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Full text of "Waves – Project Physics Programmed Instruction"

it^ 




The Project Physics Course 

Programmed Instruction 



Waves 1 





Waves 2 



# 



The Kinetic-Molecular Theory of Gases 




INTRODUCTION 

You are about to use a programmed text. 
You should try to use this booklet where there 
are no distractions — a quiet classroom or o study 
area at home, for instance. Do not hesitate to 
seek help if you do not understand some problem. 
Programmed texts require your active participa- 
tion and are designed to challenge you to some 
degree. Their sole purpose is to teach, not to 
quiz you. 

In this book there are three separate pro- 
grams. The first. Waves 1, proceeds from left to 
right across the top port of the book. Waves 2 
parallels it, starting at the front of the book 
again, and using the middle portion of each 
page. The third program, Kinetic-Moleculor 
Theory, takes up the bottom part of each page. 
It con be studied either before or after the two 
wave programs. 



This publication is one of the many instructional material* 
developed for fh<! Project Physics Course. These mo- 
terials include Texts, Handbooks, Teocher Resource Books, 
Readers, Progrommed Instruction Booklets, Film Loops, 
Transparencies, 16mm films ond laboratory equipment. 
Development of the course has profited from the help of 
many colleagues listed in the text units. 



Directors of Project Physics 

Gerald Holton, Department of Physics, Harvard University 

F. James Rutherford, Chairman, Department of Science 

Education, New York University 
Fletcher G. Watson, Harvard Graduate School of Education 



Copyright© 1974, Project Physics 
Copyright (C) 1971, Project Physics 
All Rights Reserved 
ISBN: 0-03-089643-6 
(H2 (!()« 9K76 

Project Physics is a registered trademark 




A Component of the 
Proiect Physics Course 

Distributed by 

Holt, Rinehort ond Winston, Inc. 

New York — Toronto 



Cover Art by Andrew Ahlgren 



Waves 1 The Superposition Principle 



Knowledge of the behavior of waves is 
of basic importance in physics, in this pro- 
gram you will learn something about brief 
wave disturbances — pulses: how they travel 
and what happens when two pulses pass 
through the same region at the same time. 



Waves 2 Periodic Waves 



When the seme wave shape is repeated 
over and over again, the wove is called a 
periodic wave. In this program you will learn 
the relationships among the frequency, period/ 
wavelength, and speed of a periodic wove. 



Kinetic-Molecular Theory of Gases 

This program consists of a series of problems 
that will help you to understand the kinetic mole- 
cular theory end the behavior of gases. Following 
most problems is a hint (printed upside down below 
the dashed line). In the answer frames the solutions 
to the problems are worked out in detail. To derive 
the most benefit from this program, make a reason- 
able effort to solve each problem without help. If 
success does not come, read the hint. If you still 
hove trouble, look at the solution, then go back 
and work through the problem yourself. 

The problem sequence used in this 
program is adapted from: 

Physics For the Enquiring Mind 
Eric M. Rogers 
Princeton University Press, 1960 



INSTRUCTIONS 

1. Frames: Each frame contains a question. Answer the question by writing in the blank space next to the frome. 

Fromes ore numbered 1, 2, 3, ... 

2. Answer Blocks: To find an answer to o frame, turn the page. Answer blocks are numbered Al, A2, A3, . . . 

This booklet is designed so that you can compare your answer with the given answer by folding 
back the page, like this: 





1 







"Tu^ 


2 






;__ 
















1 




— -■■■ 







3. Always write your answer before you look at the given answer. 

4. If you get the right answers to the sample questions, you do not hove to complete the program. 



INSTRUCTIONS: Same as for Waves 1 aboN 



INSTRUCTIONS: Same as for Waves 1 abov 



Sample Question A 



The disturbance travels from P^ to P2 in a time At. 

If At = 1.0 seconds, what is the speed ot wl, h the 
disturbance travels? 



I ■ I H 



Lmetfit.- 



I I I li 



i ! r- i, II I ^ 



-U»- ' ♦ 



,,,:'P.:;! 



^ 



^■ 



t-^ 



1 1 ' 1 1 



I M I I I 






£ 



:^ 



t > ■ I I — . . I . . . i I I I I H ' ! M ! T ) 



' I I ) I I ' ' — ' ' t t r p t ' ■ ' I ' 



Sample Question A 



The waves on the diagram were produced in 1.5 seconds. 



What is the frequency f in cycles per second? 



Turn page to begin the program 

The Kinetic-Molecular Theory of Gases 



Answer to A 
Ad 

" = a7 

\6 = 3.0 m 
At = 1.0 sec 
3.0 m 



3.0 m, sec 



Answer to A 



6.0 cycles 
1.5 sec 



4.0 cycles sec 



In developing the kinetic-molecular theory of gases, we use elastic collisions of 
bolls as a model, and find results that correspond to the observed behavior of gases. 
So, for that set of phenomeno, we consider invisible molecules as though they were per- 
fectly elastic spheres. The theory may seem difficult at first, but if you carry it through 
once, thinking about it corefully, it will soon begin to moke good sense. 



Sample Question B 

The diagram shows two pulses traveling towards each other 
along a rope. An instant later the pulses arrive at Point P, the 
center of the rope. What is the maximum displacement of the 
rope at that instant? 



Sample Question B 



The frequency of a wave is the number of 

(i) that pass a point 

per (ii) , and is equal 

to the inverse of the (iii) . 



A ball of moss 2.0 kg moving 12 m/sec to the right hits 
a massive wall head-on and stops dead. 



The momentum of the 
ball before impact is . 



("nits) t^vV 



'X|i30|3A sauji; ssoiu sjonba ujn;u9LuoiiJ joij; |io33>j uutu 



An! 



to B 



a + b 



Maximum displacement is a > b. 
If you missed this or Somple 
Question A, you can profit by 
working through this progrom 
booklet, beginning with Question 
1. if you got Sample Questions 
A and B right, go on to Sample 
Question C. 



Anss 



to B 



(i) cycles 

(complete waves) 

(ii) second 

(iii) period 



A1 



p = mv 



m = 2.C kg 
w « -^12 w sec 

T = (2.0 kg; (♦12 m sec) 



1 p - *24-3' — 



Sample Question C 

A rope is hung as in the diagram, one end fixed 
and the other end free. At point A the rope is moved 
sideways and back suddenly, creating similar pulses 
that travel towards the ends. 

Sketch what the pulses will look like after re- 
flection from the ends. 




Fixe<i 



aaj-j 



Sample Question C 

If waves of frequency f = 20 cycles per 
second travel with speed v = 40 meters per 
second in a given medium, 

(i) what is the wavelength of the waves in 
the medium? 

(ii) what is the period of the waves? 



Since the ball stops dead when it strikes the wall, the 

momentum of the boll after impact is 

(units) 



'0J3Z SI IDOduJI J3|^D 
{|Dq 3L|4 jO Xil30|3A 9L|4 'UOISI{|OD DliSD|3UI Sll^4 U| ■iU}\-{ 



Answer to C 



^^- 

^ 



>^ 



l( you got thrs question right also, you 
need not complete Waves 1. Turn instead 
to the Proiect Physics Programmed lr>- 
struction booklet Waves 2. If you missed 
Somple Question C (but got A and B right) 
begin this progrom at question 16. 



Answer to C 



(i) A =1 
40 



20 cycles sec 
A = 2.0 m cycle 



(ii) T = — = — sec cycle 
f 20 ^ 

T = .05 sec cycle 



A2 



p = m V 



m' = 2.0 kg 
v' = 0.0 m sec 

■p - (2.0 kg) (0.0 m sec) 



T = 0.0 



kg ■ m 



Two people hold opposite ends of a rope. The sender, 
s, snaps his end of the rope, creating a disturbance that 
travels along the rope towards the receiver, r. 




1 

T 



—A 



A short time (At) later, the disturbance passes point P. 
Draw the disturbance as it passes P. 



Sample Question D 



The figure shows an attentuated (domped) periodic wave. 




Of the following, which property is changing? 
speed of wave, wavelength, amplitude, 
frequency, period. 



If the momentum of the ball before impact was 

-^24 leg • m/ sec, and the momentum after impact was 

0.0 kg- m/ sec, then the change o{ momentum that 

the ball underwent is . 

(units) 



•d - ,d = d\7 

'looduji sjoisq luntusujouj snujiu ^Doduii 
j^^D ujoiudujouj Sjonba ujniusujouj ^o sSuoi^ij) -^^!H 



Al 



A 

P 



Answer to D 

Only amplitude is changing. 



If you were able to answer al 
of the sample questions cor- 
rectly, the rest of the prograr 
is optionol. 



A3 

Ap" = p' - "p 

P -- 0.0 - — 

sec 

— ^ »9' •" 



_-. _ kg • rr 

Ap (0.0 - 24 ;- 



The disturbance below travels through the rope from P^ 
to Pj in time At. 

If At = 1.0 seconds, what is the speed at which the dis- 
turbance travels in the rope? 

^^ 1 mete-- ^ ^ ■ 

»' < » t I — I — I — I — ( — I — I— ♦• • 




-• r 



P^2 



-•r 



Waves are shaken on a rope by moving point P up and down. 

Crests are generated when the displacement of P is positive, 
and troughs are made by negative displacements of P. 

Label the crests and troughs on the diagram. 




Newton's third law applies to this case, therefore we can 
say that the change of momentum of the wall must be 



(units; 



•Das/iu.6)| t^j - 

SI {{0<^ am 10 uin^usuiouj ^o 36uoij3 dij; ;ol|| mou)| 

3M puo 'OJ3Z s: mn^usujouj lo s36uol|3 9L|4 ^o luns 3L| i 

';33|ao J3L|io 9L|; Xq p3u;o6 ujn(U3UJoiu 91^4 0{ spn; 
-luboLLi ui lonoa si uois||iod d ui i33iqo suo Xq jsoj oin; 
-U3UJOUJ joq; S3|0|S ujn^udiuouj jO uojiOAjasuoD ^o mdi 9i^j_ -t^lH 



A2 



Ad 

At 

Ad = 3.0 m 
At = 1.0 sec. 
3.0 



1.0 sec 



= 3.0 m/sec 



A1 









c 


r« 


;s 


t 


s 




















' 1 








% 










r 


\ 










i 














i 














S-T.. 








i 


' 




\ 








1 






1 








J 






Ik 






I 






i , 








- n 






I 




r 






JK-- J 














1 




/ 






1 TL 
















\ 




J 










-W^ 


















^ 


f 










-3^ 




















\ 










*i 




















K 




^ . 












^ 


t 






troughs 



A4 



ATwoII -^ -^boll = 



i5iPwoll = --^Pbo 

= - (-24 



leg* m 



APwoii » +24 



kg- m 





3 [ 

A brief disturbance that moves through a medium is | 
called a pulse. | 

In the cose of the previous two frames, the medium ' 
is (i) , and the source of ' 
the pulse is (ii) 






2 1 

A complete waveform is a cycle. | ^ 


■ I 1 
Draw 3 cycles between points P and P'. . " • 




5 1 

Now suppose that the wall is hit by a stream of such bolls, ' 
each of mass 2.0 kg and velocity of +12 m sec. Suppose that ' 
1000 balls hit the wall head-on during 10 seconds. ' 

The total change of momentum of the wall in that 10- second l 
Deriod is i 


(units) j 


•uBis uj aijsoddo inq 'S||Dq qqq[ ai^; 40 ' 

96u0LfD UJn4U3UJ0UJ 31^; 04 |Dnb3 SI IJOM 3lj4 ^0 36U0U3 1 

ujniusiuoiu am 3jo^3J3ij; 'p3aj3suo3 5 1 ujn4U3UJoyy ' 

•S||D<| QQQ[ JO^ luniUSUJOlU 40 36uDgD 31^4 34D|nD|03 :*ui|^ 1 





A3 

(i) the rope 

(ii) the hand movenient of the sender' 



A2 




A5 

~m. kg • m 

APwoll = -24—— 



ATiooo - -24000 



eg • m 



ball) 



APwoll • Apiooo = 
balls 

ATwoII " - Apiooo 
bolU 

= - ( - 24000 ^^^-^ ) 



ATwoII - * 24000 



kg- m 



A small object is fastened to the rope. 
Draw the object and pulse when the pulse is 
centered at point P. 



^ 



In one second., point P makes 6 cycles, and each cycle 
contains a crest and a trough. 

(i) How many cycles per second pass an observer at point S? 
(11) How many troughs pass the observer at S in one second? 



crest 




trough 



Note: Actually, *he changes of momentum occur in bumps, one 
bump when each boil strikes the wall. You can still cal- 
culate the total change ot momentum, then use this total 
change to calculate an overoge force. Just forget that 
the momenta changes occur in bumps, and divide the 
total change cf momentum by the total time (ten seconds 

in this case) and obtain overage force, F 

^ ' av 

The following frame is en illustration of this point. 



A4 




A3 



(i) 6 cycles 
(ii) 6 troughs 





^ 1 

We know that the water wove below contains energy ' 
because | 


^ 1 


^ ^^ ^"^ 1 


J^ i 






4 1 

The number of cycles per second is the frequency • 
of the wave. (In the last question, the frequency was • 
6 cycles per second.) j 

If 12 cycles are produced in 3.0 ■ 
seconds, the frequency of the i 
waves is 1 


(units) 1 




6 j 

The average force on the wall, during the ten second period, ' 

due to all 1000 bolls losing momentum, is given by applying ' 

Newton's second law to the whole collection of balls. ' 

^ _ A^ 1 

F = ma and a = "T" • 1 

At 1 

Hence ' 
— Av" m A v" A (niv) Ap" 

~ "" At " At " At " At ■ ! 

Average force, Favf o" ^he wall must be . , 


(units) 1 





A5 



For one thing, the wave lifts 
the boat. The increase in the 
boots gravitotiorvjl potential 
energy must hove cor»>e from 
the wove. 



A4 



f = 



12 cycles 
3.0 sec 



= 4.0 cycles sec 



A6 



Fw 



^w 



oil 



oil = 



\t 



A^Pwall = ^24000 



9 • m 



At - 10 sec 
F'woII • 24000 kg • m/sec 



2400 



10 sec 

kg • m 
sec^ 



Fqv " +2400n«wton$ 



If you lift a weight against gravity, you do work, and 
to do work requires energy. 

A pulse passes a small weight attached to the rope. 



JX 



-•r 



s«- 



-i- 



5*- 



/M 



From this experiment it can be seen that the pulse 

transmits , because it lifted the 

weight as it passed. 



The waves on the diagram were produced in 1.5 seconds. 




What is the frequency f in cycles per second? 



In the previous frames we dealt with inelastic collisions; 
the balls struck the wall and stopped dead. Suppose 1000 hard 
steel balls, each of mass 2.0 kg, hit a massive wall head-on in 
the course of 10 seconds; but this time they arrive with a velo- 
city of +12 m/sec, bounce straight back with equal speed, 
12 m/sec, but in the opposite (negative) direction. 

The momentum of each ball before impact is 



vunits) 



A6 



energy 



A5 



6 cycles 
1.5 sec 



= 4.0 cycles sec 



A7 

p = m V 

m = 2.0 kg 
V = ♦12m sec 

P = (2.0 leg) (+12 m sec) 

kg- m 



Notice that momentum is o vector 
quantity and the oirection ot the 
momentum vector i$ the direction 
of the velocity vector. 



A pulse is a disturbance moving through 

(i) , and 

transmitting (ii) . 



The quantity T represents the time required to 
generate one cycle (complete wave). The time interval 
(T) is called the period of the wave. 

When f = 1 cycle/sec; T = 1 sec/cycle 

f = 2 cycles/sec; T = 1/2 sec/cycle 

f = 3 cycles/sec; T = 1/3 sec/cycle 

In general: 

f = X cycles/sec; T = . 



8 



The momentum of each ball after impact is 
(units) 



i^ 



•43od 
-lUI (a) Jdi^o puD 

(a) SJO^Sq X4j30|3A 
31^; ^O UO!;33Jip 31^4 
SMOl^S UIOjSdjP 31^^ 



Hint: This is easy, but watch the sign indicating the direction 
of the vector. 



A7 

(i) a medium 
(ii) energy 



A6 



1 



T - — sec/cycle. 



A8 



p = mv 

m = 2.0 kg 

v" = - 12 m sec 

?' = (2.0 kg) (-12 m sec) 



■p' = -24 



The minus sign before the 
onswer meons thof the mo- 
mentum of each ball after 
collision IS directed in the 
opposite direction from the 
momentum before collision. 



A disturbance can be described as a displacement of 
particles in a medium from their normal positions. 

Using the information below, complete the table. 




t-rr+- 



Displacement 











Point 


A 


B 


C 


D 



From frame 6 we see that the period is related to the 
frequency of a wave. 

1 , t ^ 

T - — , and t = — . 

What is the period of a wave whose frequency is 
10 cycles/sec? 



The change of momentum of one ball is 



(units) 



-OJ3Z 40U SI J3MSUD 9l^J_ -^"!|-| 



A8 



Displacement 





2 


1 





Point 


A 


B 


C 


D 



A7 



T = r 



= — sec/cycle 
10 

= 0.1 sec/cycle 



A9 



Ap = "p • - "p 



P" = -24 
■p = *24 



kg 



sec 

kg- m 



AT = -24 



kg 



kg- m 



_^ kg* m 

Ap=-48^^ 



Draw a possible pulse that has the following displacements: 



Displacement 





-1 





+2 


.2 








At Point 


A 


B 


C 


D 


E 


F 


G 



8 



The frequency of a wave is the number 

of (i) that pass a point 

per (ii) , and is equal to 

the inverse of the (iii) . 



10 



For each collision the change of momentum of the wall 



(units) 



-|u;od SjLji uo ^uaiunSjD aiji 
MO{|o^ u|o6d puo p siuDjj o; ujn^s^ ^pssn^uo^ 

•Q = "°''dv + ll^^dy 

'00| SUOISIIJOS 3I|SD|3 Ul pSAJSSUOD Sj LUn;U3lUO^ -^'^!H 



A9 



one possibility: 




A8 

(i) cycles 
(ii) second 
(iii) period 



A10 



Apwoii + Ap^boii = 

APwoll = "Apboll 

APbol. = -48^ 



APwoii = -(-48 '^ll^) 
sec 



APwoll - -^48 



Kg- m 



10 



Two pulses are shaken onto the rope from opposite ends, 
and the pulses pass at mid-rope. A "movie camera" series 
shows the interaction of the two pulses. 

Picture 12 3 4 S 



^W 



•■--^ 




ii: 



-i-^-i— 



:^3g; 



Has there been any change in the shape of the pulses as 
a result of the interaction? 



Points marked P' on the diagram correspond to the 
point P because they are in the same part of a complete 
cycle. 

On the diagram, mark points H that correspond to 
the point H. 




11 



When each ball collides with the wall, the wal 
undergoes a momentum change of +48 kg- m/sec. 



If, during 10 seconds, 1000 balls strike the wall 

and rebound thus, the total change in momentum of the 

wa 1 1 is . 

(units) 



A10 



A9 



-^ Ul.l . ■ 




tiUtil^ln. t-'fmt^ 



All 



Ap, - lOOOAp 



kg 



Sp, =- (1000) (.48 ^^ ) 
^ sec 



^p^ - > 48000 



kg ■ m 



11 



When two waves pass through the same region of a medium, the displacement of each point is 
the sum of the displacements that each wave would cause by itself. For example, as two pulses 
of displacements a and b pass a point P on a rope, the displacement of point P will be a + b: 



a*b 



10 



On the wave diagram, mark as A', B' and C 
all points which correspond respectively to points 
marked A, B and C. 



12 



Average force on the wall is 



(units) 



IV +V 



*V 



iv {i^)s/ ^v"" -i 

:md| puooas s^uo+MatNj uioj^ uoi^onba 9^ jsqiuauiay :iu\y\ 




AID 




A12 



F -^P 

' av - -7— 
At 



Ap = ♦ 48000 



kg- m 



At = 10 sec 
h 48000 kgm/s«c 



h4800 



10 sec 

kg- m 
sec2 



^4800 rtewtons 



12 



Sketch what the displacement of the rope will be 
as the two pulses arrive at the center of the rope: 



n 



The Greek letter lambda (A) is used to represent 
wavelength. 

The wavelength is the distance between corres- 
ponding points of adjacent cycles. For example, 
points A and A' are one A apart. 

On the diagram mark points whose separation is 
A from points B and C. 



4l . |A . I A' 

Mi 



13 



The balls are directed at a patch of wall whose 
dimensions are 2.0 m high by 3.0 m wide. 

The area in which the balls impact is 



(units) 



A12 



All 




A13 

areo » height x width 

height < 2.0 
width . 3.0 

area = (2.0m) (3.0 m) 



orea = 6.0 m* 



13 



Sketch what the displacement of the rope will be 
as the two pulses arrive at the center of the rope. 




12 



Higher frequencies in the same medium 
produce shorter wavelengths. 




Mathematically speaking, the wavelength are 
frequency are proportional. 



14 



Pressure is usually defined as the ratio of perpendicular 

force to the area, and written as 

force 

pressure = -^ . 

area 

The pressure on the patch of wall due to the impocting 

balls is , ^. 

(units) 



pUD 3UJ Q'9 = D3JD 'SUJ3|qOJd SnoiAdjd lUOJ^ ■^''!H 



A13 




I 1 



A12 



inversely 



A14 



pressure = 



for 



pressure - 



force = + 4800 N 

orea ^ 6.0 m' 

♦ 4800N 
6.0 ni2 



pressure - 800 N/m' 



14 




















1 






in 
























































































5 
















































































" 


^ 



































' 


1 

























i 




















1 




















-5 























1 1 






j 




.1. . 



e 



The two pulses 
in the top dia- 
gram are shown 
superposed in 
the lower dia- 
gram. Complete 
the displace- 
ment table for 
the labeled 
points. 



Displacement 














Point 


A 


B 


C 


D 


E 


F 



13 



The top diagram shows a wave moving away from source s, and 
the bottom diagram shows the wave one complete cycle later. 




(i) 



(ii) 



During the time interval from one diagram to the other, the 

elapsed time is equal to the (i) of the wave, and the 

wave had moved a distance of (ii) . 



Before we moke the change from bouncing balls to bouncing 
molecules, we must look at the behavior of things moving inside 
a closed box. Suppose we have an oblong box, 4.0 meters long 
from end to end, with only one ball in it. The ball moves from 
end to end with a speed of 12 m/sec. The ball hits head-on and 
rebounds elastically with a speed of 12 m/sec toward the other 
end. The same ball will hit the front end of the box many times 
in 10 seconds. Instead of 
using the number of balls 
hitting the end, we must 
calculate and use the num- 
ber of hits made by this one 
ball. To find the force on 
one end, we use the hits on 
that end only. 




A14 



Displacement 





5 


8 


2 


5 





Point 


A 


B 


C 


D 


E 


F 



A13 



(i) period (T) 

(ii) one wovelength (A) 



15 



Complete the drawings, showing superposition of the 
two pulses when the pulse centers coincide. 



n 



u 



u 



n 



14 



If a wave moves one wavelength A in one period of time T, 
wave speed can be calculated. 



V (wave speed) = 



distar 



time 



Find the speed of a wave whose wavelength A is 1.5 cm, 
and whose period T is 0.1 seconds. 



15 



Between successive hits on the front end of the box, the 
ball travels one "round trip." It travels the whole length of 
the box from the front end to the opposite end and back to the 
front end again. 

The distance traveled by the ball during one round-trip 
is meters. 



A15 



tn 



net = 



At the exoct moment they cross 
the center the displacement is 
zero. 



A14 



" = T 



A = 1.5 cm 

T = 0.1 sec 

1 .5 cm 



0.1 sec 
= 15 cm sec 



A15 



8.0 meters 



16 

A pulse on a rope approaches a point that is 
attached to a wall. 



7\ 






m 



Experiments show thot the wave does not continue 
past the fixed point, but is reflected back in the oppo- 
site direction, and the puise appears on the opposite 
side of the rope. 

Draw the pulse after reflection. 



15 



1 A 

Recall that f = — (frame 7) and v = — (frame 14). 

What is fhe speed of o wove written in terms of wavelength A and 
frequency f? 



16 



The speed of the boll is 12 m/sec. The total distance 
traveled by the ball in 10 seconds is meters. 



-9UJ{4 S8UJJ4 p33dS 51 aDU04SIQ UUJI^ 



A16 



A15 



fA 



A16 

d = vf 

V - 12 m sec 

t = 10 sec 
d = (12 nv sec)(10 sec) 



d - 120 m 



17 



Draw the pulse after reflection. 



A 



\ A_ 



16 



!f waves of frequency f = 20 cycles per second travel in a 
particular medium with speed v = 40 meters per second, 

(i) what is the wavelength of the waves in the medium, 

(ii) what is the period of the waves? 



17 



The number of round trips made by the ball in 10 
seconds is . 



SDUD^Sip dlJ| punOJ 9L|4 pUO 'lU Q3[ S| D9S Q[ 
Ul p3|dADJ| 93UD4SJP 'SlUOJ^ snoiASJd 9L|4 UIOJ^ -^"IH 



A17 




A16 



(i) A = - 



= 40 m/sec 



20 cycles/sec 
= 2.0 m/cycle 

Since a cycle is always implied, 
wavelength is usually written in 
terms of distance units only. In 
this example the answer would 
be reported os A = 2.0 m. 

(..)T = -j- = 20 

T = .05 sec (per cycle) 



A17 



120 
8.0 



15 



15 round trips 



18 



A rope is suspended in a stairwell, and a 
pulse is shaken on the rope from the top. 

Experiments show that the wave is reflected 
from the free end but on the same side of the 
rope. 

Draw the pulse after reflection. 




>\ 



17 



Actually, it is very difficult to produce a perfectly periodic 
wave. One reason is the dissipation of energy which causes 
waves to be "damped" or "attenuated." 

The figures shows an attenuated (damped) periodic wave. 




Of the following, which property is changing? 
speed of wave, wavelength, amplitude, frequency, period 



18 



The number of times that the ball strikes the front end of 
the box in 10 seconds is (times). 



•diJi i^Doa aDuo xoq aij; jo pua ^uoj^ aL|; sajjiJ^s ||Dq aL|j^ :;u!|_| 



A18 




i=i 



A17 

On Only amplitude is changing. 
Amplitude refers to the heights of 
the crests and troughs above end 
below the undisturbed position. 
As the figure indicates, the ompli- 
tude of the damped wove is 
decreasing. 



A18 



15 times 



19 



A rope is hung in the diagram, one end fixed and 
the other end free. At point A the rope is moved side- 
ways and back suddenly, creating similar pulses that 
travel towards the ends. 

Sketch what the pulses will look like after reflec- 
tion from the ends. 




38J^ 




This ends the two programs on waves. To further your knowledge of waves 
you should refer to some of the other Project Physics materials, especially 
the Unit 3 laboratory activities, some of the articles in Reader 3 (as for 
example, "What is a Wave?" by Albert Einstein and Leopold Infeld), and 
the following Film Loops: 

Superposition 

Standing Waves on a String 

Standing Waves in a Gas 

Vibrations of a Rubber Hose 

Vibrations of a Wire 

Vibrations of a Drum 

Vibrations of a Metal Plate 



19 












1 

1 
1 

1 




If we 


want to have 1000 hits 


on the front wall in 10 


1 
seconds, ' 


as 


we had before w 


ith the stream 


of bolls, we would have to 


hove ' 


more than 


one boll 


in the box. In 


fact, we would need a 


aout 




ba 


Is, all 


moving back and forth between the ends. 






■M 


30|3A 


>UD SSDUJ 


3UJDS am 3AOL| 

•xoq 


s||Dq {|D 4DL|; SLunssy 
3L|| ui S||oq jo jaqiunu 








am Oi 


ouoi;jodojd si S|il| ^o jaqiunu 3L|; ;di^; siunssy 


uui 


H [ 

















A19 




r 



A19 



1000 
15 



66.7 



67 bails 





20 [ 

What will the rope look like when the two pulses cross? 1 







Pressure i 


s g 


ven as fo 


rce/area; if yo 


1 

u know the area of i 


the 


front of the 


bo 


K you can 


calcula 


te the average pressure on | | 


the 


end caused 


by 


repeated 


mpacts 


of the 


balls. 1 




If we have 


on 


y 67 ball. 


in the 


box, a 


1 

calculation of the | 


"pressure" does not seem to 


be use 


ful. But if the box contains | | 


thousands of m 


1 lions of mol 


ecules 


moving 


at high speeds, it l 


bee 


omes worthwhi 


e to compute on 


average 


result for the in- i 


vis 


ble particles. 

























A20 




They cancel 





21 j 

After onof/ier reflection the pulses again cross. What will | 
the rope look like at that instant? 1 





20 j 

So let's now do a similar calculation using gas molecules in a box.' 

A metal box, 4.0 meters long, with ends 3.0 meters wide by 2.0 | 
meters high, contains one gas molecule which moves to and fro along | 
the length of the box with a speed of 500 meters per second. The | 
molecule bounces elasticolly from each end, so the speed remains | 
constant at 500 m/sec. The mass of the molecule is approximately 1 
5.0 X 10-26 kilograms.* 1 

The momentum of the molecule before impact with the front end , 

is 1 


(units) j 
*This is approximately the moss of on average molecule of air. ' 





A21 




Both pulses will be on the left 
side of the rope and so the dis- 
placement will be twice that of 
either pulse alone. 



A20 



m = 5.0 . 10-^« kg 
V = 500 m sec 

- 5.0 ^ 10' m sec 
p - (5.0 X 10-" kg) (5.0 y 10' 

p - *25 X io-»* i^i:^ I 



sec) 





This is the end of Waves 1. Now that you understand the nature of wave | 
pulses, you can study the behavior of trains of wave pulses. This is | 
dealt with in Waves 2 Periodic Waves which begins just below this pro- | 
gram in the front of the book. | 



21 



The momentum of the molecule after impact (p') with 
the front end of the box is 



(units) 



A21 

p' = mv' 

m . 5.0 . 10-" leg 

v' = -500 m sec 

r -5.0 . 10^ msec 
p' = (5.0 . 10-" kg) (-5.0 . 10' m sec) 



p' = -25 - 10 



-24 ^JJl 



22 

The total change of momentum of the molecule, therefore 



IS 



(units) 



•d - ,d = dy 'ajo^aq sy uui^ 



A22 



Ap p' -"p 



p = .25 . 10 



P = -25. 10 



.2.^ 



-2. ^i: 



Ap = (-25 . 10-2*) _ (.25 . 10 



-J4v ^g 



Ap -50 - 10 



-24 ^9- 



23 j 

In 10 seconds the molecule can make trips to 1 
end fro, and so can moke this number of impacts on the front end. I 


•sja^aiu o'8 ^' 33UD4Sip dui-punoj a^ \ 
puD 'sjaiauj OOOS ^^°^ p|noM a|noa|ouj am spuooas q[ u| 4ui|_| | 





A23 



5000 



8.0 m per trip 



625 trips 



24 j 

In 10 seconds, the molecule mokes impacts | 


with the front end of the box. | 


-aiuoj^ snoiAdid 3i|; u\ uousanb 3L|4 poe^ -^"!H 1 





A24 



625 impacts 



(one per trip) 



25 1 

The change of momentum of the molecule per impact was | 
calculated to be -50 x 10~^* kg • m/sec. Therefore, the total | 
change of momentum of the front wall of the box in 10 seconds 1 
is ' 


(units) 1 


•a|nDa|ouj 3^\ jo lun^uaoiouj ^o aBuoi^D 31^4 ot uou | 
-DSJip ui a;isoddo si ||dm 3l|4 ^0 luniusuiouj ^0 aSuoi^D 3l|j_ ^^^IH I 





A25 



A?wall ' -50 . 10-" ^l- per ,mpoct 



\Pwoii = '625) (.50 . 10 



.24 kg- m 
sec 



31250. 10-" ^1— 



APwoii - -3.12 . 10- 



kg 



26 j 

The average force on the front of the box, during the 10 1 
second period, is . 1 


(units) 1 


•Xj^ jslhoud sil^ aAi6 '41 Buimsia . 
-aj ja+^D puD 9 aujojj o; >|3Dq oq ^a|qnoj^ 6u|adl| \\\l^ 1 

•jDAjaiui 3UJU 31^; Aq papiAip oiniusiuouj ui sBudijd ' 
3L|; so p9;Din3|Do som 3djo^ aSojaAO aij; |DL|4 jsqiusiud^ '^^!H 





A26 

P _ ACmv) ^ Ap 
" At ° aT 

Ap^„„ = .3.12.10- 

At = 10 sec 

* ♦3.12-10-" 

r = newtons 

10 



F = .3.12 . 10-*' newtons 



27 

The front end wall has dimensions of 2.0 meters by 3.0 
The nvernge pressure on the end wall is 


meters. 1 


(units) 


-sSjOj 9<^ o^ djnsssjd silj^ psdxs i^uoQ 
•pa;DejD poij sm ^d^ mnnooA ai^; ^o pnojd a^inb aq p|noM sm 
'it ui 3|nos|OLU auo ^snl i^;im xoq d a>jDLu o; 3\<^d sjsm sm j| 

•D3JD ijun jad aojoj si ajnssajj 


UUJH 







A27 



pressure = force orea 

area = 6.0 



pressure 



Torce = 3.12 . 10-'' N 

3.12 > 10-'' N 
6.0 m2 



! oressore = 5.0 - 10"'' N m* i 



Now suppose that this box contains 6.0 x 10^^ molecules (that is, 600,000,000,000,000,000, 
000,000,000 molecules). This figure is roughly the number of molecules in such o box if it were 
filled with air at atmospheric pressure. In reality these molecules would be moving about in all 
directions at random; but in order to simplify the calculation, suppose that they are sorted out 
into three groups. One lot moves up and down, another lot moves to and fro along the length, 
and the third lot moves back and forth across the width. Symmetry considerations suggest we 
should have the molecules equally divided between the three groups. 

(If you wish a more rigorous explanation, think of dividing the velocity of each molecule 
into three components Vx, Vy and v^. This cannot be done individually for each molecule, so 
it is done statistically for the sample of randomly-moving molecules. The results are the same 
as given in the first paragraph.) 




ouDuose we nove three 
groups, 2.0 . IC^' in 
each, having some ove*'- 
oge velocity parallel tc 
one edge of the box. 



There ore 6.0 ■ 10^^ molecules 
in the box moving in random 
directions. 



iTaj 






I 



^ 



^. 






The pressure on an end of the box will be due only to impacts 
of molecules moving to and fro along the length. We will now cal- 
culote the pressure on the end of the box. The pressure is caused 
by 2.0 X 10^^ molecules, moving ot a speed of 500 m sec, and col- 
liding elostically with the front end of the box. 



Assume that the pressure 
on the end face is aue to the 
impacts of the molecules that 
are moving parallel to the 
length of the box. 



2m 





28 



The average pressure on the end of the box will be 



^units; 



-XOO 3U4 p pu3 9^ 
L|4iM 5u!pi||0D sainD3joui g^Ql * 3 ^^°^ ^'^ asoD sjlji u: 

z^ H 22-01 " C? 5DM a|nDa|oui 
9U0 04 9np djnssajd dSojdAO 3lj; ;di^| ^3 auioj^ ujoj^ jiooa^ -^"!H 



A28 



(5.0 ■ 10-" N m2) (2.0 . 10^*) 
= 10 - 10* N fn2 



pressure = 1.0 • 10^ N m^ 



The values used for the moss 
of a molecule and the number of 
molecules m a box of that sue 
ore roughly correct for ordirxiry 
oir in o room. 

Standard atmospheric pres- 
sure IS 1.0132 ■ lO'^ N m'. 

Thus your calculated resT;lt 
approximates the standard atmo- 
spheric pressure quite closely. 



29 



Suppose that we reduce the size of the box slowly to one-half 
of the original length (that is, to 2.0 meters). Assume that we do 
not change the speed of the molecules or the size of the end wall. 

The average pressure of the end of the box will now be 
(units) 



-SJOjaq so ujouisj suoi;o|n3|03 am |o 4S9J aL|j[ 

'spodiui uasM^sq jsadj^ 04 33UDj 
-sip JdjJOLjS D aAOL| S3|no9|oui am 9DUIS '||0M pus am uo 
S||L| jo jaquinu 9i{; 3^0|nD|0D3j noX loiji sjmbaj ||im sii|j[_ '^"!H 



A29 

number of impacts per molecule 
in 10 seconds is 1250 

change of momentum per irrtpact 

is .50 . 10-'* ''?■"' (per molecule) 
sec 

2.0 « 10^' molecules strike the face 



pressure^^ = 2.0 - 10*K m' 



This is what you should expect 
since twice as many impacts per 
unit of time should result m twice 
the pressure. 



30 1 

If the box is reduced in size to one-half the original | 
length, what effect does this have on the volume of the 1 
box? 1 





A30 



The volume is reduced to 
half its originol size. 



31 1 

How did this change in volume of one-half affect the | 
pressure exerted by the gas? | 





A31 



The pressure of the gas 
was doubled. 



32 

From the calculations made in this program, how 
pressure of a gas related to its volume? 


is the j 


-S3|n39|OUJ ^O SUOISJIIOD OU pUO 'S3|n03|OUJ pUD ||DM 

uddMjaq suo|Si||OD 3uso|3 XjiosjJdd puo s3|n33|ouj 
jo jdqiunu slugs 'spssds JO|n33|Oui jonbs diunssy 


:+"!H 







A32 



The pressure of a gcs is 
inversely prcportionol to 
the volume it occupies. 



Having now completed this program, it should be much easier to follow 
the more complete discussion of The Kinetic-Molecular Theory of Gases 
to be found in Chapter II of the Project Physics Text, and in other 
physics books. Also you will now be in a position to enjoy many of the 
Project Physics Reader 3 articles, such as: 

The Great Molecular Theory of Cases by Eric M. Rogers 

Entropy and the Second Low of Thermodynamics by Kenneth W. Ford 

The Low of Disorder by George Gamow 

The Law by Robert M. Coates 

The Arrow of Time by Jacob Bronowski 
Also in the same Reader, you will find a brief biography of James Clerk 
Maxwell, the great scientist, who was a key figure in the development of 
kinetic theory. 



0-03-089643-6