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WAVES C. A. Coulson, M.A., Ph.D. 

INTEGRATION . . . . . R. P. G-illespie, Ph.D. 


E. L. Inco, D.Sc. 

E. G. Phillips, M.A., M.Sc. 

VECTOR METHODS . . D. E. Rutherford, Dr. Math. 

THEORY OF EQUATIONS . Prof. H. W. Turnbull, F.R.S. 

Other volumes in preparation 




C. A. COULSON, M.A., Ph.D. 


With 29 Figures 







THE object of this book is to consider from an elementary 
standpoint as many different types of wave motion as 
possible. In almost every case the fundamental problem 
is the same, since it consists in solving the standard equation 
of wave motion ; the various applications differ chiefly 
in the conditions imposed upon these solutidn.jj For this 
reason it is desirable that the subject of waves should be 
treated as one whole, rather than in several distinct parts ; 
the present tendency is in this direction. 

It is presupposed that the reader is familiar with the 
elements of vector analysis, the simpler results of which 
are freely quoted. In a sense this present volume may 
be regarded as a sequel to Rutherford's Vector Methods, 
published in this series^. 

In a volume of this size, it is not possible to deal 
thoroughly with any one branch of the subject : nor 
indeed is this desirable in a book which is intended as 
an introduction to the more specialised and elaborate 
treatises necessary to the specialist. This book is intended 
for University students covering a general course of Applied 
Mathematics or Natural Philosophy in the final year of 
their honours degree. A few topics, such as elastic waves 
in continuous media, or at the common boundary of two 
media, and radiation from aerials, have unavoidably had 
to be omitted for lack of space. The reader is referred to 
any of the standard works on elasticity and wireless for 
a discussion of these problems, 


This book would not be complete without a reference 
of gratitude to my friends Dr D. E. Rutherford and 
Dr G. S. Rushbrooke, who have read the proofs, checked 
most of the examples and contributed in no small way 
to the clarity of my arguments. My thanks are also 
offered to my wife for her share in the preparation of 
the manuscript. 

C. A. 0. 

January 1941. 





Introductory 1 

General Form of Progressive Waves ..... 1 

Harmonic Waves 2 

Plane Waves 4 

The Equation of Wave Motion 5 

Principle of Superposition ....... 5 

Special Types of Solution 6 

List of Solutions ........ 13 

Equation of Telegraphy 15 

Exponential Form of Harmonic Waves . . . .16 

A Solved Example 17 

Examples 19 


The Differential Equation 21 

Kinetic and Potential Energies ...... 23 

Inclusion of Initial Conditions ...... 25 

Reflection at a Change of Density 25 

Reflection at a Concentrated Load 28 

Alternative Solutions 29 

Strings of Finite Length, Normal Modes .... 30 
String Plucked at its Mid-point , . . . , .31 

The Energies of the Normal Modes 33 

Normal Co-ordinates ....... 36 

String with Load at its Mid-point . . . . .37 

Damped Vibrations ........ 39 

Method of Reduction to a Steady Wave .... 40 

Examples 41 





The Differential Equation ....... 43 

Solution for a Rectangular Membrane .... 44 

Normal Co-ordinates for Rectangular Membrane ... 47 

Circular Membrane ........ 48 

Examples ......... 50 



Differential Equation for Waves along a Bar . . .51 
Free Vibrations of a Finite Bar ...... 53 

Vibrations of a Clamped Bar . . . . . .53 

Normal Co-ordinates ........ 53 

Case of a Bar in a state of Tension ..... 54 

Vibrations of a Loaded Spring ...... 55 

Examples . . . . . . . . .59 


Summary of Hydrodynamical Formulae .... 60 
Tidal Waves and Surface Waves ..... 62 

Tidal Waves, General Conditions 63 

Tidal Waves in a Straight Channel ..... 64 

Tidal Waves on Lakes and Tanks 67 

Tidal Waves on Rectangular and Circular Tanks ... 70 

Paths of the Particles 70 

Method of Reduction to a Steady Wave . . . .71 
Surface Waves, the Velocity Potential . . . .72 

Surface Waves on a Long Rectangular Tank ... 74 
Surface Waves in Two Dimensions . . . . .75 

Paths of the Particles 77 

The Kinetic and Potential Energies . * . . .78 



Rate of Transmission of Energy ..... 79 

Inclusion of Surface Tension, General Formulae . . .81 
Capillary Waves in One Dimension ..... 83 
Examples ......... 84 



Relation between Pressure and Density .... 87 

Differential Equation in Terms of Condensation ... 87 

Boundary Conditions ....... 90 

Solutions for a Pipe of Finite Length ..... 90 

Normal Modes in a Tube with Movable Boundary . . 91 

The Velocity Potential, General Formulae .... 92 

The Differential Equation of Wave Motion .... 93 

Stationary Waves in a Tube of Finite Length ... 95 

Spherical Symmetry ........ 95 

The Kinetic and Potential Energies ..... 96 

Progressive Waves in a Tube of Varying Section ... 97 

Examples 100 



Maxwell's Equations . . . . . . . .102 

Non-conducting Media, the Equation of Wave Motion . .105 
Electric and Magnetic Potentials . . . . .106 

Plane Polarised Waves in a Dielectric Medium . . .109 
Rate of Transmission of Energy in Plane Waves . . .111 
Reflection and Refraction of Light Waves . . . .113 

Internal Reflection . . . . . . . .118 

Partially Conducting Media, Plane Waves . . . .119 

Reflection from a Metal 122 

Radiation Pressure . .124 

Skin Effect 125 

Examples ......... 125 





Doppler Effect 128 

Beats 130 

Amplitude Modulation . . . . . . .132 

Group Velocity 132 

Motion of Wave Packets 135 

Kirchhoff's Solution of the Equation of Wave Motion . . 138 

Fresnel's Principle ........ 142 

Fraunhofer Diffraction Theory . . . . . .145 

Retarded Potential Theory 149 

Examples ......... 151 

Index 154 



1. We are all familiar with the idea of a wave ; thus, 
when a pebble is dropped into a pond, water waves travel 
radially outwards ; when a piano is played, the wires 
vibrate and sound waves spread through the room ; when 
a wireless station is transmitting, electric waves move 
through the ether. These are all examples of wave motion, 
and they have two important properties in common : 
firstly, energy is propagated to distant points ; and 
secondly, the disturbance travels through the medium 
without giving the medium as a whole any permanent 
displacement. Thus the ripples spread outwards over a 
pond carrying energy with them, but as we can see by 
watching the motion of a small floating body, the water 
of the pond itself does not move with the waves. In the 
following chapters we shall find that whatever the nature 
of the medium which transmits the waves, whether it be 
air, a stretched string, a liquid, an electric cable or the 
ether, these two properties which are common to all these 
types of wave motion, will enable us to relate them 
together. They are all governed by a certain differential 
equation, the Equation of Wave Motion (see 5), and 
the mathematical part of each separate problem merely 
consists in solving this equation with the right boundary 
conditions, and then interpreting the solution appropriately. 

2. Consider a disturbance </> which is propagated 
along the x axis with velocity c. There is no need to 
l A 


state explicitly what < refers to ; it may be the elevation 
of a water wave or the magnitude of a fluctuating electric 
field. Then, since the disturbance is moving, <f>.will depend 
on x and t. When t = 0, <f> will be some function of x 
which we may call f(x). f(x) is the wave profile, since 
if we plot the disturbance cf> against x, and " photograph " 
the wave at t 0, the curve obtained will be < ~f(x). 
If we suppose that the wave is propagated without change 
of shape, then a photograph taken at a later time t will 
be identical with that at t = 0, except that the wave 
profile has moved a distance ct in the positive direction 
of the x axis. If we took a new origin at the point x ct, 
and let distances measured from this origin be called X, 
so that x = X-\-ct, then the equation of the wave profile 
referred to this new origin would be 

Referred to the original fixed origin, this means that 

<f>=f(x-ct) . . . . (1) 

This equation is the most general expression of a wave 
moving with constant velocity c and without change of 
shape, along the positive direction of x. If the wave is 
travelling in the negative direction its form is given by 
(1) with the sign of c changed, i.e. 


3. The simplest example of a wave of this kind is the 
harmonic wave, in which the wave profile is a sine or 
cosine curve. Thus if the wave profile at t = is 

(<f>)t=o = # cos mx, 
then at time t y the displacement, or disturbance, is 

= a cos m(xct) ... (3) 

The maximum value of the disturbance, viz. a, is called 
the amplitude. The wave profile repeats itself at regular 


distances 27r/m. This is known as the wavelength A. 
Equation (3) could therefore be written 

(f) a cos ~(x~ ct) .... (4) 

The time taken for one complete wave to pass any point 
is called the period r of the wave. It follows from (4) that 


-^-(x--ct) must pass through a complete cycle of values 

as t is increased by r. Thus 


T = 277 ' 

i.e. r = A/c . . . (5) 

The frequency n of the wave is the number of waves 
passing a fixed observer in unit time. Clearly 

n=l/r . . . . (6) 
so that c = n\, .... (7) 

and equation (4) may be written in either of the equivalent 

</) == a cos27r{^ J (8) 

eft = a cos 27rl^ nt\ ... (9) 

Sometimes it is useful to introduce the wave number k, 
which is the number of waves in unit distance. Then 

*=1/A, . - (10) 

and we may write equation (9) 

(f) = a cos 27r(kxnt) . . . (11) 


If wo compare two similar waves 
<^ a cos 27r(kxnt), 

we see that </>% is the same as <^ except that it is displaced 
a distance e/27r&, i.e. eA/277. is called the phase of <f> 2 
relative to <f> v If e = 27r, 4?!, ... then the displacement 
is exactly one, two, ... wavelengths, and we say that the 
waves are in phase ; if e = TT, STT, . . . then the two waves 
are exactly out of phase. 

Even if a wave is not a harmonic wave, but the wave 
profile consists of a regularly repeating pattern, the 
definitions of wavelength, period, frequency and wave 
number still apply, and equations (5), (6), (7) and (10) 
are still valid. 

4. It is possible to generalise equation (1) to deal 
with the case of plane waves in three dimensions. A 
plane wave is one in which the disturbance is constant 
over all points of a plane drawn perpendicular to the 
direction of propagation. Such a plane is called a wave- 
front, and the wavefront moves perpendicular to itself 
with the velocity of propagation c. If the direction of 
propagation is x : y : z = I : m : n, where Z, m, n are the 
direction cosines of the normal to each wavefront, then the 
equation of the wavefronts is 

Ix -\-my-\-nz = const., . . . (12) 

and at any moment t, <f> is to be constant for all x, y, z 
satisfying (12). It is clear that 

<f)=^f{lx+my+nzct) . . . (13) 

is a function which fulfils all these requirements and 
therefore represents a plane wave travelling with velocity 
c in the direction I : m : n without change of form. 


5. The expression (13) is a particular solution of the 
equation of wave motion referred to on p. 1. Since 
I, m, n are direction cosines, Z 2 +m 2 +i& 2 = 1, and it is 
easily verified that <f> satisfies the differential equation * 

This is the equation of wave motion. j It is one of the 
most important differential equations in the whole of 
mathematics, since it represents all types of wave motion 
in which the velocity is constant. The expressions in 
(1), (2), (8), (9), (11) and (13) are all particular solutions 
of this equation. We shall find, as we investigate different 
types of wave motion ^subsequent chapters, that equation 
(14) invariably appears, and it will be our task to select 
the solution that is appropriate to our particular problem. 
There are certain, types of solution that occur often, and 
we shall discuss some of them in the rest of this chapter, 
but before doing so, there is one important property of 
the fundamental equation that must be explained. 

6. The equation of wave motion is linear. That is 
to say, (f> and its differential coefficients never occur in 
any form other than that of the first degree. Consequently, 
if < x and <f> 2 are any two solutions of (14), a 1 <^ 1 +a 2 <^ 2 is 
also a solution, a x and 2 being two arbitrary constants. 
This is an illustration of the principle of superposition, 
which states that, when all the relevant equations are 
linear, we may superpose any number of individual 
solutions to form new functions which are themselves also 
solutions. We shall often have occasion to do this. 

A particular instance of this superposition, which is 
important in many problems, comes by adding together 

* This equation has a close resemblance to Laplace's Equation 
which is discussed in Rutherford, Vector Methods, Chapter VII. 

t Sometimes called the wave equation, but we do not use this 
phrase to avoid confusion with modern wave mechanics. 


two harmonic waves going in different directions with the 
same amplitude and velocity. Thus, with two waves 
similar to (11) in opposite directions, we obtain 

<f) = a cos k 2n(kxnt)-\-a cos 2tr(Tcx-\-nt) 

= 2a cos 2nkx cos 2irnt (15) 

This is known as a stationary wave, to distinguish it from 
the earlier progressive waves. It owes its name to the 
fact that the wave profile does not move forward. In fact, 
<f) always vanishes at the points for which cos 2irkx = 0, 


viz. x i~7> T7> db-r> These points are called the 
4& 4& 4fc 

nodes, and the intermediate points, where the amplitude 
of c/> (i.e. 2a cos 2irkx) is greatest, are called antinodes. 
The distance between successive nodes, or successive 
antinodes, is l/2k, which, by (10), is half a wavelength. 

Using harmonic wave functions similar to (13), we find 
stationary waves in three dimensions, given by 

(f) = a cos ~Y- (Ix+my+nzct) + a cos (lx-\-my+nz+ct) 
A A 

= 2a cos ~r- (Ix+my+nz) cos ct . . . (16) 
A A 

In this case < always vanishes on the planes Ix -\-my-\-nz 

A 3A 

= "-> db 9 > an d these are known as nodal planes. 
4 4 

7. We shall now obtain some special types of solution 
of the equation of wave motion ; we shall then be able to 
apply them to specific problems in later chapters. We 
may divide our solutions into two main types, representing 
stationary and progressive waves. 

We have already 'dealt with progressive waves in one 
dimension. The equation to be solved is 


Its most general solution may be obtained by a 
method due to JD'Alembert. We change to new variables 
u = x ct, and v = x+ct. Then it is easily verified that 

d<f> d< ty ty d<(> 3<f> 

~- transforms to + , TT transforms to c +c 
dx du dv dt du dv 

o2 J 

so that the equation becomes = ; the most general 


solution of this is 

/ and g being arbitrary functions. In the original variables 
this is 

. . . (17) 

The harmonic waves of 2 are special cases of this, in 
which / and g are cosine functions. The waves / and g 
travel with velocity c, in opposite directions. 

In two dimensions the equation of wave motion is 

c d 

and the most general solution involving only plane * waves 
is </> = f(to+my ct) +g(lx+my+ct), . (19) 

where, as before, / and g are arbitrary functions and 
p+m* - 1. 

In three dimensions the differential equation is 

?V , ^.^ = i^ /9m 

dx*^~dy*^ dz 2 c* dt* ' ' l " ; 

and the most general solution involving only plane waves is 
<f) ~f(lx+my+nz~ct)+g(lx+my+nz+ct) . (21) 
in which l*+m*+ri* = 1. 

* Strictly these should be called line waves, since at any 
moment < is constant along the lines Ix -}- }ny = const. 


There are, however, other solutions of progressive type, 
not involving plane waves. For suppose that we transform 
(20) to spherical polar coordinates r, 0, $* The equation 
of wave motion becomes 

o t 9 __ t\ on i "" * on I ' 


If we are interested in solutions possessing spherical 
symmetry (i.e. independent of 8 and ifj) we shall have to 
solve the simpler equation 

This may be written 

1 8 2 

showing (cf. eq. (17)) that it has solutions 
^ =*f(r-ct)+g(r+ct), 

f and g again being arbitrary functions. We see, therefore, 
that there are progressive type solutions 

<f> = f(r~ct) + *g(r+ct) . . (24) 

Let us now turn to solutions of stationary type. These 
may all be obtained by the method known as the separation 
of variables. In one dimension we have to solve 

Let us try to find a solution of the form 
= X(x)T(t), 
* See e.g. Rutherford, p. 62, equation 20. 


X and T being functions of x and t respectively, whose 
form is still to be discovered. Substituting this value of 
</> in the differential equation and dividing both sides 
by X(x)T(t) we obtain 


X dx* c 2 T dt* ' ' * V ; 

The left-hand side is independent of t, being only a function 
of x, and the right-hand side is independent of x. Since the 
two sides are identically equal, this implies that each is 
independent both of a: and t, and must therefore be constant. 
Putting this constant equal to p 2 , we find 

X'+p^X^Q, T"+c 2 p 2 T = Q. . . (26) 

These equations give, apart from arbitrary constants 

~ cos m cos A ._,_. 

X = . px , T = . opt . . (27) 

sm^ sin * ' 

A typical solution therefore is a cos px cos cpt, in which 
p is arbitrary. In this expression we could replace either 
or both of the cosines by sines, and by the principle of 
superposition the complete solution is the sum of any 
number of terms of this kind with different values of p. 

The constant p 2 which we introduced, is known as 
the separation constant. We were able to introduce it in 
(25) because the variables x and t had been completely 
separated from each other and were in fact on opposite sides 
of the equation. There was no reason why the separation 
constant should have had a negative value of p 2 except 
that this enabled us to obtain harmonic solutions (27). 
If we had put each side of (25) equal to -\-p 2 , the solutions 
would have been 

X = ev* , T = e c & . . . (28) 

and our complete solution should therefore include terms 
of both types (27) and (28). The same distinction between 
the harmonic and exponential types of solution will occur 


This method of separation of variables can be extended 
to any number of dimensions. Thus in two dimensions a 
typical solution of (18) is 

, cos cos cos 

d> = . px . ay . ret , . . (29) 
^ sin^ sin sm ' 

in which p 2j rq 2 = f 2 , p and # being allowed arbitrary 
values. An alternative version of (29), in which one of 
the functions is hyperbolic, is 

= COS pxe G S rct. . . (30) 
T sm^ sm ' 

in which p 2 q z r 2 . 

It is easy to see that there is a variety of forms similar 
to (30) in which one or more of the functions is altered 
from a harmonic to a hyperbolic or exponential term. 

In three dimensions we have solutions of the same type, 
two typical examples being 

. COS COS COS COS , 0,0,0 

* = sin * sin W Bin " Hm** ' * +<1 +* = * 

^ e rz SC , -p'-f+r* = ^ (32) 
sin sin x a 

There are two other examples of solution in three 
dimensions that we shall discuss. In the first case we put 
x = r cos 0, y = r sin 0, and we use r, # and z as cylindrical 
coordinates. The equation of wave motion becomes * 

A solution can be found of the form 

$ = B(r)@(6)Z(z)T(t), . . . (33) 

* See Rutherford, p. 63. 


where, by the method of separation of variables, R, @, Z, 
T satisfy the equations 

_ o. 

r dr r* 

2 2 02 (VA\ 

jp - -r- > n -p q . . (M) 

The only difficult equation is the first, and this * is just 
Bessel's equation of order m, with solutions J m (nr) and 
Y m (nr). J m is finite and Y m is infinite when r = 0, so 
that we shall usually require only the J m solutions. The 
final form of is therefore 

, J m/ .cos n cos cos A /0 _ 

6 = (nr) . m6 . qz . cpt . . (35) 
1 Jr TO sin sm sin 

If (f> is to be single valued, m must be an integer ; but n, 
q and p may be arbitrary provided that n 2 = p*q 2 . 
Hyperbolic modifications of (35) are possible, similar in 
all respects to (31) and (32). 

Our final solution is one in spherical polar coordinates 
r, Q, ifj. The equation of wave motion (22) has a solution 


* See Ince, Integration of Ordinary Differential Equations, p. 127. 


ra, n and p are arbitrary constants, but if W(ifj) is to be 
single valued, ra must be integral. The first two of these 
equations present no difficulties. The 0-equation is the 
generalised Legendre's Equation * with solution 

= P n (cos 0), 

and if & is to be finite everywhere, n must be a positive 
integer. When m = and n is integral, P n m (cos 9) 
reduces to a polynomial in cos of degree n y known 
as the Legendre's polynomial P n (cos 9). For other 
integral values of m, P n m (cos 9) is defined by the equation 

A few values of P n (cos 9) and P n m (cos 9) are given 
below, for small integral values of n and m. When m>n, 
P n m (cos 9) vanishes identically. 

P (cos 9) - 1 

P! (cos 0) = cos 

P 2 (cos0) = i (3 cos 2 01) 

P 3 (cos 0) = $ (5 cos 3 0-3 cos 0) 

P 4 (cos 0) | (35 cos 4 0-30 cos 2 0+3) 
P^ (cos 0) = sin 
Pa 1 (cos 0) = 3 sin cos 
P 3 X (cos 0) = I sin (5 cos 2 0-1) 
P 2 2 (cos 0) = 3 sin 2 0. 

To solve the JS-equation put R(r) = r"" 1 / 2 /S(r), and we 
find that the equation for S(r) is just Bessel's equation 

Therefore S(r) = e/ n+1/2 (pr) or 

* See Ince, Integration of Ordinary Differential Equations, p. 119, 
for the case m = 0. 


Collecting the various terms, the complete solution, apart 
from hyperbolic modifications, is seen to be 

^ = f -1/8 +1/2 (:pr) p^ (cos 6} mi/1 cpt . (36) 

* n 4- 1/2 Slli felli 

If ( has axial symmetry, we must only take functions 
with ra = 0, and if it has spherical symmetry, terms with 
m = n = 0. Now ' J^(z) = \/(2/7rz) sin z, and also 
Y l!2 (z) = <\/(2ir/z) cos 2;, so that this becomes 

A solution finite at the origin is obtained by omitting the 
cos pr term. 

8. We shall now gather together for future reference 
the solutions obtained in the preceding pages, 

Progressive -waves 
1 dimension 

<l>=f(x-ct)+g(x+ct) . . (17) 

2 dimensions 

d*<f> dty __ 1 3 2 <f> 

fa* + %~ 2 ^ ^ M* 
<f> =f(lz+my-ct)+g(lx+my+ct), l 2 +m* ^ 1 . (19) 

3 dimensions 

3 dimensions, spherical symmetry 



Stationary waves 
1 dimension 

2 dimensions 


sill & x sin 


cos cos cos 
sin P x sin ^ sin '* 


3 dimensions 



= r z 



TJ ^ ^ *<*> 


sm * J sin sin 



cos cos 

Plane Polar Coordinates (r, 6) 

r dr 

Cylindrical polar coordinates (r, 9, z) 



sin sin cn ' 

i J w COS COS COS 22<> 

V ^ Y m (nr > sin m ^ sin ^sin ^ n = p ~-q~ 

and other hyperbolic modifications. 
Spherical Polar Coordinates (r, #, iff) 

a 2 ^ 2 a<^ i 

n+1 / 2 

i a 2 <j i a 2 <^> 

' ..9 , ' Q /I 0/9 ^2 P)/2 


. COS . COS 

M w (cos C7) . mib . ( 
sin ^ sm 


Spherical symmetry 

d*J> 2 3d) I d 2 (h ' , cos cos 


In solving problems, we shall more often require progressive 
type solutions in cases where the variables x, y, z are allowed 
an infinite range of values, and stationary type solutions 
when their allowed range is finite. 

9. There is an important modification of the equation 
of wave motion which arises when friction, or some other 
dissipative force, produces a damping. The damping effect 
is usually allowed for (see e.g. Chapter II and elsewhere) 

O I 

by a term of the form k 9 which will arise when the 

damping force is proportional to the velocity of the vibra- 
tions. The revised form of the fundamental equation, 
known as the equation of telegraphy, is 

/-\2 i 

If we omit the term ~~ this equation is the same as that 

occurring in the flow of heat. If we put < ue~ Jct/2 ) we 
obtain an equation for u of the form 


Very often k is so small that we may neglect & 2 , and then 
(39) is in the standard form which we have discussed in 
8, and the solutions given there will apply. In such a 
case the presence of the dissipative term is shown by a 
decay factor e~ ktl2 . If this is written in the form e~ tft , 
then t (= 2/k) is called the modulus of decay. When 
the term in fc 2 may not be neglected, we have to solve 
(38) and the method of separation of variables usually 
enables a satisfactory solution to be obtained without 
much difficulty. 


There is an alternative solution to the equation of 

telegraphy that is sometimes useful. Taking the case of 

one dimension, and supposing that k is so small that 
k 2 may be neglected, we have shown that the solution of 
(38) may be written in the form 

ct), .... (40) 

where / is any function. Since / is arbitrary, we can put 


f(x-ct) = e'*!***' g(x-ct), 

and g is now an arbitrary function. Substituting this in 
(40) we get 

ct) ..... (41) 

This expression resembles (40) except that the exponential 
factor varies with x instead of with t. 

10. Most of the waves with which we shall be 
concerned in later chapters will be harmonic. This is 
partly because, as we have seen in 8, harmonic functions 
arise very naturally when we try to solve the equation of 
wave motion ; it is also due to the fact that by means of 
a Fourier analysis, any function may be split into harmonic 
components, and hence by the principle of superposition, 
any wave may be regarded as the resultant of a set of 
harmonic waves. 

When dealing with progressive waves of harmonic type 
there is one simplification that is often useful and which 
is especially important in the electromagnetic theory of 
light waves. We have seen in (11) that a progressive 
harmonic wave in one dimension can be represented by 
= a cos 2m(kxnt). If we allow for a phase e, it 
will be written <f> = a cos {%rr(kxnt)-{-}. Now this 
latter function may be regarded as the real part of the 
complex quantity a e*(2flrtf -<)+}. It is most convenient 
for our subsequent work if we choose the minus sign and 


also absorb the phase and the amplitude a into one 
complex number A. We shall then write 

< = A e^^-kx) } A = a e~ i . . . (42) 

This complex quantity is itself a solution of the equation 
of wave motion, as can easily be seen by substitution, and 
consequently both its real and imaginary parts are also 
solutions. Since all our equations in <j> are linear, it is 
possible to use (42) itself as a solution of the equation 
of wave motion, instead of its real part. In any equation 
in which (f> appears to the first degree, we can, if we wish, 
use the function (42) and assume that we always refer to 
the real part, or we can just use (42) as it stands, without 
reference to its real or imaginary parts. In such a case the 
apparent amplitude A is usually complex, and since 
A = a e~~ i , we can say that |^4| is the true amplitude, 
and arg A is the true phase. The velocity, of course, 
as given by (7) and (10), is n/k. 

We can extend this representation of <f> to cover waves 
travelling in the opposite direction by using in such a case 

<f> = A e W*+W ..... (43) 

There is obviously no reason why we should not extend 
this to two or three dimensions. For instance, in three 

& A e 27Ti { nt ~( pX+ w +rz V . . . (44) 

would represent a harmonic wave with amplitude A 
moving with velocity 7&/y'(# 2 +g 2 +r 2 ) in the direction 
x : y : z = p : q : r. 

11. We shall conclude this chapter with an example. 

Let us find a solution of ~ + -~ = -~ ~-~ such that <b 

dx 2 dy 2 c 2 dt 2 ^ 

vanishes on the lines # = 0, x = a, y = 0, y = b. Since 


the lines x 0, a, and y 0, b are nodal lines, our 
solution must be of the stationary type. Referring to 8, 
equation (29), we see that possible solutions are 

. COS COS COS A , 0,0 9 

= . px . qy . ret , where 0"+<r = r . 
r sin sin sm *- * 

Since < is identically zero at x 0, and y = 0, we shall 
have to take the sine rather than the cosine in the first 
two factors. Further, since at x = a, </> for all 
values of /, therefore 

sin pa 0. 
Similarly, sin qb = 0. 

Hence p = ?/wr/a, and # = ^77/6, m and n being integers. 
A solution satisfying all the conditions is therefore 

, . nrnx . mni cos 
^ = sin sm-gi- ^frt, 

where r 2 = 7r 2 (m 2 /a 2 +n 2 /6 2 ) . 

The most general solution is the sum of an arbitrary 
number of such terms, e.g. 

sin - {C mn cos rct-\-D mn sin re/}. (45) 

m,n & 

At = 0, this gives 

. mry 
sm , 

rr r^ 

o = 27rcD mn 8m sin . 

By suitable choice of the constants C mn and D wn we can 
make (f> and ^ have any chosen form at t = 0, The value 
at any subsequent time is then given by (45). 


12. Examples 

(1) Show that (/> = f(x cos Q-\-y sin. ct) represents a 
wave in two dimensions, the direction of propagation making 
an angle 6 with the axis of x. 

(2) Show that <f> = a cos (Ix -\-my-ct) is a wave in two 
dimensions and find its wavelength. 

(3) What is the wavelength and velocity of the system of 
plane waves <f> = a sin (Ax+By + CzDt) ? 

(4) Show that three equivalent harmonic waves with 120 
phase between each pair have zero sum. 

(5) Show that (/> r~~ l t' 2 cos %0f(r-j~ct) is a progressive type 
wave in two dimensions, r and 6 being plane polar coordinates, 
and / being an arbitrary function. By superposing two of 
these waves in which / is a harmonic function, obtain a 
stationary wave, and draw its nodal lines. Note that this is 
not a single-valued function unless we put restrictions upon the 
allowed range of 6. 

(6) By taking the special case of f(x) = g(x) = sin px in 
equation (24), show that it reduces to the result of equation 
(36) in which m n = 0. Use the relation 

(7) Find a solution of ~4 + - = 0, such that < = 

dx 2 c 2 dt* 

when t = oo, and <j> = when x = 0. 

d 2 <I> 1 2 < 

(8) Find a solution of ,-^ - ^ such that < = when 

c)x c ot 

x +00 or^= +00. 

d*z d*z 

(9) Solve the equation - = c 2 - given that z is never 

ot ox 

infinite for real values of x and t, and z = when x = 0, or 
when t 0. 

(10) Solve ~ '- given that F = when t ~ co and 

when x = 0, and when x I. 


(11) x, y, z are given in terms of the three quantities 
> *?? by th e equations 

x = a sinh sin 77 cos 
y = a sinh sin 77 sin 
z = a cosh cos 77 

3V 3V ^V J 0V 
Show that the equation - -f- 7 + ^ = - - is of the 

dx 2 dy 2 dz 2 c 2 dt 2 

correct form for solution by the method of separation of 
variables, when , 77, are used as the independent variables. 
Write down the subsidiary equations into which the whole 
equation breaks down. 

12. Show that the equation of telegraphy (38) in one 
dimension has solutions of the form 

where m and p are constants satisfying the equation 

"[ANSWERS : 2. 2rr/(l* + m 2 )* ; 3. A = 27r/(^. 2 + B* -f C a )*, 
vel. = AD/27r; 7. ^L sin naje~ cni ; 8. Ae~ n(x + ct ^ ; 9. ^ sin ^ 
sin cp^ ; 10. u4e~^ 8< sin px , p = *r/l , %ir/l 9 ; 11- Show 
that ^ = const., 77 = const., const, form an orthogonal 
system of coordinates, and transform y 2 ^ in terms of , 17, f 
as in Rutherford, Vector Methods, 47. The result is 
$ ~ X(g)Y(7))Z( > )T(t), where m, p and 7 are arbitrary con- 
stants, and 

sinh { -- X + p 2 

sinh ^ cff dg smh 2 f 

_ s i n ~ -- ^4-^23^2 Y = -g 2 F, 
- 2 

szn 77 d-t] dtj sin 2 ?/ 



13. In this chapter we shall discuss the transverse 
vibrations of a heavy string of mass p per unit length. By 
transverse vibrations we mean vibrations in which the 
displacement of each particle of the string is in a direction 
perpendicular to the length. When the displacement is 
in the same direction as the string, we call the waves 
longitudinal ; these waves will be discussed in Chapter IV. 
We shall neglect the effect of gravity ; in practice this 
may be achieved by supposing that the whole motion takes 
place on a smooth horizontal plane. 

In order that a wave may travel along the string, 
it is necessary that the string should be at least slightly 
extensible ; in our calculations, however, we shall assume 
that the tension does not change appreciably from its 
normal value F. The condition for this (see 14) is that 
the wave disturbance is not too large. 

Let us consider the motion of a small element of the 
string PQ (fig. 1) of length ds. Suppose that in the 
equilibrium state the string lies along the axis of x, and 
that PQ is originally at P Q Q . Let the displacement of 
PQ from the x axis be denoted by y. Then we shall obtain 
an equation for the motion of PQ in terms of the tension 
and density of the string. The forces acting on this 
element, when the string is vibrating, are merely the two 
tensions F acting along the tangents at P and Q as shown 
in the figure ; let if/ and \fj-\-difs be the angles made by these 
two tangents with the x axis. We can easily write down 



the equation of motion of the element PQ in the y direction ; 
for the resultant force acting parallel to the y axis is 
F sin (\fj+d\ls)F sin $. Neglecting squares of small 
quantities, this is F cos ifj di/j. The equation of motion is 

F cos i/j difj pds 


FIG. 1 

Now tan 
from (1) 

= , so that 


= 7~ 2 dx, and so, 


. (2) 

f /%\ 2 1 ~ l 
But cos 2 A ~ 4 1 -f- 1 - J I ,80 that if the displacements 

I \&*y J 


small enough for us to neglect I - 1 compared with 

unity, we may write (2) in the standard form for wave 
motion * (Chapter I, 5), viz., 


It follows from Chapter I, equation (17) that the general 
solution of this equation may be put in the form 

y=f(x-ct)+g(x+ct), ... (4) 

/ and g being arbitrary functions. f(xct) represents a 
progressive wave travelling in the positive direction of 
the x axis with velocity c, and g(x-\-ct) represents a 
progressive wave with the same velocity in the negative 
direction of x. Thus waves of any shape can travel in 
either direction with velocity c = -\/(F/p), and without 
change of form. A more complete discussion, in which 
we did not neglect terms of the second order, would show 
us that the velocity was not quite independent of the 
shape, and indeed, that the wave profile would change 
slowly with the time. These corrections are difficult to 
apply, and we shall be content with (4), which is, indeed, 
an excellent approximation except where there is a sudden 

" kink " in y, in which case we cannot neglect I 

14. Since the velocity of any point of the string is y, 
we can soon determine the kinetic energy of vibration. It is 

T = Jj pfdx .... (5) 

* The student who is interested in geometry will be able to 
prove that the two tensions at P and Q are equivalent to a single 
force of magnitude FdsfR, where JR is the radius of curvature of 
the string. This force acts perpendicularly to PQ. Putting 

E = l + , and neglecting , we obtain (3). 


The potential energy F is found by considering the increase 
of length of the element PQ. This element has increased 
its length from dx to ds. We have therefore done an 
amount of work F(dsdx). Summing for all the elements 
of the string, we obtain the formula 

V= |V(<fo-efc) = lF 

J J 

r/%\ 2 

= J-F I I J dx, approximately. . (6) 

The integrations in (5) and (6) are both taken over the 
length of the string. 

With a progressive wave y =f(xct), these equations 

T = J lf**(f')*dx = W /(/')* - (7) 

Thus the kinetic and potential energies are equal. The 
same result applies to the progressive wave y = g(x-\-ct) 9 
but it does not, in general, apply to the stationary type 
waves y f(xct) +g(x+ct). 

We can now decide whether our initial assumption is 
correct, that the tension remains effectively constant. 
If the string is elastic, the change in tension will be pro- 
portional to the change in length. We have seen in (6) 

that the change in length of an element dx is - I -^ I dx. 

2 \dxj 

Thus, provided that is of the first order of small 


quantities, the change of tension is of the second order, 
and may safely be neglected. This assumption is equivalent 
to asserting that the wave profile does not have any large 
" kinks," but has a relatively gradual variation with x. 


15. The functions / and g of (4) are arbitrary. But 
they may be fixed by a knowledge of the initial conditions. 
Thus, with a string of unlimited length, such that 
^ =0 = </>(#), y t== Q = 0(aj),* we must have, from (4), 

Integrating this last equation we have 

and so 

The displacement at any subsequent time t is therefore 

if i r x ~ ct i r x+ct } 

= 5 1 t(x-ct) + </>(x+ct) - - $(x)dx + - i/,(x)dx \ 

z I C J C J J 
I f x + ct } 

- t(x)dx\. . . (9) 

C J x-ct J 

16. The discussion above applies specifically to 
strings of infinite length. Before we discuss strings of 
finite length, we shall solve two problems of reflection 
of waves from a discontinuity in the string. The first is 
when two strings of different densities are joined together, 
and the second is when a mass is concentrated at a point 
of the string. In each case we shall find that an incident 
wave gives rise to a reflected and a transmitted wave. 

Consider first, then, the case of two semi-infinite 
strings 1 and 2 joined at the origin (fig. 2). Let the 

* This function \f/(x) must be distinguished from the angle ^ in 



densities of the two strings be p and p 2 . Denote the dis- 
placements in the two strings by y and y 2 . Let us suppose 
that a train of harmonic waves is incident from the negative 
direction of x. When these waves meet the change of 
wire, they will suffer partial reflection and partial trans- 
mission. If we choose the exponential functions of 10 
to represent each of these waves, we may write 


= 2/incidcnt + 2/reflccted 
== ^/transmitted 

?/ . __ A eZTTidl 

^/incident -^-^ 

?V * i 7? 

//reflected ^l 

... . yJ 

transmitted ^2 



FIG. 2 

A 1 is real, but ^4 2 and B l may be complex. According to 
10 equation (42), the arguments of A% and S l will give 
their phases relative to the incident wave. All three waves 
in (11) must have the same frequency n, but since the 
velocities in the two wires are different, they will have 
different wavelengths 1/& X and l/& 2 . The reflected wave 
must, of course, have the same wavelength as the incident 
wave. Since the velocities of the two types of wave are 


n/k-L and n/k 2 (Chapter I, equations (7) and (10)), and we 
have shown in (3) that c 2 = F/p, therefore 

*lW = ft/ft .... (12) 

In order to determine A 2 and B l we use what are known 
as the boundary conditions. These are the conditions 
which must hold at the boundary point x 0. Since the two 
strings are continuous, we must have y = y 2 identically 
for all values of t, and also the two slopes must be the same, 

so that -~ = -~ for all t. If this latter condition were 

ox ox 

not satisfied, we should have a finite force acting on an 
infinitesimal piece of wire at the common point, thus 
giving it infinite acceleration. We shall often meet 
boundary conditions in other parts of this book ; their 
precise form will depend of course upon the particular 
problem under discussion. In the present case, the two 
boundary conditions give 

A 1 +B 1 = A 2 , 
27ri(k l A l ~\-k l B l ) = 27ri(k 2 A 2 ). 

These equations have a solution 

_ *"" 2 * 

A J/t I 1/t A J/> l_ T* 

"i A^I~| /I/O "^^1 i / i T'* / 2 

Since k lt k 2 and A 1 are real, this shows that B 1 and A 2 
are both real. A 2 is positive for all k and & 2 , but jBj is 
positive if Tc^>Tc^ and negative if k^k^ Thus the 
transmitted wave is always in phase with the incident 
wave, but the reflected wave is in phase only when the 
incident wave is in the denser medium ; otherwise it is 
exactly out of phase. 

The coefficient of reflection R is defined to be the 

k k 
ratio \BtlAi\ 9 i.e. 7^ r- 2 , which, by (12), we may write 



Similarly, the coefficient of transmission T is equal 

to \A 2 fA 1 1 , i.e. 

o. /^ 

. . . . (15) 

17. A similar discussion can be given for the case of 
a mass M concentrated at a point of the string. Let us 
take the equilibrium position of the mass to be the origin 
(fig. 3) and suppose that the string is identical on the two 

sides. Then if the incident wave comes from the negative 
side of the origin, we may write, just as in (10) and (11) : 

2/1 = 2/incident + 2/reflected 
2/2 == 2/transmitted 





The boundary conditions are that for all values of t 
(i) foiL-o = M*=o .... 




The first equation expresses the continuity of the string 
and the second is the equation of motion of the mass M . 
We can see this as follows : the net force on M is the 
difference of the components of F on either side, so that 
if ifj^ and if/ 2 are the angles made with the x axis, we have 

~ a -. 

L ~ _U=o 

Since ^ and ifj 2 are small, we may put sin */r 2 = 

tan i/r 2 = , sin ^ = , and (18) is then obtained. 

ox fix 

Substituting from (16) into (17) and (18), and cancelling 
the term e 27rint , which is common to both sides, we find 

= A, 

Let us write Trn 2 M/kF = p . . . . (19) 

A solution of the equations then gives 

A, l+ip l+p* ' ' 

. . . (21) 

A 2 I lip 

A l l+ip 

In this problem, unlike the last, B t and A z are complex, 
so that there are phase changes. These phases (according 
to 10) are given by the arguments of (20) and (21). 
They are therefore tan -1 (p) and tan~ 1 ( l/p) respectively. 
The coefficient of reflection R is IBJA^, which equals 
) 1/2 > and the coefficient of transmission T is 
i.e. l/(l+p 2 ) 1/2 . If we write p = tan 0, where 
then we find that the phase changes are 
and 77/2+0, and also R = sin 0, T = cos 6. 

18. The two problems in 16, 17 could be solved 
quite easily by taking a real form for each of the waves 


instead of the complex forms (11) and (16). The student 
is advised to solve these problems in this way, taking, for 
example, in 17, the forms 

^incident = a l cos %ir(nt kx) 
^reflected = b l GOS{2rrr(nt+kx)+} 
2/transmiitcd = 2 cos {^(nt-Jcx) +rj} . (22) 

In most cases of progressive waves, however, the complex 
form is the easier to handle ; the reason for this is that 
exponentials are simpler than harmonic functions, and 
also the amplitude and phase are represented by one 
complex quantity rather than by two separate terms. 

19. So far we have been dealing with strings of 
infinite length. When we deal with strings of finite length 
it is easier to use stationary type waves instead of progres- 
sive type. Let us now consider waves on a string of length 
Z, fastened at the ends where x = 0, I. We have to find a 

d 2 y 1 3 2 y 

solution of the equation (3), viz. - , subject to 

dx 2 c* dt* 

the boundary conditions T/ 0, at x 0, I, for all t. 
Now by Chapter I, 8, we see that suitable solutions are 
of the type 

cos cos 

sin ** sin c ^' 

It is clear that the cosine term in x will not satisfy the 
boundary condition at x = 0, and we may therefore write 
the solution 

y = sin px (a cos cpt-}-b sin cpt). 

The constants a, b and p are arbitrary, but we have still 
to make y = at x = 1. This implies that sin pi = 0, 
i.e. pi = TT, 277, STT. . . . It follows that the solution is 

. TTTX t TTTCt . T7TCt\ 

y = sin la cos - -- 1- b sin r- , r=^l,2,3, . . . (23) 
I \ I IJ 


Each of the solutions (23) in which r may have any positive 
integral value, is known as a normal mode of vibration. 
The most general solution is the sum of any number of 
terms similar to (23) and may therefore be written 

. TTTX\ TTTCt . T7TCt\ 

y == Z sin \ a r cos + b r sin \ . (24) 
r l \ i l } 

The values of a r and b r are determined from the initial 
conditions ; thus, when t = 0, 

y tssQ = S a r sin ' . . . . (25) 


T7TC . T7TX 

y t=s{) = E b r -ysm ... (26) 

If we are told the initial velocity and shape of the string, 
then each a r and b r is found from (25) and (26), and hence 
the full solution is obtained. We shall write down the 
results for reference. If we suppose that when t = 0, 
y = <f)(x) 9 y = *fj(x) 9 then the Fourier analysis represented 
by (25) and (26) gives 

2 C J ., . TTTX 
a r = ~ (f)(x) sm dx 

'Jo I 

2 f l 

b r MX) si 
me J o 


sin ~dx . . (27) 

In particular, if the string is released from rest when 
t = 0, every b r 0. 

20. As an illustration of the theory of the last section, 
let us consider the case of a plucked string of length I 
released from rest when the midpoint is drawn aside 
through a distance h (fig. 4). In accordance with (25) 
and (26) we can assume that 

t TTTX met 
= L a sin -- cos r . 



When t = 0, this reduces to 2a r sin , and the coefficients 

r ^ 

a r have to be chosen so that this is identical with 

y = ~x 

y= (l-x) , 

FIG. 4 


If we multiply both sides of the equation y = 27a r sin 

by sin -y- , and integrate from x = to a: ?, as in the 

method of Fourier analysis, all the terms except one will 
disappear on the right-hand side, and we shall obtain 

1 C l l 2 2h . TTTXj C l 2Jl (1 . T7TX J 

-a r = x sin -y- dx+ -=- (lx) sin ax. 

2 J o ^ ^ J 112, I * 

Sh . TTT 
a = - sin when r is odd, 
22 2 


= when r is even. 

So the full solution, giving the value of y at all subsequent 
times, is 

Sh 1 

; Sin 






Thus the value of y is the result of superposing certain 
normal modes with their appropriate amplitudes. These 
are known as the partial amplitudes. The partial 
amplitude of any selected normal mode (the rth for 
example), is just the coefficient a r . In this example, a r 
vanishes except when r is odd, and then a r is proportional 
to 1/r 2 , so that the amplitude of the higher modes is 
relatively small. 

21. The rth normal mode (23) has a frequency re/21. 
Also, there are zero values of y (i.e. nodes) at the points 
x = 0, Z/r, 21 /r .... (r 1)1 /r, I. If the string is plucked 
with the finger lightly resting on the point l/r it will be 
found that this mode of vibration is excited. With even 
order vibrations (r even) the mid- point is a node, and with 
odd order vibrations it is an antinode. 

We can find the energy associated with this mode of 
vibration most conveniently by rewriting (23) in the form 

. . TTTX . ,__ x 

y = A sin cos 4 + e \ . . (29) 
/ { I j 

Here A is the amplitude and e is the phase. According 
to (5) the kinetic energy is 

T = P f dx = A* sin* + .. . (30) 

Similarly, by (6) the potential energy is 

V* f r&)*te^A^ + <. (31) 

2 J o\dxj 4Z \ I j 

Now by (3) F/p = c 2 , and so the two coefficients in (30) 
and (31) are equal. The total energy of this vibration 
is therefore 


The total energy is thus proportional to the square of the 
amplitude and also to the square of the frequency. This 
is a result that we shall often find as we investigate various 
types of wave motion. 

As a rule, however, there are several normal modes 
present at the same time, and we can then write the dis- 
placement (24) in the more convenient form 

* . TTTX (met { 
y = 2M r sin cos^ - + r \. . (33 
r=i I \ I } 

A r is the amplitude, and c r is the phase, of the rth 
normal mode. When we evaluate the kinetic energy 
as in (30) we find that the " cross- terms " vanish, since 

sin sin -j- dx = 0, if r ^ s. Consequently the total 

I i 

kinetic energy is just 


J c 

and in a precisely similar way tho total potential energy is 

By addition we find that the total energy of vibration is 
^2>M r *. . . . (34) 

This formula is important. It shows that the total energy 
is merely the sum of the energies obtained separately for 
each normal mode. It is due to this simple fact, which 
arises because there are no cross-terms involving A r A s , 
that the separate modes of vibration are called normal 
modes. It should be observed that this result holds for 
both the kinetic and potential energies separately as well 
as for their sum. 


We have already seen that when a string vibrates more 
than one mode is usually excited. The lowest frequency, 
viz. c/2Z, is called the ground note, or fundamental, and 
the others, with frequencies rc/2Z, are harmonics or over- 
tones. The frequency of the fundamental varies directly 
as the square root of the tension and inversely as the 
length and square root of the density. This is known as 
Mersenne's law. The tone, or quality, of a vibration is 
governed by the proportion of energy in each of the 
harmonics, and it is this that is characteristic of each 
musical instrument. The tone must be carefully distin- 
guished from the pitch, which is merely the frequency of 
the fundamental. 

We can use the results of (34) to determine the total 
energy in each normal mode of the vibrating string which 
we discussed in 20. According to (28) and (33) A 2n = 0, 

A A 8A 1 . (271 + 1)77 ^ ^ ^. 

and A 2n+l = sin . Consequently, the 

7T (^72~p~JL) Zi 

total energy of the even modes is zero, and the energy 
of the (2n+l)th mode is 16c 2 /^>/(2^+l) 2 7r 2 /. Tlu's shows 
us that the main part of the energy is associated with the 
normal modes of low order. We can check these formulae 
for the energies in this example quite easily. For the total 
energy of the whole vibration is the sum of the energies of 
each normal mode separately : i.e. 

total energy = -~- 2 

It is shown in books on algebra that the sum of the series 
l/P+l/3 2 +l/5 2 + ... is77 2 /8. Hence the total energy is 
2c% 2 p/, i.e. 2Fh 2 /l. But the string was drawn aside and 
released from rest in the position of fig. 4, and at that 
moment the whole energy was in the form of potential 
energy. This potential energy is just F times the increase 
in length, i.e. 2jF{(Z 2 /4+/& 2 ) 1 / 2 -Z/2}. A simple calculation 
shows that if we neglect powers of h above the second, 


as we have already done in our formulation of the equation 
of wave motion, this becomes 2Fh 2 /l, thus verifying our 
earlier result. 

This particular example corresponds quite closely to 
the case of a violin string bowed at its mid-point. A listener 
would thus hear not only the fundamental, but also a 
variety of other frequencies, simply related to the funda- 
mental numerically. This would not therefore be a pure 
note, though the small amount of the higher harmonics 
makes it much purer than that of many musical instru- 
ments, particularly a piano. 

If the string had been bowed at some other point than 
its centre, the partial amplitudes would have been different, 
and thus the tone would be changed. By choosing the 
point properly any desired harmonic may be emphasised 
or diminished, a fact well known to musicians. 

22. We have seen in 21 that it is most convenient 
to analyse the motion of a string of finite length in terms 
of its normal modes. According to (33) the rth mode is 

. . TTTX (met , ) 
y r = A r sin cos j + e r k 
i \ i } 

We often write this 

y r = <f> r sin ... (35) 

The expressions (/> r are known as the normal coordinates 
for the string. There are an infinite number of these 
coordinates, since there are an infinite number of degrees 
of freedom in a vibrating string. The advantage of using 
these coordinates can be seen from (30) and (31) ; if the 
displacement of the string is 

y - S<{> r sin . . . (36) 

r-l * 







The reason why we call (f> r a normal coordinate is 
now clear ; for in .mechanics the normal coordinates 
(?i> #2 $n are suitable combinations of the original 
variables so that the kinetic and potential energies can be 
written in the form 

V - 


The similarity between (37) and (38) is obvious. Further, 
it can be shown, though we shall not reproduce the analysis 
here, that Lagrange's equations of motion apply with the 
set of coordinates <^ r in just the same way as with the 
coordinates q r in ordinary mechanics. 

23. We shall next discuss the normal modes of a 
string of length I when a mass M is tied to its mid-point 
(fig. 5). Now we have already seen in 21 that in the 

FIG. 5 

normal vibrations of an unloaded string the normal modes 
of even order have a node at the mid-point. In such a 


vibration there is no motion at this point, and it is clearly 
irrelevant whether there is or is not a mass concentrated 
there. Accordingly, the normal modes of even order are 
unaffected by the presence of the mass, and our discussion 
will apply to the odd normal modes. 

Just as in the calculations*' of 16, 17, in which there 
was a discontinuity in the string, we shall have two 
separate expressions y l and y 2 valid in the regions 0^x^.1/2 
and 1/2^x^1. It is obvious that the two expressions 
must be such that y is symmetrical about the mid-point 
of the string. y l must vanish at x and y 2 at x = I. 
Consequently, we may try the solutions 

y^ = a sin px cos (cpt-\-e) 

y 2 = a sin p(lx) cos (cpt~\-) . . (39) 

We have already satisfied the boundary condition y l = y 2 
at x = 1/2. There is still the other boundary condition 
which arises from the motion of M. Just as in (18) we 
may write this 

Substituting the values of y and y 2 as given by (39) and 
using the relation F c 2 p, we find 

pi pi pi 

The quantity pl/2 is therefore any one of the roots of the 
equation x tan x = pl/M. If we draw the curves y = tan x, 
y = pl/Mx, we can see that these roots lie in the regions 
to 7T/2, TT to 3?7/2, 2?r to 57T/2, etc. If we call the roots 
x l9 x 2 ... then the frequencies cp/27T become cx r /7rl. If M 
is zero so that the string is unloaded, x r = (r+l/2)7r, 
so the presence of M has the effect of decreasing the 
frequencies of odd order. 

If we write n for the frequency of a normal mode, 
then, since n = cpj27r, it follows that (40) can be written 


in the form of an equation to determine n directly ; viz., 
x tan x = pljM, where x = (irl^n . . (41) 

This equation is called the period equation. Its solutions 
are the various permitted frequencies (and hence periods) 
of the normal modes. Period equations occur very fre- 
quently, especially when we have stationary type waves, 
and we shall often meet them in later chapters. This 
particular period equation is a transcendental equation 
with an infinite number of roots, ^j 

24. In the previous paragraphs we have assumed that 
there was no frictional resistance, so that the vibrations were 
undamped. In practice, however, the air does provide 
a resistance to motion ; this is roughly proportional to the 
velocity. Let us therefore discuss the motion of a string 
of length I fixed at its ends but subject to a resistance 
proportional to the velocity. The fundamental equation 
of wave motion (3) has to be supplemented by a term in 

and it becomes 

ex* c* 

A solution by the method of separation of variables (cf . 9) 
is easily made, and we find 

Since y is to vanish at tho two ends, we must have, as 
before, sin pi = 0, and hence p = TTT/I, r = 1, 2, 3 .... 
The normal modes of vibration are therefore 

y = A r e~ w sin -y- cos (_#+e r ) . . (43) 




The exponential term erW represents a decaying amplitude 
with modulus (see 9) equal to 2/k. The frequency 
q/27T is slightly less than when there is no frictional resist- 
ance. However, Jc is usually small, so that this decrease 
in frequency is often so small that it may be neglected. 

25. There is another interesting method of obtaining 
the velocity of propagation of waves along a string, which 
we shall now describe and which is known as the method 
of reduction to a steady wave. Suppose that a wave is 
moving from left to right in fig. 6 with velocity c. Then, 

FIG. 6 

if we superimpose on the whole motion a uniform velocity 
c the wave profile itself will be reduced to rest, and 
the string will everywhere be moving with velocity c, 
keeping all the time to a fixed curve (the wave profile). 
We are thus led to a different problem from our original 
one ; for now the string is moving and the wave profile 
is at rest, whereas originally the wave profile was moving 
and the string as a whole was at rest. Consider the motion 
of the small element PQ of length ds situated at the top 
of the hump of a wave. If R is the radius of curvature at 
this top point, and we suppose, as in 13, that the string 
is almost inextensible, then the acceleration of the element 
PQ is c 2 /R downwards. Consequently, the forces acting on 
it must reduce to (c*/R) pds. But these forces are merely the 
two tensions F at P and Q, and just as in 13 (especially 


note at foot of p. 23), they give a resultant Fds/R downwards. 
Equating the two expressions, we have 

This is, naturally, the same result as found before. The 
disadvantage of this method is that it does not describe 
in detail the propagation of the wave, nor does it deal 
with stationary waves, so that we cannot use it to get 
the equation of wave motion, etc. It is, however, very 
useful if we are only concerned with the wave velocity, 
and we shall see later that this simple artifice of reducing the 
wave to rest can be used in other problems as well. 

26. Examples 

( 1 ) Find the velocity of waves along a string whoso density 
is 4 gms. per cm. when stretched to a tension 90000 dynes. 

(2) A string of unlimited length is pulled into a harmonic 
shape y = a cos kx, and at time t = it is released. Show 
that if F is the tension and p the density of the string, its 
shape at any subsequent time t is y ~ a cos kx cos kct, where 
c 2 = F/p. Find the mean kinetic and potential energies per 
unit length of string. 

(3) Find the reflect ion coefficient for two strings which 
are joined together and whose densities are 25 gms. per cm. 
and 9 gms. per cm. 

(4) An infinite string lies along the x axis. At t = that 
part of it between x = a i s given a transverse velocity 
a z x 2 . Describe, with the help of equation (9) the subsequent 
motion of the string, the velocity of wave motion being c. 

(5) Investigate the same problem as in question (4) except 
that the string is finite and of length 2a, fastened at the 
points x = ia. 

(6) What is the total energy of the various normal modes 
in question (5) ? Verify, by smnmatioii over all the normal 
modes, that this is equal to the initial kinetic energy. 

(7) The two ends of a uniform stretched string are fastened 
to light rings that can slide freely on two fixed parallel wires 
a distance I apart. Find the normal modes of vibration. 


(8) A uniform string of length 3Z fastened at its ends, is 
plucked a distance a at a point of trisection. It is then 
released from rest. Find the energy in each of the normal 
modes and verify that the sum is indeed equal to the work 
done in plucking the string originally. 

(9) Discuss fully the period equation (41) in 23. Show in 
particular that successive values of x approximate to rrr, 
and that a closer approximation is x = TTT -f- pl/Mrn. 

(10) Show that the total energy of vibration (43) is 
%plA r *e- kt {q* + kq cos (qt+c r ) sin (#+ f ,)+Jfca cos* (qt+c r )} 9 

and hence prove that the rate of dissipation of energy is 
%kplA r 2 e~ kt {2q sin (qt+c r )+k cos (qt+ r )}*. 

(11) Two uniform wires of densities p L and p a and of equal 
length are fastened together at one end and the other two ends 
are tied to two fixed points a distance 21 apart. The tension 
is F. Find the normal periods of vibration. 

(12) The density of a stretched string is m/x*. The end- 
points are at x = a, 2a, and the tension is F. Show that the 
normal vibrations are given by the expression 

y=A sin [6 log, WY'" * pt ' 

' ' 

Show that the period equation is 6 log e 2 = nir, n ~ 1, 2, .... 

(13) A heavy uniform chain of length I hangs freely from 
one end, and performs small lateral vibrations. Show that 
the normal vibrations are given by the expression 

y = A J (2p\/{x/g}) cos (pt+e), 
where J represents Bessel's function ( 7) of order zero. 

Deduce that the period equation is J^(2p^/{l/g}) = 0, x 
being measured from the lower end. 


1. 150 cms./sec. ; 2. %Fa 2 k 2 siu*kct, %Fa 2 k* cos 2 kct ; 3. 1/4; 

5. y = Sb r cos ^i^ sin ( I*>^, b r = (- 

Ct d 


6. 8pa 5 /15; 7. y~a r cos cos I + r l ; 8. energy in rth 
normal mode = sin 2 - ; sum = 3c 2 a 2 />/4Z ; 11. 2n/p 

4t7T T O 

where c 1 ta>n(pl/c l ) = c 2 tan(pZ/c a ), c 1 2 =j^ T /p 1 , c a 2 = F/ p 2 -] 



27. The vibrations of a plane membrane stretched to a 
uniform tension T may be discussed in a manner very 
similar to that which we have used in Chapter II for 
strings. When we say that the tension is T we mean that 
if a line of unit length is drawn in the surface of the 
membrane, then the material on one side of this line 



exerts a force T on the material on the other side, and this 
force is perpendicular to the line we have drawn. Let us 
consider the vibrations of such a membrane ; we shall 
suppose that its thickness may be neglected. If its 
equilibrium position is taken as the xy plane, then we are 
concerned with displacements z(xy) perpendicular to this 
plane. Consider a small rectangular element ABCD 
(fig. 7) of sides Sx, y. When this is vibrating the forces 



on it are (a) two forces T$x perpendicular to AB and CD, 
and (b) two forces "TSy perpendicular to AD and BC. 
These four forces act in the four tangent planes through 
the edges of the element. An argument precisely similar 
to that used in Chapter IT, 13, shows that the forces (a) 

d 2 z 
give a resultant TSo: . 8y perpendicular to the plate. 

Similarly, the forces (b) reduce to a force TSy . $x. Let 


the mass of the plate be p per unit area ; then, neglecting 
gravity, its equation of motion is 

c) 2 z r) 2 z f) z 

T SxSy+J SxSy = pSxSy , 
dy 2 dx 2 r dt 2 

(8*z <Pz\ __ 8% 
[dx 2 'dy 2 ) ~~ P ~dt 2 ' 

This may be put in the standard form 

L ^ n \ 

where c 2 = "T/p .... (2) 

Thus we have reduced our problem to the solution of the 
standard equation of wave motion, and shown that the 
velocity of waves along such membranes is c \S(l"/p). 

28. Let us apply these equations to a discussion of the 
transverse vibrations of a rectangular membrane ABCD 
(fig. 8) of sides a and 6. Take AB and AD as axes of x 
and y. Then we have to solve (1) subject to certain 
boundary conditions. These are that z = at the boundary 
of the membrane, for all t. With our problem this means 
that z = when x = 0, x = a, y = 0, y = 6, independent 



of the time. The most suitable solution of the equation 
of wave motion is that of 8, equation (29). It is 

cos cos cos , 
z = . px . qy . ret , 
sin sin sin 

If z is to vanish at x = 0, y = 0, we shall have to reject 
the cosines in the first two factors. Further, if z vanishes 

FIG. 8 

at x = a, then sin pa 0, so that p = m7r/a, and similarly 
q = nir/b, m and n being positive integers. Thus the 
normal modes of vibration may be written 

. . mrrx . mry 

z = A sin sin - cos 

a b 



We may call this the (m, n) normal mode. Its frequency 
is cr/27r, i.e. 

6 2 


If &>&, the fundamental vibration is the (1, 0) mode, for 
which the frequency is c/2a. The overtones (4) are not 
related in any simple numerical way to the fundamental, 
and for this reason the sound of a vibrating plate, in 
which as a rule several modes are excited together, is 
much less musical to the ear than a string, where the 
harmonics are all simply related to the fundamental. 



In the (m, n) mode of (3) there are nodal lines x = 0, 

a/m, 2a/m, a, and y = 0, 6/w, 26/w, ... 6. On opposite 

sides of any nodal line the displacement has opposite sign. 
A few normal modes are shown in fig. 9, in which the 
shaded parts are displaced oppositely to the unshaded. 





FIG. 9 


The complete solution is the sum of any number of 
terms siich as (3), with the constants chosen to give any 
assigned shape when t 0. The method of choosing 
these constants is very similar to that of 19, except that 
there are now two variables x and y instead of one, and 
consequently we have double integrations corresponding 
to (27). 

According to (4) the frequencies of vibration depend 
upon the two variables m and n. As a result it may 
happen that there are several different modes having the 
same frequency. Thus, for a square plate, the (4, 7), 
(7, 4), (1, 8) and (8, 1) modes have the same frequency ; 
and for a plate for which a = 36, the (3, 3) and (9, 1) 
modes have the same frequency. When we have two or 
more modes with the same frequency, we call it a 
degenerate case. It is clear that any linear combination 
of these modes gives another vibration with the same 


29. We can introduce normal coordinates as in the 
case of a vibrating string (cf. 22). According to (3) 
the full expression for z is 

A / j , v nir y /K \ 

z = 27 A mn cos (rct-\~ r ) sin sm - . (5) 

m,n a b 

We write this 

'. . WITX . ft7n/ 

^ 270^ sm sin , . . (6) 

m t n a u 

where </> mn are the normal coordinates. The kinetic 
energy is 

. . . (7) 

and this is easily shown to be 

T - 2 \ patymn ... (8) 

m, n 8 

The potential energy may be calculated in a manner 
similar to 14. Referring to fig. 7 we see that in the 
displacement to the bent position, the two tensions T8y 
have done work T8y . (arcAB 8x). As in 14, this 

reduces to approximately - T I I 8x8y. The other two 

2 \dxj 

tensions TSo; have done work - T I ~- I 8x8 u. The total 

2 \dyj 
potential energy is therefore 

In the case of the rectangular membrane this reduces to 

F = Z \ p6cVVl (10) 


It will be seen that T and F are both expressed in the form 
of Chapter II, equation (38), typical of normal coordinates 
in mechanical problems. 


30. With a circular membrane such as a drum of 
radius a, we have to use plane polar coordinates r,9, 
instead of Cartesians, and the solution of equation (1), 
apart from an arbitrary amplitude, is given in 8, 
equation (35a). It is 


z = J m (nr) g - n m9 cos net. 

We have neglected the Y m (nr) term since this is not finite 
at r 0. If we choose the origin of 6 properly, this normal 
mode may be written 

z == J m (nr) cos m9 cos net. . . (11) 

If z is to be single-valued, m must be a positive integer. 
The boundary condition at r = a is that for all values of 
6 and t, J m (na) cos m9 cos net equals zero, i.e., J m (na) 0. 
For any assigned value of m this equation has an infinite 
number of real roots, each one of which determines a 
corresponding value of n. These roots may be found 
from tables of Bessel functions. If we call them n m , ly 
n m, 2> n m, fc> > then the frequency of (11) is nc/27r, 
i.e. cn m ^ fc/27r, and we may call it the (m, k) mode. The 
allowed values of m are 0, 1, 2, ... and of k are 1, 2, 3, ... . 
There are nodal lines which consist of circles and radii 
vectores. Fig. 10 shows a few of these modes of vibration, 
shaded parts being displaced in an opposite direction to 

The nodal lines obtained in figs. 9 and 10 are known 
as Chladni's figures. A full solution of a vibrating mem- 
brane is obtained by superposing any number of these 
normal modes, and if nodal lines exist at all, they will 
not usually be of the simple patterns shown in these 
figures. As in the case of the rectangular membrane so 
also in the case of the circular membrane, the overtones 
bear no simple numerical relation to the fundamental 
frequency, and thus the sound of a drum is not very 



musical. A vibrating bell, however, is of very similar 
type, but it can be shown,* that some of the more important 
overtones bear a simple numerical relation to the funda- 
mental ; this would explain the pleasant sound of a well- 
constructed bell. But it is a little difficult to see why 
the ear so readily rejects some of the other overtones 



whose frequencies are not simply related to the fundamental. 
A possible explanation ) is that the mode of striking may 
be in some degree unfavourable to these discordant 
frequencies. In any case, we can easily understand why 
a bell whose shape differs slightly from the conventional, 
will usually sound unpleasant. 

* See Slater and Frank, Introduction to Theoretical Physics, 
1933, p. 161. 

t Lamb, Dynamical Theory of Sound (Arnold), 1910, p. 155. 


31. Examples 

(1) Find two normal modes which are degenerate (28) 
for a rectangular membrane of sides 6 and 3. 

(2) Obtain expressions for the kinetic and potential 
energies of a vibrating circular membrane. Perform the in- 
tegrations over the ^-coordinate for the case of the normal mode 

z = A J m (nr) cos m6 cos net. 

(3) A rectangular drum is 10 cm. X 20 cm. It is stretched 
to a tension of 5 kgm., and its mass is 20 gm. What is the 
fundamental frequency ? 

(4) A square membrane bounded by x = 0, a and y = 0, a 

is distorted into the shape z = A sin - sin - and then 

a a 

released. What is the resulting motion ? 

(5) A rectangular membrane of sides a and b is stretched 
unevenly so that the tension in the x direction is Tj_ and in 
the y direction is T 2 . Show that the equation of motion is 

d 2 z 8 2 z d 2 z 

T-i +T 2 = p -. Show that this can be brought into 
dx 2 dy 2 dt 2 

the standard form by changing to now variables x/ VT~ , 
2//VT" ' anc ^ nence fiftd. the normal modes. 

(6) Show that the number of normal modes for the 
rectangular membrane of 28 whose frequency is less than N is 
approximately equal to the area of a quadrant of the ellipse 

vj2 /i2 A * 

-- 1 -- = =r N 2 . Hence show that the number is roughly 
a 2 b 2 T 

[ANSWERS: 1. (2, 0) and (0, 1): in general (2m, n) and 

(2n, m) ; 2. T = J7rpn 2 c 2 ^ 2 sin 2 net f {J m (nr)}*rdr, 

J o 

V = %7T P 2 2 A 2 cos 2 net (* [n 2 {J rn / (nr)} 2 +m 2 { J m(^)} 2 / r2 ] r dr > 
which becomes, after integration by parts 

V = %7rpn 2 c 2 A 2 cos 2 nctj a {J m (nr)} 2 rdr; 3. 175-1 ; 

4. z = A sin (27rx/a) sin (^nyja) cos (^13-rTCt/a) ; 

5. z A sin (mnx/a) sin (rnry/b) cos npt, 



32. The vibrations which we have so far considered have 
all been transverse, so that the displacement has been 
perpendicular to the direction of wave propagation. We 
must now consider longitudinal waves, in which the 
displacement is in the same direction as the wave. Sup- 
pose that AB (fig. 11) is a bar of uniform section and 

P 1 Q 1 

FIG. 11 

mass p per unit length. The passage of a longitudinal 
wave along the bar will be represented by the vibrations 
of each element along the rod, instead of perpendicular 
to it. Consider a small element PQ of length Sx, such 
that AP = x, and let us calculate the forces on this element, 
and hence its equation of motion, when it is dis- 
placed to a new position P'Q' '. If the displacement of 
P to P' is f , then that of Q to Q' will be f +8f , so that 
P'Q' = 8x +8. We must first evaluate the tension at 
P' . We can do this by imagining Sx to shrink to zero. 



Then the infinitesimally small element around P' will be 
in a state of tension T where, by Hooke's Law, 

Tp> A . 

orig. length 
= A Lim - 

= Ag (1) 

Returning to the element P'Q', we see that its mass is the 
same as that of PQ, i.e. px, and its acceleration is ~. 



~*~'W ^ T ^'~ T ^' 

Thus the equation of motion for these longitudinal waves 
reduces to the usual equation of wave motion 

The velocity of waves along a rod is therefore \/(A//>), 
a result similar in form to the velocity of transverse 
oscillations of a string. 

The full solution of (2) is soon found if we know the 
boundary conditions. 

(i) At a free end the tension must vanish, and thus, 


from (1), = 0, but the displacement will not, in general, 


vanish as well. 

(ii) At a fixed end the displacement must vanish, 
but the tension will not, in general, vanish also. 


33. If wo are interested in the free vibrations of a 
bar of length Z, we shall use stationary type solutions of 
(2) as in 8, equation (27). Thus 

= (a cos px-\-b sin px) cos {cpt-^-e}. 

If we take the origin at one end, then by (i) fig/dx has to 
vanish at x = and x = I. This means that 6 = 0, and 
sin pi = 0. i.e. pi mr, where n = 1, 2, ... . The free 
modes are therefore described by the functions 

co l~T~ +e 4 ' * (3) 

This normal mode has frequency nc/2Z, so that the funda- 
mental frequency is c/2Z, and the harmonics are simply 
related to it. There are nodes in (3) at the points x = l/2n, 
3l/2n, 5l/2n, .... (2nl)l/2n ; and there are antinodes ( 6) 
at x = 0, 2l/2n, 4Z/2/1 .... Z. From (1) it follows that 
these positions are interchanged for the tension, nodes of 
motion being antinodes of tension and vice versa. We 
shall meet this phenomenon again in Chapter VI. 

34. The case of a rod rigidly clamped at its two ends 
is similarly solved. The boundary conditions are now 
that | = at x 0, and at x = Z. The appropriate 
solution of (2) is thus 

. mrx (mrct } 

,sm - cos \ - - +}-. . . (4) 

This solution has the same form as that found in Chapter II, 
19, for the transverse vibrations of a string. 

35. We may introduce normal coordinates for these 
vibrations, just as in 22 and 29. Taking, for 
example, the case of 34, we should write 

, , . nm 

g = S n sin , . . . (5) 

w = l l 

where . (nrrct , 

tn == a n COS -j + 


The kinetic energy of the element PQ is |pS# . | 2 , so 
that the total kinetic energy is 



... (6) 

1 o n 

The potential energy stored up in P'Q' is approximately 
equal to one-half of the tension multiplied by the increase 

1 P 

in length : i.e. - A . S. Thus the total potential energy is 

36. The results of 33, 34 for longitudinal vibrations 
of a bar need slight revision if the bar is initially in a state 
of tension. We shall discuss the vibrations of a bar of 
natural length Z stretched to a length Z, so that its equili- 
brium tension T is , , 

T =A J -p .... (8) 


Referring to fig. 11, the unstretched length of P'Q' is not 
Sx but -2$x, so that the tension at P f is not given by (1), 


but by the modified relation 

8x+$- l j Sx 
J P . = A Lim - - - 

^To+' using (8). . (9) 

The mass of PQ is p(l /l)8x where p refers to the unstretched 
bar, so the equation of motion is 


We have again arrived at the standard equation of wave 

It follows that c = (Z// )c , where c is the velocity under 
no permanent tension. Appropriate solutions of (10) are 
soon seen to be 

.. . UTTX (nrrct } , rt 

| = a n sin -y- cos J -r- + e w }- , tt=l,2 . . . (11) 
I ^ J 

The fundamental frequency is c/2l, which, from (10), can 
be written c /2 . Thus with a given bar, the frequency is 
independent of the amount of stretching. 

The normal mode (11) has nodes where x = 0, l/n, 
2l/n, . . ,1. A complete solution of (10) is obtained by 
superposition of separate solutions of type (11). 

37. We shall conclude this chapter with a discussion 
of the vibrations of a spring suspended from its top end 
and carrying a load M at its bottom end. When we 
neglect the mass of the spring it is easy to show that 
the lower mass M (fig. 12) executes Simple Harmonic 
Motion in a vertical line. Let us, however, consider the 
possible vibrations when we allow for the mass m of the 
spring. Put m = pi, where p is the unstretched mass per 
unit length and I is the unstretched length. We may 
consider the spring in three stages. In stage (a) we have 
the unstretched spring of length 1. The element PP' of 
length Sx is at a distance x from the top point A. In stage 
(6) we have the equilibrium position when the spring is 
stretched due to its own weight and the load at the bottom. 
The element PP' is now displaced to QQ'. P is displaced 
a distance X downwards and P' a distance X+SX. Lastly, 
in stage (c) we suppose that the spring is vibrating anc 
the element QQ' is displaced to RR'. The displace! 
of Q and Q' from their equilibrium positions are f ar 



The new length EE' is therefore 8x +SJ?+S. The mass of 
the element is the same as the mass of PP' } viz. p8x, and 
is of course the same in all three stages. 

We are now in a position to determine the equation 
of motion of EE' . The forces acting on it are its weight 

A 1 





FIG. 12 


downwards and the two tensions at E and E '. The 
tension TR may be found from Hooke's Law, by assuming 
that 8x is made infinitesimally small. Then, as in 36, 

= A. 


orig. length 

A Lim 



So the equation of motion of RR' is 

pSx - = resultant force downwards 

r ' 

s , * 
== gp + dx 

Dividing by p&x and using (12), this becomes 

a 2 g A 

This last equation must be satisfied by 0, since this is 
merely the position of equilibrium (6) . So 


= 9 + 

By subtraction we discover once more the standard equation 
of wave motion 

x c pm 

This result is very similar to that of 36. However, 
before we can solve (13) we must discuss the boundary 
conditions. There are two of these. Firstly, when x = 0, 
we must have = for all t. Secondly, when x = I, 
(i.e. the position of the mass M ) we must satisfy the law 
of motion 

Using (12), this becomes 


As before, this equation must be satisfied by = 0, since 
this is just the equilibrium stage (6). Thus 

So, by subtraction we obtain the final form of the second 
boundary condition 



The appropriate solution of (13) is 

a sin px cos {pc+e}. . . (15) 
This gives = when # = 0, and therefore satisfies the 
first boundary condition. It also satisfies the other 
boundary condition (14) if 

plt&npl = mlM. . . . (16) 
By plotting the curves y = tan x, y = (m/M)/x, we see 
that there are solutions of (16) giving values of 2>Z in the 
ranges to vr/2, TT to 37T/2, .... The solutions become 
progressively nearer to nrr as n increases. 

We are generally interested in the fundamental, or 
lowest, frequency, since this represents the natural vibra- 
tions of M at the end of the spring. The harmonics 
represent standing waves in the spring itself, and may be 
excited by gently stroking the spring downwards when in 
stage (6). If m/M is small, the lowest root of (16) is 
small ; writing pi = z, we may expand tan z and get 

z(z+z 3 /3+...) =m/M. 

z 2 (l+z 2 /3) =m/M. 

We may put z 2 in the term in brackets equal to the first 
order approximation z 2 = m/M 9 and then we find for the 
second order approximation 

. 1+m/BM' 


The period of the lowest frequency in (15) is 2-jTJpc, i.e., 
27rl/cz. Using the fact that c 2 = A?/w, this becomes 

277 A/ 1 . If the mass of the spring m had been 

' A 

neglected we should have obtained the result 2?r\/(Of/A). 
It thus appears that the effect of the mass of the spring 
is equivalent, in a close approximation, to adding a mass 
one -third as great to the bottom of the spring. 

38. Examples 

(1) Find the velocity of longitudinal waves along a bar 
whose mass is 2-25 gms. per cm. and for which the modulus 
is 9-0 . 1C 10 dynes. 

(2) Two semi -infinite bars are joined to form an infinite 
rod. Their moduli are Aj and A 2 and the densities are p 
and p 2 . Investigate the reflection coefficient (see 16) and 
the phase change on reflection, when harmonic waves in the 
first medium meet the join of the bars. 

(3) Investigate the normal modes of a bar rigidly fastened 
at one end and free to move longitudinally at the other. 

(4) A uniform bar of length I is hanging freely from one 
end. Show that the frequencies of the normal longitudinal 
vibrations are (n-j-J) c/2Z, where c is the velocity of longi- 
tudinal waves in the bar. 

(5) The modulus of a spring is 7-2 . 10 3 dynes. Its mass 
is 10 gms. and its unstretched length is 12 cms. A mass 
40 gms. is hanging on the lowest point, and the top point is 
fixed. Calculate to an accuracy of 1 per cent, the periods of 
the lowest two vibrations. 

(6) Investigate the vertical vibrations of a spring of un- 
stretched length 21 and mass 2m, supported at its top end 
and carrying loads M at the mid-point and the bottom. 

[ANSWEKS : 1. 2 km. per sec. ; 2. B = 

/(r+i)wc \ 

( +4 : 6 " 

3. e = A r sin v ' ' cos | v ' *'" +e r l ; 5. 1-690 sees., 

0-252 sees. ; 6. Period = 2ir/nc where k 3 Skcotnl-}- cot 2 nl = 1, 
k = Mln/m.] 



39. In this chapter we shall discuss wave motion in 
liquids. We shall assume that the liquid is incompressible, 
with constant density p. This condition is very nearly 
satisfied by most liquids, and the case of a compressible 
fluid is dealt with in Chapter VI. We shall further assume 
that the motion is irrotational. This is equivalent to 
neglecting viscosity and assuming that all the motions 
have started from rest due to the influence of natural 
forces such as wind, gravity, or pressure of certain bound- 
aries. If the motion is irrotational, wo may assume 
the existence of a velocity potential <j> if we desire it. 
It will be convenient to summarise the formulae which 
we shall need in this work. 

(i) If the vector u j with components (u, v, w) J 
represents the velocity of any part of the fluid, then from 
the definition of (f> 

u = -\7< == grad <f>, . . (I) 

so that in particular u = ~8<f>/dx, v = d(f)/dy, 
w = fy/dz. 

(ii) On a fixed boundary the velocity has no normal 

* Before reading this chapter the student is advised to read 
Rutherford's Vector Methods, Chapter VI, from which several 
results will be quoted. 

f Using Clarendon type for vectors. 

J Many writers use (u x u y u z ) for the velocity components. We 
shall find (u, v, w) more convenient for our purposes. It is necessary, 
however, to distinguish u, which is a vector representing the velocity 
and u, which is just the x component of the velocity. 


component, and hence if 8/8 v denotes differentiation along 
the normal 

= ..... (2) 

(iii) Since no liquid will be supposed to be created or 
annihilated, the equation of continuity must express the 
conservation of mass ; it is 

v-"r + ? + 7r-<> (3) 

dx dy dz 
Combining (1) and (3), we obtain Laplace's equation 

- < 

(iv) If H(x, y, z, t) is any property of a particle of the 


fluid, such as its velocity, pressure or density, then 


is the variation of H at a particular point in space, and 


- is the variation of H when we keep to the same particle 


of fluid. is known as the total differential coefficient, 

and it can be shown * that 

DH dH , 

-fjr - ^T+ 

Dt dt ,-* 

DH OH dH , dH , dH ' ( } 

{ Q - -- L u - 4- v -- U w - 

Dt dt ^ dx 8y ^ 8z 

(v) If the external forces acting on unit mass of liquid 
can be represented by a vector F, then the equation of 
motion of the liquid may be expressed in vector form 

Du 1 

* See Rutherford, 66. 


In Cartesian form this is 

8u 8u , 3u 8u _ 1 dp /a . 

+ U+V--+W-- =F x ---f, . (6) 
dt dx % dz p ox 

with two similar equations for v and w. 

(vi) An important integral of the equations of motion 
can be found in cases where the external force F has a 
potential V, so that F = yF. The integral in question 
is known as Bernoulli's Equation : 

- + l* + v -jl = c > < 7 > 

p 2 dt 

where C is an arbitrary function of the time. Now 
according to (1), addition of a function of t to <j> does 
not affect the velocity distribution given by <f> ; it is often 

convenient, therefore, to absorb C into the term ~- and 


(7) can then be written 

A particular illustration of (8) which we shall require later 
occurs at the surface of water waves ; here the pressure 
must equal the atmospheric pressure and is hence constant. 
Thus at the surface of the waves (sometimes called the 
free surface) 

\ u 2 + V - ^ = constant. . . (9) 

40. We may divide the types of wave motion in 
liquids into two groups ; the one group has been called 
tidal waves,* and arises when the wavelength of the 
oscillations is much greater than the depth of the liquid. 
Another name for these waves is long waves in shallow 
water. With waves of this type the vertical acceleration 
* Lamb, Hydrodynamics, Chapter VIII. 


of the liquid is neglected in comparison with the horizontal 
acceleration, and we shall be able to show that liquid 
originally in a vertical plane remains in a vertical plane 
throughout the vibrations ; thus each plane of liquid 
moves as a whole. The second group may be called 
surface waves, and in these the disturbance does not 
extend far below the surface. The vertical acceleration 
is no longer negligible and the wavelength is much less 
than the depth of the liquid. To this group belong most 
wind waves and surface tension waves. We shall consider 
the two types separately, though it will be recognised that 
Tidal Waves represent an approximation and the results 
for these waves may often be obtained from the formulse 
of Surface Waves by introducing certain restrictions. 


41. We shall deal with Tidal Waves first. Here we 
assume that the vertical accelerations may be neglected. 
One important result follows immediately. If we draw 
the z axis vertically upwards (as we shall continue to do 
throughout this chapter), then the equation of motion in 
the z direction as given by (6), is 

Dw 1 dp 

___ = __0____. 

We are to neglect and thus 


= gp, i.e. p = gpz -(- constant. 


Let us take our xy plane in the undisturbed free surface, 
and write (#, y, t) for the elevation of the water above 
the point (x, y, 0). Then, if the atmospheric pressure is 
p Q , we must have p = p Q when z = . So the equation 
for the pressure becomes 

)- (10) 


We can put this value of p into the two equations of 
horizontal motion, and we obtain 

Du dt> Dv d 

~Dt~~ ~ g dx y ~Dt = ~ g dy 


The right-hand sides of these equations are independent 
of z, and we deduce therefore that in this type of motion 
the horizontal acceleration is the same at all depths. 
Consequently, as we stated earlier without proof, on still 
water the velocity does not vary with the depth, and 
the liquid moves as a whole, in such a way that particles 
originally in a vertical plane, remain so, although this 
vertical plane may move as a whole. 

42. Let us now apply the results of the last section 
to discuss tidal waves along a straight horizontal channel 
whose depth is constant, but whose cross-section A varies 


FIG. 13 

from place to place. We shall suppose that the waves 
move in the x direction only (extension to two dimensions 
will come later). Consider the liquid in a small volume 
(fig. 13) bounded by the vertical planes x, x~\-dx at P 
and Q. The liquid in the vertical plane through P is all 
moving with the same horizontal velocity u(x) independent 
of the depth. We can suppose that A varies sufficiently 
slowly for us to neglect motion in the y direction. We 


have two equations with which to obtain the details of 
the motion. The first is (11) and may be written 

du , du du dt, 

+ u +w = -0^-. 
dt dx 8z dx 

Since u is independent of 2, = 0. Further, since we 


shall suppose that the velocity of any element of fluid 

is small, we may neglect u which is of the second order, 


and rewrite this equation 

dU --a 8 ^ (12) 

8* ~ 9 te (") 

The second equation is the equation of continuity. Equa- 
tion (3) is not convenient for this problem, but a suitable 
equation can be found by considering the volume of liquid 
between the planes at P and Q, in fig. 13. Let b(x) be the 
breadth of the water surface at P. Then the area of the 
plane P which is covered with water is [A -\~b] P ; therefore 
the amount of liquid flowing into the volume per unit 
time is [(A~}-b)u]p. Similarly, the amount flowing out 
per unit time at Q is [(A +&)^]Q. The difference between 
these is compensated by the rate at which the level is 
rising inside the volume, and thus 


Since bQu, is of the second order of small quantities, we may 
neglect this term and the equation of continuity becomes 


Eliminating u between (12) and (13) gives us the equation 

In the case in which A is constant, this reduces to the 
standard form 

This is the familiar equation of wave motion in one 
dimension, and we deduce that waves travel with velocity 
^/(Ag/b). If the cross-section of the channel is rectangular, 
so that A = bh, h being the depth, 

c=V(fl*) .... (16) 
With an unlimited channel, there are no boundary 

conditions involving #, and to our degree of approximation 

waves with any profile will travel in either direction. 

With a limited channel, there will be boundary conditions. 

Thus, if the ends are vertical, u = at each of them. 

We may apply this to a rectangular basin of length I, 

whose two ends are at x = 0, 1. Possible solutions of (15) 

are given in 8, equation (27). They are 

= (a cos px+f3 sin px) cos (cpt+e). 

Then, using (13) and also the fact that A = bh, we find 

8u cp ~ 

= -=- (a Gospx+p sinpx) sin (cpt+e). 

OX fl 


u - (a sin pxj3 cos px) sin (cpt+c). 

The boundary conditions u = at x = 0, Z, imply that 
j8 = 0, and sin pi = 0. So 

, r == 1,2,3,... (17) 


It will be noticed that nodes of u and do not occur at 
the same points. 

The vertical velocity may be found from the general 
form of the equation of continuity (3). Applied to our 
case, this is 

8u 8w _ 

T; h ~T~ == 0- 
- ox GZ 

Now u is independent of z and w = on the bottom of 
the liquid where z h. Consequently, on integrating 


7rra r c TTTX . /ryrc^ \ 

(z+A cos sin I -7- +* r f 

Ih I \ I ) 

We may use this last equation to deduce under what 
conditions our original assumption that the vertical 
acceleration could be neglected, is valid. For the vertical 

Dw . dw . 

acceleration is effectively , i.e. 

The maximum value of this is TrWaJZ 2 , and may be com- 
pared with the maximum horizontal acceleration Trrc 2 a r /lh. 
The ratio of the two is rirh/1, i.e. 277-A/A, since, from (17) 
A = 2l/r. We have therefore confirmed the condition 
which we stated as typical of these long waves, viz. that 
the vertical acceleration may be neglected if the wavelength 
is much greater than the depth of water. 

43. We shall now remove the restriction imposed in 
the last section to waves in one dimension. Let us use 
the same axes as before and consider the rate of flow of 
liquid into a vertical prism bounded by the planes 
x, x-\-dx, y, y+dy. In fig. 14, ABCD is the undisturbed 
surface, EFGH is the bottom of the liquid, and PQES is 
the moving surface at height (x } y) above ABCD. The 



rate of flow into the prism across the face PEHS is 
[u(h+)dy] a , and the rate of flow out across RQFO is 
The net result from these two planes is 

FIG. 14 


a gain {ufy-^Qftdxdy. Similarly, from the other two 



vertical planes there is a gain - {v(h-{-^)}dxdy. The 

total gain is balanced by the rising of the level inside the 
prism, and thus 

- ~{u(h+)}dxdy - j^{v(h+)}dxdy - ^ . dxdy. 

As in 42, we may neglect terms such as ut, and vt, and 
write the above equation of continuity 

d(hu) d(hv) __ __a^ 
dx ~ + ~~dy ~8t ' 



We have to combine this equation with the two equations 
of motion (11), which yield, after neglecting square terms 
in the velocities 

8u dv d 

Eliminating u and v gives us the standard equation 


If h is constant (tank of constant depth) this becomes 

This is the usual equation of wave motion in two dimensions 
and shows that the velocity is \/(gh). If we are concerned 
with waves in one dimension, so that is independent of 

y (as in 42) we put = and retrieve (15). 

We have therefore to solve the equation of wave 
motion subject to the boundary conditions 

(i) w = at z = h, 

(ii) = at a boundary parallel to the y axis, and 


at a boundary parallel to the x axis, 

f\Y r\ 

(iii) at any fixed boundary, where denotes 
ov ov 

differentiation along the normal to the boundary. This 
latter condition, of which (ii) is a particular case, can be 
seen as follows. If Ix -{-my = 1 is the fixed boundary, 
then the component of the velocity perpendicular to this 
line has to vanish. That is, lu -\-rnv = 0. By differentiating 
partially with respect to t and using (21), the condition (iii) 
is obtained. 


44. We shall apply these formulae to two cases ; first, 
a rectangular tank, and, second, a circular one, both of 
constant depth. 

Rectangular tank. Let the sides be x = 0, a and y = 0, 6. 
Then a suitable solution of (23) satisfying all the boundary 
conditions (i) and (ii) would be 

= A cos cos cos (rTTCt+e), . (24) 
a b 

where p = 0, 1, 2 ... , gr - 0, 1, 2, ... , and r 2 = p*/a*+q*lb*. 

This solution closely resembles that fora vibrating membrane 
in Chapter III, 28, and the nodal lines are of the same 
general type. The student will recognise how closely the 
solution (24) resembles a " choppy sea." 

Circular tank. If the centre of the tank is origin and 
its radius is a, then the boundary condition (iii) reduces to 


at r = a. Suitable solutions of (23) in polar 


coordinates have been given in Chapter I, equation (35a). 
We have 

= A cos mO J m (nr) cos (cnt-\-e) . (25) 

We have rejected the Y m solution since it is infinite at 
r = 0, and we have chosen the zero of 6 so that there is 
no term in sin m9. This expression satisfies all the condi- 
tions except the boundary condition (iii) at r ~ a. This 
requires that J m '(na) 0. For a given value of m (which 
must be integral) this condition determines an infinite 
number of values of n, whose magnitudes may be found 
from tables of Bessel Functions. The nodal lines are 
concentric circles and radii from the origin, very similar 
to those in fig. 10 for a vibrating membrane. The period 
of this motion is 2?r/ 'en. \ 

45. It is possible to determine the actual paths of 
individual particles in, many of these problems. Thus, 


referring to the rectangular tank of 42, the velocities 
u and w are given by (18) and (19). We see that 

w ~Trr(z-\-h) TTTX 

= cot 7. 

u I I 

This quantity is independent of the time and thus any 
particle of the liquid executes simple harmonic motion 
along a line whose slope is given by the above value of 
w/u. For particles at a fixed depth, this direction changes 
from purely horizontal beneath the nodes to purely vertical 
beneath the antinodes. 

46. We shall conclude our discussion of tidal waves 
by applying the method of reduction to a steady wave, 
already described in 25, to the case of waves in a channel 
of constant cross-section A and breadth of water-line b. 
This is the problem of 42 with A constant. Let c be the 
velocity of propagation of a wave profile. Then super- 
impose a velocity ~c on the whole system, so that the 
wave profile becomes stationary and the liquid flows under 
it with mean velocity c. The actual velocity at any point 
will differ from c since the cross-sectional area of the liquid 
is not constant. This area is A-\-b,, and varies with . 
Let the velocity be c~\-6 at sections where the elevation 
is . Since no liquid is piling up, the volume of liquid 
crossing any plane perpendicular to the direction of flow 
is constant, i.e. 

(A +&) (c +0) = constant = Ac. . (26) 

Wo have still to use the fact that the pressure at the free 
surface is always atmospheric. In Bcrnouilli's equation 
at the free surface (9) we may put d</>/dt = since the 
motion is now steady motion ; also V = g at the free 
surface. So, neglecting squares of the vertical velocity, 
this gives 

= const. = |c 2 . 


Eliminating between this equation and (26), we have 



. _ . (27) 
l ' 

If is small, so that we may neglect compared with -4/6, 
then this equation gives the same result as (16), viz. 
c 2 = gA/b. We can, however, deduce more than this 
simple result. For if >0, the right-hand side of (27) is 
greater than gA/b, and if <0, it is less than gA/b. Thus 
an elevation travels slightly faster than a depression and 
so it is impossible for a long wave to be propagated 
without change of shape. Further, since the tops of waves 
travel faster than the troughs, we have an explanation of 
why waves break on the sea-shore when they reach shallow 


47. We now consider Surface Waves, in which the 
restriction is removed that the wavelength is much greater 
than the depth. In these waves the disturbance is only 
appreciable over a finite depth of the liquid. We shall 
solve this problem by means of the velocity potential </>. 
<f> must satisfy Laplace's equation (4) and at any fixed 
boundary d<f>/dv = 0, by (2). There are, however, two 
other conditions imposed on <j> at the free surface. The 
first arises from Bernoulli's equation (9). If the velocity 
is so small that u 2 may be neglected, and if the only 
external forces are the external pressure and gravity, we 


may put u 2 = and F = gr in this equation, which 

free surface 

The second condition can be seen as follows. A particle 
of fluid originally on the free surface will remain so always. 
Now the equation of the free surface, where z = (#, y, t) 
may be written 

= /(3, y, z, t) = (&, 2/, Q z. 

Consequently, / is a function which is always zero for a 
particle on the, free surface. We may therefore use (5) 
with H put equal to /, and we find 

A Df . , 0f 
^j^ == ^ + u ^ +v -^~ w ' 

Dt dt dx 8y 

Now from (28) - = - -- I - I = -- TT- on the surface. 
v } dx g di \dx) g 8i 


Thus ~- is a small quantity of order of magnitude not 

f\Y ?\Y 

greater than u\ consequently u and v -, being of 

ex cy 

order of magnitude not greater than u 2 , may be neglected. 
We are left with the new boundary condition 

% - w - - * (29) 

dt ~ W ~~ 8z (M) 

Combining (28) and (29) we obtain an alternative relation 

We summarise the conditions satisfied by <f> as follows : 
(i) Laplace's equation y 2 < = in the liquid . (2) 
(ii) d<f>l&v = on a fixed boundary ... (4) 


1 firh 

(iii) = - -? on the free surface . . . (28) 
(7 d 

oJ" o J 

(iv) = on the free surface . . . (29) 

vt dz 

( v ) .? + g, f!r = on the free surface . . (30) 

Gv 0% 

Only two of the last three conditions are independent. 

48. Let us apply these equations to the case of a 
liquid of depth h in an infinitely long rectangular tank, 
supposing that the motion takes place along the length 
of the tank, which we take as the x direction. The axes 
of x and y lie, as usual, in the undisturbed free surface. 
Condition (i) above gives an equation which may be 
solved by the method of separation of variables (see 7), 
and if we want our solution to represent a progressive 
wave with velocity c, a suitable form of the solution would 

<f> = (Ae mz + Be~ mz ) cos m(xct). 

A, B, m and c are to be determined from the other condi- 
tions (ii)-(v). At the bottom of the tank (ii) gives d</>/dz~0, 
i.e. Ae~ mh Be mh = 0. So Ae~ mh = Be mh = W, say, and 

(/) C cosh m(z-\-h) cos m(xct). . (31) 

Condition (v) applies at the free surface where, if the 
disturbance is not too large, we may put z = ; after 
some reduction it becomes 

c 2 = (g/m) tanh mh. 

Since m = 27T/A, where A is the wavelength, we can write 

. . . (32) 


Condition (iii) gives us the appropriate form of ; it is 

Y mcO . 

- cosh mft sin m(xct). 

This expression becomes more convenient if we write a 

for the amplitude of ; i.e., a = - cosh mh. Then 


= a sin m(xct), .... (33) 

gfl coshw(2+&) , .. /0 , x 

A = * -- ^ ! i cosm( c). . (34) 
me cosh mh 

If the water is very deep so that tanh (27rA/A) = 1, then 
(32) becomes c 2 0A/277, and if it is very shallow so 
that tanh (2?r^/A) = 27rh/X, we retrieve the formula of 42 
for long waves in shallow water, viz. c 2 = gh. 

We have seen in Chapter I that stationary waves result 
from superposition of two opposite progressive harmonic 
waves. Thus we could have stationary waves analogous 
to (33) and (34) defined by 

= a sin mx cos met, .... (35) 

. ga . . .__ 

6 = -- - - sin mx sin met. . (36) 
me cosh mh 

We could use these last two equations to discuss stationary 
waves in a rectangular tank of finite length. 

49. We shall now discuss surface waves in two dimen- 
sions, considering two cases in particular. 

Rectangular tank. With a rectangular tank bounded 
by the planes x = 0, a and y = 0, 6, it is easily verified 
that all the conditions of 47 are satisfied by 

PTTX qiry 

= A cos - cos ~- cos ret , 
a o 

. a A . 

A =- * -- v cos ^L_ cos 1^! sm 

rc cosh rA a b 



p = 1, 2, ... ; q = 1, 2, .... ; r 2 = 7r 2 (p 2 /a 2 + 2 /& 2 ) and 

c 2 = (gr/r) tanh r/L . . . (37) 

Circular tank. Suppose that the tank is of radius a 
and depth h. Then choosing the centre as origin and 
using cylindrical polar coordinates r, 0, z, Laplace's equation 
(cf. Chapter I, 7) becomes 

$ 2 cA 1 dJ> 1 ffij) d^d) 

' .. 1- ' | > [ r ___ f\ /QC\ 

A suitable solution can be found from Chapter I, equation 
(35a), which gives us a solution of the similar equation 

in the form 

. J m cos cos 
' ~ Y m ^ nr ' sin sin 

In this equation let us make a change of variable, writing 
ct = iz> where i 2 = 1. We then get Laplace's equation 
(38) and its solutions are therefore 

In our problem we must discard the Y solution as Y m (r) 
is infinite when r 0. So, choosing our zero of 6 suitably, 
we can write 

< = J m (nr) cos m9 (A cosh nz+B sinh nz). 
At the bottom of the tank condition (ii) gives, as in 48, 
A sinh rih = B cosh rih,, so that 

</) = (7 J m (nr) cos m0 cosh n(z+A). 

The constants m and w are not independent, since we 
have to satisfy the boundary condition at r = a. This 
gives J m '(na) = 0, so that for any selected m, r& is restricted 


to have one of a certain set of values, determined from 
the roots of the above equation. The function C above 
will involve the time, and in fact if we are interested in 
waves whose frequency is /, we shall try C oc sin 2irft. 
Putting C = D sin Sir/I, where D is now a constant 
independent of r, 6, z or t, we have 

$ = DJ m (nr) cos m6 cosh n(z +h) sin 27rft. (39) 

The boundary condition 47 (iii) now enables us to find ; 
it is 

- i J m (nr) cos m9 cosh nh cos 2-rrft . (40) 


The remaining boundary condition 47 (iv) gives us the 
period equation ; it is 

47T 2 / 2 D J m (nr) cos m9 cosh nh sin 2^ 

-{-gnD J m (nr) cos m# sinh n& sin 2rrft = 0. 

i.e. 4rr 2 / 2 == grn tanh TiA. . . . (41) 

For waves with a selected value of m (which must be 
integral) n is found and hence, from (41) / is found. We 
conclude that only certain frequencies are allowed. Apart 
from an arbitrary multiplying constant, the nature of the 
waves is now completely determined. 

50. In 48 we discussed the progressive wave motion 
in an infinite straight channel. It is possible to determine 
from (34) the actual paths of the particles of fluid in this 
motion. For if X , Z denote the displacements of a particle 
whose mean position is (x, z) we have 

8<h ga Goshm(z+h) 
X = TT = - -- r T- 1 

ex c cosh mh 

84> qa 
Z = - - = - 

dz c cosn mh 


in which we have neglected terms of the second order of 
small quantities. Thus 

X = - - - - - - cosra(x-~c), 

me 2 cosh m h 

gasmhm(z+h) . 
Z = * _ - . - L smm(x-ct). 
me 1 cosh mh 

Eliminating t, we find for the required path 

* (42) 

^ ' 

These paths are ellipses in a vertical plane with a constant 
distance (2ga/mc z ) sech mh between their foci. A similar 
discussion could be given for the other types of wave 
motion which we have solved in other paragraphs. 

51 . The Kinetic and Potential energies of these waves 
are easily determined. Thus, if we measure the P.E. 
relative to the undisturbed state, then, since (#, y) is the 
elevation, the mass of liquid standing above a base dA 
in the xy plane is p dA. Its centre of mass is at a height 
, and thus the total P.E. is 

4, .... (43) 

the integral being taken over the undisturbed area of 
surface. Likewise the K.E. of a small element is J pu 2 dr, 
dr being the element of volume of the liquid, so that the 
total K.E. is 

... (44) 

the integral being taken over the whole liquid, which may, 
within our approximation, be taken to be the undisturbed 

With the progressive waves of 48, and <f> are given 



by (33) and (34), and a simple integration shows that the 
K.E. and P.E. in one wavelength (27r/m) are equal, and 
per unit width of stream, have the value 

..... (45) 

In evaluating (44) it is often convenient to use Green's 
Theorem in the form * 


\\dxj l 

The latter integral is taken over the surface S which 
bounds the original volume, and d/dv represents differen- 
tiation along the outward normal to this volume. Since 
d</)/dv = on a fixed boundary, some of the contributions 
to T will generally vanish. Also, on the free surface, if 
is small, we may put d0/dz instead of d(f)/dv. 

52. We shall next calculate the rate at which energy 
is transmitted in one of these surface waves. We can 



FIG. 15 

illustrate the method by considering the problem discussed 
in 48, i.e. progressive waves in a rectangular tank of 
depth h. Let AA' (fig. 15) be an imaginary plane fixed 
in the liquid perpendicular to the direction of wave 
* See Rutherford, Chapter VI, p. 66 (ii). 


propagation. We shall calculate the rate at which the 
liquid on the left of A A' is doing work upon the liquid on 
the right. This will represent the rate at which the energy 
is being transmitted. Suppose that the tank is of unit 
width and consider that part of A A' which lies between 
the two lines z, z+dz (shown as PQ in the figure). At 
all points of this area the pressure is p, and the velocity 
is u. The rate at which work is being done is therefore 

pudz. Thus the total rate is I pudz. We use Bernouilli's 

J -h 

equation (8) to give us p ; since u 2 may be neglected, 
and F = gz, therefore 

p = 

Now, according to (1) u = d(f>/dx and from (34), 

ga . 

6 = -- - - cosm(x-ct). 
me cosh mh 

Putting these various values in the required integral we 

f ga cosh m(z+h) 

Binm(xct) -- - - (p Q gpz)dz 
J ~h c coshwA r 

/ A f P0 2 a 2 
(x ct) 
J ^ h c 

2 / 
Bin 2 m( 

This expression fluctuates with the time, and we are 
concerned with its mean value. The mean value of 
sin m(xct) is zero, and of sin 2 m(xct) is J. Thus the 
mean rate at which work is being done is 


sech 2 m& cosh 2 m(z -\-Ti)dz. 

After some reduction this becomes 

%gpa?c (I + 2mh cosech 2mh). 


In terms of the wavelength A = 27r/m, this is 

1 ^Jl + ^eosech^l . . (46) 

4 I A A J 

Now from (45) we see that the total energy with a stream 
of unit width is %gpa 2 per unit length. Thus the velocity 
of energy flow is 

C ( , . 4:7Th , 4:7Th} /Am ^ 

_|l + _cosech_| . . (47) 

We shall see in a later chapter that this velocity is an 
important quantity known as the Group Velocity. 

53. In the preceding paragraphs we have assumed 
that surface tension could be neglected. However, with 
short waves this is not satisfactory and we must now 
investigate the effect of allowing for it. When we say 
that the surface tension is T, we mean that if a line of 
unit length is drawn in the surface of the liquid, then 
the liquid on one side of this line exerts a pull on the 
liquid on the other side, of magnitude T. Thus the effect 
of Surface Tension is similar to that of a membrane 
everywhere stretched to a tension T (as in Chapter III, 
27), placed on the surface of the liquid. We showed in 
Chapter III that when the membrane was bent there was 
a downward force per unit area approximately equal to 


T -i 

g Thus in fig. 16, the pressure p 1 just inside 

the liquid does not equal the atmospheric pressure 
but rather 


The reader who is familiar with hydrostatics will 
recognise that the excess pressure inside a stretched film 
(as in a soap bubble) is ^(l/E^+l/B^), where R and B 2 
are the radii of curvature in any pair of perpendicular 


planes through the normal to the surface. We may put 
E l = d^/dx* and E 2 = d 2 /% 2 to the first order of 
small quantities, and then (48) follows immediately. 

Thus, instead of making p = p Q at the free surface of 

fd* 3 2 ) 

the liquid, the correct condition is that p -j-~H + - n r 

(dx z cy*) 

is constant and equal to p Q . We may combine this with 
Bernoulli's equation (9), in which we neglect u 2 and put 
V = gz. Then the new boundary condition which replaces 
47 (iii) is 


p x y 

We still have the boundary condition 47 (iv) holding, 
since this is not affected fey any sudden change in pressure 
at the surface. By combining (29) and (49) we find the 
new condition that replaces 47 (v). It is 

8z p 

We may collect these formulae together ; thus, with surface 

(i) y2^ = o in the body of the liquid . . (4) 
(ii) 3<f>/dv = en all fixed boundaries . . (2) 

Qt T f P2^* Q"* }"^\ 

(m) - ^ + ~ + = on the free surface 

(iv) d/dt = d<j>ldz on the free surface . . (29) 

( y ) - + ~-^+^l ^ on the free 

3t z dz p \dx* dy*\ 3z ^^ 

^ ^ y J surface . (50) 

Only two of the last three equations are independent. 


54. Waves of the kind in which surface tension is 
important are known as capillary waves. We shall 
discuss one case which will illustrate the conditions (i)-(v). 
Let us consider progressive type waves on an unlimited 
sheet of water of depth h, assuming that the motion takes 
place exclusively in the direction of x. Then, by analogy 
with (31) we shall try 

C cosh m(z-\-Ji) cos m(xct). . . (51) 

This satisfies (i) and (ii), (iv^ ffives the form of , which is 

= (C/c) sinh mh sin m(x >cl). . . (52) 

We have only one more condition to satisfy ; if we choose 
(v) this gives 

m 2 c 2 (7 cosh mil cos m(xct)~{-mCg sinh mh cos m(xct) 

H m 3 (7 sinh mh cos m(xct) = 0, 

i.e. c 2 = (g/m-}-Tm/p) taiih mh. . . . (53) 

Tliis equation is really the modified version of (32) when 
allowance is made for the surface tension ; if T = 0, it 
reduces to (32). 

When h is large, tanh. mh = 1, and if we write m = 277/A, 
we have 

'-+ <. 

The curve of c against A is shown in fig. 17, from which 
it can be seen that there is a minimum velocity which 
occurs when A 2 ^Tr^T/gp. Waves shorter than this, in 
which surface tension is dominant, are called ripples, and 
it is seen that for any velocity greater than the minimum 
there are two possible types of progressive wave, one of 
which is a ripple. The minimum velocity is (4</T//>) 1/4 , 
and if, as in water, T = 75, p 1*00 and g = 981 c.g.s. 
units, this critical velocity is about 23 cms. per sec., and 



the critical wavelength is about 17 cms. Curves of c against 
A for other values of the depth h are very similar to fig. 17. 

FIG. 17 



(1) Find the Potential and Kinetic energies for tidal 
waves in a tank of length I, using the notation of 42. 

(2) Find the velocity of any particle of liquid in the 
problem of tidal waves in a circular tank of radius a ( 44). 
Show that when m = in (25), particles originally on a vertical 
cylinder of radius r coaxial with the tank, remain on a coaxial 
cylinder whose radius fluctuates ; find an expression for the 
amplitude of oscillation of this radius in terms of r. 

(3) Tidal waves are occurring in a square tank of depth h 
and side a. Find the normal modes, and calculate the Kinetic 
and Potential energies for each of them. Show that when 
more than one such mode is present, the total energy is just 
the sum of the separate energies of each normal mode. 

(4) What are the paths of the particles of the fluid in the 
preceding question ? 

(5) A channel of unit width is of depth h, where h = kx, 
k being a constant. Show that tidal waves are possible with 
frequency p/27r, for which 

f = AJ Q (ax l l 2 ) cos pt, 


where a 2 = 4p z /kg, and J Q is BesseFs function of order zero. 
It is known that the distance between successive zeros of 
J (x) tends to TT when x is large. Hence show that the wave- 
length of these stationary waves increases with increasing 
values of x (This is the problem of a shelving beach.) 

(6) At the end of a shallow tank, we have x = 0, and the 
depth of water h is h h x zm . Also the breadth of the tank 
b is given by 6 = b x n . Show that tidal waves of frequency 
p/2ir are possible, for which 

= Ax u J Q (rx s ) cos pt, 
s ~ 1 m, a 2 = p 2 /gho, r = a/s, 2u l 2m~n arid q = | u/s |. 

Use the fact that J m (x) satisfies the equation 

d*J 1 dJ / m 2 \ __ 

dx 2 xdx \ x*\ 


(7) Prove directly from the conditions (i)-(v) in 47 
without using the results of 48 that the velocity of surface 
waves in a rectangular channel of infinite depth is ^(gXjZn). 

(8) Find the paths of particles of fluid in the case of surface 
waves on an infinitely deep circular tank of radius a. 

(9) A tank of depth h is in the form of a sector of a circle 
of radius a and angle 72. What are the allowed normal modes 
for surface waves ? 

(10) If X, Y, Z denotes the displacement of a particle of 
fluid from its mean position x, y, z in a rectangular tank of 
sides a and b when surface waves given by equation (37) are 
occurring, prove that the path of the particle is the straight line 

d WTTX b Qirij 1 

cot X = cot ^-^ Y = - coth r(z+h) Z. 
PIT a qir b r 

(11) Show that in surface waves on a cylindrical tank of 
radius a and depth h, the energies given by the normal modes 
(39) are 

V = J ~J- cosWnh cos z 2<rrft J m 2 (nr) r dr, and 


i r a 

T = - n?rpD 2 sin 2 27r/ cosh nh siiih nh I J m 2 (nr) r dr. 
2 Jo 


Use the fact that the total energy must be independent of the 
time to deduce from this that the period equation is 

47T 2 / 2 = gn tarih nh. 

(12) Show that when we use cylindrical polar coordinates 
to describe the capillary waves of 53, the pressure condition 
at the free surface 53 (iii) is 


dt p (dr* r'dr ?- 2 d0 a J 

Use this result to show that waves of this nature on a 
circular basin of infinite depth are described by 
<f> = C J m (nr) cos mB e nz cos 2irft, 

m , 

where J m '(na) = and 4?r 2 / 2 = gn +~Tn 3 /p. 

(13) Show that capillary waves on a rectangular basin of 
sides a, b and depth h are given by 

cosh r(z-{-h) mrrx niry 
A ~ A - : - cos - cos -^ cos ZTrft, 
smh rh a b 

v rA mirx n-ny . 

4 = cos - cos - - sin 2<nft, 
ATTJ a o 

where m = 0, 1, 2, ... ; w = 0, 1, 2 ... ; r a = 7r 2 (m 2 /a 2 +n 2 /6 2 ), 
and the period equation is 

47T 2 / 2 = (gr+Tr 3 //)) tanh rh. 

Verify, that when n = 0, this is equivalent to the result of 
54, equation (53). 


(r-nct \ ir-jTct \ 

~Y + *r 1, i ffpfar* Sin 8 I -y- + r 1 ; 

(2) radial vol. is (gA/c) cos m6 J m '(nr) sin (cnt + t), trans- 
verse velocity is (gAm/cnr) sin m^ J m ( nr ) B i n ( cn ^ ~f~ e), 
(gA/c) J '(nr) ; (3) J = A cos (p-nxfa) cos (qiry/a) cos (met /a), 

sin 2 (rnctja), P.E. = - 

cos 2 (t-Trttf/a) ; (4) = tan cot ; (8) X : F : Z = 

r g a a 

nrJ m '(nr) : mJ m (nr) tan m^ : nr J m (nr) ; (9) Same as in 
a. (39)-(41) except that m = 5A;/2, where k = 0, 1, 2 ---- ] 



56. Throughout Chapter V we assumed that the liquid 
was incompressible. Ail important class of problems is 
that of waves in a compressible fluid, such as a gas. In 
this chapter we shall discuss such waves, of which sound 
waves are particular examples. The passage of a sound 
wave through a gas is accompanied by oscillatory motion 
of particles of the gas in the direction of wave propagation. 
These waves are therefore longitudinal. Since the density 
p is not constant, but varies with the pressure p, we require 
to know the relation between p and p. If the compressions 
and rarefactions that compose the wave succeed each other 
so slowly that the temperature remains constant (an 
isothermal change) this relation is p kp. But normally 
this is not the case and no flow of heat, which would be 
needed to preserve the temperature constant, is possible ; 
in such cases (adiabatic changes) 

P = W, (i) 

where Jc and y are constants depending on the particular 
gas used. We shall use (1) when it is required, rather 
than the isothermal relation. 

57. There are several problems in the propagation 
of sound waves that can be solved without using the 
apparatus of velocity potential <f> in the form in which 
we used it in Chapter V, 47-54 ; we shall therefore 
discuss some of these before giving the general development 
of the subject. 



Our first problem is that of waves along a uniform 
straight tube, or pipe, and we shall be able to solve this 
problem in a manner closely akin to that of Chapter IV, 
32, where wo discussed the longitudinal vibrations of a 
rod. We can suppose that the motion of the gas particles 
is entirely in the direction of the tube, and that the velocity 
and displacement are the same for all points of the same 

Suppose for convenience that the tube is of unit cross- 
sectional area, and let us consider the motion of that 
part of the gas originally confined between parallel planes 
at P and Q a distance Ax apart (fig. 18). The plane P 

p Q 

dx I 
+ d| 

P' Q 1 
FIG. 18 

is distant x from some fixed origin in the tube. During 
the vibration let PQ move to P'Q', in which P is displaced 
a distance from its mean position, and Q a distance 
g+dg. The length P'Q' is therefore dx+dg. We shall 
find the equation of motion of the gas at P'Q' . For this 
purpose we shall require to know its mass and the 
pressure at its two ends. Its mass is the same as the 
mass of the undisturbed element PQ, viz. p dx, where p 
is the normal average density. To get the pressure at P' 
we imagine the element dx to shrink to zero ; this gives 
the local density p, from which, by (1), we calculate the 
pressure. We have 

p = Lim Pd tx/(dx+dg) = P Jl - ^V 
dx-^o \ cx) 



if we may neglect powers of d/dx higher than the first. 
The quantity (p p )//> w ^ often occur in this chapter ; 
it is called the condensation s. Thus 

s= -dfldx, p = Po (l+s). . . (3) 

The net force acting on the element P'Q' is p t p /9 and 
hence the equation of motion is 

We may rewrite (4) in the form 

8^__dp8p_ dp8*t 

po dt*~ dpte- p( >dpdx* ilom (2) ' 

It appears then that satisfies the familiar equation of 
wave motion 

g-^.'-** <> 

This equation shows that waves of any shape will be 
transmitted in either direction with velocity \/(dp/dp). 
In the case of ordinary air at C., using (1) as the relation 
between p and p, we find that the velocity is c = 332 
metres per sec., which agrees with experiment. Newton, 
who made this calculation originally, took the isothermal 
relation between p and p and, naturally, obtained an 
incorrect value for the velocity of sound. 

A more accurate calculation of the equation of motion 
can be made, in which powers of d$/dx are not neglected, 
as follows. From (2) we have the accurate result 

p = p = 



So, now using (4) in which no approximations have been 

Equation (5) is found from (6) by neglecting 8g/dx compared 
with unity. A complete solution of (6) is, however, beyond 
the scope of this book. It is easy to see that, since (6) 
is not in the standard form of a wave equation, the velocity 
of transmission depends upon the frequency, and hence 
that a wave is not, in general, transmitted without change 
of shape. 

58. We must now discuss the boundary conditions. 
With an infinite tube, of course, there are no such condi- 
tions, but with a tube rigidly closed at x = X Q , we must 
have = at x X Q , since at a fixed boundary the gas 
particles cannot move. 

Another common type of boundary condition occurs 
when a tube has one or more ends open to the atmosphere. 
At this end, the pressure must have the normal atmospheric 
value, and thus, from (1) and (2), dg/dx = 0. 

To summarise : 

02 I 02 

(i) ^ = -Qp in the tube, and c 2 = dp/dp . (5) 
(ii) = at a closed end (7) 


(iii) ~- = s = at an open end. ... (8) 

59. We shall apply these equations to find the normal 
modes of vibration of gas in a tube of length L These 
waves will iiaturally be of stationary type. 

(a) Closed at both ends x = 0, 1. This problem is the 
same mathematically, as the transverse vibrations of a 


string of length Z, fixed at its ends (cf. Chapter II, 19). 
Conditions (i) and (ii) of 58 give for the normal modes 

.. . . TTTX (rirct 1 

f - 4, am -T- cos { + f | , r = 1, 2, .... (9) 

(6) Closed atx = 0, opera aZ # = Z (a " stopped tube "). 
Here conditions (ii) and (iii) give = at x 0, and 


= at a; = I. The normal modes are 


\ "^ I i r\ i 9 nn\ 

'\' ' 2/ Z W "(V '"2; Z +*rj>r-U,l,Z,... l">) 

(c) Opera 6o/fc eratfo # = 0, Z. We have to satisfy 
the boundary condition (iii) d/dx = at x = 0, Z. So 
the normal modes are 

r *//* ^ 

r = l,2,... (11) 

In each case the full solution would be the superposition 
of any number of terms of the appropriate type with 
different r. The fundamental frequencies in the three 
cases are 2l/c, 4-l/c, and 2Z/c respectively. The harmonics 
bear a simple numerical relationship to the fundamental, 
which explains the pleasant sound of an organ pipe. 

60. We shall now solve a more complicated problem. 
We are to find the normal modes of a tube of unit sectional 
area, closed at one end by a rigid boundary and at the 

FIG. 19 

other by a mass M free to move along the tube. Let 
the fixed boundary be taken as x 0, and the normal 
equilibrium position of the moveable mass be at x I 


(fig. 19). Then we have to solve the standard equation 
of wave motion with the boundary conditions that when 
x = 0, (ii) gives = 0, and that when x I the excess 
pressure inside, pp Q , must be responsible for the 
acceleration of the mass M. This implies that 

d 2 
PPQ = M-Qp when x = I- 

The first condition is satisfied by the function 

= A sin nx cos (nct-\-e) . . . (12) 
To satisfy the second condition, we observe that 

p-Po - (dp/dp)(p-p ) = -~c* Po d/dx, from (3). 
So this condition becomes 

M i* = -*' **=' 

Using (12) this gives, after a little reduction, 
ril tan ril = IpJM. 

The allowed values of n are the roots of this equation. 
There is an infinite number of them, and when M = 0, so 
that the tube is effectively open to the air at one end, 
we obtain equation (10) ; when M oo, so that the tube 
is closed at each end, we obtain equation (9). 

61. So far we have developed our solutions in terms 
of , the displacement of any particle of the gas from its 
mean position. It is possible, however, to use the method 
of the velocity potential <f>. Many of the conditions which 
(f> must satisfy are the same as in Chapter V, but a few 
of them are changed to allow for the variation in density. 
It is convenient to gather these various formulae together 

(i) If the motion is irrotational, as we shall assume, 
u = V</>, (cf. Chapter V, equation (1)) . (13) 


(ii) At any fixed boundary, dfydv = (cf. Chapter V, 
equation (2)y ....... (14) 

(iii) The equation of Continuity (cf . Chapter V, equation 
(3)) is slightly altered, and it is * 

(iv) The equations of motion are unchanged ; if F is 
the external force on unit mass, in vector form, 
they are 

Du 1 

j~ = F -- yp (cf. Chapter V, equation (6)) . (16) 

JL/Z p 

(v) In cases where the external forces have a potential 
F, we obtain Bernouillf s equation (cf. Chapter V, 
equation (8)) 

+ -|u 2 +F - = const. . . (17) 
p ct 

in which we have absorbed an arbitrary function of the 
time into the term 8(f>/dt (cf. Chapter V, equation (8)). 

62. In sound waves we may neglect all external 
forces except such as occur at boundaries, and thus we 
may put F = in (17). Also we may suppose that the 
velocities are small and neglect u 2 in this equation. With 
these approximations Bernouilli's equation becomes 

dp dcf> 

= const. 

p ct 

We can simplify the first term ; for = I -f- ] , 

J P J \ d p' P 
and if the variations in density are small, dp/dp may be 

* Rutherford, 67. 


taken as constant, and equal to c 2 as in (5). Thus 

Cdp Cdp 

= c 2 M- = c*log e p = C 2 {log,(l+s)+log,p }. So 
J p J p 

-P = c 2 s+const., if s is small. If we absorb this constant 
in <, then Bernouilli's equation takes its final form 

Laplace's equation for <f> does not hold because of the 
changed equation of continuity. But if u, v, w and s 
are small, (15) can be written in a simpler form by the 
aid of (13) ; viz., 

This is effectively the same as 

| = v* - - - do) 

Now let us eliminate s between (18) and (19), and 
we shall find the standard equation of wave motion 

w-ig . - . < 

This shows that c is indeed the velocity of wave propaga- 
tion, but before we can use this technique for solving 
problems, we must first obtain the boundary conditions 
for (/>. At a fixed boundary, by (ii) d(f)/dv = 0. At an 
open end of a tube, the pressure must be atmospheric, 
and hence s = 0. Thus, from (18), 

8^/0* = (21) 

This completes the development of the method of the 
velocity potential, and we can choose in any particular 
problem whether we solve by means of the displacement 
| or the potential <. It is possible to pass from one to 
the other, since from (3) and (18) 


63. We shall illustrate these equations by solving the 
problem of stationary waves in a tube of length I, closed 
at one end (x 0) and open at the other (x = I). This 
is the problem already dealt with in 59 (6), and with 

,, , , . . , , . c &<(> 1 &<$> 

the same notation, we require a solution of ~ = _ r 

H dx* c 2 dt 2 

subject to the conditions 

d<j>/dx = at x = 0, 

It is easily seen that 

<f) = a cos mx cos (cmt-\-e) 

satisfies all these conditions provided that cos ml = 0, 
i.e. ml 77/2, 877/2, .... (r+l/2)7r/2 .... So the normal 
modes are 

and from this expression all the other properties of these 
waves may easily be obtained. The student is advised 
to treat the problems of 59 (a) and (c) in a similar manner. 

64. Our next application of the equations of 62 
will be to problems where there is spherical symmetry 
about the origin. The fundamental equation of wave 
motion then becomes (see Chapter I, equation (23)) 

a 2 ^ 2 ty __ i d^ 

~3r^ + r Or c 2 ~dt? ' 
with solutions of progressive type 

< = lf( r -ct)+lg(r+ct). 

There are solutions of stationary type (see Chapter I, 
equation (37)) 

cos cos 

<A = (l/r) . mr . cmt. 
r v ; ; sm sin 


If the gas is contained inside a fixed sphere of radius a, 
then we must have </> finite when r = 0, and dcf>/dr = 
when r = a. This means that 

with the condition 

tan ma = ma .... (23) 

This period equation has an infinite number of roots which 
approximate to ma = (n+l/tyn when n is large. So for 
its higher frequencies the system behaves very like a 
uniform pipe of length a open at one end and closed at 
the other. 

This analysis would evidently equally well apply to 
describe waves in a conical pipe. 

65. We shall now calculate the energy in a sound 

r i 
wave. The Kinetic energy is clearly - p u 2 ^F, where 

dV is an element of volume. In terms of the velocity 
potential this may be written 

^ (24) 

The last expression follows from Green's theorem just as 
in Chapter V, 51, and the surface integral is taken over 
the boundary of the gas. There is also Potential energy 
because each small volume of gas is compressed or rarified, 
and work is stored up in the process. To calculate it, 
consider a small volume F , which during the passage 
of a wave is changed to F x . If s is the corresponding 
value of the condensation, then from (3), we have, to the 
first degree in s l9 

F^FoU-^) . . . (25) 

Further, suppose that during the process of compression, 
F and s are simultaneous intermediary values, Then we 


can write the work done in compressing the volume from 

' f Fl 
F to F! in the form I p dV. But, just as in (25), 

J Fo 

V = F (l s), and henco 

dV = -V Q ds. 

We may also write p pQ+(dp/dp)(pp ) 

Thus the potential energy may be written 



J o 

This is the contribution to the P.E. which arises from the 
volume F . The total P.E. may be found by integration. 
The first term will vanish in this process since it merely 
represents the total change in volume of the gas, which 
we may suppose to be zero. We conclude, therefore, that 


f i 

the Potential Energy is - c 2 p s 2 dV ... 

It can easily be shown that with a progressive wave 
the K.E. and P.E. are equal ; this does not hold for 
stationary waves, for which their sum remains constant. 

66. We conclude this chapter with a discussion of 
the propagation of waves along a pipe whose cross-sectional 
area A varies along its length. Our discussion is similar 
in many respects to the analysis in 57. 

Consider the pipe shown in fig. 20, and let us measure 
distances x along the central line. It will be approximately 
true to say that the velocity u is constant across any 
section perpendicular to the x axis. Suppose that the 
gas originally confined between the two planes P, Q at 




distances x, x+dx is displaced during the passage of a 
wave, to P'Q', the displacement of P being | and of Q 
being g+dg. Consider the motion of a small prism of gas 

FIG. 20 

such as that shaded in the figure ; its equation of motion 
may be found as in 57, and it is 

PO "ol? ~oT 

dt* dx 


We must therefore find the pressure in terms of . This 
may be obtained from the equation of continuity, which 


expresses the fact that the mass of gas in P'Q' is the same 
as that in PQ. Thus, if p is the density, 

p Q A(x) dx = p A(x 

Neglecting small quantities, this yields 

\ dx A dx) 

Eliminating p between (27) and (28) we find 

8'*__zp2e_. 1/I1 
P W~ dp dx ~ c PQ dx [A dx 

where, as usual, c 2 = . 


So the equation of motion is 

In the case in which A is constant this reduces to the former 
equation (5). An important example when A is not 
constant is the so-called exponential horn used on the 
best acoustic gramophones ; here the tube is approximately 
symmetrical about its central line and the area varies with 
the distance according to the law A = A Q e Zax , where a and 
A are constants. 

With this form of A, (29) reduces to 

100 WAVES 

A solution is possible by the method of separation of 
variables (see 7). We soon find 

where m l and m 2 are given by a\/(a 2 n 2 ). In most 
exponential horns ri 2 is considerably larger than a 2 in the 
range of audible frequencies, so that m 1 and m z may be 
written a^in. Thus 

+*>} . . (30) 

The first term represents a wave going outwards and the 
second a wave coming inwards. We conclude from this 
that waves can be sent outwards along the horn with a 
velocity c which is approximately independent of the 
frequency, and with an attenuation factor e~ ax which is 
also independent of the frequency. It is this double 
independence which allows good reproduction of whatever 
waves are generated at the narrow end of the horn, and 
which is responsible for this choice of shape in the best 
gramophones. Other forms of A will not, in general, give 
rise to the same behaviour. 

67. Examples 

(1) Use the method of 58 to investigate sound waves 
in a closed rectangular box of sides a l9 a 2 and a 3 . Show 
that the number of such waves for which the frequency is less 
than n is approximately equal to one-eighth of the volume 
of the quadric x^/a^+y^/a^+z 2 /^ = 4n 2 /c 2 . Hence show 
that this number is approximately 47m 3 a 1 a 2 a 3 /3c 3 . 

(2) Investigate the reflection and transmission of a train 
of harmonic waves in a uniform straight tube at a point 
where a smooth piston of mass M just fits into the tube and 
is free to move. 

(3) Show that the kinetic and potential energies of a plane 
progressive wave are equal. 

(4) Show that the kinetic and potential energies of 
stationary waves in a rectangular box have a constant sum. 

(5) Find an equation for the normal modes of a gas which 


is confined between two rigid concentric spheres of radii 
a and b. 

(6) Show that a closer approximation to the roots of 
equation (23) is ma = (n + l/2)7r l/{(n-fl/2)ff}. 

(7) Find numerically the fundamental frequency of a 
conical pipe of radius 1 metre open at its wide end. 

(8) The cross-sectional area of a closed tube varies with the 
distance along its central line according to the law A = A Q x n . 
Show that if its two ends are x = 0, and x = I, then standing 
waves can exist in the tube for which the displacement is 
given by the formula 

f = x^ l ~ n ^J m (qx/c) cos {qct + c}, 

where m ~ (n-f l)/2 and J m (ql/c) = 0. 

Use the fact that J m (x) satisfies the equation 

<PJ I dJ 

[ANSWERS: 2. reflection coefft. = { 
transmission coefft. = {l+^^V^Po 2 }" 1 ^ 2 > ^ period = Sir/ 
where (abp*-{-I) sin p(ba) ~ p(b a) cos p(b+a) ; 7. 166.] 



68. Before we discuss the propagation of electric waves, 
we shall summarise the most important equations that we 
shall require. These are known as Max-well's equations. 
Let the vectors E (components E x , E v , E z ) and H (com- 
ponents H x , H y , Hy) denote the electric and magnetic 
field strengths. These are defined | as the forces on a 
unit charge or pole respectively when placed inside a 
small needle-shaped cavity, the direction of the cavity 
being the same as the direction in which we wish to measure 
the component of E or H. We shall suppose that all our 
media are isotropic with no ferromagnetism or permanent 
polarisation ; thus, if we write e for the dielectric constant, 
and fj, for the permeability, then the related vectors, 
viz. the magnetic induction B and the dielectric dis- 
placement D are given by the equations B = /zH, D = eE. 
Further, let j (components j x , j y , j z ) denote the current 
density vector, and p the charge density. Then, if we 
measure j, B and H in electromagnetic units, E and D 
in electrostatic units, writing c for the ratio between the 
two sets of units,}; Maxwell's equations may be summarised 
in vector form as follows : 

div D = 47rp (1) 

div B = (2) 

* Before reading this chapter, the student is advised to 
familiarise himself with the equations of electromagnet ism, as 
found in text books such as those by Jeans, Pidduck, or 
Abraham - B ecker . 

f See, e.g., Abraham-Becker, Chapters IV, VII. 

J This system is known as the Mixed System. If we had used 
entirely e.s.u., or entirely e.m.u., the powers of c would have been 
different. Particular care is required in discussing the units in 
(3) and (7). In this chapter c will always denote the ratio of the 
two sets of units. 





D = cE ..... (5) 
B r^ pM ...... (6) 

To these equations wo must add the relation between j 
and E. If a is the conductivity, which is the inverse of 
the specific resistance, this relation is 

J = orE ..... (7) 

For conductors a is large, and for insulators it is small. 

The above equations have been written in vector form ; 
until the student has acquired familiarity with the use of 
the vector notation and operation, he is advised to verify 
the various calculations of this chapter, using the equations 
in Cartesian form as well as vector form. This will soon 
show how much simpler the vector treatment is, in nearly 
every case. If we wish to write these equations in their 
full Cartesian form, we have to remember that 

__ dD x dD y 

dx dy 


- and that 


(orr ) ZJ 3T7 ^ U 3 II 3TJ \ 

Ctlz Vtly VFLx Cflz Oily Cl2x\ 

dy dz ' dz dx dx dy / 

The preceding equations then become 

dx dy 


dy dz 
dHx dHz 


ldD x } 

c ct 
1 dD s 

' dz 

dE 2 

dx dy 

c dt 
idB s 


104 WAVES 

D x = E X , D y = E V ,D Z = E, . . (5 7 ) 
B x = nH x , B y = fjiH y , B z = fjiH z . . (6') 
j x = o^ , j y = <7^ , j, = aE z . . (7') 

Equations (l)-(4) are sometimes called Maxwell's 
Equations and equations (5)- (7) constitutive relations. 
Simple physical bases can easily be given for (l)-(4). 
Thus, (1) represents Gauss' Theorem, and follows from the 
law of force between two charges ; (2) represents the fact 
that isolated magnetic poles cannot be obtained ; (3) is 
Ampfere's Rule that the work done in carrying a unit pole 
round a closed circuit equals 4rr times the total current 
enclosed in the circuit ; part of this current is the conduc- 
tion current j and part is Maxwell's displacement current 

1 3D 

- -- ~- ; (4) is Lenz's law of induction. 

4:7TC Ot 

These seven equations represent the basis of our 
subsequent work. They need to be supplemented by a 
statement of the boundary conditions that hold at a change 
of medium. If suffix n denotes the component normal to 
the boundary of the two media, and suffix s denotes the 
component in any direction in the boundary plane, then 
on passing from the one medium to the other 

D n , B n , E s and H a are continuous . . (8) 

In cases where there is a current sheet (i.e. a finite 
current flowing in an indefinitely thin surface layer) some 
of these conditions need modification, but we shall not 
discuss any such cases in this chapter. 

There are two other important results that we shall 
use. First, we may suppose that the electromagnetic 
field stores energy, and the density of this energy per unit 
volume of the medium is 

. . . (9) 

Second, there is a vector, known as the Poynting 


vector, which is concerned with the rate at which energy 
is flowing. This vector/ whose magnitude and direction 
are given by 

(E X H), . . . . (10) 

represents the amount of energy which flows in unit time 
across unit area drawn perpendicular to it. E and H are 
generally rapidly varying quantities and in such cases it is 
the mean value of (10) that has physical significance. 

69. We shall first deal with non-conducting media, 
such as glass, so that we may put a = in (7) ; we suppose 
that the medium is homogeneous, i.e. and p, are constants. 
If, as usually happens, there is no residual charge, we may 
also put p = in (1), and with these simplifications, 
Maxwell's equations may be written 

div E == , div H = 0, 


ur H = -- 

c dt 

ir. ITT e l 

curl E = - , curl H = -- 

c dtj 

These equations lead immediately to the standard equation 
of wave motion, for we know * that 

curl curl H = grad div H y 2 H. 

Consequently, from the fourth of the equations in (11), 
we find 

, ,. o,,. . as a . 

grad div H \7 2 H = - curl = --- curl E. 
6 v c dt cdt 

Substituting for div H and curl E, we discover the standard 


* Rutherford, Vector Methods, p. 59, equation (10). 

106 WAVES 

Eliminating H instead of E we find the same equation for E : 

According to our discussion of this equation in Chapter I, 
this shows that waves can be propagated in such a medium, 
and that their velocity is c/y^cju,). In free space, where 
e = ju, = 1, this velocity is just c. Now c, which was 
defined as the ratio of the two sets of electrical units, 
has the dimensions of a velocity, and its magnitude can 
be obtained experimentally; it is approximately 2*998 . 10 10 
cms. per sec. But it is known that the velocity of light 
in free space has exactly this same value. We are thus 
led to the conviction that light waves are electromagnetic 
in nature, a view that has subsequently received complete 
verification. X-rays, y-rays, ultra-violet waves, infra-red 
waves and wireless waves are also electromagnetic, and 
differ only in the order of magnitude of their wavelengths. 
We shall be able to show later, in 71, that these waves 
are transverse. 

In non-conducting dielectric media, like glass, e is not 
equal to unity ; also JJL depends on the frequency of the 
waves, but for light waves in the visible region we may 
put fj, = 1. The velocity of light is therefore c/\/e. Now 
in a medium whose refractive index is K, it is known 
experimentally that the velocity of light is c/K. Hence, 
if our original assumptions are valid, = K 2 . This is 
known as Maxwell's relation. It holds good for many 
substances, but fails because it does not take sufficiently 
detailed account of the atomic structure of the dielectric. 
It applies better for long waves (low frequency) than for 
short waves (high frequency). 

70. A somewhat different discussion of (11) can be 
given in terms of the electric and magnetic potentials. 
Since div H = 0, it follows that we can write 

H = curl A, .... (14) 


where A is a vector yet to be determined. This equation 
does not define A completely, since if t/r is any scalar, 
curl (A+grad $) curl A. Thus A is undefined to the 
extent of addition of the gradient of any scalar, and we 
may accordingly impose one further condition upon it. 


If H curl A, and curl E = - , it follows, by 

c ct 

elimination of H, that 

where (j> is any arbitrary function, 

i.e. E=-grad^- . . (15) 

In cases where there is no variation with the time, this 
becomes E = grad </>, showing that <f> is the analogue 
of the electrostatic potential. 

Eliminating H from the relations H = curl A, 

curl H = -- , and using (15) to eliminate E, we find 
c ct 

Let us now introduce the extra allowed condition upon A, 
and write g / 

divA+-^-0 . . . (16) 
c ct 

Then A satisfies the standard equation of wave motion 


108 WAVES 

Further, taking the divergence of (15), we obtain, by (16) 

Thus <f> also satisfies the standard equation 

**-? " 8 > 

A similar analysis can be carried through when p and j 
are not put equal to zero, and wo find 

H = curl A .... (14') 
~ . . (15') 

-. . . (16') 
c dt 

<f) and A are known as the electric potential and magnetic 
or vector potential respectively. It is open to our 
choice whether we solve problems in terms of A and <f>, 
or of E and H. The relations (14') -(18') enable us to pass 
from the one system to the other. The boundary condi- 
tions for <f> and A may easily be obtained from (8), but 
since we shall always adopt the E, H type of solution, 
which is usually the simpler, there is no need to write 
them down here. 

There is one other general deduction that can be made 
here. If we use (3), (5) and (7) we can write, for 
homogeneous media, 

curl H = 47TcrE -) --- . 
c dt 


Taking the divergence of each side, and noting, from (1), 
that div E = 47r/>/e, we find 


Thus, on integration, 

p = /) e~^, where 6 = e/47rcrc . . (19) 

6 is called the time of relaxation. It follows from (19) 
that any original distribution of charge decays exponentially 
at a rate quite independent of any other electromagnetic 
disturbances that may be taking place simultaneously, 
and it justifies us in putting p = in most of our 
problems. With metals such as copper, 6 is of the order 
of 10~ 13 sees., and is beyond measurement ; but with 
dielectrics such as water is large enough to be deter- 
mined experimentally. 

71. We next discuss plane waves in a uniform non- 
conducting medium, and show that they are of transverse 
type, E and H being perpendicular to the direction of 
propagation. Let us consider plane waves travelling with 
velocity V in a direction I, ra, n. Then E and H must be 
functions of a new variable 

u ES Ix+my+nzVt . . . (20) 

When we say that a vector such as E is a function of u, 
we mean that each of its three components separately 
is a function of u, though the three functions need not 
be the same. Consider the fourth equation of (11). Its 
^-component (see (3')) is 

en, _ afly __ as* 

~dy dz ~ c dt * 


If dashes denote differentiation with respect to u, this is 

mH z '~nH y f = ~~~ 


Integrating with respect to u, this becomes 

- -E 

y ^ 

in which we have put the constant of integration equal 
to zero, since we are concerned with fluctuating fields 
whose mean value is zero. There are two similar equations 
to the above, for E y and E g , and we may write them as 
one vector equation. If we let n denote the unit vector 
in the direction of propagation, so that n = (I, m, n), we 

nxH- - E (21) 


Exactly similar treatment is possible for the third equation 
of (11) ; we get 


Equation (21) shows that E is perpendicular to n and H, 
and (22) shows that H is perpendicular to n and E. In 
other words, both E and H are perpendicular to the direc- 
tion of propagation, so that the waves are transverse, and 
in addition, E and H are themselves perpendicular, E, 
H and n forming a right-handed set of axes. If we 
eliminate H from (21) and (22) and use the fact that 

n x [n x E] = (n . E)n (n . n)E = E, 

since n is perpendicular to E and n is a unit vector, we 
discover that F 2 = c 2 /e/>t, showing again that the velocity 
of these waves is indeed c/\/(eju.). 

It is worth while writing down the particular cases of 
(21) and (22) that correspond to plane harmonic waves 


in the direction of the z axis, and with the E vector in the 
x or y directions. The solutions are 

J0 = H a = -V(*lriM ip(t ~ zlV) 

E y = aeW-W H y = (23) 

E 9 = H z = 0. 

E x = 

E y =:0 H y = 4V(*//*) 6e '' x '~ z/7) (24) 

E 8 ==Q H 2 = 0. 

In accordance with 10, a and b may be complex, the 
arguments giving the two phases. It is the general 
convention * to call the plane containing H and n the 
plane of polarisation. Thus (23) is a wave polarised 
in the xz plane, and (24) a wave polarised in. the yz plane. 
By the principle of superposition ( 6) we may superpose 
solutions of types (23) and (24). If the two phases are 
different, we obtain elliptically polarised light, in which 
the end-point of the vector E describes an ellipse in the 
xy plane. If the phases are the same, we obtain plane 
polarised light, polarised in the plane y/x b/a. If 
the phases differ by ?r/2, and the amplitudes are equal, 
we obtain circularly polarised light, which, in real form, 
may be written 

E x ^=a cos p(tz/V) H x = V( e /j^) a sin Pttz/V) 
E y = a sin p(tz/V) H y = +V r ( //*) a cos P(t*IV) 
E z = H z = 0. (25) 

The end-points of the vectors E and H each describe 
circles in the xy plane. 

In all three cases (23) -(25), when we are dealing with 
free space (e === /z = 1) the magnitudes of E and H are 

72. By the use of (10) we can easily write down the 
rate at which energy is transmitted in these waves. Thus, 

* To which, unfortunately, not all writers conform. 

112 WAVES 

(CdP I \ 
0, 0, TIA/ / 

This vector is in the direction of the positive z-axis, showing 
that energy is propagated with the waves. According to 
(9), the total energy per unit volume is 

877 1 ' r J 477 

From these two expressions we can deduce the velocity 
with which the energy flows ; for this velocity is merely 
the ratio of the total flow across unit area in unit time 
divided by the energy per unit volume. This is c/v'fc/x), 
so that the energy flows with the same velocity as the 
wave. This does not hold with all types of wave 
motion ; an exception has already occurred in liquids 

When we calculate the Poynting Vector for the waves 

(23) and (24), we must remember that ExH is not a 

linear function and consequently (see 10) we must choose 
either the real or the imaginary parts of E and H. Taking, 
for example, the real part of (23), the Poynting Vector 
lies in the z direction, with magnitude 


' 4^ 

This is a fluctuating quantity whose mean value with 

ca 2 /e 

respect to the time is A/ The energy density, from 
877 \ jj, 


(9), is cos 2 jp( z/F), with a corresponding mean value 
ea 2 /877. Once again the velocity of transmission of energy 
is -r-A/ r = c/-v/(eu), which is the same as the 

077 V jU, 077 

wave velocity. 



73. We shall next discuss the reflection and refraction 
of plane harmonic light waves. This reflection will be 
supposed to take place at a plane surface separating two 
non-conducting dielectric media whose refractive indices are 
KI and K 2 . Since we may put ^ = {JL 2 1, the velocities 
in the two media are c/K^ cjK^. In fig. 21 let Oz be the 

FIG. 21 

direction of the common normal to the two media, and let 
AO, OB, OC be the directions of the incident, reflected 
and refracted (or transmitted) waves. We have not yet 
shown that these all lie in a plane ; let us suppose that 
they make angles 0, 0' and <f> with the z axis, OA being 
in the plane of the paper, and let us take the plane of 
incidence (i.e. the plane containing OA and Oz) to be the 
xz plane. The y axis is then perpendicular to the plane 
of the paper. 

114 WAVES 

Since the angle of incidence is 0, then as in (20), each 
of the three components of E and H will be proportional to 

gip{ct KI(X sin 6 -f z cos 0)} 

Let the reflected and transmitted rays move in directions 
(I l9 m v n ) and (1 2 , w 2 , n 2 ) respectively. Then the corres- 
ponding components of E and H for these rays will be 
proportional to 

Thus, considering the E x components, we may write the 
incident, reflected and transmitted values 

A $&& ~ R i(% sin -f z cos 6)} ^ e ip{ft - Ki(l& -f m\y + n^)} an( J 
e ip{ct - E z (l s x + m a y + n^ 

These functions all satisfy the standard equation of wave 
motion and they have the same frequency, a condition 
which is necessary from the very nature of the problem. 

We shall first show that the reflected and transmitted 
waves lie in the plane of incidence. This follows from the 
boundary condition (8) that E x must be continuous on the 
plane z 0, i.e. for all x, y, t, 

This identity is only possible if the indices of all three 
terms are identical : i.e. 

ctKiX sin 9 = ct K^^x-^-m^) ~ ct K 2 (l 2 x-{~m 2 y). 

Thus #! sin = K^ = 

= K l m 1 = K 2 m 2 . 

The second of these relations shows that m = m 2 = 0, 
so that the reflected and transmitted rays OB, OG lie in 
the plane of incidence xOz. The first relation shows that 
Z x sin 0, i.e. that the angle of reflection 0' is equal to 
the angle of incidence 0, and also that 

KI sin = KZ sin < . . . (26) 



This well-known relationship between the angles of 
incidence and refraction is known as Snell's law. 

Our discussion so far has merely concerned itself with 
directions, and we must now pass to the amplitudes of 
the waves. There are two main cases to consider, according 
as the incident light is polarised in the plane of incidence, 
or perpendicular to it. 

Incident light polarised in the plane of incidence. The 
incident ray AO has its magnetic vector in the xz plane, 
directed perpendicular to AO. To express this vector in 
terms of x, y, z it is convenient to use intermediary axes 
> ??> (see fig- 22, where 7? is not shown, however, as it 

FIG. 22 

is parallel to Oy and perpendicular to the plane of the 
paper). is in the direction of propagation, and is in 
the plane of incidence. Referred to these new axes, H 

116 WAVES 

lies entirely in the direction, and E in the 77 direction. 
We may use (23) and write 

J0f = JSff = , Jff,, = ateW*-*^ 

H r) ==H^==0 ) H^ = -K^e^-K^. 
Now = x sin 6-{-z cos 0, and so it follows that : 
incident wave 

E x = , H x = -#!% COS 9 e M* 

E a l e ip ^ ct ~~ :K ^ xaln ^ +zco& ^ , H 
E z = , #z = #!% sm 0e ? '^~^ 

Similar analysis for the reflected and refracted waves 
enables us to write 
reflected wave 

E x = , fls = #A COS eMrf-tfiteBinfl-scosfl)^ 

y 3= & ie &** - *<* sin ^ -^os 19)} } H y = Q, 
E z = , # 2 - EA sin e^W- 
refracted wave 

E x = , #3. = ^2 cos < e^^ 

JB y = a^cf-^^sin^-fzcos^)} ? H y = 0, 

E z = , # z = # 2 a 2 sin <^ e*2^c 

We may write the boundary conditions in the form that 
E x , E y , K 2 Ey, H x> H y and H z are continuous at z = 0. 
These six conditions reduce to two independent relations, 
which we may take to be those due to E y and H x : 

K-fl^ cos 0+K^ cos = K 2 a% cos <f>. 

KI cos 6+E" 2 cos K 1 cos 0~K 2 cos ^ ~~ 2^ cos ^ ' 

Using Snell's law (26) in the form K l : K 2 = sin ^ : sin 0, 
this gives 

a " 


am(6-<f>) 2 sin ^ cos 6' 


Equation (27) gives the ratio of the reflected and refracted 
amplitudes. If medium 2 is denser than medium 1 , K 2 > K 1? 
so that 6>(f>, and thus b l /a l is negative ; so there is a 
phase change of 77 in the electric field when reflection takes 
place in the lighter medium. There is no phase change on 
reflection in a denser medium, nor in the refracted wave. 

Incident light polarised perpendicular to the plane of 
incidence. A similar discussion can be given when the 
incident light is polarised perpendicular to the plane of 
incidence ; in this case the roles of E and H are practically 
interchanged, H y for example being the only non-vanishing 
component of H. It is not necessary to repeat the analysis 
in full. With the same notation for the amplitudes of the 
incident, reflected and refracted waves, we have 


_ . _ 
sin 29+ sin 2</> sin 20 sin 2<f> 4 cos 9 sin </> ' 

If reflection takes place in the lighter medium, 7 1 
9>(f>, and there is no phase change in E at reflection ; 
if K^>K^ then there is a phase change of TT. 

It follows from (28) that the reflected ray vanishes if 
sin 29 = sin 2<. Since 9 ^ ^, this implies that 9-\-</) 77/2, 
and then Snell's law gives 

K l sin 9 = K% sin </> = K 2 cos 0, 

tan 9 = K^Ki = V(*a/*i) - - (29) 

With this angle of incidence, known as Brewster's angle, 
there is no reflected ray. 

In general, of course, the incident light is composed of 
waves polarised in all possible directions. Equations (27) 
and (28) show that if the original amplitudes in the two 
main directions are equal, the reflected amplitudes will 
not be equal, so that the light becomes partly polarised 
on reflection. When the angle of incidence is given by 
(29) it is completely polarised on reflection. This angle is 
therefore sometimes known as the polarising angle. 

118 WAVES 

74. An interesting possibility arises in the discussion 
of 73, which gives rise to the phenomenon known as 
total or internal reflection. It arises when reflection 
takes place in the denser medium so that <>#. If we 
suppose 9 to be steadily increased from zero, then <f> also 
increases and when sin 6 = K^K^ , <f> = rr/2. If 6 is 
increased beyond this critical value, <f> is imaginary. 
There is nothing to disturb us in this fact provided that 
we interpret the analysis of 73 correctly, for we never 
had occasion to suppose that the coefficients were real. 
We can easily make the necessary adjustment in this 
case. Take for simplicity the case of incident light 
polarised in the plane of incidence. Then the incident 
and reflected waves are just as in our previous calculations. 
The refracted wave has the same form also, but in the 
exponential term, K 2 sin c/> = K l sin 0, and is therefore 
real, whereas 

KZ cos < = <S(K 2 *-KJ sin 2 

and is imaginary, since we are supposing that internal 
reflection is taking place and therefore K t sin 6>K Z . We 
may therefore write K 2 cos <f> = -j-iq, where q is real. 
Thus the refracted wave has the form 

J5J __ a e 

= a 

For reasons of finiteness at infinity, we have to choose 
the negative sign, so that it appears that the wave is 
attenuated as it proceeds into the less dense medium. 
For normal light waves it appears that the penetration 
is only a few wavelengths, and this justifies the title 
of total reflection. The decay factor is 

e -pqz __ e -p^(KSm\*e-K&Z f 

This factor increases with the frequency so that light of 
great frequency hardly penetrates at all. In actual 
physical problems, the refractive index does not change 


from K l to K 2 abruptly, as we have imagined ; however, 
Drude has shown that if we suppose that there is a thin 
surface layer, of thickness approximately equal to one 
atomic diameter, in which the change takes place smoothly, 
the results of this and the preceding paragraphs are hardly 
affected . 

75. In our previous calculations we have assumed 
that the medium was non-conducting, so that we could 
put a 0. When we remove this restriction, keeping 
always to homogeneous media, equations (l)-(7) give us 

div E = 0, 
div H = 0, 

curl H = 47TC7E+- , 
c dt 

, LJL rH 

curl E = ~~*~ . 

c 'dt 

Now curl curl E = grad div E y 2 E = y 2 E, so that 

dt c dt c dt c* dt 2 ' 

Ua 2 E 477(7LldE 

V 2 E=-|^ + -/- . . (30) 

A similar equation holds for H. Equation (30) is the 
well-known equation of telegraphy (see 9). The first 
term on the right-hand side may be called the displacement 

I )T\ 

term, since it arises from the displacement current -- 

477-c dt 

and the second is the conduction term, since it arises from 
the conduction current j. If we are dealing with waves 
whose frequency is p/2-rr, E will be proportional to e*** ; 
the ratio of these two terms is therefore ep/47rccr. Since 
e is generally of the order of unity, this means that if 
pl^TT is much greater than ccr, only the displacement term 
matters (this is the case of light waves in a non-conducting 

120 WAVES 

dielectric) ; but if p/2?r is much less than cor, only the 
conduction term matters (this is the case of long waves 
in a good metallic conductor). In the intermediate region 
both terms must bo retained. With most metals, if p<10 7 
we can neglect the first term, and if >>10 15 we can neglect 
the second term. 

Let us discuss the solutions of (30) which apply to 
plane harmonic waves propagated in the z direction, such 
that only E x and H y are non- vanishing (as in (24)). We 
may suppose that each of these components is pro- 
portional to 

eW-rt .... (31) 

where pfZir is the frequency and q is still to be determined. 
This expression satisfies the equation (30) if 

q is therefore complex, and we may write it 

q = a-ifi, 

The " velocity " of (31) is l/q ; but we have seen in 73 
that in a medium of refractive index K the velocity is c/JfiT. 
So the effective refractive index is cq which is complex. 
Complex refractive indices occur quite frequently and are 
associated with absorption of the waves ; for, combining 
(31) and (33) we have the result that E x and H v are 
proportional to 

e -ppz e ip(t-) m . t (34) 

This shows that a plane wave cannot be propagated in 
such a medium without absorption. The decay factor may 


be written e~ kz where k pp. k is called the absorbtion 
coefficient. In the case where 4wac/p is small compared 
with unity (the case of most metals), k is approximately 
equal to 27rcr\/(iJL/). Now the wavelength in (34) is 
A = 27T/ap, so that in one wavelength the amplitude 
decays by a factor e , approximately ^^a^P AS 
we are making the assumption that ccr/ep is small, the 
decay is gradual, and can only be noticed after many 
wavelengths. The distance travelled before the amplitude 
is reduced to l/e times its original value is l/k, which is 
of the same order as a. 

The velocity of propagation of (34) is I/a, and thus 
varies with the frequency. With our usual approximation 
that ccr/ep is small, this velocity is 


We can show that in waves of this character E and H 
are out of phase with each other. For if, in accordance 
with (31), we write 

then the y- component of the vector relation 

i*. P * 

curl E = ^ , 

c dt 

gives us the connection between a and b. It is 
8E X ___ p dH y 

~dz ~~ "~~ c ~dT ' 

i.e. qa = ^ 6 . . . . (36) 


Thus b/a is equal to (c///,)g. Now q is complex and hence 
there is a phase difference between E 9 and H v equal to 
the argument of q. This is tan~ 1 (j3/a), and with the same 
approximation as in (35), this is just tan~ 1 (27ro i c/e^>), which 
is effectively 27rac/p. 



76. It is interesting to discuss in more detail the 
case in which the conductivity is so great that we may 
completely neglect the displacement term in (30). Let 
us consider the case of a beam of light falling normally 
on an infinite metallic conductor bounded by the plane 
z 0. Let us suppose (fig. 23) that the incident waves 



FIQ. 23 

come from the negative direction of z, in free space, for 
which e = fju = 1, and are polarised in the yz plane. Then, 
according to (24) they are defined by : 
incident wave 

E x = a l e^- 2 / c > , H v = a l e^-^. 

reflected wave 

E x = b l eW+*M , H v = - 


In the metal itself we may write, according to (31) and (36), 
E x = a 2 e ^ (t ~^ , H y -= - q a 2 e^-^. 

These values will satisfy the equation of telegraphy (30) 
in which we have neglected the displacement term, if 

02 _ 

where y 2 = Sira^/pc. Thus 

= y(i-) ....... (37) 

Inside the metal, E and H have a 7r/4 phase difference, 
since, as we have shown in (36), this phase difference is 
merely the argument of q. 

The boundary conditions are that E x and H y are 
continuous at z = 0. This gives two equations 





Since g is complex, all three electric vectors have phase 
differences. The ratio R of reflected to incident energy 
is !&!/&! | 2 , which reduces to 

(cy-/*) 2 + 

In the case of non-ferromagnetic metals, cy is much larger 
than JLC, so that approximately 

124 WAVES 

This formula has been checked excellently by the experi- 
ments of Hagen and Rubens, using wavelengths in the 
region of 10~ 5 cms. 

It is an easy matter to generalise these results to apply 
to the case when we include both the displacement and 
conduction terms in (30). 

We can use (38) to calculate the loss of energy in the 
metal. If we consider unit area of the surface of the metal, 
the rate of arrival of energy is given by the Poynting Vector. 

This is \a 1 | 2 . Similarly the rate of reflection of energy is 


So the rate of dissipation is -j | a^ | 2 -| b l | 2 \ . 

STT 877 

This must be the same as the Joule heat loss. In our 
units, this loss is ccrE 2 per unit volume per unit time. 
If we take the mean value of E x 2 in the metal, it is an 


easy matter to show that I ccrE x 2 dz is indeed exactly 

J o 
equal to this rate of dissipation. 

77. When the radiation falls on the metal of 76, it 
exerts a pressure. We may calculate this, if we use the 
experimental law that when a current j is in the presence 
of a magnetic field H there is a force JLCJ X H acting on 
it. In our problem, there is, in the metal, an alternating 
field E, and a corresponding current aE. The force on the 
current is therefore ju,aE x H, and this force, being perpen- 
dicular to E and H, lies in the z direction. The force on 
the charges that compose the current is transmitted by 
them to the metal as a whole. Now both E and H are 
proportional to e~ p Y e (see equation 37) so that the force 
falls off according to the relation e~~ 2p Y z . To calculate the 
total force on unit area of the metal surface, we must 
integrate juaExH from z = to z = oo. ExHisa 
fluctuating quantity, and so we shall have to take its mean 


value with respect to the time. The pressure is then 

c f 30 1 

- Kl 2 
P J o z 


Using (38) this may be expressed in the form 

78. There is another application of the theory of 
76 which is important. Suppose that we have a straight 
wire of circular section, and a rapidly alternating e.m.f. 
is applied at its two ends. We have seen in 76 that 
with an infinite sheet of metal the current falls off as we 
penetrate the metal according to the law e~ p Y z . If py 
is small, there is little diminution as we go down a distance 
equal to the radius of the wire, and clearly the current 
will be almost constant for all parts of any section (see, 
however, question (12) in 79). But if py is large, then 
the current will bo carried mainly near the surface of the 
wire, and it will not make a great deal of difference whether 
the metal is infinite in extent, as we supposed in 76, or 
whether it has a cross-section in the form of a circle ; in 
this case the current density falls off approximately 
according to the law e~ p Y f as we go down a distance r 
from the surface. This phenomenon is known as the 
skin effect ; it is more pronounced at very high frequencies. 

We could of course solve the problem of the wire quite 
rigorously, using cylindrical polar coordinates. The 
formulae are rather complicated, but the result is 
essentially the same. 

79. Examples 

(1) Prove the equations (17') and (18') in 70. 

(2) Find the value of H when E x = E y = 0, and E z = 
A cos nx cos net . It is given that H = when t 0, and also 
6 = ^=1, p = cr = 0. Show that there is no mean flux of 
energy in this problem. 

126 WAVES 

(3) Prove the equation (28) in 73 for reflection and 
refraction of light polarised perpendicular to the plane of 

(4) Show that the polarising angle is less than the critical 
angle for internal reflection. Calculate the two values if 
KI - 6, K 2 = 1. 

(5) Show that the reflection coefficient from glass to air at 
normal incidence is the same as from air to glass, but that 
the two phase changes are different. 

(6) Light falls normally on the plane face which separates 
two media K! , K 2 . Show that a fraction R of the energy is 
reflected, and T is transmitted, whore 

R = / 

K t +K t 

Hence prove that if light falls normally on a slab of dielectric, 
bounded by two parallel faces, the total fraction of energy 

reflected is ^ -T-, and transmitted is r o l * . It is 

necessary to take account of the multiple reflections that take 
place at each boundary. 

(7) Light passes normally through the two parallel faces 
of a piece of plate glass, for which K = 1-5. Find the fraction 
of incident energy transmitted, taking account of reflection 
at the faces. 

(8) Show that when internal reflection ( 74) is taking 
place, there is a phase change in the reflected beam. Evaluate 
this numerically for the case of a beam falling at an angle 
of 60 to the normal when K l = 1*6, K 2 = 1, the light being 
polarised in the plane of incidence. 

(9) Show that if we assume /A = 1, then the reflection 
coefficient with metals (76) may be written in the form 
R = 1 2/V(c<7/i>), where v is the frequency. If a is 1-6 . 10 7 
(in our mixed units), calculate R for A = 10~ 3 cms. and 
A = 10-* cms. 

(10) A current flows in a straight wire whose cross -section 
is a circle of radius a. The conduction current j depends 
only on r the radial distance from the centre of the wire, 
and the time t. Assuming that the displacement current 
can be neglected, prove that H is directed perpendicular to the 


radius vector. If j(r, t) and H(r, t) represent the magnitudes 
of j and H, prove that 

^ iTT \ 4 ' ty 1 L<J 3H 

(rLr) = 47rn , = 

dr v ' J ' dr c ct 

(11) Use the results of question (10) to prove that j" satisfies 
the differential equation 

1 d I dj\ 4m7/z dj 
r dr \ dr/ c dt ' 

Show also that H satisfies the equation 

12 tT 1 n ~ET TT A 1 TJT 

G H J. (JJLL JLJ. TrTTLtO" ufl 

Use the method of separation of variables to prove that 
there is a solution of the ^ -equation of the form j =/(r)e^, 

d z f I df 

dr 2 r dr 

Hence show that / is a combination of Bessel functions of 
order zero and complex argument. 

(12) If a (in question (1 1)) is small, show that an approximate 
solution of the current equation is j = A(l -\-iar* Ja 2 r 4 )e^, 
where A is a constant. Hence show that the total current 
fluctuates between iJ", where, neglecting powers of a above 
the second, J = 7Ta 2 u4(l-j-a 4 a 2 /24). Use this result to show 
that the heat developed in unit length of the wire in unit 

time is o (l+aV/12). (Questions (10), (11) arid (12) 
are the problem of the skin effect at low frequencies.) 

[ ANSWERS : 2. H x H t = , H y A sin nx sin net ; 
4. 9 28', 9 36' ; 7. 12/13 of the incident energy is trans- 
mitted; 8. 100 20'; 9, 0-984,0-950.] 



80. The speed at which waves travel in a medium is 
usually independent of the velocity of the source ; thus, if a 
pebble is thrown into a pond with a horizontal velocity, the 
waves travel radially outwards from the centre of disturb- 
ance in the form of concentric circles, with a speed which is 
independent of the velocity of the pebble that caused them. 

When we have a moving source, sending out waves 
continuously as it moves, the velocity of the waves is 
often unchanged,* but the wavelength and frequency, as 
noted by a stationary observer, may be altered. 

Thus, consider a source of waves moving towards an 
observer with velocity u. Then, since the source is moving, 


f\ j^\ ^\. _i^\ uXTV -- '- 


'Vi^^' ^^^^ 'V^' ~^^r ^^r 




ut ntX 1 

U /\ /\ S\ /\ 


A 1 B 1 

FIG. 24 

(a) Waves when source is stationary. 

(b) Waves when source is moving. 

the waves which are between the source and the observer 

will be crowded into a smaller distance than if the source 

had been at rest. This is shown in fig. 24, where the waves 

are drawn both for a stationary and a moving source. If 

the frequency is n, then in time t the source emits nt waves. 

* It is changed slightly when there is dispersion ,* see 83. 



If the source had been at rest, these waves would have 
occupied a length AB. But due to its motion the source 
has covered a distance ut, and hence these nt waves are 
compressed into a length A'B', where ABA'B' = ut. 

ntXntX' = ut, 
i.e. A' = X-ufn = A(l w/c), .... (1) 

if c is the wave velocity. If the corresponding frequencies 
measured by the fixed observer are n and n', then, since 
n\ = c = n'X' > therefore 

' = ^- . . . . (2) 

If the source is moving towards the observer the frequency 
is increased ; if it moves away from him, the frequency is 
decreased. This explains the sudden change of pitch 
noticed by a stationary observer when a motor-car passes 
him. The actual change in this case is from hc/(c u) to 
nc/(c+u), so that 

Arc = 2ncu/(c*--u*). ... (3) 

This phenomenon of the change of frequency when a source 
is moving is known as the Doppler effect. It applies 
equally well if the observer is moving instead of the source, 
or if both are moving. 

For, consider the case of the observer moving with 
velocity v away from the source, which is supposed to be 
at rest. Let us superimpose upon the whole motion, 
observer, source and waves, a velocity v. We shall 
then have a situation in which the observer is at rest, 
the source has a velocity v y and the waves travel with 
a speed cv. We may apply equation (2) which will then 
give the appropriate frequency as registered by the observer ; 
if this is n*, then 

n(c-v) n(c-v) 

- - -- - (4) 

130 WAVES 

To deal with the case in which both source and observer 
are moving, with velocities u and v respectively, in the 
same direction, we superimpose again a velocity v upon 
the whole motion. Then in the new problem 3 the observer 
is at rest, the source has a velocity uv, and the waves 
travel with velocity c v. Again, we may apply (2) and 
if the frequency registered by the observer is n'" 9 we have 

(cv) (uv) cu 


These considerations are of importance in acoustic and 
optical problems ; it is not difficult to extend them to 
deal with cases in which the various velocities are not in 
the same line, but we shall not discuss such problems here. 

81. We have shown in Chapter I, 6 that we may 
superpose any number of separate solutions of the wave 
equation. Suppose that we have two harmonic solutions 
(Chapter I, equation (11)) with equal amplitudes and nearly 
equal frequencies. Then the total disturbance is 

(f) = a cos 27r(k 1 x--n 1 t)-\-a cos 

[2 2 J [2 2 

The first cosine factor represents a wave very similar to 
the original waves, whose frequency and wavelength are 
the average of the two initial values, and which moves 

with a velocity ^ ~. This is practically the same as the 

velocity of the original waves, and is indeed exactly the 
same if %/ij = 7i 2 /& 2 - But ^ e secon d cosine factor, which 
changes much more slowly both with respect to x and t, 
may be regarded as a varying amplitude. Thus, for the 
resultant of the two original waves, we have a wave of 
approximately the same wavelength and frequency, but 
with an amplitude that changes both with time and distance. 


We may represent this graphically, as in fig. 25. The 
outer solid profile is the curve 

y = 2a cos 2* x- - 2 *. 

The other profile curve is the reflection of this in the x 
axis. The actual disturbance (f> lies between these two 
boundaries, cutting the axis of x at regular intervals, 
and touching alternately the upper and lower profile 
curves. If the velocities of the two component waves 
are the same, so that n^fk^ = n^k^ then the wave system 
shown in fig. 25 moves steadily forward without change 

FIG. 25 

of shape. The case when njk^ is not equal to n 2 /k z is 
dealt with in 83. 

Suppose that refers to sound waves. Then we shall 
hear a resultant wave whose frequency is the mean of 
the two original frequencies, but whose intensity fluctuates 
with a frequency twice that of the solid profile curve. 
This fluctuating intensity is known as beats ; its frequency, 
which is known as the beat frequency, is just n l ^n 29 
that is, the difference of the component frequencies. We 
can detect beats very easily with a piano slightly out of 
tune, or with two equal tuning-forks on the prongs of 
one of which we have put a little sealing wax to decrease 
its frequency. Determination of the beat frequency 
between a standard tuning-fork and an unknown frequency 

132 WAVES 

is one of the best methods of determining the unknown 
frequency. Beats of low frequency are unpleasant to 
the ear. 

82. There is another phenomenon closely related to 
beats. Let us suppose that we have a harmonic wave 
<f> = A cos 27r(nt fcr), with amplitude A and frequency n. 
Suppose further that the amplitude A is made to vary 
with the time in such a way that A = a-\-b cos 2irpt. 
This is known as amplitude modulation. The result is 

<f> = (a +6 cos %7rpt) cos 27r(ntkx) 

= acos27r(n kx) 
+~-|cos 27r[(n+p)t kx] + cos 

The effect of modulating, or varying, the amplitude, is to 
introduce two new frequencies as well as the original one ; 
these new frequencies np are known as combination 

83. If the velocities of 81 are not the same (n^k^ 
not equal to n 2 /k 2 ), then the profile curves in fig. 25 move 
with a speed (n l n 2 )/(k l k 2 ) ) which is different from that 
of the more rapidly oscillating part, whose speed is 
(n 1 +n 2 )l(k 1 +k 2 ). In other words, the individual waves 
in fig. 25 advance through the profile, gradually increasing 
and then decreasing their amplitude, as they give place to 
other succeeding waves. This explains why, on the sea- 
shore, a wave which looks very large when it is some 
distance away from the shore, gradually reduces in height 
as it moves in, and may even disappear before it is 
sufficiently close to break. 

This situation arises whenever the velocity of the 
waves, i.e. their wave velocity F, is not constant, but 
depends on the frequency. This phenomenon is known 
as dispersion. We deduce that in a dispersive system 
the only wave profile that can be transmitted without 


change of shape is a single harmonic wave train ; any 
other wave profile, which may be analysed into two or 
more harmonic wave trains, will change as it is propagated. 
The actual velocity of the profile curves hi fig. 25 is known 
as the group velocity U. We see from (6) that if the 
two components are not very different, F = n/k, and 

U = (% Wa)/^ fc a ) == dn/dk. . . (7) 
In terms of the wavelength A, we have k = I/A, so that 

We could equally well write this 

dk dk dk dX 

We have already met several cases in which the wave 
velocity depends on the frequency ; we shall calculate 
the group velocity for three of them. 

Surface waves on a liquid of depth h : 

The analysis of Chapter V, equation (32) shows that the 
velocity of surface waves on a liquid of depth h is given by 

According to (9) therefore, the group velocity is F \dVjd\, 

TT 1 T r f\ , ^Trft , &nh} 
i.e. 2 I T" C ~A~ I ' 

When h is small, the two velocities are almost the same, 
but when h is large, U = F/2, so that the group velocity 
for deep sea waves is one-half of the wave velocity. 
Equation (10) is the same as the expression obtained hi 
52, equation (47), for the rate of transmission of energy 
in these surface waves. Thus the energy is transmitted 
with the group velocity. 

134 WAVES 

Electric waves in a dielectric medium : 
The analysis in Chapter VII, 69, shows that the wave 
velocity in a dielectric medium is given by 

F 2 = c a //A. 

We may put fju = 1 for waves in the visible region. Now 
the dielectric constant 6 is not independent of the frequency, 
and so F depends on A. The group velocity follows from 
(9) ; it is 


In most regions, especially when A is long, e decreases 
with A so that U is less than F. For certain wavelengths, 
however, particularly those in the neighbourhood of a 
natural frequency of the atoms of the dielectric, there is 
anomalous dispersion, and U may exceed F. When A 
is large, we have the approximate formula 

It then appears from (11) that 

Electric waves in a conducting medium : 
The analysis in Chapter VII, 76, shows that the electric 
vector is propagated with an exponential term e ip ^~^ z \ 

1 pc 

where y 2 = 27rcrfJi/pc. Thus F 2 = = ~ . According 

y 2 27T0y/. 

to (7), the group velocity is 

77 _ dP _ fa 

~~ d(py) ~ 

If we suppose that a and JJL remain constant for all 
frequencies, then this reduces to 

U = 2/y = 2V. . . . (12) 


The group velocity here is actually greater than the wave 

84. We shall now extend this discussion of group 
velocity to deal with the case of more than two component 
waves. We shall suppose that the wave profile is split 
up into an infinite number of harmonic waves of the type 

e 2m(kx-fU) 9 .... (13) 

in which the wave number k has all possible values ; we 
can suppose that the wave velocity depends on the 
frequency, so that n is a function of k. If the amplitude 
of the component wave (13) is a(k) per unit range of k, 
then the full disturbance is 

(, t) = I a(k) . eW**-nt)dk . 

k - -00 

This collection of superposed waves is known as a wave 
packet. The most interesting wave packets are those in 
which the amplitude is largest for a certain value of &, 
say & , and is vanishingly small if k~k Q is large. Then 
the component waves mostly resemble e 27r ^ k x ~ not \ and 
there are not many waves which differ greatly from this. 
We shall discuss in detail the case in which 

a(k)=A e-^*-*')'. . . . (15) 

This is known as a Gaussian wave packet, after the 
mathematician Gauss, who used the exponential function 
(15) in many of his investigations of other problems. 
A, a and k Q are, of course, constants for any one packet. 

Let us first determine the shape of the wave profile at 
t = 0. The integral in (14) is much simplified because 
the term in n disappears. In fact, 


= J 



On account of the term e~ a(k ~ k ^, the only range of k 
which contributes significantly to this integral lies around 
fc ; since when & & l/V a ^is term becomes er\ and 
for larger values of kk it becomes rapidly smaller, this 
range of k is of order of magnitude A& l/\/a. 
[n order to evaluate the integral, we use the result * 

4, (17) 

This enables us to integrate at once, and we find that 

fL/ T f\\ A I ajT*z*/cr p 27rikoX (1R\ 

Y'V* </ j "/ *-* A / t-* O . \JHJI 

V a 

The term e 27n '^ represents a harmonic wave, whose 
wavelength A = l/& , and the other factors give a varying 

mplitude A J- e~^^ G . If we take the real part of (18), 

FIG. 26 

>(#, 0) has the general shape shown in fig. 26. The outer 
urves in this figure are the two Gaussian curves 

nd </>(%, Q) oscillates between them. Our wave packet 
* Gillespie, Integration, p. 88. 


(14) represents, at t = 0, one large pulse containing several 
oscillations. If we define a half -width, as the value of x 
that reduces the amplitude to l/e times its maximum 
value, then the half-width of this pulse is (\/cr)/7T. 

At later times, >0, we have to integrate (14) as it 
stands. To do this we require a detailed knowledge of n 
as a function of k. If we expand according to Taylor's 
theorem, we can write 

n = n Q +a(k-k Q )+p(k-k Q )*j2+... 

a = (dn/dk) , j3 = (d*nldk z ) , ..... (19) 

As a rule the first two terms are the most important, and 
if we neglect succeeding terms, we may integrate, using 
(17). The result is 

+ 00 

= f 


When t 0, it is seen that this does reduce to (18), thus 
providing a check upon our calculations. The last term 
in (20) shows that the individual waves move with a 
wave velocity n /k Qt but their boundary amplitude is given 

by the first part of the expression, viz. A ^- e~ 7rl( 

Now this expression is exactly the same as in (18), drawn 
in fig. 26, except that it is displaced a distance at to the 
right. We conclude, therefore, that the group as a whole 
moves with velocity a = (dn/dk) Q , but that individual 
waves within the group have the wave velocity n /& . 
The velocity of the group as a whole is just what we have 
previously called the group velocity (7). 

If we take one more term in (19) and integrate to 
obtain </>(%, t) we find that <f> has the same form as in (20) 

138 WAVES 

except that a is replaced by or -\-7rj3it. The effect of this 
is twofold ; in the first place it introduces a variable 
phase into the term e 27ri ^ k x ~ nQt \ and in the second place it 
changes the exponential term in the boundary amplitude 
curve to the form 

This is still a Gaussian curve, but its half- width is increased 
t0 {(<j*+ir*p*P)l<m*} l l* . . . . (21) 

We notice therefore that the wave packet moves with 
the wave velocity ft /& , and group velocity (dn/dk)^ 
spreading out as it goes in such a way that its half- width 
at time t is given by (21). 

The importance of the group velocity lies mainly in 
the fact that in most problems where dispersion occurs, 
the group velocity is the velocity with which the energy 
is propagated. We have already met this in previous 

85. We shall next give a general discussion of the 
standard equation of wave motion y 2 ^ ~ , in which 

C Ov 

c is constant. We shall show that the value of </> at any 
point P (which may, without loss of generality be taken 
to be the origin) may be obtained from a knowledge of 

of o / 

the values of <t, ~~ and -~ on any given closed surface S, 
dn dt 

which may or may not surround P ; the values of cf> and 
its derivatives on S have to be associated with times 
which differ somewhat from the time at which wo wish 
to determine <f> P . 

Let us analyse <j> into components with different 
frequencies ; each component itself must satisfy the 
equation of wave motion, and by the principle of super- 
position, which holds when c is constant, we can add the 



various components together to obtain the full solution. 
Let us consider first that part of </> which is of frequency 
p ; we may write it in the form 

t(x, y, z) e<** 9 . . . (22) 

where k = Z-npIc (23) 

i/j is the space part of the disturbance, and it satisfies the 
Poisson equation 

( v *+*')lA = 0. . . . (24) 

This last equation may be solved by using Green's theorem.* 
This theorem states that if ifj 1 and t/r 2 are any two functions, 
and S is any closed surface, which may consist of two or 
more parts, such that i/^ and 2 have no singularities inside 
it, then 


The volume integral on the left-hand side is taken over 
the whole volume bounded by S, and d/dn denotes 
differentiation along the outward normal to dS. 

FIG. 27 

In this equation ^ 1 and i/r 2 are arbitrary, so we may 

/? tfcf 
put I/TJ, equal to i/j 9 the solution of (24), and ?/r 2 = , 

* See Rutherford, p. 65, equation (29). 

140 WAVES 

r being measured radially from the origin P. We take 
the volume through which we integrate to be the whole 
volume contained between the given closed surface 8 
(fig. 27) and a small sphere 2 around the origin. We 
have to exclude the origin because *Jj 2 becomes infinite at 
that point. Fig. 27 is drawn for the case of P within S j 
the analysis holds just as well if P lies outside S. 

Now it can easily be verified that V 2 <// 2 ~& 2 <A2> 

so that the left-hand side of (25) becomes I ^ 2 (\7 2 +& 2 )0 dr, 

and this vanishes, since (V 2 +& 2 )*A by (24). The 
right-hand side of (25) consists of two parts, representing 
integrations over $ and 2. On 2 the outward normal 
is directed towards P and hence this part of the full 
expression is 

When we make the radius of Z tend to 0, only one term 
remains ; it is 

where da) is an element of solid angle round P. Taking 
the limit as r tends to zero, this gives us a contribution 
4:7Tift P . Equation (25) may therefore be written 

r(e-*d<lt . ijcr 3 /1\ , . 7 .e^ 
~ J -- r e -' fcr ( - 1 +tM - 

J\rdn r 8n\rJ Y r 



Since by definition <f> = \jj(xyz)e ikct 9 we can write this 
last equation in the form 

. . . (26) 



e ik(ct-r) 

X = 

_4 e <rf-r>l A\ 

dn V dn\r) 


= A B +C, say. 

We may rewrite X in a simpler form ; for on account of 
the time variation of <, iff e ik(ct ~ r} is the same as <, taken, 
not at time t y but at time tr/c. If we write this symboli- 

r\ /T \ 

cally [6]t-ric> ^en # = ^- I- 1 [<l-r/c- I 11 a similar way, 
(77i \7y 

. 1 r^l i ^ 1 0P f^l i. r 

A = I I , and O = I I , wnere, lor 

r l^ n \t-rlc cr dn L lt-r/c 

example, ~ means that we evaluate dc/>/dn as a 

L^ n ]t-r{c 

function of x, y y z, t and then replace t by tr/c. We 
call tr/c the retarded time. We have therefore proved 

1 f 
</>P = - \ 

X dS, where 

So far we have been dealing with waves of one definite 
frequency. But there is nothing in (27) which depends 
upon the frequency, and hence, by summation over all 
the components for each frequency present in our complete 
wave, we obtain a result exactly the same as (27) but 
without the restriction to a single frequency. 

This theorem, which is due to Kirchhoff, is of great 
theoretical importance ; for it implies (a) that the value 
of < may be regarded as the sum of contributions X/far 
from each element of area of S ; this may be called the 
law of addition of small elements, and is familiar in a 
slightly different form in optics as Huygen's Principle ; 
and (6) that the contribution of dS depends on the value 
01 <f>, not at time t t but at time trfc. Now r/c is the 

142 WAVES 

time that a signal would take to get from dS to the point 
P, so that the contribution made by dS depends not on 
the present value of <f> at dS, but on its value at that 
particular previous moment when it was necessary for a 
signal to leave dS in order that it should just have arrived 
at P. This is the justification for the title of retarded 
time, and for this reason also, [^>]^_ r / c is Sometimes known 
as a retarded potential. 

It is not difficult to verify that we could have obtained 
a solution exactly similar to the above, but involving 
t-\-r/c instead of tr/c ; we should have taken if/ 2 in the 

previous work to be - instead of - . In this way 

we should have obtained advanced potentials, [0]j +r / c , 
and advanced times, instead of retarded potentials and 
retarded times. More generally, too, we could have 
superposed the two types of solution, but we shall not 
discuss this matter further. 

In the case in which c = oo, so that signals have an 
infinite velocity, the fundamental equation reduces to 
Laplace's equation,* y 2 ^ = 0, and the question of time 
variation does not arise. Our equation (27) reduces to 
the standard solution for problems of electrostatics. 

86. We shall apply this theory to the case of a source 
sending out spherical harmonic waves, and we shall 
take S to be a closed surface surrounding the point P 
at which we want to calculate <, as shown in fig. 28. 
Consider a small element of dS at Q ; the outward normal 
makes angles O l and with QO and PQ, and these two 
distances are r x and r. The value of </> at Q is given by 
the form appropriate to a spherical wave (see Chapter I, 
equation (24)) : 

(j) Q = - cos m(&r^ . . (28) 

r i 
* See Rutherford, p. 67, equation (33). 



Thus = ~ cos 0. 

em >!_ 

fl m . . 1 

a cos t/ t -I cosm(c r t ) smm(a r.) V. 

(r^ r x J 

Now A = 277/m, so that if r x is much greater than A, which 

FIG." 28 

will almost always happen in practical problems, we may 


ma cos 0, . 

sm m i c i r \ ^ 



f - = -Acos^ 

a^ \r/ 


The retarded values are easily found, and in fact, from (27), 

ma cos 9, . 
v c.i^ 

_j C os 9 cosm(ci~-[r+r 1 ]) cos 9 sin ra(c [r+r,]). 

r 2 n cr r n 

144 WAVES 

We may neglect the second term on the right if r is much 
greater than A, and so 

ma . * /n . 

X = (cos 0+cos X ) smm(d [r+rj) . (29) 

Combining (29) with (26) it follows that 

1 fraa 

<h p = I (cos 6 + cos cM sin m(ct[r -\-rJ\dS 

477-J rr x 

= r (cos 6 -f- cos 0j) sinm(c^ [r+rj])^ . (30) 

2AJ rri 

If, instead of a spherical wave, we had had a plane 
wave coming from the direction of 0, we should write 

r x now being measured from some plane perpendicular to 
OQ, and (30) would be changed to 

<h p = -^ I - (cos + cos 0J sin m(ct[r +r l ])dS. (31) 
2A J r 

We may interpret (30) and (31) as follows. The effect 
at P is the same as if each element dS sends out a wave 

- -4 /cos 0+cos 0A 
of amplitude y- I \dS, A beuig the amplitude 

of the incident wave at dS ; further, these waves are 
a quarter of a period in advance of the incident wave, 
as is shown by the term sin w(c [r-f-fi]) instead of 

cQ$m(ct r x ). ~ (cos 0+cos gj { s called the inclination 


factor and if, as often happens, only small values of 
and t occur significantly, it has the value unity. This 
interpretation of (30) and (31) is known as Fresnel's 

The presence of this inclination factor removes a 
difficulty which was inherent in Huygen's principle ; this 


principle is usually stated in the form that each element 
of a wave -front emits wavelets in all directions, and these 
combine to form the observed progressive wave-front. In 
such a statement there is nothing to show why the wave 
does not progress backwards as well as forwards, since 
the wavelets should combine equally in either direction. 
The explanation is, of course, that for points behind the 
wave -front cos 6 is negative with a value either exactly 
or approximately equal to cos 6 l9 and so the inclination 
factor is small. Each wavelet is therefore propagated 
almost entirely in the forward direction. 

Now let us suppose that some screens are introduced, 
and that they cover part of the surface of S. If we assume 
that the distribution of <f> at any point Q near the screens 
is the same as it would have been if the screens were not 
present, we have merely to integrate (30) or (31) over 
those parts of S which are not covered. This approxi- 
mation, which is known as St Venant's principle, is not 
rigorously correct, for there will be distortions in the 
value of (f)Q extending over several wavelengths from the 
edges of each screen. It is, however, an excellent approxi- 
mation for most optical problems, where A is small ; 
indeed (30) and (31) form the basis of the whole theory 
of the diffraction of light. With sound waves, on the other 
hand, in which A is often of the same order of magnitude 
as the size of the screen, it is only roughly correct. 

87. We conclude this discussion with an example of 
the analysis summarised in (31). Consider an infinite 
screen (fig. 29) which we may take to be the xy plane. 
A small part of this screen (large compared with the wave- 
length of the waves but small compared with other distances 
involved) is cut away, leaving a hole through which waves 
may pass. We suppose that a set of plane harmonic waves 
is travelling in the positive z direction, and falls on the 
screen ; we want to find the resulting disturbance at a 
point P behind the screen. 




In accordance with 86 we take the surface S to be 
the infinite xy plane, completed by the infinite hemisphere 
on the positive side of the xy plane. We may divide the 

P (x, y, z) 

Fia. 29 

contributions to (31) into three parts. The first part 
arises from the aperture, the second part arises from the 
rest of the screen, and the third part arises from the 


If the incident harmonic waves are represented by 
= a cos m(ct z) this first contribution amounts to 

a Cl 

2Aj ; 

(1+ cos 6) smm(ctr)d8. 

We have put 9 l = in this expression since the waves 
fall normally on to the xy plane. We shall only be con- 
cerned here with points P which lie behind, or nearly 
behind, the aperture, so that we may also put cos 0=1 
without loss of accuracy. This contribution is then 

T - smm(ctr)dS . . (32) 

The second part, which comes from the remainder of 
the xy plane, vanishes, since no waves penetrate the 
screen and thus there are no secondary waves starting there. 

The third part, from the infinite hemisphere, also 
vanishes, because the only waves that can reach this part 
of $ are those that came from the aperture, and when 
these waves reach the hemisphere their inclination factor 
is zero. Thus (32) is in actual fact the only non-zero con- 
tribution and we may write 

- -s 
Aj r 

mim(ctr)d8 . . (33) 

Let P be the point (.E, y, z) and consider the contribution 
to (33) that arises from a small element of the aperture at 
Q (, ??, 0). If OP = /, and QP = r, we have 

77 2 . . (34) 

Let us make the assumption that the aperture is so small 
that | 2 and 17 2 may be neglected. Then to this approxi- 
mation (34) shows us that 


148 WAVES 



Again without loss of accuracy, to the approximation to 
which we are working, we may put l/r I//, and then 
we obtain 

</>P = A sin {m(ct /)+}, 


A* = <7 2 + 2 , tan e = S/C, 


c(x, y) = ^> I c s^(z+^?)d<fy, 

. . (35) 

Once we know the shape of the aperture it is an easy 
matter to evaluate these integrals. Thus, if we consider 
the case of a rectangular aperture bounded by the lines 
= a, T? = j3, we soon verify that 8 = 0, and that 


= Xf J J C S A? 

___ ^ ^ ^ 

A/ px py 

where j9 = 2?r/A/. If we are dealing with light waves, 
then the intensity is proportional to O 2 and the diffraction 
pattern thus observed in the plane z = / consists of a 
grill network, with zero intensity corresponding to the 
values of x and y satisfying either sin pax = 0, or 
sin pfty = 0. 

The theory of this paragraph is known as Fraunhofer 
Diffraction Theory. 


88. We conclude this chapter with a discussion of 
the equation 

-... (37) 

where p is some given function of x, y y z and t. When p = 
this is the standard equation of wave motion, whose solution 
was discussed in 85. Equation (37) has already occurred 
in the propagation of electric waves when charges were 
present (Chapter VIII, equations (17') and (18')). We 
may solve this equation in a manner very similar to that 
used in 85. Thus, suppose that p(x, y, z, t) is expressed 
in the form of a Fourier series with respect to t, viz., 

p(*,y,*> t) = Za k (x, y, z)e ikct . . . (38) 


There may be a finite, or an infinite, number of different 
values of &, and instead of a summation over discrete 
values of k we could, if we desired, include also an integra- 
tion over a continuous range of values. We shall discuss 
here the case of discrete values of k ; the student will 
easily adapt our method of solution to deal with a 

Suppose that <^(#, y, z, t) is itself analysed into com- 
ponents similar to (38), and let us write, similarly to (22), 

t(x, y, z, t) = 2fa(x, y, z)e*** . . . (39) 

the values of k being the same as in (38). If we substitute 
(38) and (39) into (37), and then equate coefficients of 
e ikct , we obtain an equation for \f/ k . It is 

This equation may be solved just as in 85. Using 
Green's theorem as in (25), we put ^ = *l* k (x 9 y, z), 

e -ikr 

^ a . - 1 taking H and 8 to be the same as in fig. 27. 

150 WAVES 

With these values, it is easily seen that the left-hand side 
of (25) no longer vanishes, but has the value 

y> ^-ttr,*- t t e (41) 

the integral being taken over the space between S and S. 
The right-hand side may be treated exactly as in 85, 
and gives two terms, one due to integration over , and 
the other to integration over S. The first of these is 

y p9 Z P ) .... (42) 

The second may be calculated just as on p. 140. Gathering 
the various terms together, we obtain 

_| I J JL* ih k e~ ikr I- 1 -\-ikdfk \dS . (43) 

477 J ( r dn dn\r/ r dn] 

Combining (38), (39) and (43) we can soon verify that our 
solution can be written in the form 

^(^p? UP* Z P) = t "~ r!c dr-{-- X d8, . . (44) 
J J 

where X is defined by (27). This solution reduces to (27) 
in the case where p = 0, while it reduces to the well-known 
solution of electrostatics in the case where c = oo. 

We have now obtained the required solution of (37). 
Often, however, there will be conditions imposed by the 
physical nature of our problem that allow us to simplify 
(44). Thus, if p(x, y, z, t) is finite in extent, and has only 
had non-zero values for a finite time t>t Q) we can make 
X = by taking 8 to be the sphere at infinity. 'This 
follows because X is measured at the retarded time trjc, 
and if r is large enough, we shall have tr/c<t Q , so that 


[(/>]t~ric and its derivatives will be identically zero on 8. 
In such a case we have the simple result 

%^r, . . . (45) 

the integration being taken over the whole of space. 
Retarded potentials calculated in this way are very 
important in the Classical theory of electrons. 

89. Examples 

(1) Aii observer who is at rest notices that the frequency 
of a car appears to drop from 272 to 256 per second as the 
car passes him. Show that the speed of the car is 
approximately 20 m.p.h. How fast must he travel in the 
direction of the car for the apparent frequency to rise to 
280 per second, and what would it drop to in that case ? 

(2) Show that in the Dopplor effect, when the source and 
observer are not moving in the same direction, the formula) 
of 80 are valid to give the various changes in frequency, 
provided that u and v denote, not the actual velocities, but 
the components of the two velocities along the direction in 
which the waves reach the observer. 

(3) The amplitude A of a harmonic wave A cos 27r(nt kx) 
is modulated so that A a-\~b cos 2?rpt-{-c cos 2 2irpt. Show 
that combination tones of frequencies np, n2p appear, 
and calculate their partial amplitudes. 

(4) The dielectric constant of a certain gas varies with the 
wavelength according to the law e = A-\-B/\ 2 (7A 2 , where 
A, B and C are constants. Show that the group velocity U 
of electromagnetic waves is given in. terms of the wave velocity 
V by the formula 

(5) In a region of anomalous dispersion ( 83) the dielectric 

A A 2 

constant obeys the approximate law e = 1-f- . A more 


152 WAVES 

j4A 2 (A 2 _ A 2 ) 

accurate expression is e = 1 H -- - - - ~ , where A, B and 

(A AQ ) -j-x>A 

A are constants. Find the group velocity of electric waves 
in these two cases. 

(6) Calculate the group velocity for ripples on an infinitely 
deep lake. ( 55, equation (54).) 

(7) Investigate the motion of a wavepacket ( 84) for 
which the amplitude a is given in terms of the wave number k 
by the relation 

a(k) = 1 if \k-kfl <kt 

k and k l being constants. Assume that only the first two 
terms of the Taylor expansion of n in terms of k are required. 
Show that at time t the disturbance is 


TT(X at) 

where a = (dn/dk) . Verify that the wavepacket moves as 
a whole with the velocity a. 

(8) Show that when dS is normal to the incident light 

( 86), the inclination factor is - . Plot this function 


against 6, and thus show that each little element dS of a 
wave gives zero amplitude immediately behind the direction 
of wave motion. Using the fact that the energy is proportional 
to the square of the amplitude of <f> t show that each small 
element sends out 7/8 of its energy forwards in front of the 
wave, and only 1/8 backwards. 

(9) A plane wave falls normally on a small circular 
aperture of radius b. Discuss the pattern observed at a large 
distance / behind the aperture. Show that with the formulae 
of 87, if the incident wave is ^ = a cos m(ct z), then 
S = 0, and if P is the point (x, 0, /), then 

C I \/(^ 2 "~ 2 ) cos Pfdg where p = 

= p f cos (pb cos 0) sin 2 dO. 
A / Jo 


Expand cos (pb cos 0) in a power series in cos 0, and hence 
show that 


~ A/ \ 

where & = pb/2 = nbx/Xf. Since the system is symmetrical 
around the z axis, this gives the disturbance at any point 
in the plane z =/. It can be shown that the infinite series 
is in fact a Bessel function of order unity. It gives rise to 
diffraction rings of diminishing intensity for large values of x. 
(10) The total charge q on a conducting sphere of radius a 
is made to vary so that q = 47ra 2 cr, where a = for <0, and 
a = cr sin pt for t>0. Show that if = ^ = 1, ( 70 eq. (18')) 
the electric potential ^ at a distance E from the centre of the 
sphere is given by 

ct<Ra, (f> = 0, 

Ra<ct<R+a 9 </> = -^^{lcosp(t -- 

#/t i \ c 

47racor ?a / JR\ 

R+a<ct, <f> = - -sin siiipU -- ). 
p,R c \ c/ 

[ANSWERS : 1. c/34, where c = vel. of sound, 248-5 ; 
3. a+c, 6/2, c/4; 

i, A7 iv 

where v = wave v oclty; 

6. U - JF.] 


The Numbers refer to pages 

Absorption coefficient, 121 

Adiabatic, 87 

Advanced potentials, 142 

times, 142 
Ampere's Rule, 104 
Amplitude, 2 

modulation, 132 
Amplitudes, partial, 33 
Anomalous dispersion, 134 
Antinodes, 6 

Bars and springs, longitudinal 

waves in, 51-59 
Basins, tides in, 66, 70, 74-76 
Beats, 131 

frequency, 131 
Bell, vibrations of, 49 
Boundary conditions, 1, 27, 30, 

38, 52, 69, 73, 82, 90, 94 
Brewster's angle, 117 

Capillary waves, 81-84 
Chladni's figures, 48 
Circularly polarised light, 111 
Combination tones, 132 
Compressible fluid, 87 
Condensation, 89 
Conductivity, 103 
Conical pipe, sound waves in, 96 
Constant of separation, 9 
Constitutive relations, 104 
Coordinates, normal, 36, 47, 53 

D'Alembert, 7 
Damping, 15, 39 
Decay, modulus of, 15 
Degenerate vibrations, 46 
Dielectric displacement, 102 
Diffraction of light, 145 

theory, Fraunhofer, 148 
Dispersion, 132 

anomalous, 134 

Displacement current, Maxwell's, 

104, 119, 122 
Doppler effect, 129 
Drude, 119 

Electric and magnetic field 

strengths, 102 
waves, 102-127 

Elliptically polarised light, 111 
Energy, kinetic, 23, 33, 47, 54, 

78, 96 
loss of, 124 

potential, 24, 33, 47, 54, 78, 96 
rate transmitted, 79, 111, 133 
Equation of telegraphy, 15 
wave motion, 1-20, 5 
wave motion, complex solu- 
tions, 16-17 
Exponential horn, 99 

Field strengths, electric and 

magnetic, 102 
Fraunhofer diffraction theory, 


Free surface, 62, 82 
Frequency, 3 
Fresnol's principle, 144 
Fundamental, 35, 49, 58, 91 

Gaussian wave packet, 135 
Gauss' theorem, 104 
General considerations, 128-151 
Ground note, 35 
Group velocity, 81, 133, 135, 
137, 138 

Hagen and Rubens, 124 
Half-width, 137 
Harmonic wave, 2, 16-17 
Horn, 96, 99 
Huygen's Principle, 141, 144 

Inclination factor, 144 
Incompressible liquid, 60 



Index, refractive, 106, 118 
Induction, magnetic, 102 
Intensity, 148 

Internal or total reflection, 118 
Isothermal, 87 

Joule heat, 124 

Kinetic energy in bars, f>4 

liquids, 78 

membranes, 47 

sound, 96 

strings, 23, 33 
Kirchhoff, 141 

Lenz's law of induction, 104 
Light, velocity of, 106 
Liquids, waves in, 60-86 
Longitudinal waves, 21 

in bars and springs, 51-59, 87 
Long waves in shallow water, 62 
Lowest frequency, 35 

Magnetic and electric tield 

strengths, 102 
Maxwell's displacement current, 

104, 119, 122 
equations, 102 
relation, 106 

Membranes, waves in, 43-50 
Mersenne's law, 35 
Mode, normal, 30, 33, 35, 37, 

39, 45, 48, 55, 91, 95 
Modulation, amplitude, 132 
Modulus of decay, 15 

Nodal planes, 6 
Nodes, 6 

Non-conducting media, 105 
Normal coordinates, 36, 47, 53 
Normal modes in bars, 55 
circular membranes, 48 
rectangular membranes, 48 
sound waves in pipes, 91, 95 
strings, 30-33, 35, 37, 39 

Observer, moving, 129 
Organ pipe, 91 
Overtones, 35, 49 

Packet, wave, 135 
Partial amplitudes, 33 
Paths of particles, 70, 77 
Period, 3 

equation, 39, 77, 96 
Phase, 4, 16, 17 
Pipes, sound waves in, 90-96 
Pitch, 35 
Plane of polarisation, 1 1 1 

polarised light, 111 

nodal, 6 

wave, 4 

Polarisation, piano of. 111 
Polarising angle, 1 1 7 
Potential, advanced, 112 

electric, 106, 108 

energy in bars, 54 

oiiorgy in liquids, 78 

energy in rnembraiio;, 47 

onergy in sound, 96 

energy in strings, 24, 33 

magnetic or vector, 106, 108 

retarded, 142, 151 

velocity, 60, 72, 87, 92 
Poyiit ing vector, 104, 112, 124 
Pressure, radiation, 124 
Principle of superposition, 5, 130, 

135, 138 
Profile, wave, 2 

Progressive waves, 6, 13, 23, 26, 
28, 30, 40, 66, 71, 74, 77, 
83, 95, 100, 109-125 

Reduction to a steady wave, 40, 


Reflection coefficient, 27, 117, 

of light waves, 113 

total or internal, 118 
Refraction of light waves, 113 
Refractive index, 106, 118 

complex, 120 
Relaxation, time of, 109 
Resistance, specific, 103 
Retarded potential, 142, 151 

time, 141 
Ripples, 83 

Screen, 145, 146 
Separation constant, 9 
Skin effect, 125 



Snell's law, 115 
Sound, velocity of, 89 

waves, 87-101 
Source, moving, 128 
Springs and bars, longitudinal 
waves in, 51-59 

vibration of, 55 
Stationary waves, 6, 32, 38, 45, 

48, 53, 75, 95 
Strings, normal modes, 3 1 

waves on, 21-42 
St Venant's Principle, 145 
Superposition, principle of, 5, 

130, 135, 138 
Surface, free, 62, 82 

tension, 63, 81 

waves in liquids, 63, 72-81 

Telegraphy, equation of, 15, 119 
Tidal waves, 62, 63-72 
Time of relaxation, 109 
Tone, 35 

combination, 132 
Total or internal reflection, 118 
Transmission coefficient, 28, 1 17, 

Transverse waves, 21, 44, 109 

Vector, Poynting, 104, 112, 124 
Velocity, group, 81, 133, 135, 

137, 138 
of light, 106 

Velocity, of sound, 89 
potential, 60, 72, 87, 92 
wave, 132 

Vibrations, degenerate, 46 

Wave, capillary, 81-84 
electric, 102-127 
harmonic, 2, 16-17 
in bars and springs, 51-59 
in liquids, 60-86 
in membranes, 43-50 
long, in shallow wa-ter, 62 
longitudinal, 21, 51-59, 87 
motion, equation of, 1-20, < 
number, 3 
on strings, 21-42 
packet, 135 

plane, 4 
profile, 2 
30, 40, ' 

45, 4S, 

6, 13, 23, 26, 28, 
71, 74, 77, 83, 
95, 100, 109-125 
reduction to a steady, 40, 7 1 
sound, 87-101 
stationary, 6, 32, 

53, 75/95 
surface, 63, 72-81 
tidal, 62, 63-72 
transverse, 21, 44, 
velocity, 132 
Wave-front, 4 
Wavelength. 3