Instructor Solution Manual - Santa Rosa Junior College

SUPERPOSITION

21

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

m  4. (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) m  2 (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 v  Ts / and v   f  Assume the linear mass density is the same in all parts of this question. Use primes T  v2  ( f )2 f 2 (2 f0 )2      4 Changing the tension is how one tunes a guitar; it T0 v02  ( f0 )2 f02 f02 would be difficult to increase the tension by a factor of 4 without breaking the string, so only small changes are made this way. L   v /f  v /(2 f0 ) 1 (b)     so the length must be decreased by a factor of 2. Guitar players effectively L0  0 v /f0 v /(f0 ) 2 for new quantities. (a)

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

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21-1

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 (f  2 Hz), 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 t  1.0 s differs from the graph t  0.0 s 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 t  2.0, 3.0, and 4.0 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines.

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


21-4

Chapter 21

The graph at t  1.0 s differs from the graph at t  0.0 s 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 t  2.0, 3.0, and 4.0 s are a superposition of the left- and the right-moving waves. The overlapping parts of the two waves are shown by the dotted lines.

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

At t  4 s the shorter pulses overlap and cancel. At t 6 s 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

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

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

(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 t  2 s differs from the graph at t  0 s 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 t  4 s. At t  6 s, 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 t  8 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 v   f  030 m100 Hz  30 m/s

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   52 (20 m)  080 m. The frequency of the standing wave is

f 

v





40 m/s  50 Hz 080 m

21.8. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: (a) When the frequency is doubled ( f   2 f0 ), the wavelength is halved (  12  0 ). 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

v

T



 v 

T





4T



 2v

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

v  2v 

f   2 f0 0 

f  0  2 f0 0 

f   2 f0

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 2L m  m  1, 2, 3,  m The three longest wavelengths for standing waves will therefore correspond to m  1, 2, and 3. Thus, 2(24 m) 2(24 m) 2(24 m) 1   48 m  2   24 m  3   16 m 1 2 3 (b) Because the wave speed on the string is unchanged from one m value to the other, f 22 (50 Hz)(24 m) f 22  f33  f3    75 Hz 3 16 m

21.10. Model: A string fixed at both ends supports standing waves. Solve: (a) We have fa  36 Hz  mf1 where f1 is the fundamental frequency that corresponds to m  1. The next successive frequency is fb  48 Hz  (m  1) f1. Thus,

f b (m  1) f1 m  1 48 Hz 4 4m      m 1  fa mf1 m 36 Hz 3 3

 m3 

f1 

36 Hz  12 Hz 3

The wave speed is

2L f1  (20 m)(12 Hz)  24 m/s 1 (b) The frequency of the fourth harmonic is 48 Hz. For m  4, the wavelength is 2L 2(10 m) 1 m    m m 4 2 v  1 f1 

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 vwire  f   f (2L)  (80 Hz)(2  090 m)  144 m/s and is also equal to

TS / , where



mass 50 103 kg   5555  103 kg/m length 090 m


21-8

Chapter 21

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

vwire 

Mg



 M

2 vwire

g



(5555 103 kg/m)(144 m/s)2 98 m/s2

 12 kg

21.12. Model: The laser light forms a standing wave inside the cavity. Solve: The wavelength of the laser beam is 2L 2(05300 m) m   100,000   1060 m m 100,000 The frequency is f100,000 

c

100,000



29979  108 m/s 1060 106 m

 2828  1013 Hz

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

The three longest wavelengths are 2(121 m) 1   242 m 1 (b) In the case of an open-closed tube,

The three longest wavelengths are 4(121 m) 1   484 m 1

m 

2L m

2 

2(121 m)  121 m 2

m 

4L m

2 

4(121 m)  161 m 3

m  1, 2, 3, 

3 

2(121 m)  0807 m 3

3 

4(121 m)  0968 m 5

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 m  2 mode and the wavelength of the standing wave is equal to the length of the tube, that is,   0.80 m. Since the frequency f  500 Hz, the speed of sound in this case is v  f   (500 Hz)(0.80 m)  400 m/s. 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 f1  1500 Hz when the tube is filled with helium gas at 0C. Using  m  2L/m, f1 helium 

vhelium

1

Similarly, when the tube is filled with air, v 331 m/s f1 air 331 m/s f1 air  air    1 2L f1 helium 970 m/s



970 m/s 2L



 331 m/s  f1 air    (1500 Hz)  512 Hz  970 m/s 

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: fm  m

v v  mf1  L  4L 4 f1


Superposition

21-9

for a fundamental frequency of f1  250 Hz and a speed of sound of v  350 m/s, the length of the vocal tract must be approximately 350 m/s L  35 cm 4(250 Hz)

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 2L m   1  2L  2(190 m)  380 m m The wave speed along the string is v  f11  (27.5 Hz)(3.80 m)  104.5 m/s. The tension in the wire can be found as follows:

v

TS



 mass  2  0400 kg  2  TS  v 2   v    (1045 m/s)  2180 N length 2  00 m    

21.18. Model: A string fixed at both ends forms standing waves. Solve: A simple string sounds the fundamental frequency f1  v /(2L). Initially, when the string is of length

LA  30 cm, the note has the frequency f1A  v /(2LA ). For a different length, f1B  v /(2LB ). Taking the ratio of each side of these two equations gives f1A v /(2LA ) LB f    LB  1A LA f1B v /(2LB ) LA f1B We know that the second frequency is desired to be f1B  523 Hz. The string length must be

440 Hz (30 cm) 252 cm 523 Hz 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. LB 

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 (x)1  20 cm but out of phase when

(x)2  60 cm Thus,



   2(60 cm  20 cm)  80 cm 2 (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. (x)2  x 1 

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

m  0 for the smallest d and the condition for destructive interference, we get x   2   0  2 m  12  rad, m  0, 1, 2, 3, 








21-10

Chapter 21



1 v     2 2 f (b) When the speakers are out of phase,  0    Using m  1 for the smallest 2

x

d

  0   rad  2

 0 rad   rad  d 

 1  343 m/s      25 cm  2  686 Hz  d and the condition for constructive

interference, we get

  2 2

d



   2

x



  0  2m ,

 d

m  0, 1, 2, 3, 



1  v  1  343 m/s        25 cm 2 2  f  2  686 Hz 

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:  Cm (600  109 m)(1) 2nd C   d   216 nm m 2n (2)(139) where we have used m  1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair  nfilm  nglass .

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:  Cm (500  109 m)(1) 2nd C   d   200 nm m 2n (2)(125) where we have used m  1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair  noil  nwater .

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, r1  3 and r2  4. Thus, r  r2  r1  4  3  . The phase difference is

 

2r





2 ( )



 2

This corresponds to constructive interference. For the point Q, r1  72  and r2  2. The phase difference is

 

2r





2

 32    3 

This corresponds to destructive interference. For the point R, r1  52  and r2  72 . The phase difference is

 

2 ( )



 2

This corresponds to constructive interference.



r 

C/D C

2

3  2

D

7  2



C

r1

r2

P

3

Q

7  2

R

5  2


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, r1  2 and r2  3. The phase difference is

 

2r



2 (3  2 )

  0 



   3

This corresponds to destructive interference. For the point Q, r1  3 and r2  32   The phase difference is

 

2

 32      4 

This corresponds to constructive interference. For the point R, r1  52  and r2  3. The phase difference is

 

2

 12      2 

This corresponds to constructive interference.

r1

r2

P

2

3

r 

C/D D

Q

3

3  2

3  2

C

R

5  2

3

1 2

C

Assess: Note that it is not r1 or r2 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,

r2  r12  (20 m)2  (40 m)2  (20 m) 2  4472 m so r  r2  r1  4472 m  40 m  0472 m The phase difference between the sources is  0  0 rad and the wavelength of the sound waves is

v 340 m/s   01889 m f 1800 Hz Thus, the phase difference of the waves at the point 4.0 m in front of one source is




21-12

Chapter 21

2 (0472 m)  0 rad  5 rad  52 (2 rad) 01889 m This is a half-integer multiple of 2π rad, so the interference is perfect destructive.   2

r



  0 

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, r1  800 m and r2  (800 m)2  (600 m)2  1000 m

So, r  r2  r1  200 m and  0   rad The wavelength of the waves is

c 30  108 m/s   100 m f 30 106 Hz Thus, the phase difference of the waves at the point (300 m, 800 m) is r 2 (200 m)   2   0    rad  5 rad  52 (2 rad)  100 m 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 fbeat  f1  f2

 3 Hz  f1  200 Hz



f1  203 Hz

f1 is larger than f 2 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 fbeat  f1  f 2  100 MHz where we have used the fact that the 1   2 so f1  f 2 . The frequency of emitter 1 is f1  c /1, where

1  1250 102 m. The wavelength of emitter 2 is  2  c /f 2 

c c (300 108 m/s)    126 cm. f1  100 MHz c /1  100 MHz (300  108 m/s)/(1250 102 m)  100 MHz


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 x  1 m at t  0.050 s, was thus located at x  6 m at t  0 s. This helps us draw the snapshot of the wave pulse moving left at t  0 s (shown as a dashed line). Subtracting this wave snapshot from the resultant at t  0 s (shown as a solid line) yields the right-traveling wave’s snapshot at t  0 s (shown as a dotted line). Finally, the snapshot graph of the wave pulse moving right at t  0.050 s is the same as at t  0 s (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 L  2.0 m, so   L  2.0 m. This means the wave number is

2   rad/m 20 m According to Equation 21.5, the displacement of a medium when two sinusoidal waves superpose to give a standing wave is D( x,t )  A( x)cos t , where A( x)  2a sin kx  Amax sin kx The amplitude function gives the amplitude of oscillation from point to point in the medium. For x  10 cm, k

2





A( x  10 cm)  (20 cm)sin[( rad/m)(010 m)]  062 cm Similarly, A( x  20 cm)  12 cm, A( x  30 cm)  16 cm, A( x  40 cm)  19 cm, and A( x  50 cm)  20 cm Assess: Consistent with the above figure, the amplitude of oscillation is a maximum at x  0.50 m.

21.32. Model: The wavelength of the standing wave on a string is  m  2L/m, where m  1, 2, 3,  We assume that 30 cm is the first place from the left end of the string where A  Amax /2 Visualize:

Solve: The amplitude of oscillation on the string is A( x)  Amax sin kx. Since the string is vibrating in the third harmonic, the wave number is 2 2  k  3  (2 L /3) L


21-14

Chapter 21

Substituting into the equation for the amplitude gives 1 3   3   3  1 Amax  Amax sin  (030 m)   sin  (030 m)    (030 m)  rad  L  54 m 2 L 6 L  L  2

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 A( x)  2a sin kx, where a is the amplitude of the traveling wave and the wave number is 2 2  k    2L L Substituting k into the above equation gives    L    1  A( x  14 L)  20 cm  2a sin      10 cm  a    a  2 cm  14 cm L 4     2  

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

f 

Thus T  2(025 s)  050 s, and so

1 T 2

1 1 v 30 m/s   20 Hz      15 m T 050 s f 20 Hz

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   1100 kg/m3 and   90 mm2 . The velocity of waves on this string is v  T/  T/( ). The allowed wavelengths for standing waves are   2L/m (Equation 21.13), and the fundamental frequency corresponds to m  1 (see Equation 21.15). Thus, the fundamental frequency is f m1 

mv 1 T 1 (500 N)    177.7 Hz  180 Hz 2L 2 L  2(0.20 m) (1100 kg/m3 )(90 106 m2 )

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   1300 kg/m3 and    r 2   (10 106 m)2 

314 1010 m2. To have a fundamental frequency (i.e., m  1 ) at 100 Hz, Equation 21.14 gives f1 

v 1 T   T  4L2 f12   4(0.12 m)2 (100 Hz) 2 (1300 kg/m3 )(3.14 10 10 m 2 )  2.4  104 N 2L 2L 

21.37. Model: The wave on a stretched string with both ends fixed is a standing wave. For a vibration at its fundamental frequency,   2L. Solve: The wavelength of the wave reaching your ear is 39.1 cm  0.391 m, so the frequency of the sound wave is v 344 m/s f  air   8798 Hz  0391 m This is also the frequency emitted by the wave on the string. Thus,

8798 Hz 

vstring





1 TS







1

150 N  00006 kg/m

   0568 m

L  12   0284 m  284 cm

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

f 

vsound

 sound



343 m/s  8575 Hz 040 m


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  2L  060 m  vstring   string f  (060 m)(8575 Hz)  5145 m/s The tension is the string is found as follows:

vstring 

TS



 TS   (vstring )2  (0001 kg/m)(5145 m/s)2  260 N

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  3  23 L and the tension TS in the wire is TS  k (0.080 m), where k is the spring constant. For this tension,

vwire 

TS





f 3 

TS





f 

3 k (008 m) 2L 

We will let the stretching of the spring be x when the standing wave on the wire displays two antinodes. This means 2  L and TS  kx. For the tension TS,

vwire 

TS





f 2 

TS





f 

1 k x 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 k x 3 k (0080 m)  L  2L 

 x  018 m  18 cm

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

5.00  103 kg  2.00  103 kg/m 2.50 m The length of the vibrating part of the string is L  2.00 m and the tension on it is T  Mg , where M  4.00 kg. Applying Equation 21.14 gives  m2 M  mv m T m Mg fm     f m2  g  2   4L   2L 2L  2L   



so a plot of f m2 versus

m2M 4 L2 

should give a straight line with slope g.

The fit to the data gives g  8.42. © Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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  L  180 m The wave speed on the cord is

vcord  f   (20 Hz)(180 m)  36 m/s The linear density of the cord is vcord  Ts / . The tension TS 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,

vcord 

Ts





k L mv 2 (0.075 kg)(36 m/s)2  k  cord   90 N/m m/L LL (1.80 m)(0.60 m)

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:

f1 

1 T0 , 2L 

f2 

2 T0 2L 

For the second-harmonic frequency to be the same as the fundamental, we must have f1  f 2 , which gives

1 T0 2 T0  2L  2L 

 T0 

T0 4

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 f5  f3 and M  15 kg. We also know the density of water is   1000 kg/m3 In the third mode before the sphere is submerged L  32     23 L Likewise, after the sphere is submerged L  52     52 L The tension in the string before the sphere is submerged is Ts  Mg , but after the sphere is submerged, according to Archimedes’ principle, it is reduced by the weight of the water displaced by the sphere: Ts  Mg  Vg , where V  43  R3 Solve: We are looking for R so solve Ts  Mg  Vg for Vg and later we will isolate R from that.

Vg  Mg  Ts Solve v  Ts / for Ts Also substitute for V.

4  

   R3 g  Mg  v2 3 Now use v   f 

 4  

   R3 g  Mg   (  f5)2 3 © Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Superposition

21-17

Recall that f5  f3 and   52 L 2

 4  

2 

 

 4  

2 

v

    R3 g  Mg    Lf3  3 5 Substitute f3  v /

    R3 g  Mg    L  3 5  Now use  

2L 3

2



and v  Ts / 

2

4  

Ts /   2L  3 

    R3 g  Mg    L 5 3 

2

The 2’s, ’ s, and L’s cancel.

 4  

3 

4  

 3  

 

    R3 g  Mg   Ts  3 5

2

Recall that Ts  Mg 2

    R3 g  Mg    Mg 3 5 Cancel g and factor M out on the right side.



 4  

2  3    

 

9  

 16   

   R3  M 1      M 1    M   3 5 25 25 

Now solve for R.

 3  16  M R3      4  25   R3

12 M 12 (15 kg) 3  61 cm 25  25  (1000 kg/m3 )

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. L  20 2 m Now   M /L  (0075 kg)/(20 2 m)  002652 kg/m Also, in the fundamental mode   2L; 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 mgd mg /2  Mg (4.0 kg)/2+8.0 kg    Tsd sin  Mgd  0 N  Ts   (9.8 m/s 2 )  138.6 N 2 sin sin(45) Use these values for  ,  , and Ts to find f.

Ts /

(1386 N)/(002652 kg/m)  13 Hz 5657 m Assess: This seems like a reasonable frequency for a mechanical system like this. f 

v









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 (Ts )L  (Ts )R  Ts  We also know f L  f R  f and LL  LR  L Notice from the diagram that  L  L and

 R  23 L


21-18

Chapter 21

Solve: From v   f and v  Ts / eliminate v and solve for :   T /( f )2  Take the ratio of the linear densities to find  R : 2

R (Ts )R /(R f R )2 Ts /( 32 Lf ) 1 9 9       R  0 2 2 2 0 (Ts )L /(L f L ) 4 Ts /(Lf ) 2 4 3

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 vlight c 30 108 m/s fm  m m m  m(15  109 Hz)  15m GHz 2L 2L 2(010 m) 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

f m (GHz)

7 8 9 10 11 12 13

10.5 12.0 13.5 15.0 16.5 18.0 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 f1 open closed  f3 open open  3 f1 open open

vair

1 open closed

3

vair

 1 open open

Lopen closed 



2Lopen open 12



1 3  4Lopen closed 2Lopen open

2(780 cm)  130 cm 12

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 390 Hz  3 130 Hz = 3f1, 520 Hz  4f1, and 650 Hz = 5f1. 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 openopen tube because fm = mf1 with m  1, 2, 3, 4,  For an open-closed tube m has only odd values. (b) Knowing f1, we can now find the length of the tube:

L

v 343 m/s   13 m 2 f1 2(130 Hz)

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 750 m/s 750 m/s f m 1  (270 Hz)  580 Hz, f m 2  (2300 Hz)  4.9 kHz 350 m/s 350 m/s


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 cm      820 cm 3 2 Because the piston is driven at a frequency of 400 Hz, the speed of the sound wave in oxygen is v  f   (400 Hz)(0820 m)  328 m/s 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  2L/m. Or, equivalently, the length of the tube that generates the mth mode is L  m( /2). 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 m  1 requires a length change

L  (m  1)( /2) m /2   /2 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 L  567 cm  425 cm  142 cm, then by L  70.9 cm  56.7 cm  14.2 cm. Thus  /2  14.2 cm and thus   28.4 cm  0.284 m. Therefore the frequency of the tuning fork is v 343 m/s f    1208 Hz  121 kHz  0284 m

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 f m  m(v /4L), where v is the speed of sound in air and m is an odd integer: m  1, 3, 5,  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 m(343 m/s) L (4)(580 Hz)


21-20

Chapter 21

We can place the values of L, and corresponding values of h  1.0 m  L, in a table: h  1.0 m  L m L 0.852 m  85.2 cm 1 0.148 m 0.556 m  55.6 cm 3 0.444 m 0.261 m  26.1 cm 5 0.739 m 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 m  3 standing wave at h  56 cm.

21.53. Model: A stretched wire, which is fixed at both ends, creates a standing wave whose fundamental frequency is f1 wire . The second vibrational mode of an open-closed tube is f3 open closed . 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

f3 open closed 

3vair 3(340 m/s)   300 Hz 4Ltube 4(085 cm)

f1 wire  300 Hz 

vwire 1 TS  2Lwire 2Lwire 

TS  (300 Hz)2 (2Lwire )2   (300 Hz)2 (2  025 m)2 (0020 kg/m)  450 N

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  2L  2(2.0 m)  4.0 m. Using the speed of sound in aluminum, the frequency is

f1 

vAl

1



6420 m/s  1605 Hz  16 kHz 40 m

21.55. Model: Model the tunnel as an open-closed tube. v (m  odd) to find L, but we need to know 4L m first. Since m takes on only odd values for the open-closed tube the next resonance after m is m  2 We are given f m  45 Hz and f m 2  63 Hz Solve: v (m  2) fm 2 2 2 4 L  m  2  m  f m  2   m  2  m  f m  2  1  2  m    5     v f 6  3 Hz fm m m 2  1  fm   fm  ( m) 1 4L fm 45 Hz Visualize: We are given v  335 m/s We would like to use f m  m


Superposition

21-21

Now that we know m we can find the length L of the tunnel. v v 335 m/s fm  m  Lm  (5)  93 m 4L 4 fm 4(45 Hz) 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

1 4

of a wavelength, so the tube’s length is L  14  The next mode is

3 4

of a wavelength.

The tube’s length hasn’t changed, so in this mode L  34   The next mode is now slightly more than a wavelength: L  54   The next mode is

7 4

of a wavelength, so L  74   We see that there is a pattern. The length of the tube and

the possible standing-wave wavelengths are related by m L where m  1, 3, 5, 7,   odd integers 4 Solving for , we find that the wavelengths and frequencies of standing waves in an open-closed tube are 4L  m   m  m  1,3,5,7,   odd integers  v v  fm  m m 4L  

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 A  2a cos

 12   , where a is the amplitude of each wave and   2x/   0 is the phase difference

between the two waves. The speakers are emitting identical waves so they have identical phase constants, so  0  0. Thus,

  x  1  15  A  15a  2a cos    x  cos        2  © Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21-22

Chapter 21

The wavelength of a 1000 Hz tone is   vsound /f  0343 m Thus the separation must be

x 

0343 m



cos1(075)  00789 m  79 cm

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 x  12  , which is equivalent to a phase change of  rad. Since x  40 cm, we find   80 cm. (b) Destructive interference at x  10 cm requires

3  10 cm    0  2  rad  rad   0   rad   0  4  80 cm  (c) When side by side, with x  0, the phase difference is    0  3 /4 rad. The amplitude of the superposition 2

x



of the two waves is

3    a  2a cos   077a   2a cos 2 8  

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 x   2   0



We have a maximum intensity when x  0.50 m and x  0.90 m. This means (050 m)  090 m  2   0  2m rad 2     0  2(m  1) rad    


Superposition

21-23

Taking the difference of these two equations gives vsound 340 m/s  040 m  2    850 Hz   2    040 m  f   040 m    (b) Using again the equations that correspond to constructive interference, we find   050 m  2     0  2m rad   0   20  10   rad 0  40 m 2   We have taken m  1 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). m  1 gives  12  rad (or equivalently, m  2 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: 2nd 2(142)(500 nm) (1420 nm) m    m m m Thus, 1  1420 nm,  2  12 (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, 2nd 2(142)(500 nm) 1420 nm    1 m 2 m  12 m  12

Thus, 1  2 1420 nm  2840 nm,  2  23 (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: x   2   0  2 m  12  rad







The wavelength of the sound is



v 343 m/s   02858 m f 1200 Hz

Let d be the separation between the mesh and the wall. Substituting  0  0 rad, x  2d , m  0, and the above value for the wavelength,

2 (2d ) 02858 m  0 rad  rad  d   00715 m  72 cm 02858 m 4

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 nfilm  nair , 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  kx1  10   rad  2  kx2   20  0 rad

   2  1  k ( x2  x1)  ( 20  10 )   rad  k x   0   rad  k x   rad

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


21-24

Chapter 21

 2    2m rad  k x   rad  2m rad    (2d )  (2m  1) rad   film   

 film 

C n



2d m  12

 C 

2nd 266d  m  12 m  12

m  0, 1, 2, 3,

(b) For m  0 the wavelength for constructive interference is (266)(390 nm) C   2075 nm

 12 

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

750 MHz = 7.50 108 Hz wave has a wavelength   c /f  040 m Thus, the antenna spacing d  2.0 m is exactly 5. The maximum possible intensity is on the line connecting the antennas, where r  d  5 So this is a line of maximum intensity. Similarly, the line that bisects the two antennas is the r  0 line of maximum intensity. In between, in each of the four quadrants, are four lines of maximum intensity with r  , 2, 3, and 4. 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 r  m  12 . We assume  0  0 rad for two speakers playing “exactly the same” tone. The wavelength of the sound is   vsound /f  (343 m/s)/686 Hz  0500 m. Consider a point that is a distance x in front of the top speaker. Let r1 be the distance from the top speaker to the point and r2 the distance from the bottom speaker to the point. We have r1  x r2  x2  30 m2 Destructive interference occurs at distances x such that





r  x2  90 m2  x  m  12 


Superposition

21-25

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





2







x  90 m   x  m  12    x 2  2 m  12  x  m  12   2



2



2

 x



90 m  m  12







2

2

2 m  12 

Evaluating x for different values of m: m 0 1 2 3

x (m) 17.88 5.62 2.98 1.79

Because you start at x  2.5 m and walk away from the speakers, you will only hear minima for values x  2.5 m. 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, 1 2

 r  r2  r1  (r1)2  (50 m)2  r1  (120 m)2  (50 m)2  120 m  10 m

  20 m 

f 

v





340 m/s  170 Hz 20 m

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


21-26

Chapter 21

Solve: The amplitude of the sound wave is A  2a cos

 12    With  0  0 rad, the phase difference between the

waves is

 r  A  2a cos    20 m  At the coordinates (0.0 m, 0.0 m), r  0 m, so A  2.0a. At the coordinates (0.0 m, 0.5 m),   2  1  2

r



 2

r 20 m



r  (30 m)2  (25 m)2  (30 m)2  (15 m)2  0551 m 

A  2a cos

(0551 m)  13a 20 m

At the coordinates (0.0 m, 1.0 m),

 (108 m)  A  2a cos    025a  20 m  At the coordinates (0.0 m, 1.5 m), r  1568 m and A  156a At the coordinates (0.0 m, 2.0 m), r  20 m and A  20a r  (30 m)2  (30 m)2  (30 m)2  (10 m) 2  108 m 

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

Solve: The amplitude of the sound wave is A  2a cos

 12   . We are to find the minimum ratio x/

at which

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

   1 a  2a cos      2cos  2 

 12   2 x



x





1



cos1

 12   13

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 r   2   0 r  r2  r1  (800 m)2  (650 m)2  (800 m) 2  (550 m) 2  59.96 m



The wavelength of the radio wave is



c 30  108 m/s   100 m f 30  106 m


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 m    2    0 rad 12 rad  100 m 





(b) Since   120  06(2 ), which is neither m2π nor m  12 2 , the interference at P is somewhere in between maximum constructive and perfect destructive. (c) At a point 10 m further north we have r  (800 m)2  (660 m)2  (800 m)2  (560 m)2  6058 m

 6058 m    2    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   m  12 2 with m  0 toward a constructive interference condition   m(2 ) with m  1. 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 r  12  , 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 A  3a. (c) The maximum intensity is I max  CA2  9Ca 2 . The ratio of the intensity to the intensity of a single speaker is I max Isingle speaker



9Ca 2 Ca 2

9

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 f3A  3 f1A  3(440 Hz)  1320 Hz f 2E  2 f1E  2(659 Hz)  1318 Hz The beat frequency is therefore f3A  f2E  1320 Hz  1318 Hz  2 Hz (b) The beat frequency between the first harmonics is f1E  f1A  659 Hz  440 Hz  219 Hz The beat frequency between the second harmonics is f2E  f2A  1318 Hz  880 Hz  438 Hz The beat frequency between f3A and f 2E 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 f 2E  1320 Hz  4 Hz  1316 Hz   f1E  12 (1316 Hz) 658 Hz Note that the second harmonic frequency of the E string could also be f2E  1320 Hz  4 Hz  1324 Hz   f1E 662 Hz This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that

vstring   f 

T





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/s   077727 m 440 Hz © Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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/s f  445 Hz 077727 m Thus the flutist will hear beats at the following frequency: f   f  445 Hz  440 Hz  5 beats/s 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 L  12   12 (077727 m)  03886 m The new length to eliminate beats needs to be



1  v  1  346 m/s        03932 m 2  f  2  440 Hz  Thus, she will have to extend the “tuning joint” of her flute by L 

2

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 n

 12  

n 2vY   t t  1 Note that 2  is the distance between two consecutive antinodes, and n is the number of such half wavelengths that vY 

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  v   v  f   f 1  Y   f   f 1  Y  v  v    The expression for the beat frequency is v vv 2v  v   v  f   f   f 1  Y   f 1  Y   2 f Y  2 Y  Y 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 f  400 Hz  8 Hz  392 Hz This frequency is

f 

f0 1  vS /v

 1

vS 400 Hz   1020408  vS  70 m/s 343 m/s 392 Hz

That is, the velocity of the source vS 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

vstring 

T



 f 

f 

1 T

 

The wavelength  cannot change if the length of the string does not change. So, df 1 1 1 1 1 1 1 T  1 f T  (T )1/2   f      2 dT   2 T 2T     2T f 2T (b) Since there are 5 beats per second, f T T 10 Hz 10 Hz f  5 Hz       0020  20, 2 T T f 500 Hz That is, an increase of 2.0% in the tension of one of the strings will cause 5 beats per second. © Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 v f  280 Hz m m  1,2,3, 2L 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 s  80 cm, plus a semicircular bend of length 12 (2 r ). The radius is r  12 (10 cm)  50 cm. The tube’s total length is

L  2(s  80 cm)  12 (2  50 cm)  1757 cm  2s  1757 m  2s A standing-wave resonance will be created if

 v  343 m/s [ L  1757  2s]   m m  06125m  2 f 2(280 Hz)   s  03063m  08785 meters We can tabulate the different extensions s that correspond to standing-wave modes m  1, m  2, m  3, and so on. m

s

1 2 3 4 5 6

0.572 m 0.266 m 0.040 m  4.0 cm 0.347 m  34.7 cm 0.653 m  65.3 cm 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, L 

 12  l 

Solve: The wave speed on a stretched wire is

vwire 

Ts



 f


21-30

Chapter 21

The frequency f  100 Hz and the wavelength   12 l because it is a second harmonic wave. The tension Ts  (1.25 kg) g because the hanging mass is in static equilibrium and   1.00 103 kg/m. Substituting in these values, 3 l  l  (100 10 kg/m)  g  (100 Hz) 2    (200 m 1s 2 )l 2 2 (125 kg) 2 2

(125 kg) g (100 103 kg/m)

 (100 Hz)

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

T  2

l g

 l

T2

g 2

(314 s/100)2

4 4 Substituting this expression for l into the equation for g, we get

g  (2000 m1s 2 )(0025 s 2 )2 g 2

2

g   0250 s 2  g

 (0125 m 1s 2 ) g 2  g  0

(0125 m1s 2 ) g  1 g  0  g  800 m/s 2   Assess: A value of 8.0 m/s2 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 Ts 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 f 2  200 Hz, f 2  245 Hz, and M   M  10 kg Apply v   f and v  Ts /  Solve: Taking the ratio of the frequencies allows gives T / M g ( M  10 kg) g f 2 v/ M  10 kg   s    f 2 v / M Ts / Mg Mg 2

 f 2  M  10 kg    f M  2 2  f    M  2   1  10 kg  f 2     10 kg 10 kg M   20 kg 2 2  f 2   245 Hz     1  200 Hz   1    f2  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 f 2 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 f0  f0 (1  v0 /v). Second, as this moving object reflects (or acts as a “source” of ultrasound waves), the frequency fecho as observed by the original source So is fecho  f0 (1  v0 /v)1. Combining these two equations gives f 0 f (1  v0 /v) v  v0 fecho   0  f0 1  v0 /v 1  v0 /v v  v0 (b) If v0

v, then  v  v  fecho  f0 1  0 1  0  v  v 

1

f beat

 v  v   2v   f0 1  0 1  0   f0 1  0  v  v v     2v0  fecho  f0  f0 v

(c) Using part (b) for the beat frequency,  2v0  6 65 Hz    (240 10 Hz)  v0  209 cm/s 1540 m/s   (d) Assuming the heart rate is 90 beats per minute the angular frequency is

  2 f  2 (15 beats/s)  9425 rad/s Using v0  vmax   A,

A

v0





209 cm/s 90 min 1



209 cm/s  22 mm 9425 rad/s

21.79. Solve: (a) The wavelengths of the standing-wave modes are 2L m  1, 2, 3, m 2(100 m) 2(100 m) 2(100 m) 1   200 m 2   100 m 3   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

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

v1 

g1 2



(98 m/s2 )(200 m)  56 m/s 2


21-32

Chapter 21

Likewise, v2  40 m/s and v3  32 m/s. (c) We have

v

gm  f mm 2



fm 

g mg  2 m 4 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

f1 

(1)(98 m/s 2 ) 1  0279 Hz T1   36 s 4 (100 m) f1

Likewise, T2  25 s and T3  21 s.

21.80. Model: The overlap of the waves causes interference. Solve: (a) The waves traveling to the left are   x t    ( x  L) t   D1  a sin  2      D2  a sin  2       20   T    T     The phase difference between the waves on the left side of the antenna is thus L  ( x  L) t   x t  L   2  1  2       20  2      2   20  T  T      On the right side of the antennas, where x1  x2  L, the two waves are

  x t    ( x  L) t   D1  a sin  2     D2  a sin  2      20   T  T        Thus, the phase difference between the waves on the right is 2 L  ( x  L) t  x t   R   2  1  2      20  2        20 T     T  We want to have destructive interference on the country side or on the left and constructive interference on the right. This requires 2 L 2 L  L   20   2 m  12  R   20   2 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 ,

4 L





 2 m  12  n





 L  m  12  n

 2

The smallest value of L that works is for n  m, in which case L  14  (b) From the  R equation,

 

 2 14  2 L    20    20    20    2 n   20   2 n rad    2    Adding integer multiples of 2π to the phase constant doesn’t really change the wave, so the physically significant phase constant is for n  0, namely  20  12  rad. (c) We have  20  12  rad  14 (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 be achieved by delaying the wave by t 

1 (2 ) 4

rad can

1 T. 4

(d) A wave with frequency f  1000 kHz  100 106 Hz has a period T  100 106 s  100 s and wavelength   c/f  300 m. So this broadcast scheme will work if the antennas are spaced L  75 m apart and if the broadcast from antenna 2 is delayed by t  0.25 s  250 ns. © Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Instructor Solution Manual - Santa Rosa Junior College

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