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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