Instructor Solution Manual - Santa Rosa Junior Collegelwillia2/41/41hw4keych21.pdf · 21.6. The...

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21-1

Conceptual Questions

21.1. (a) When a string is fixed at both ends the number of antinodes is the mode (m-value) of the standing wave; so

4.m (b) When the frequency is doubled the wavelength is halved. This halving of the wavelength will increase the

number of antinodes to eight.

21.2. When the string is flat most of the segments of the string (except at the nodes) are moving (quite rapidly at the

antinodes), so the energy is in the kinetic energy of the moving string.

21.3. (a) 2m (b) horizontally, because sound is a longitudinal wave (c) 0 cm, 16 cm, 32 cm (d) 8 cm and 24 cm

because the pressure antinodes occur where there are displacement nodes.

21.4. The speed of sound in air is related to the temperature by / ,v RT M so speed increases with increasing

temperature. Some organ pipes are open at both ends and some are open-closed. However, in both cases the

frequency increases with increasing speed. So frequency increases as speed increases, and speed increases as

temperature increases. As a result, as the air in the organ pipe warms up, the frequency of the fundamental and hence

all the harmonics will increase slightly.

21.5. The sound you hear is the vibration of the glass; it is set into motion by the disturbance of the flowing,

sloshing liquid. The disturbances are of many frequencies but the natural frequency of the glass resonates and is

amplified by the glass while other frequencies are quickly damped out. The reason the pitch rises as the glass fills is

that the natural frequency of the glass changes; as the glass fills the resonance frequency rises.

21.6. The length of the flute is the same regardless of the gas it is filled with; consequently, the fundamental

wavelength is the same too. But because helium atoms have a smaller atomic mass than air molecules, helium atoms

move faster on average at the same temperature. So the speed of sound is greater in helium than in air at the same

temperature. In v f if is unchanged but v is increased, then f must increase also; this is perceived as a higher

pitch. The same explanation applies to the voices of people who have inhaled a breath of helium.

21.7. Use s /v T and v f Assume the linear mass density is the same in all parts of this question. Use primes

for new quantities. (a) 2 2 2 2

02 2 2 2

0 0 0 0 0

( ) (2 )4

( )

T v f f f

T v f f f

Changing the tension is how one tunes a guitar; it

would be difficult to increase the tension by a factor of 4 without breaking the string, so only small changes are made

this way.

(b) 0

0 0 0 0

/ /(2 ) 1

/ /( ) 2

L v f v f

L v f v f

so the length must be decreased by a factor of 2. Guitar players effectively

decrease the length by putting their fingers down on the string at various frets to create a node there.

SUPERPOSITION

21

21-2 Chapter 21


21.8. (a) 4 mm (b) 0 mm (c) 4 mm. This is easily seen by adding the contributions from Wave 1 and Wave 2 at

each point.

21.9. At point a, two crests are arriving at the same place at the same time from in-phase sources, so it is a point of

constructive interference. At point b, a crest from source 2 is arriving at the same time as a trough is arriving from

source 1, so it is a point of destructive interference. At point c, two troughs are arriving at the same place at the same

time from in-phase sources, so it is a point of constructive interference.

21.10. Pulling out the tuning valve lengthens the instrument and the wavelength of the standing waves. This lowers

the frequency. If she pulled out the valve slightly to go from a beat frequency of 5 Hz to 3 Hz ( 2 Hz),f then she

was at 445 Hz and dropped 2 Hz to 443 Hz. Had she been at 435 Hz, the beat frequency would have increased from

by f from 5 to 7 Hz.

Exercises and Problems

Section 21.1 The Principle of Superposition

21.1. Model: The principle of superposition comes into play whenever the waves overlap.

Visualize:

Superposition 21-3


The snapshot graph at 1.0 st differs from the graph 0.0 st in that the left wave has moved to the right by 1.0 m

and the right wave has moved to the left by 1.0 m. This is because the distance covered by each wave in 1.0 s is

1.0 m. The snapshot graphs at 2.0, 3.0, and 4.0 st are a superposition of the left and the right moving waves. The

overlapping parts of the two waves are shown by the dotted lines.


Visualize:

21-4 Chapter 21


The graph at 1.0 st differs from the graph at 0.0 st in that the left wave has moved to the right by 1.0 m and the

right wave has moved to the left by 1.0 m. This is because the distance covered by the wave pulse in 1.0 s is 1.0 m.

The snapshot graphs at 2.0, 3.0, and 4.0 st are a superposition of the left- and the right-moving waves. The

overlapping parts of the two waves are shown by the dotted lines.


Visualize:

At 4 st the shorter pulses overlap and cancel. At 6 st the longer pulses overlap and cancel. The dashed line

shows where the waves would be in the absence of the other wave.

Superposition 21-5



Solve: (a) As graphically illustrated in the figure below, the snapshot graph of Figure EX21.4b was taken at 4 s.t

(b)

Section 21.2 Standing Waves

Section 21.3 Standing Waves on a String

21.5. Model: A wave pulse reflected from the string-wall boundary is inverted and its amplitude is unchanged.

Visualize:

21-6 Chapter 21


The graph at 2 st differs from the graph at 0 st in that both waves have moved to the right by 2 m. This is

because the distance covered by the wave pulse in 2 s is 2 m. The shorter pulse wave encounters the boundary wall at

2 s and is inverted upon reflection. This reflected pulse wave overlaps with the broader pulse wave, as shown in the

snapshot graph at 4 s.t At 6 s,t only half of the broad pulse is reflected and hence inverted; the shorter pulse

wave continues to move to the left with a speed of 1 m/s. Finally, at 8t s both the reflected pulse waves are

inverted and they are both moving to the left.

21.6. Model: Reflections at the string boundaries cause a standing wave on the string.

Solve: Figure EX21.6 indicates two full wavelengths on the string. Hence 12

(60 cm) 30 cm 0 30 m Thus

0 30 m 100 Hz 30 m/sv f

21.7. Model: Reflections at both ends of the string cause the formation of a standing wave.

Solve: Figure EX21.7 indicates 5/2 wavelengths on the 2.0-m-long string. Thus, the wavelength of the standing

wave is 25

(2 0 m) 0 80 m. The frequency of the standing wave is

40 m/s50 Hz

0 80 m

vf

21.8. Model: Reflections at the string boundaries cause a standing wave on the string.

Solve: (a) When the frequency is doubled 0( 2 ),f f the wavelength is halved 102

( ). This halving of the

wavelength will increase the number of antinodes from 4 to 8.

(b) Increasing the tension by a factor of 4 means

42

T T Tv v v

For the string to continue to oscillate as a standing wave with four antinodes means 0. Hence,

0 0 0 0 0 02 2 2 2v v f f f f f f

That is, the new frequency is twice the original frequency.

Superposition 21-7


21.9. Model: A string fixed at both ends supports standing waves.

Solve: (a) A standing wave can exist on the string only if its wavelength is

2 1, 2, 3, m

Lm

m

The three longest wavelengths for standing waves will therefore correspond to 1, 2, and 3.m Thus,

1 2 3

2(2 4 m) 2(2 4 m) 2(2 4 m)4 8 m 2 4 m 1 6 m

1 2 3

(b) Because the wave speed on the string is unchanged from one m value to the other,

2 22 2 3 3 3

3

(50 Hz)(2 4 m)75 Hz

1 6 m

ff f f

21.10. Model: A string fixed at both ends supports standing waves.

Solve: (a) We have a 136 Hzf mf where 1f is the fundamental frequency that corresponds to 1.m The next

successive frequency is b 148 Hz ( 1) .f m f Thus,

b 11

a 1

( 1) 1 48 Hz 4 4 36 Hz1 3 12 Hz

36 Hz 3 3 3

f m f m mm m f

f mf m

The wave speed is

1 1 1

2(2 0 m)(12 Hz) 24 m/s

1

Lv f f

(b) The frequency of the fourth harmonic is 48 Hz. For 4,m the wavelength is

2 2(1 0 m) 1 m

4 2m

L

m

21.11. Model: For the stretched wire vibrating at its fundamental frequency, the wavelength of the standing wave

is 1 2L

Visualize:

Solve: The wave speed on the steel wire is

wire (2 ) (80 Hz)(2 0 90 m) 144 m/sv f f L

and is also equal to S / ,T where

33mass 5 0 10 kg

5 555 10 kg/mlength 0 90 m

21-8 Chapter 21


The tension ST in the wire equals the weight of the sculpture or Mg. Thus,

2 3 2wire

wire 2

(5 555 10 kg/m)(144 m/s)12 kg

9 8 m/s

Mg vv M

g

21.12. Model: The laser light forms a standing wave inside the cavity.

Solve: The wavelength of the laser beam is

100,000

2 2(0 5300 m)10 60 m

100,000m

L

m

The frequency is 8

13100,000 6

100,000

2 9979 10 m/s2 828 10 Hz

10 60 10 m

cf

Section 21.4 Standing Sound Waves and Musical Acoustics

21.13. Solve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possible

wavelengths for this case are

2 1, 2, 3, m

Lm

m

The three longest wavelengths are

1 2 3

2(1 21 m) 2(1 21 m) 2(1 21 m)2 42 m 1 21 m 0 807 m

1 2 3

(b) In the case of an open-closed tube,

4 1, 3, 5, m

Lm

m

The three longest wavelengths are

1 2 3

4(1 21 m) 4(1 21 m) 4(1 21 m)4 84 m 1 61 m 0 968 m

1 3 5

21.14. Model: We have an open-open tube that forms standing sound waves.

Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends.

There is another antinode in the middle of the tube. Thus, this is the 2m mode and the wavelength of the standing

wave is equal to the length of the tube, that is, 0.80 m. Since the frequency 500 Hz,f the speed of sound in

this case is (500 Hz)(0.80 m) 400 m/s.v f

Assess: The experiment yields a reasonable value for the speed of sound.

21.15. Solve: For the open-open tube, the fundamental frequency of the standing wave is 1 1500 Hzf when the

tube is filled with helium gas at 0 C. Using 2 / ,m L m

helium1 helium

1

970 m/s

2

vf

L

Similarly, when the tube is filled with air,

air 1 air1 air 1 air

1 1 helium

331 m/s 331 m/s 331 m/s(1500 Hz) 512 Hz

2 970 m/s 970 m/s

v ff f

L f

Assess: Note that the length of the tube is one-half the wavelength whether the tube is filled with helium or air.

21.16. Solve: For the open-closed tube, Equation 21.18 gives the possible frequencies:

114 4

m

v vf m mf L

L f

Superposition 21-9


for a fundamental frequency of 1 250 Hzf and a speed of sound of 350 m/s,v the length of the vocal tract must

be approximately

350 m/s35 cm

4(250 Hz)L

21.17. Model: Reflections at the string boundaries cause a standing wave on a stretched string.

Solve: Because the vibrating section of the string is 1.9 m long, the two ends of this vibrating wire are fixed, and the

string is vibrating in the fundamental harmonic. The wavelength is

1

22 2(1 90 m) 3 80 mm

LL

m

The wave speed along the string is 1 1 (27.5 Hz)(3.80 m) 104.5 m/s.v f The tension in the wire can be found as

follows:

2 2 2SS

mass 0 400 kg(104 5 m/s) 2180 N

length 2 00 m

Tv T v v

21.18. Model: A string fixed at both ends forms standing waves.

Solve: A simple string sounds the fundamental frequency 1 /(2 ).f v L Initially, when the string is of length

A 30 cm,L the note has the frequency 1A A/(2 ).f v L For a different length, 1B B/(2 ).f v L Taking the ratio of

each side of these two equations gives

1A A B 1AB A

1B B A 1B

/(2 )

/(2 )

f v L L fL L

f v L L f

We know that the second frequency is desired to be 1B 523 Hz.f The string length must be

B

440 Hz(30 cm) 25 2 cm

523 HzL

The question is not how long the string must be, but where must the violinist place his finger. The full string is 30 cm

long, so the violinist must place his finger 4.8 cm from the end.

Assess: A fingering distance of 4.8 cm from the end is reasonable.

Section 21.5 Interference in One Dimension

Section 21.6 The Mathematics of Interference

21.19. Model: Interference occurs according to the difference between the phases ( ) of the two waves.

Solve: (a) A separation of 20 cm between the speakers leads to maximum intensity on the x-axis, but a separation of

60 cm leads to zero intensity. That is, the waves are in phase when 1( ) 20 cmx but out of phase when

2( ) 60 cmx Thus,

2 1( ) 2(60 cm 20 cm) 80 cm2

x x

(b) If the distance between the speakers continues to increase, the intensity will again be a maximum when the

separation between the speakers that produced a maximum has increased by one wavelength. That is, when the

separation between the speakers is 20 cm 80 cm 100 cm.

21.20. Model: The interference of two waves depends on the difference between the phases ( ) of the two

waves.

Solve: (a) Because the speakers are in phase, 0 0 rad Let d represent the path-length difference. Using

0m for the smallest d and the condition for destructive interference, we get

10 2

2 2 rad, 0, 1, 2, 3, x

m m

21-10 Chapter 21


0

1 1 343 m/s2 rad 2 0 rad rad 25 cm

2 2 2 686 Hz

x d vd

f

(b) When the speakers are out of phase, 0 Using 1m for the smallest d and the condition for constructive

interference, we get

02 2 , 0, 1, 2, 3, x

m m

1 1 343 m/s2 2 25 cm

2 2 2 686 Hz

d vd

f

21.21. Model: Reflection is maximized if the two reflected waves interfere constructively.

Solve: The film thickness that causes constructive interference at wavelength is given by Equation 21.32: 9

CC

2 (600 10 m)(1)216 nm

2 (2)(1 39)

mndd

m n

where we have used 1m to calculate the thinnest film.

Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable

because air film glass.n n n

21.22. Model: Reflection is maximized if the two reflected waves interfere constructively.

Solve: The film thickness that causes constructive interference at wavelength is given by Equation 21.32: 9

CC

2 (500 10 m)(1)200 nm

2 (2)(1 25)

mndd

m n

where we have used 1m to calculate the thinnest film.

Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable

because air oil water.n n n

Section 21.7 Interference in Two and Three Dimensions

21.23. Solve: (a) The circular wave fronts emitted by the two sources show that the two sources are in phase

because the wave fronts of each source have moved the same distance from their sources.

(b) Let us label the top source as 1 and the bottom source as 2. Since the sources are in phase, 0 0 rad. For the

point P, 1 23 and 4 .r r Thus, 2 1 4 3 .r r r The phase difference is

2 2 ( )2

r

This corresponds to constructive interference.

For the point Q, 71 2r and 2 2 .r The phase difference is

32

223

r

This corresponds to destructive interference.

For the point R, 51 2r and 7

2 2.r The phase difference is

2 ( )2


1r 2r r C/D

P 3 C

Q 72 2 3

2 D

R 52 7

2 C

Superposition 21-11


21.24. Solve: (a) The circular wave fronts emitted by the two sources indicate the sources are out of phase. This is

because the wave fronts of each source have not moved the same distance from their sources.

(b) Let us label the top source as 1 and the bottom source as 2. Because the wave front closest to source 2 has moved

only half of a wavelength, whereas the wave front of source 1 has moved one wavelength, the phase difference

between the sources is 0 . For the point P, 1 22 and 3 .r r The phase difference is

0

2 2 (3 2 )3

r

This corresponds to destructive interference.

For the point Q, 1 3 r and 32 2

r The phase difference is

32

24


For the point R, 51 2r and 2 3 .r The phase difference is

12

22


1r 2r r C/D

P 2 3 D

Q 3 32 3

2

C

R 52 3

12

C

Assess: Note that it is not 1r or 2r that matters, but the difference r between them.

21.25. Model: The two speakers are identical, and so they are emitting circular waves in phase. The overlap of

these waves causes interference.

Visualize:

Solve: From the geometry of the figure,

2 2 2 22 1 (2 0 m) (4 0 m) (2 0 m) 4 472 mr r

so 2 1 4 472 m 4 0 m 0 472 mr r r The phase difference between the sources is 0 0 rad and the

wavelength of the sound waves is

340 m/s0 1889 m

1800 Hz

v

f

Thus, the phase difference of the waves at the point 4.0 m in front of one source is

21-12 Chapter 21


50 2

2 (0 472 m)2 0 rad 5 rad (2 rad)

0 1889 m

r

This is a half-integer multiple of 2π rad, so the interference is perfect destructive.

21.26. Model: The two radio antennas are emitting out-of-phase, circular waves. The overlap of these waves

causes interference.

Visualize:

Solve: From the geometry of the figure, 1 800 mr and

2 22 (800 m) (600 m) 1000 mr

So, 2 1 200 mr r r and 0 rad The wavelength of the waves is

8

6

3 0 10 m/s100 m

3 0 10 Hz

c

f

Thus, the phase difference of the waves at the point (300 m, 800 m) is

50 2

2 (200 m)2 rad 5 rad (2 rad)

100 m

r

This is a half-integer multiple of 2π rad, so the interference is perfect destructive.

Section 21.8 Beats

21.27. Solve: The beat frequency is

beat 1 2 1 13 Hz 200 Hz 203 Hzf f f f f

1f is larger than 2f because the increased tension increases the wave speed and hence the frequency.

21.28. Solve: The flute player’s initial frequency is either 523 Hz 4 Hz 527 Hz or 523 Hz 4 Hz 519 Hz.

Since she matches the tuning fork’s frequency by lengthening her flute, she is increasing the wavelength of the

standing wave in the flute. A wavelength increase means a decrease of frequency because .v f Thus, her initial

frequency was 527 Hz.

21.29. Solve: The beat frequency is

beat 1 2 100 MHzf f f

where we have used the fact that the 1 2 so 1 2.f f The frequency of emitter 1 is 1 1/ ,f c where

21 1 250 10 m. The wavelength of emitter 2 is

8

2 2 8 21 1

(3 00 10 m/s)/ 1 26 cm.

100 MHz / 100 MHz (3 00 10 m/s)/(1 250 10 m) 100 MHz

c cc f

f c

Superposition 21-13


21.30. Model: The principle of superposition applies to overlapping waves.

Visualize:

Solve: Because the wave pulses travel along the string at a speed of 100 m/s, they move a distance of d vt

(100 m/s)(0.05 s) 5 m in 0.050 s. The front of the wave pulse moving left, which is located at 1x m at

0.050 s,t was thus located at 6 m at 0 s.x t This helps us draw the snapshot of the wave pulse moving left at 0 st

(shown as a dashed line). Subtracting this wave snapshot from the resultant at 0 st (shown as a solid line) yields the

right-traveling wave’s snapshot at 0 st (shown as a dotted line). Finally, the snapshot graph of the wave pulse moving

right at 0.050 st is the same as at 0 st (shown as a dotted line) except that it is shifted to the right by 5 m.

21.31. Model: The wavelength of the standing wave on a string vibrating at its second-harmonic frequency is equal

to the string’s length.

Visualize:

Solve: The length of the string 2.0 m,L so 2.0 m.L This means the wave number is

2 2 rad/m

2 0 mk

According to Equation 21.5, the displacement of a medium when two sinusoidal waves superpose to give a standing

wave is ( , ) ( )cos ,D x t A x t where max( ) 2 sin sinA x a kx A kx The amplitude function gives the amplitude of

oscillation from point to point in the medium. For 10 cm,x

( 10 cm) (2 0 cm)sin[( rad/m)(0 10 m)] 0 62 cmA x

Similarly, ( 20 cm) 1 2 cm,A x ( 30 cm) 1 6 cm,A x ( 40 cm) 1 9 cm,A x and ( 50 cm) 2 0 cmA x

Assess: Consistent with the above figure, the amplitude of oscillation is a maximum at 0.50 m.x

21.32. Model: The wavelength of the standing wave on a string is 2 / ,m L m where 1, 2, 3, m We assume

that 30 cm is the first place from the left end of the string where max /2A A

Visualize:

Solve: The amplitude of oscillation on the string is max( ) sin .A x A kx Since the string is vibrating in the third

harmonic, the wave number is

2 23

(2 /3)k

L L

21-14 Chapter 21


Substituting into the equation for the amplitude gives

max max

1 3 3 1 3sin (0 30 m) sin (0 30 m) (0 30 m) rad 5 4 m

2 2 6A A L

L L L

21.33. Model: The wavelength of the standing wave on a string vibrating at its fundamental frequency is equal to 2L.

Solve: The amplitude of oscillation on the string is ( ) 2 sin ,A x a kx where a is the amplitude of the traveling wave

and the wave number is

2 2

2k

L L

Substituting k into the above equation gives

14

1( ) 2 0 cm 2 sin 1 0 cm 2 cm 1 4 cm

4 2

LA x L a a a

L

21.34. Solve: You can see in Figure 21.4 that the time between two successive instants when the antinodes are at

maximum height is half the period, or 12

T Thus 2(0 25 s) 0 50 s,T and so

1 1 3 0 m/s2 0 Hz 1 5 m

0 50 s 2 0 Hz

vf

T f

21.35. Model: Model the tendon as a string. The wavelength of the standing wave on a string vibrating at its

fundamental frequency is equal to 2L.

Solve: The linear density of the string is , where

31100 kg/m and

290 mm . The velocity of waves on

this string is / /( ).v T T The allowed wavelengths for standing waves are 2 /L m (Equation 21.13), and

the fundamental frequency corresponds to 1m (see Equation 21.15). Thus, the fundamental frequency is

1 3 6 2

1 1 (500 N)177.7 Hz 180 Hz

2 2 2(0.20 m) (1100 kg/m )(90 10 m )m

mv Tf

L L

21.36. Model: Model the spider’s silk as a string. The wavelength of the standing wave on a string vibrating at its

equal to 2L.

Solve: The linear density of the string is , where 31300 kg/m and 2 6 2(10 10 m)r

10 23 14 10 m . To have a fundamental frequency (i.e., 1m ) at 100 Hz, Equation 21.14 gives

2 2 2 2 3 10 2 41 1

14 4(0.12 m) (100 Hz) (1300 kg/m )(3.14 10 m ) 2.4 10 N

2 2

v Tf T L f

L L

21.37. Model: The wave on a stretched string with both ends fixed is a standing wave. For a vibration at its

fundamental frequency, 2 .L

Solve: The wavelength of the wave reaching your ear is 39.1 cm 0.391 m, so the frequency of the sound wave is

air 344 m/s879 8 Hz

0 391 m

vf

This is also the frequency emitted by the wave on the string. Thus,

string S1 1 150 N879 8 Hz 0 568 m

0 0006 kg/m

v T

12

0 284 m 28 4 cmL

21.38. Model: The wave on a stretched string with both ends fixed is a standing wave.

Solve: We must distinguish between the sound wave in the air and the wave on the string. The listener hears a sound

wave of wavelength sound 40 cm 0.40 m. Thus, the frequency is

sound

sound

343 m/s857 5 Hz

0 40 m

vf

Superposition 21-15


The violin string oscillates at the same frequency, because each oscillation of the string causes one oscillation of the

air. But the wavelength of the standing wave on the string is very different because the wave speed on the string is

not the same as the wave speed in air. Bowing a string produces sound at the string’s fundamental frequency, so the

wavelength of the string is

string 1 string string2 0 60 m (0 60 m)(857 5 Hz) 514 5 m/sL v f

The tension is the string is found as follows:

2 2Sstring S string( ) (0 001 kg/m)(514 5 m/s) 260 N

Tv T v

21.39. Model: The steel wire is under tension and it vibrates with three antinodes.

Solve: When the spring is stretched 8.0 cm, the standing wave on the wire has three antinodes. This means 23 3

L

and the tension ST in the wire is S (0.080 m),T k where k is the spring constant. For this tension,

S Swire 3

3 (0 08 m)

2

T T kv f f

L

We will let the stretching of the spring be x when the standing wave on the wire displays two antinodes. This means

2 S and L T kx. For the tension S,T

S Swire 2

1T T k xv f f

L

The frequency f is the same in the above two situations because the wire is driven by the same oscillating magnetic

field. Now, equating the two frequency equations gives

1 3 (0 080 m)0 18 m 18 cm

2

k x kx

L L

21.40. Model: The wave on a stretched string with both ends fixed is a standing wave. Assume the pulley is

frictionless and ignore the effect on the string tension that is due to the mass of the string.

Solve: The linear density of the string is 3

35.00 10 kg2.00 10 kg/m

2.50 m

The length of the vibrating part of the string is 2.00 mL and the tension on it is ,T Mg where 4.00 kg.M

Applying Equation 21.14 gives

22

22 2 2 4m m

mv m T m Mg m Mf f g

L L L L

so a plot of 2mf versus

2

24

m M

L should give a straight line with slope g.

The fit to the data gives 8.42.g

21-16 Chapter 21


21.41. Model: The stretched bungee cord that forms a standing wave with two antinodes is vibrating at the second

harmonic frequency.

Visualize:

Solve: Because the vibrating cord has two antinodes, 2 1 80 mL The wave speed on the cord is

cord (20 Hz)(1 80 m) 36 m/sv f

The linear density of the cord is cord s/ .v T The tension ST in the cord is equal to ,k L where k is the bungee’s

spring constant and L is the 0.60 m the bungee has been stretched. The linear density has to be calculated at the

stretched length of 1.8 m where it is now vibrating. Thus,

2 2s cord

cord

(0.075 kg)(36 m/s)90 N/m

/ (1.80 m)(0.60 m)

T k L mvv k

m L L L

21.42. Model: The wave on a stretched string with both ends fixed is a standing wave.

Solve: Use the fact that the wave speed v is /v T and apply Equation 21.14 to express both the fundamental

frequency and the second-harmonic frequency:

0 01 2

1 2,

2 2

T Tf f

L L

For the second-harmonic frequency to be the same as the fundamental, we must have 1 2,f f which gives

0 0 00

1 2

2 2 4

T T TT

L L

Thus, the tension must be one-fourth its original value.

21.43. Visualize: Use primed quantities for when the sphere is submerged. We are given 5 3f f and 1 5 kg.M We

also know the density of water is

31000 kg/m In the third mode before the sphere is submerged 3 22 3

L L

Likewise, after the sphere is submerged 5 22 5

L L The tension in the string before the sphere is submerged

is s ,T Mg but after the sphere is submerged, according to Archimedes’ principle, it is reduced by the weight of the

water displaced by the sphere: s ,T Mg Vg where 343

V R

Solve: We are looking for R so solve sT Mg Vg for Vg and later we will isolate R from that.

sVg Mg T

Solve s /v T for sT Also substitute for V.

3 24

3R g Mg v

Now use v f

3 25

4( )

3R g Mg f

Superposition 21-17


Recall that 5 3f f and 25

L

23

3

4 2

3 5R g Mg Lf

Substitute 3 /f v

234 2

3 5

vR g Mg L

Now use 23

L and s /v T

2

s3

23

/4 2

3 5

TR g Mg L

L

The 2’s, ’ s, and L’s cancel.

23

s

4 3

3 5R g Mg T

Recall that sT Mg

234 3

3 5R g Mg Mg

Cancel g and factor M out on the right side.

234 3 9 16

1 13 5 25 25

R M M M

Now solve for R.

3 3 16

4 25

MR

3 33

12 12 (1 5 kg)6 1 cm

25 25 (1000 kg/m )

MR

Assess: The density of the sphere turns out to be about 1.5 times the density of water, which means it sinks and is in

a reasonable range for densities.

21.44. Visualize: First compute the length L of the wire from the Pythagorean theorem. 2 0 2 mL Now /M L

(0 075 kg)/(2 0 2 m) 0 02652 kg/m Also, in the fundamental mode 2 ;L here 4,0 2 m 5 657 m

Solve: To find the tension in the cable, recall that the net torque about any point in the system must be zero in a

static situation. Taking the sum of the torques about the point where the horizontal bar joins the wall gives

2s s

/2 (4.0 kg)/2+8.0 kgsin 0 N (9.8 m/s ) 138.6 N

2 sin sin(45 )

mgd mg MgT d Mgd T

Use these values for , , and sT to find f.

s / (138 6 N)/(0 02652 kg/m)13 Hz

5 657 m

Tvf

Assess: This seems like a reasonable frequency for a mechanical system like this.

21.45. Model: Assume that the spring provides the same tension to both strings and also acts as a fixed point for

the end of each string so an integral number of half wavelengths fit in each string.

Visualize: Use a subscript L for the left string and R for the right string. From the assumption above we know

s L s R s( ) ( )T T T We also know L Rf f f and L RL L L Notice from the diagram that L L and

2R 3

L

21-18 Chapter 21


Solve: From v f and s /v T eliminate v and solve for 2: /( )T f Take the ratio of the linear densities

to find R:

222sR s R R R 3

R 02 2 220 s L L L s3

/( )( ) /( ) 1 9 9

4 4( ) /( ) /( )

T LfT f

T f T Lf

Assess: We expect a slower wave speed in the right string to correspond to a larger mass density.

21.46. Model: The microwave forms a standing wave between the two reflectors.

Solve: (a) There are reflectors at both ends, so the electromagnetic standing wave acts just like the standing wave on

a string that is tied at both ends. The frequencies of the standing waves are 8

light 93 0 10 m/s(1 5 10 Hz) 1 5 GHz

2 2 2(0 10 m)m

v cf m m m m m

L L

where we have used the fact that electromagnetic waves of all frequencies travel at the speed of light c. The generator

can produce standing waves at any frequency between 10 GHz and 20 GHz. These are

m (GHz)mf

7 10.5

8 12.0

9 13.5

10 15.0

11 16.5

12 18.0

13 19.5

21.47. Model: The fundamental wavelength of an open-open tube is 2L and that of an open-closed tube is 4L.

Solve: We are given that

1 open closed 3 open open 1 open open3f f f

air air

1 open closed 1 open open open closed open open

1 33

4 2

v v

L L

open openopen closed

2 2(78 0 cm)13 0 cm

12 12

LL

21.48. Model: A tube forms standing waves.

Solve: (a) The fundamental frequency cannot be 390 Hz because 520 Hz and 650 Hz are not integer multiples of it.

But we note that the difference between 390 Hz and 520 Hz is 130 Hz as is the difference between 520 Hz and

650 Hz. We see that 1 1 1390 Hz 3 130 Hz = 3 , 520 Hz 4 , and 650 Hz = 5 .f f f So we are seeing the third, fourth,

and fifth harmonics of a tube whose fundamental frequency is 130 Hz. According to Equation 21.17, this is an open-

open tube because m 1 = with 1, 2, 3, 4, f mf m For an open-closed tube m has only odd values.

(b) Knowing 1,f we can now find the length of the tube:

1

343 m/s1 3 m

2 2(130 Hz)

vL

f

21.49. Model: Model the vocal tract as an open-closed tube that forms standing waves.

Solve: The standing-wave frequencies in an open-closed tube are proportional to the wave speed (see Equation

21.18). Therefore the new frequencies will be

1 2

750 m/s 750 m/s(270 Hz) 580 Hz, (2300 Hz) 4.9 kHz

350 m/s 350 m/sm mf f

Superposition 21-19


21.50. Model: Particles of the medium at the nodes of a standing wave have zero displacement.

Solve: The cork dust settles at the nodes of the sound wave where there is no motion of the air molecules. The

separation between the centers of two adjacent piles is 12

. Thus,

123 cm82 0 cm

3 2

Because the piston is driven at a frequency of 400 Hz, the speed of the sound wave in oxygen is

(400 Hz)(0 820 m) 328 m/sv f

Assess: A speed of 328 m/s in oxygen is close to the speed of sound in air, which is 343 m/s at 20°C.

21.51. Model: The nodes of a standing wave are spaced /2 apart. Assume there are no standing-wave modes

between those given.

Visualize:

Solve: The wavelength of the mth mode of an open-open tube is m 2 / .L m Or, equivalently, the length of the tube

that generates the mth mode is ( /2).L m Here is the same for all modes because the frequency of the tuning fork

is unchanged. Increasing the length of the tube to go from mode m to mode 1m requires a length change

( 1)( /2) /2 /2L m m

That is, lengthening the tube by /2 adds an additional antinode and creates the next standing wave. This is

consistent with the idea that the nodes of a standing wave are spaced /2 apart. This tube is first increased

56 7 cm 42 5 cm 14 2 cm,L then by 70.9 cm 56.7 cm 14.2 cm.L Thus /2 14.2 cm and thus

28.4 cm 0.284 m. Therefore the frequency of the tuning fork is

343 m/s1208 Hz 12 1 kHz

0 284 m

vf

21.52. Model: The open-closed tube forms standing waves.

Visualize:

Solve: When the air column length L is the proper length for a 580 Hz standing wave, a standing-wave resonance

will be created and the sound will be loud. From Equation 21.18, the standing-wave frequencies of an open-closed

tube are m ( /4 ),f m v L where v is the speed of sound in air and m is an odd integer: 1, 3, 5, m The frequency

is fixed at 580 Hz, but as the length L changes, 580 Hz standing waves will occur for different values of m. The

length that causes the mth standing-wave mode to be at 580 Hz is

(343 m/s)

(4)(580 Hz)

mL

21-20 Chapter 21


We can place the values of L, and corresponding values of 1.0 m ,h L in a table:

m L 1.0 mh L

1 0.148 m 0.852 m 85.2 cm

3 0.444 m 0.556 m 55.6 cm

5 0.739 m 0.261 m 26.1 cm

7 1.035 m h can’t be negative

So water heights of 26, 56, and 85 cm will cause a standing-wave resonance at 580 Hz. The figure shows the

3m standing wave at 56 cm.h

21.53. Model: A stretched wire, which is fixed at both ends, creates a standing wave whose fundamental frequency

is 1 wire.f The second vibrational mode of an open-closed tube is 3 open closed.f These two frequencies are equal

because the wire’s vibrations generate the sound wave in the open-closed tube.

Visualize:

Solve: The frequency in the tube is

air3 open closed

tube

3 3(340 m/s)300 Hz

4 4(0 85 cm)

vf

L

wire S1 wire

wire wire

1300 Hz

2 2

v Tf

L L

2 2 2 2S wire(300 Hz) (2 ) (300 Hz) (2 0 25 m) (0 020 kg/m) 450 NT L

21.54. Model: In a rod in which a longitudinal standing wave can be created, the standing wave is equivalent to a

sound standing wave in an open-open tube. Both ends of the rod are antinodes, and the rod is vibrating in the

fundamental mode.

Solve: Since the rod is in the fundamental mode, 1 2 2(2.0 m) 4.0 m.L Using the speed of sound in aluminum,

the frequency is

Al1

1

6420 m/s1605 Hz 1 6 kHz

4 0 m

vf

21.55. Model: Model the tunnel as an open-closed tube.

Visualize: We are given 335 m/sv We would like to use ( odd)4

m

vf m m

L to find L, but we need to know

m first. Since m takes on only odd values for the open-closed tube the next resonance after m is 2m We are

given 4 5 Hzmf and 2 6 3 Hzmf

Solve:

2 2 2

2

( 2)2 2 24 2 1 2 5

6 3 Hz( ) 1 1

4 4 5

m m m

mm m m

m

vm

f m f fL m m m mv ff m f f

mL f Hz

Superposition 21-21


Now that we know m we can find the length L of the tunnel.

335 m/s(5) 93 m

4 4 4(4 5 Hz)m

m

v vf m L m

L f

Assess: 93 m seems like a reasonable length for a tunnel.

21.56. Model: A standing wave in an open-closed tube must have a node at the closed end of the tube and an

antinode at the open end.

Visualize:

Solve: We first draw a series of pictures showing all the possible standing waves. By examination, we see that the

first standing wave mode is 14

of a wavelength, so the tube’s length is 14

L The next mode is 34

of a wavelength.

The tube’s length hasn’t changed, so in this mode 34

L The next mode is now slightly more than a wavelength:

54

L The next mode is 74

of a wavelength, so 74

L We see that there is a pattern. The length of the tube and

the possible standing-wave wavelengths are related by

where 1, 3, 5, 7, odd integers4

mL m

Solving for , we find that the wavelengths and frequencies of standing waves in an open-closed tube are

4

1, 3, 5, 7, odd integers

4

m

mm

L

mm

v vf m

L

21.57. Model: The amplitude is determined by the interference of the two waves.

Solve: For interference in one dimension, where the speakers are separated by a distance x, the amplitude of the net

wave is 12

2 cos ,A a where a is the amplitude of each wave and 02 /x is the phase difference

between the two waves. The speakers are emitting identical waves so they have identical phase constants,

so 0 0. Thus,

1 1 51 5 2 cos cos

2

xA a a x

21-22 Chapter 21


The wavelength of a 1000 Hz tone is sound / 0 343 mv f Thus the separation must be

10 343 mcos (0 75) 0 0789 m 7 9 cmx

It is essential to note that the argument of the arccosine is in radians, not in degrees.

21.58. Model: Constructive or destructive interference occurs according to the phases of the two waves.

Visualize:

Solve: (a) To go from destructive to constructive interference requires moving the speaker 12

,x which is

equivalent to a phase change of rad. Since 40 cm,x we find 80 cm.

(b) Destructive interference at 10 cmx requires

0 0 0

10 cm 32 2 rad rad rad

80 cm 4

x

(c) When side by side, with 0,x the phase difference is 0 3 /4 rad. The amplitude of the superposition

of the two waves is

32 cos 2 cos 0 77

2 8a a a a

21.59. Model: Interference occurs according to the difference between the phases of the two waves.

Visualize:

Solve: (a) The phase difference between the sound waves from the two speakers is

02x

We have a maximum intensity when 0.50 mx and 0.90 m.x This means

0 0

(0 50 m) 0 90 m2 2 rad 2 2( 1) radm m

Superposition 21-23


Taking the difference of these two equations gives

sound0 40 m 340 m/s2 2 0 40 m 850 Hz

0 40 m

vf

(b) Using again the equations that correspond to constructive interference, we find

0 0 20 10

0 50 m2 2 rad rad

0 40 m 2m

We have taken 1m in the last equation. This is because we always specify phase constants in the range rad to

π rad (or 0 rad to 2π rad). 1m gives 12 rad (or equivalently, 2m will give 3

2 rad)

21.60. Model: Reflection is maximized for constructive interference of the two reflected waves, but minimized for

destructive interference.

Solve: (a) Constructive interference of the reflected waves occurs for wavelengths given by Equation 21.32:

2 2(1 42)(500 nm) (1420 nm)m

nd

m m m

1Thus, 1420 nm, 12 2

(1420 nm) 710 nm, 3 473 nm, 4 355 nm, Only the wavelength of 473 nm is

in the visible range.

(b) For destructive interference of the reflected waves,

1 1 12 2 2

2 2(1 42)(500 nm) 1420 nmnd

m m m

1Thus, 2 1420 nm 2840 nm, 22 3

(1420 nm) 947 nm, 3 568 nm, 4 406 nm, The wavelengths of

406 nm and 568 nm are in the visible range.

(c) Beyond the limits 430 nm and 690 nm the eye’s sensitivity drops to about 1 percent of its maximum value. The

reflected light is enhanced in blue (473 nm). The transmitted light at mostly 568 nm will be yellowish green.

21.61. Model: Reflection is minimized when the two reflected waves interfere destructively.

Solve: Equation 21.23 gives the condition for perfect destructive interference between the two waves:

10 2

2 2 radx

m

The wavelength of the sound is

343 m/s0 2858 m

1200 Hz

v

f

Let d be the separation between the mesh and the wall. Substituting 0 0 rad, 2 ,x d 0,m and the above

value for the wavelength,

2 (2 ) 0 2858 m0 rad rad 0 0715 m 7 2 cm

0 2858 m 4

dd

21.62. Model: A light wave that reflects from a boundary at which the index of refraction increases has a phase

shift of π rad.

Solve: (a) Because film air ,n n the wave reflected from the outer surface of the film (called 1) is inverted due to the

phase shift of π rad. The second reflected wave does not go through any phase shift of π rad because the index of

refraction decreases at the boundary where this wave is reflected, which is on the inside of the soap film. We can

write for the phases

1 1 10 2 2 20 rad 0 radkx kx

2 1 2 1 20 10 0( ) ( ) rad rad radk x x k x k x

0 0 rad because the sources are identical. For constructive interference,

21-24 Chapter 21


film

22 rad rad 2 rad (2 ) (2 1) radm k x m d m

Cfilm C1 1 1

2 2 2

2 2 2 660, 1, 2, 3,

d nd dm

n m m m

(b) For 0m the wavelength for constructive interference is

C 1

2

(2 66)(390 nm)2075 nm

For m and 2, C 692 nm (~red) and C 415 nm (~violet). Red and violet together give a purplish color.

21.63. Model: The two radio antennas are sources of in-phase, circular waves. The overlap of these waves causes

interference.

Visualize:

Solve: Maxima occur along lines such that the path difference to the two antennas is r m The 8750 MHz = 7.50 10 Hz wave has a wavelength / 0 40 mc f Thus, the antenna spacing 2.0 md is exactly 5.

The maximum possible intensity is on the line connecting the antennas, where 5r d So this is a line of

maximum intensity. Similarly, the line that bisects the two antennas is the 0r line of maximum intensity. In

between, in each of the four quadrants, are four lines of maximum intensity with , 2 , 3 , and 4 .r Although

we have drawn a fairly accurate picture, you do not need to know precisely where these lines are located to know that

you have to cross them if you walk all the way around the antennas. Thus, you will cross 20 lines where r m

and will detect 20 maxima.

21.64. Model: The changing sound intensity is due to the interference of two overlapped sound waves.

Solve: Minimum intensity implies destructive interference. Destructive interference occurs where the path length

difference for the two waves is 12

.r m We assume 0 0 rad for two speakers playing “exactly the same”

tone. The wavelength of the sound is sound / (343 m/s)/686 Hz 0 500 m.v f Consider a point that is a distance

x in front of the top speaker. Let 1r be the distance from the top speaker to the point and 2r the distance from the

bottom speaker to the point. We have

2 21 2 3 0 mr x r x

Destructive interference occurs at distances x such that

2 2 12

9 0 mr x x m

Superposition 21-25


To solve for x, isolate the square root on one side of the equation and then square:

2 212 2 22 2 21 1 1

2 2 2 12

9 0 m9 0 m 2

2

mx x m x m x m x

m

Evaluating x for different values of m:

m x (m)

0 17.88

1 5.62

2 2.98

3 1.79

Because you start at 2.5 mx and walk away from the speakers, you will only hear minima for values 2.5 m.x

Thus, to correct significant figures, minima will occur at distances of 3.0 m, 5.6 m, and 18 m.

21.65. Model: The changing sound intensity is due to the interference of two overlapped sound waves.

Visualize: The listener moving relative to the speakers changes the phase difference between the waves.

Solve: Initially when you are at P, equidistant from the speakers, you hear a sound of maximum intensity. This

implies that the two speakers are in phase 0( 0). However, upon moving to Q you hear a minimum of sound

intensity, which implies that the path length difference from the two speakers to Q is /2 Thus,

2 2 2 212 1 1 12

( ) (5 0 m) (12 0 m) (5 0 m) 12 0 m 1 0 mr r r r r

340 m/s2 0 m 170 Hz

2 0 m

vf


Visualize:

21-26 Chapter 21


Solve: The amplitude of the sound wave is 12

2 cosA a With 0 0 rad, the phase difference between the

waves is

2 1 2 2 2 cos2 0 m 2 0 m

r r rA a

At the coordinates (0.0 m, 0.0 m), 0 m,r so 2.0 .A a At the coordinates (0.0 m, 0.5 m),

2 2 2 2 (0 551 m)(3 0 m) (2 5 m) (3 0 m) (1 5 m) 0 551 m 2 cos 1 3

2 0 mr A a a

At the coordinates (0.0 m, 1.0 m),

2 2 2 2 (1 08 m)(3 0 m) (3 0 m) (3 0 m) (1 0 m) 1 08 m 2 cos 0 25

2 0 mr A a a

At the coordinates (0.0 m, 1.5 m), 1 568 m and 1 56r A a At the coordinates (0.0 m, 2.0 m), 2 0 mr and

2 0A a


Visualize:

Solve: The amplitude of the sound wave is 12

2 cos .A a We are to find the minimum ratio /x at which

.A a Because the speakers are identical, we may take 0 0.

1 11 12 2

1 12 cos 2cos 2 cos

2 3

x xa a

21.68. Model: The two radio antennas are sources of in-phase waves. The overlap of these waves causes

interference.

Visualize:

Solve: (a) The phase difference of the two waves at point P is given by

2 2 2 20 2 12 (800 m) (650 m) (800 m) (550 m) 59.96 m

rr r r

The wavelength of the radio wave is 8

6

3 0 10 m/s100 m

3 0 10 m

c

f

Superposition 21-27


Since the sources are identical, so 0 0 rad The phase difference at P due to the two waves is

59 96 m2 0 rad 1 2 rad

100 m

(b) Since 1 20 0 6(2 ), which is neither m2π nor 12

2 ,m the interference at P is somewhere in between

maximum constructive and perfect destructive.

(c) At a point 10 m further north we have

2 2 2 2(800 m) (660 m) (800 m) (560 m) 60 58 mr

60 58 m2 0 rad 1 21 rad (0 605)2

100 m

Because the phase difference is increasing as you move north, you are moving from a destructive interference

condition 12

2m with 0m toward a constructive interference condition (2 )m with 1.m The

signal strength will therefore increase.

21.69. Model: The amplitude is determined by the interference of the three waves.

Solve: (a) We have three identical loudspeakers as sources. r between speakers 1 and 2 is 1.0 m and 2.0 m.

Thus 12

,r which gives perfect destructive interference for in-phase sources. That is, the interference of the

waves from loudspeakers 1 and 2 is perfect destructive, leaving only the contribution due to speaker 3. Thus the

amplitude is a.

(b) If loudspeaker 2 is moved to the left by one-half of a wavelength or 1.0 m, then all three waves will reach you in

phase. The amplitude of the superposed waves will therefore be maximum and equal to 3 .A a

(c) The maximum intensity is 2 2max 9 .I CA Ca The ratio of the intensity to the intensity of a single speaker is

2max

2single speaker

99

I Ca

I Ca

21.70. Model: The superposition of two slightly different frequencies gives rise to beats.

Solve: The third harmonic of note A and the second harmonic of note E are

3A 1A 2E 1E3 3(440 Hz) 1320 Hz 2 2(659 Hz) 1318 Hzf f f f

The beat frequency is therefore

3A 2E 1320 Hz 1318 Hz 2 Hzf f

(b) The beat frequency between the first harmonics is

1E 1A 659 Hz 440 Hz 219 Hzf f

The beat frequency between the second harmonics is

2E 2A 1318 Hz 880 Hz 438 Hzf f

The beat frequency between 3Af and 2Ef is 2 Hz. Thus, the tuner must listen for a beat frequency of 2 Hz.

(c) If the beat frequency is 4 Hz, then the second harmonic frequency of the E string is 1

2E 1E 21320 Hz 4 Hz 1316 Hz (1316 Hz) 658 Hzf f

Note that the second harmonic frequency of the E string could also be

2E 1E1320 Hz 4 Hz 1324 Hz 662 Hzf f

This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that

string

Tv f f T

21.71. Model: The superposition of two slightly different frequencies creates beats.

Solve: (a) The wavelength of the sound initially created by the flutist is

342 m/s0 77727 m

440 Hz

21-28 Chapter 21


When the speed of sound inside her flute has increased due to the warming up of the air, the new frequency of the A

note is

346 m/s445 Hz

0 77727 mf

Thus the flutist will hear beats at the following frequency:

445 Hz 440 Hz 5 beats/s f f

Note that the wavelength of the A note is determined by the length of the flute rather than the temperature of air or

the increased sound speed.

(b) The initial length of the flute is

1 12 2

(0 77727 m) 0 3886 mL The new length to eliminate beats needs to be

1 1 346 m/s0 3932 m

2 2 2 440 Hz

vL

f

Thus, she will have to extend the “tuning joint” of her flute by

0 3932 m 0 3886 m 0 0046 m 4 6 mm

21.72. Solve: (a) Yvette’s speed is the width of the room divided by time. This means

12 Y

Y

2n n vv

t t

Note that 12 is the distance between two consecutive antinodes, and n is the number of such half wavelengths that

fill the entire width of the room.

(b) Yvette observes a higher frequency f of the source she is moving toward and a lower frequency f of the

source she is receding from. If v is the speed of sound and f is the sound wave’s frequency, we have

Y Y1 1v v

f f f fv v

The expression for the beat frequency is

Y Y Y Y Y21 1 2 2

v v v v v vf f f f f

v v v v

(c) The answers to part (a) and (b) are the same. Even though you and Yvette have different perspectives, you should

agree as to how many modulations per second she hears.

21.73. Model: The frequency of the loudspeaker’s sound in the back of the pick-up truck is Doppler shifted. As the

truck moves away from you, the frequency of the sound emitted by its speaker is decreased.

Solve: Because you hear 8 beats per second as the truck drives away from you, the frequency of the sound from the

speaker in the pick-up truck is 400 Hz 8 Hz 392 Hzf This frequency is

0 SS

S

400 Hz1 1 020408 7 0 m/s

1 / 343 m/s 392 Hz

f vf v

v v

That is, the velocity of the source Sv and hence the pick-up truck is 7.0 m/s.

21.74. Model: A stretched string under tension supports standing waves.

Solve: (a) The wave speed on a stretched string is

string

1T Tv f f

The wavelength cannot change if the length of the string does not change. So,

1/212

1 1 1 1 1 1 1( )

2 2 2 2

df T f TT f

dT T T f TT

(b) Since there are 5 beats per second,

10 Hz 10 Hz5 Hz 0 020 2 0

2 500 Hz

f T Tf

T T f

,

That is, an increase of 2.0% in the tension of one of the strings will cause 5 beats per second.

Superposition 21-29


21.75. Model: The microphone will detect a loud sound only if there is a standing-wave resonance in the tube. The

sound frequency does not change, but changing the length of the tube can create a standing wave.

Solve: The standing-wave condition is

280 Hz 1, 2, 3,2

vf m m

L

where L is the total length of the tube. When the slide is extended a distance s, the tube has two straight sides, each of

length 80 cm,s plus a semicircular bend of length 12

(2 ).r The radius is 12

(10 cm) 5 0 cm.r The tube’s total

length is 12

2( 80 cm) (2 5 0 cm) 175 7 cm 2 1 757 m 2L s s s

A standing-wave resonance will be created if

343 m/s[ 1 757 2 ] 0 6125

2 2(280 Hz)

vL s m m m

f

0 3063 0 8785 meterss m

We can tabulate the different extensions s that correspond to standing-wave modes 1, 2, 3,m m m and so on.

m s

1 0.572 m

2 0.266 m

3 0.040 m 4.0 cm

4 0.347 m 34.7 cm

5 0.653 m 65.3 cm

6 1.959 m

Physically, the extension must be greater than 0 cm and less than 80 cm. Thus, the three slide extensions that create a

standing-wave resonance at 280 Hz are 4.0, 35, and 65 cm to two significant figures.

21.76. Model: The stretched wire is vibrating at its second harmonic frequency.

Visualize: Let l be the full length of the wire, and L be the vibrating length of the wire. That is, 12

L l

Solve: The wave speed on a stretched wire is

swire

Tv f

21-30 Chapter 21


The frequency 100 Hzf and the wavelength

12

l because it is a second harmonic wave. The tension s (1.25 kg)T g

because the hanging mass is in static equilibrium and

31.00 10 kg/m. Substituting in these values,

2 32 1 2 2

3

(1 25 kg) (1 00 10 kg/m)(100 Hz) (100 Hz) (2 00 m s )

2 2 (1 25 kg)(1 00 10 kg/m)

g l lg l

To find l we can use the equation for the time period of a simple pendulum: 2 2

2

2 2

(314 s/100)2 0 250 s

4 4

l TT l g g g

g

Substituting this expression for l into the equation for g, we get 1 2 2 2 2 1 2 2

1 2 2

(2 000 m s )(0 025 s ) (0 125 m s ) 0

(0 125 m s ) 1 0 8 00 m/s

g g g g

g g g

Assess: A value of 28.0 m/s is reasonable for the information given in the problem.

21.77. Model: Assume that the extra kilogram doesn’t stretch the wire longer (so L stays the same) nor thinner (so

stays the same). Also assume that, because the wire is thin, its own weight is negligible, so sT is constant

throughout the wire and is equal to Mg.

Visualize: The wire is fixed at both ends so, in the second harmonic, .L We are given 2 200 Hz,f 2 245 Hz,f

and 1 0 kgM M Apply v f and s /v T

Solve: Taking the ratio of the frequencies allows gives

s2

2 s

/ ( 1 0 kg)/ 1 0 kg

/ /

T M g M gf v M

f v MT Mg Mg

2

2

2

1 0 kgf M

f M

2

2

2

1 1 0 kgf

Mf

2 2

2

2

1 0 kg 1 0 kg2 0 kg

245 Hz11

200 Hz

M

f

f

Assess: We did not expect M to be really huge or (a) it would have broken the wire, and (b) adding one more

kilogram wouldn’t have made as big a difference in 2f as it did.

21.78. Model: The frequency is Doppler shifted to higher values for a detector moving toward the source. The

frequency is also shifted to higher values for a source moving toward the detector.

Visualize:

Superposition 21-31


Solve: (a) We will derive the formula in two steps. First, the object acts like a moving detector and “observes” a

frequency that is given by 0 0 0(1 / ).f f v v Second, as this moving object reflects (or acts as a “source” of ultrasound

waves), the frequency echof as observed by the original source oS is 1echo 0 0(1 / ) .f f v v Combining these two

equations gives

0 0 0 0echo 0

0 0 0

(1 / )

1 / 1 /

f f v v v vf f

v v v v v v

(b) If 0 ,v v then

10 0 0 0 0

echo 0 0 0

21 1 1 1 1

v v v v vf f f f

v v v v v

0beat echo 0 0

2vf f f f

v

(c) Using part (b) for the beat frequency,

600

265 Hz (2 40 10 Hz) 2 09 cm/s

1540 m/s

vv

(d) Assuming the heart rate is 90 beats per minute the angular frequency is

2 2 (1 5 beats/s) 9 425 rad/sf

Using 0 max ,v v A

01

2 09 cm/s 2 09 cm/s2 2 mm

9 425 rad/s90 min

vA

21.79. Solve: (a) The wavelengths of the standing-wave modes are

21, 2, 3,m

Lm

m

1 2 3

2(10 0 m) 2(10 0 m) 2(10 0 m)20 0 m 10 0 m 6 67 m

1 2 3

The depth of the pool is 5.0 m. Clearly the standing waves with 2 and 3 are “deep water waves” because the 5.0 m

depth is larger than one-quarter of the wavelength. The wave with 1 barely qualifies to be a deep water standing

wave.

(b) The wave speed for the first standing-wave mode is

21

1

(9 8 m/s )(20 0 m)5 6 m/s

2 2

gv

21-32 Chapter 21


Likewise, 2 34 0 m/s and 3 2 m/s.v v

(c) We have

2 2 4

mm m m

m

g g mgv f f

L

Note that m is the mode and not the mass.

(d) The period of oscillation for the first standing-wave mode is calculated as follows

2

1 11

(1)(9 8 m/s ) 10 279 Hz 3 6 s

4 (10 0 m)f T

f

Likewise, 2 32 5 s and 2 1 s.T T

21.80. Model: The overlap of the waves causes interference.

Solve: (a) The waves traveling to the left are

1 2 20

( )sin 2 sin 2

x t x L tD a D a

T T

The phase difference between the waves on the left side of the antenna is thus

L 2 1 20 20

( )2 2 2

x L t x t L

T T

On the right side of the antennas, where 1 2 ,x x L the two waves are

1 2 20

( )sin 2 sin 2

x t x L tD a D a

T T

Thus, the phase difference between the waves on the right is

R 2 1 20 20

( ) 22 2

x L t x t L

T T

We want to have destructive interference on the country side or on the left and constructive interference on the right.

This requires

1L 20 R 202

2 22 2

L Lm n

These are two simultaneous equations, and we can satisfy them both if L and 20 are properly chosen. Subtracting

the second equation from the first to eliminate 20,

1 12 2

42

2

Lm n L m n

The smallest value of L that works is for ,n m in which case 14

L

(b) From the R equation,

14

20 20 20 20

222 2 rad

2

Ln n

Adding integer multiples of 2π to the phase constant doesn’t really change the wave, so the physically significant

phase constant is for 0,n namely 120 2

rad.

(c) We have 1 120 2 4

rad (2 ). If the wave from antenna 2 was delayed by one full period T, it would shift the

wave by one full cycle. We would describe this by a phase constant of 2π rad. So a phase constant of 14

(2 ) rad can

be achieved by delaying the wave by 14

.t T

(d) A wave with frequency 61000 kHz 1 00 10 Hzf has a period 61 00 10 s 1 00 sT and wavelength

/ 300 m.c f So this broadcast scheme will work if the antennas are spaced 75 mL apart and if the broadcast

from antenna 2 is delayed by 0.25 s 250 ns.t

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