(BUKU FISGEL) CHAPTER 22 WAVE MOTION

Chapter 22

Wave Motion

A PROPAGATING WAVE is a self-sustaining disturbance of a medium that travels from one point to another, carrying energy and momentum. Mechanical waves are aggregate phenomena arising from the motion of constituent particles. The wave advances, but the particles of the med- ium only oscillate in place. A wave has been generated on the string in Fig. 22-1 by the sinusoi- dal vibration of the hand at its end. The wave furnishes a record of earlier vibrations of the source. Energy is carried by the wave from the source to the right, along the string. This direc- tion, the direction of energy transport, is called the direction (or line) of propagation of the wave.

Each particle of the string (such as the one at pointC) vibrates upand down, perpendicular to the line of propagation. Any wave in which the vibration direction is perpendicular to the direction of propagation is called atransverse wave. Typical transverse waves, besides those on a string, are electro- magnetic waves ± light and radio waves. By contrast, in sound waves the vibration direction is parallel to the direction of propagation, as you will see in Chapter 23. Such a wave is called alongitudinal (or compressional)wave.

WAVE TERMINOLOGY: The period …T† of a wave is the time it takes the wave to go through one complete cycle. It is the time taken for a particle such as the one at A to move through one complete vibration or cycle, down from point A and then back to A. The period is the number of seconds per cycle. The frequency …f† of a wave is the number of cycles per second: Thus,

f ˆ1 T

IfT is in seconds, thenf is in hertz (Hz), where 1 Hzˆ1 s ¹. The period and frequency of the wave are the same as the period and frequency of the vibration.

The toppoints on the wave, such asAandC, are called wavecrests. The bottom points, such asB andD, are calledtroughs. As time goes on, the crests and troughs move to the right with speedv, the speed of the wave.

Theamplitudeof a wave is the maximum disturbance undergone during a vibration cycle, distancey₀ in Fig. 22-1.

213 Fig. 22-1

Copyright 1997, 1989, 1979, 1961, 1942, 1940, 1939, 1936 The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

The wavelength…j†is the distance along the direction of propagation between corresponding points on the wave, distanceACfor example. In a timeT, a crest moving with speedvwill move a distancejto the right. Therefore,sˆvtgives

jˆvTˆv f and

vˆfj

This relation holds for all waves, not just for waves on a string.

IN-PHASE VIBRATIONS exist at two points on a wave if those points undergo vibrations that are in the same direction, in step. For example, the particles of the string at points A and C in Fig. 22-1 vibrate in-phase, since they move up together and down together. Vibrations are in- phase if the points are a whole number of wavelengths apart. The pieces of the string at A and B vibrate opposite to each other; the vibrations there are said to be 1808, or half a cycle, out-of- phase.

THE SPEED OF A TRANSVERSE WAVE on a stretched string or wire is

vˆ



tension in string mass per unit length of string s

STANDING WAVES: At certain vibrational frequencies, a system can undergo resonance. That is to say, it can eciently absorb energy from a driving source in its environment which is oscil- lating at that frequency (Fig. 22-2). These and similar vibration patterns are called standing waves, as compared to the propagating waves considered above. These might better not be called waves at all since they do not transport energy and momentum. The stationary points (such as B and

Fig. 22-2

D) are called nodes; the points of greatest motion (such as A, C, and E) are called antinodes. The distance between adjacent nodes (or antinodes) is ¹₂j. We term the portion of the string between adjacent nodes a segment, and the length of a segment is also ¹₂j:

CONDITIONS FOR RESONANCE: A string will resonate only if the vibration wavelength has certain special values: the wavelength must be such that a whole number of wave segments (each

12j long) exactly ®t on the string. A proper ®t occurs when nodes and antinodes exist at positions demanded by the constraints on the string. In particular, the ®xed ends of the string must be nodes. Thus, as shown in Fig. 22-2, the relation between the wavelength j and the length L of the resonating string is Lˆn…¹₂j†, where n is any integer. Because jˆvT ˆv=f, the shorter the wave segments at resonance, the higher will be the resonance frequency. If we call the fundamen- tal resonance frequency f1, then Fig. 22-2 shows that the higher resonance frequencies are given by fnˆnf1.

LONGITUDINAL (COMPRESSIONAL) WAVES occur as lengthwise vibrations of air columns, solid bars, and the like. At resonance, nodes exist at ®xed points, such as the closed end of an air column in a tube, or the location of a clampon a bar. Diagrams such as Fig. 22-2 are used to display the resonance of longitudinal waves as well as transverse waves. However, for longitu- dinal waves, the diagrams are mainly schematic and are used simply to indicate the locations of nodes and antinodes. In analyzing such diagrams, we use the fact that the distance between node and adjacent antinode is ¹₄j:

Solved Problems

22.1 Suppose that Fig. 22-1 represents a 50-Hz wave on a string. Take distancey₀ to be 3.0 mm, and distance AE to be 40 cm. Find the following for the wave: (a) amplitude, (b) wavelength, (c) speed.

(a) By de®nition, the amplitude is distancey₀and is 3.0 mm.

(b) The distance between adjacent crests is the wavelength, and sojˆ20 cm.

(c) vˆjf ˆ …0:20 m†…50 s ¹† ˆ10 m=s

22.2 Measurements show that the wavelength of a sound wave in a certain material is 18.0 cm. The frequency of the wave is 1900 Hz. What is the speed of the sound wave?

FromjˆvTˆv=f, which applies to all waves,

vˆjf ˆ …0:180 m†…1900 s ¹† ˆ342 m=s

22.3 A horizontal cord 5.00 m long has a mass of 1.45 g. What must be the tension in the cord if the wavelength of a 120 Hz wave on it is to be 60.0 cm? How large a mass must be hung from its end (say, over a pulley) to give it this tension?

We know that

vˆjf ˆ …0:600 m†…120 s ¹† ˆ72:0 m=s

Further, sincevˆ 

(tension)/(mass per unit length);

p

Tensionˆ …mass per unit length†…v²† ˆ 1:4510 ³kg 5:00 m

!

…72:0 m=s†²ˆ1:50 N The tension in the cord balances the weight of the mass hung at its end. Therefore,

F_Tˆmg or mˆFT

g ˆ 1:50 N

9:81 m=s²ˆ0:153 kg

22.4 A uniform ¯exible cable is 20 m long and has a mass of 5.0 kg. It hangs vertically under its own weight and is vibrated from its upper end with a frequency of 7.0 Hz. (a) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint?

(a) We shall usevˆ 

(tension)/(mass per unit length).

p The midpoint of the cable supports half its weight, so the tension there is

FT ˆ¹₂…5:0 kg†…9:81 m=s²† ˆ24:5 N Mass per unit lengthˆ5:0 kg

20 m ˆ0:25 kg=m Further

vˆ



24:5 N 0:25 kg=m s

ˆ9:9 m=s so that

(b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore the frequency, 7.0 Hz, is the same at all points.

To ®nd the wavelength at the midpoint, we must use the speed we found for that point, 9.9 m/s.

That gives us

jˆv

f ˆ9:9 m=s 7:0 Hz ˆ1:4 m

22.5 Suppose that Fig. 22-2 shows a metal string under a tension of 88.2 N. Its length is 50.0 cm and its mass is 0.500 g. (a) Computevfor transverse waves on the string. (b) Determine the frequen- cies of its fundamental, ®rst overtone, and second overtone.

vˆ



tension mass per unit length s

ˆ



88:2 N

…5:0010 ⁴kg†=…0:500 m†

s

ˆ297 m=s …a†

(b) We recall that the length of the segment isj=2 and we usejˆv=f. For the fundamental:

jˆ1:00 m and f ˆ297 m=s

1:00 m ˆ297 Hz For the ®rst overtone:

jˆ0:500 m and f ˆ297 m=s

0:500 mˆ594 Hz For the second overtone:

jˆ0:333 m and f ˆ297 m=s

0:333 mˆ891 Hz

22.6 A string 2.0 m long is driven by a 240-Hz vibrator at its end. The string resonates in four segments. What is the speed of transverse waves on the string?

Since each segment isj=2 long, we have 4 j

2 ˆL or jˆL

2ˆ2:0 m

2 ˆ1:0 m Then, usingjˆvTˆv=f, we have

vˆfjˆ …240 s ¹†…1:0 m† ˆ0:24 km=s

22.7 A banjo string 30 cm long resonates in its fundamental to a frequency of 256 Hz. What is the tension in the string if 80 cm of the string have a mass of 0.75 g?

First we shall ®ndvand then ®nd the tension. We know that the string vibrates in one segment when f ˆ256 Hz. Therefore, from Fig. 22-2,

j

2ˆL or jˆ …0:30 m†…2† ˆ0:60 m vˆfjˆ …256 s ¹†…0:60 m† ˆ154 m=s and

The mass per unit length of the string is 0:7510 ³kg

0:80 m ˆ9:410 ⁴kg=m Then, fromvˆ 

(tension)/(mass per unit length);

p

F_Tˆ …154 m=s†²…9:410 ⁴kg=m† ˆ22 N

22.8 A string vibrates in ®ve segments to a frequency of 460 Hz. (a) What is its fundamental fre- quency? (b) What frequency will cause it to vibrate in three segments?

Detailed Method

If the string isnsegments long, then from Fig. 22-2 we haven…¹₂j† ˆL. Butjˆv=f_n, soLˆn…v=2f_n†.

Solving forfngives

f_nˆn v 2L

We are told thatf₅ˆ460 Hz, and so

460 Hzˆ5 v 2L

or v

2Lˆ92:0 Hz Substituting this in the above relation gives

f_nˆ …n†…92:0 Hz†

(a) f1ˆ92:0 Hz:

(b) f3ˆ …3†…92 Hz† ˆ276 Hz Alternative Method

We recall that for a string held at both ends,fnˆnf1. Knowing thatf5ˆ460 Hz, we ®ndf1ˆ92:0 Hz andf₃ˆ276 Hz.

22.9 A string fastened at both ends resonates at 420 Hz and 490 Hz with no resonance frequencies in between. Find its fundamental resonance frequency.

In general,fnˆnf1. We are told thatfnˆ420 Hz andfn‡1ˆ490 Hz. Therefore, 420 Hzˆnf1 and 490 Hzˆ …n‡1†f1

We subtract the ®rst equation from the second to obtainf1ˆ70:0 Hz.

22.10 A violin string resonates in its fundamental at 196 Hz. Where along the string must you place your ®nger so its fundamental becomes 440 Hz?

In the fundamental, Lˆ¹₂j. Since jˆv=f, we have f1ˆv=2L. Originally, the string of length L1

resonated to a frequency of 196 Hz, and so

196 Hzˆ v 2L1

We want it to resonate to 440 Hz, so we have

440 Hzˆ v 2L₂ We eliminatevfrom these two simultaneous equations and ®nd

L2

L₁ˆ196 Hz 440 Hzˆ0:445

To obtain the desired resonance, the ®nger must shorten the string to 0.445 of its original length.

22.11 A 60 cm long bar, clamped at its middle, is vibrated lengthwise by an alternating force at its end.

(See Fig. 22-3.) Its fundamental resonance frequency is found to be 3.0 kHz. What is the speed of longitudinal waves in the bar?

Because its ends are free, the bar must have antinodes there. The clamppoint at its center must be a node. Therefore, the fundamental resonance is as shown in Fig. 22-3. Because the distance from node to antinode is always¹₄j, we see thatLˆ2…¹₄j†. SinceLˆ0:60 m, we ®ndjˆ1:20 m.

Then, from the basic relation (p. 214)jˆv=f, we have

vˆjf ˆ …1:20 m†…3:0 kHz† ˆ3:6 km=s

22.12 Compressional waves (sound waves) are sent down an air-®lled tube 90 cm long and closed at one end. The tube resonates to several frequencies, the lowest of which is 95 Hz. Find the speed of sound waves in air.

The tube and several of its resonance forms are shown in Fig. 22-4. Recall that the distance between a node and an adjacent antinode isj=4. In our case, the top resonance form applies, since the segments are longest for it and its frequency is therefore lowest. For that form,Lˆj=4, so

jˆ4Lˆ4…0:90 m† ˆ3:6 m UsingjˆvTˆv=f gives

vˆjf ˆ …3:6 m†…95 s ¹† ˆ0:34 km=s

22.13 To what other frequencies will the tube described in Problem 22.12 resonate?

The ®rst few resonances are shown in Fig. 22-4. We see that, at resonance, Lˆn…¹₄jn†

wherenˆ1;3;5;7;. . ., is an odd integer, andjnis the resonance wavelength. Butjnˆv=fn, and so Lˆn v

4fn or fnˆn v 4Lˆnf1

where, from Problem 22.12, f₁ˆ95 Hz. The ®rst few resonance frequencies are thus 95 Hz, 0.29 kHz, 0.48 kHz,. . ..

22.14 A metal rod 40 cm long is dropped, end ®rst, onto a wooden ¯oor and rebounds into the air.

Compressional waves of many frequencies are thereby set up in the bar. If the speed of compres- sional waves in the bar is 5500 m/s, to what lowest-frequency compressional wave will the bar resonate as it rebounds?

Both ends of the bar will be free, and so antinodes will exist there. In the lowest resonance form (i.e., the one with the longest segments), only one node will exist on the bar, at its center, as shown in Fig. 22-5. We will then have

Lˆ2 j

4 or jˆ2Lˆ2…0:40 m† ˆ0:80 m Then, fromjˆvT ˆv=f,

f ˆv

jˆ5500 m=s

0:80 m ˆ6875 Hzˆ6:9 kHz

22.15 A rod 200 cm long is clamped 50 cm from one end, as shown in Fig. 22-6. It is set into longi- tudinal vibration by an electrical driving mechanism at one end. As the frequency of the driver is slowly increased from a very low value, the rod is ®rst found to resonate at 3 kHz. What is the speed of sound (compressional waves) in the rod?

Fig. 22-5 Fig. 22-6

Fig. 22-3 Fig. 22-4

The clamped point remains stationary, and so a node exists there. Since the ends of the rod are free, antinodes exist there. The lowest-frequency resonance occurs when the rod is vibrating in its longest possible segments. In Fig. 22-6 we show the mode of vibration that corresponds to this condition. Since a segment is the length from one node to the next, then the length from Ato N in the ®gure is one-half segment.

Therefore, the rod is two segments long. This resonance form satis®es our restrictions about positions of nodes and antinodes, as well as the condition that the bar vibrate in the longest segments possible. Since one segment isj=2 long,

Lˆ2…j=2† or jˆLˆ200 cm Then, fromjˆvTˆv=f,

vˆjf ˆ …2:00 m†…310³s ¹† ˆ6 km=s

22.16 (a) Determine the shortest length of pipe closed at one end that will resonate in air to a sound source of frequency 160 Hz. Take the speed of sound in air to be 340 m/s. (b) Repeat for a pipe open at both ends.

(a) Figure 22-4(a) applies in this case. The shortest pipe will bej=4 long. Therefore, Lˆ1

4jˆ1 4

v

f ˆ 340 m=s

4…160 s ¹†ˆ0:531 m

(b) In this case the pipe will have antinodes at both ends and a node at its center. Then, Lˆ2 1

4j ˆ1 2

v

f ˆ 340 m=s

2…160 s ¹†ˆ1:06 m

22.17 A pipe 90 cm long is open at both ends. How long must a second pipe, closed at one end, be if it is to have the same fundamental resonance frequency as the open pipe?

The two pipes and their fundamental resonances are shown in Fig. 22-7. As we see, L_oˆ2…¹₄j† L_cˆ¹₄j

from whichLcˆ¹₂Loˆ45 cm.

22.18 A glass tube that is 70.0 cm long is open at both ends. Find the frequencies at which it will resonate to sound waves that have a speed of 340 m/s.

A pipe that is open at both ends must have an antinode at each end. It will therefore resonate as in Fig.

22-8. We see that the resonance wavelengthsjnare given by Lˆn jn

2 or jnˆ2L

n

Fig. 22-7 Fig. 22-8

wherenis an integer. Butjnˆv=fn, so f_nˆ n

2L

…v† ˆ …n† 340 m=s 20:700 m

ˆ243nHz

Supplementary Problems

22.19 The average person can hear sound waves ranging in frequency from about 20 Hz to 20 kHz. Determine the wavelengths at these limits, taking the speed of sound to be 340 m/s. Ans. 17 m, 1.7 cm

22.20 Radio station WJR in Detroit broadcasts at 760 kHz. The speed of radio waves is 3:0010⁸m/s. What is the wavelength of WJR's waves? Ans. 395 m

22.21 Radar waves with 3.4 cm wavelength are sent out from a transmitter. Their speed is 3:0010⁸m/s. What is their frequency? Ans. 8:810⁹Hzˆ8:8 GHz

22.22 When driven by a 120 Hz vibrator, a string has transverse waves of 31 cm wavelength traveling along it. (a) What is the speed of the waves on the string? (b) If the tension in the string is 1.20 N, what is the mass of 50 cm of the string? Ans. (a) 37 m/s; (b) 0.43 g

22.23 The wave shown in Fig. 22-9 is being sent out by a 60 cycle/s vibrator. Find the following for the wave: (a) amplitude, (b) frequency, (c) wavelength, (d) speed, (e) period. Ans. (a) 3.0 mm; (b) 60 Hz; (c) 2.00 cm;

(d) 1.2 m/s; (e) 0.017 s

22.24 A copper wire 2.4 mm in diameter is 3.0 m long and is used to suspend a 2.0 kg mass from a beam. If a transverse disturbance is sent along the wire by striking it lightly with a pencil, how fast will the disturbance travel? The density of copper is 8920 kg/m³. Ans. 22 m/s

22.25 A string 180 cm long resonates in three segments to transverse waves sent down it by a 270 Hz vibrator.

What is the speed of the waves on the string? Ans. 324 m/s

22.26 A string resonates in three segments to a frequency of 165 Hz. What frequency must be used if it is to resonate in four segments? Ans. 220 Hz

22.27 A ¯exible cable, 30 m long and weighing 70 N, is stretched between two poles by a force of 2.0 kN. If the cable is struck sideways at one end, how long will it take the transverse wave to travel to the other end and return? Ans. 0.65 s

Fig. 22-9

22.28 A wire under tension vibrates with a fundamental frequency of 256 Hz. What would be the fundamental frequency if the wire were half as long, twice as thick, and under one-fourth the tension? Ans. 128 Hz 22.29 Steel and silver wires of the same diameter and same length are stretched with equal tension. Their densities are 7.80 g/cm³and 10.6 g/cm³, respectively. What is the fundamental frequency of the silver wire if that of the steel is 200 Hz? Ans. 172 Hz

22.30 A string has a mass of 3.0 gram and a length of 60 cm. What must be the tension so that when vibrating transversely its ®rst overtone has frequency 200 Hz? Ans. 72 N

22.31 (a) At what point should a stretched string be plucked to make its fundamental tone most prominent? At what point should it be plucked and then at what point touched (b) to make its ®rst overtone most prominent and (c) to make its second overtone most prominent? Ans. (a) center; (b) plucked at 1/4 of its length from one end, then touched at center; (c) plucked at 1/6 of its length from one end, then touched at 1/3 of its length from that end

22.32 What must be the length of an iron rod that has the fundamental frequency 320 Hz when clamped at its center? Assume longitudinal vibration at a speed of 5.00 km/s. Ans. 7.81 m

22.33 A rod 120 cm long is clamped at the center and is stroked in such a way as to give its ®rst overtone. Make a drawing showing the location of the nodes and antinodes, and determine at what other points the rod might be clamped and still emit the same tone. Ans. 20.0 cm from either end

22.34 A metal bar 6.0 m long, clamped at its center and vibrating longitudinally in such a manner that it gives its

®rst overtone, vibrates in unison with a tuning fork marked 1200 vibration/s. Compute the speed of sound in the metal. Ans. 4.8 km/s

22.35 Determine the length of the shortest air column in a cylindrical jar that will strongly reinforce the sound of a tuning fork having a vibration rate of 512 Hz. Use vˆ340 m/s for the speed of sound in air.

Ans 16.6 cm

22.36 A long, narrow pipe closed at one end does not resonate to a tuning fork having a frequency of 300 Hz until the length of the air column reaches 28 cm. (a) What is the speed of sound in air at the existing room temperature? (b) What is the next length of column that will resonate to the fork? Ans. (a) 0.34 km/s;

(b) 84 cm

22.37 An organ pipe closed at one end is 61.0 cm long. What are the frequencies of the ®rst three overtones ifvfor sound is 342 m/s? Ans. 420 Hz, 700 Hz, 980 Hz