Ch 12 Physics Answers

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Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 12

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CHAPTER 12

1. Because the sound travels both ways across the lake, we haveL = !vt = !(343 m/s)(1.5 s) = 2.6 102 m.

2. Because the sound travels down and back, we haveL = !vt = !(1560 m/s)(2.0 s) = 1.6 103 m = 1.6 km.

3. (a) We find the extreme wavelengths from1 = v/f1 = (343 m/s)/(20 Hz) = 17 m;2 = v/f2 = (343 m/s)/(20,000 Hz) = 1.7 102 m = 1.7 cm.

The range of wavelengths is 1.7 cm ? ? 17 m.(b) We find the wavelength from

= v/f= (343 m/s)/(10 106 Hz) = 3.4 105 m.

4. For the travel times we havetwire = d/vwire = (8.4 m)/(5100 m/s) = 0.0016 s;

tair = (d/vair) = (8.4 m)/(343 m/s) = 0.024 s ( 15twire).

5. The speed in the concrete is determined by the elastic modulus:vconcrete = (E/)1/2 = [(20 109 N/m2)/(2.3 103 kg/m3)]1/2 = 2.95 103 m/s.

For the time interval we havet = (d/vair) (d/vconcrete);1.4 s = d{[1/(343 m/s)] [1/(2.95 103 m/s)]}, which gives d = 5.4 102 m.

6. (a) For the time interval in sea water we havet = d/vwater = (1.0 103 m)/(1560 m/s) = 0.64 s.

(b) For the time interval in the air we havet = d/vair = (1.0 103 m)/(343 m/s) = 2.9 s.

7. If we let L1 represent the thickness of the top layer, the total transit time ist = (L1/v1) + [(L L1)/v2)];4.5 s = [L1/(331 m/s)] + [(1500 m L1)/(343 m/s)], which gives L1 = 1200 m.

Thus the bottom layer is 1500 m 1200 m = 300 m.

8. Because the distance is d = vt, the change in distance from the change in velocity isd = t v; so the percentage change is(d/d)100 = (v/v)100 = (343 331 m/s)(100)/(343 m/s) = 3.5%.


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9. We find the intensity of the sound from1= 10 log10(I1/I0);

120 dB = 10 log(I1/1012 W/m2), which gives I1 = 1.0 W/m

2.For a whisper we have

2 = 10 log10(I2/I0);

20 dB = 10 log(I1/1012

W/m2

), which gives I2 = 1.0

1010

W/m2

.

10. We find the intensity level from= 10 log10(I1/I0) = 10 log10[(2.0 10

6 W/m2)/(1012 W/m2)] = 63 dB.

11. We find the ratio of intensities of the sounds from= 10 log10(I2/I1);2.0 dB = 10 log10(I2/I1), which gives I2/I1 = 1.58.

Because the intensity is proportional to the square of the amplitude, we haveI2/I1 = (A2/A1)

2;

1.58 = (A2/A1)2, which givesA2/A1 = 1.3.

12. When three of the four engines are shut down, the intensity is reduced by a factor of 4, so we have2 1 = 10 log10(I2/I0) 10 log10(I1/I0) = 10 log10(I2/I1);

2 120 dB = 10 log((I1/I1), which gives 2 = 114 dB.

13. We find the ratio of intensities from= 10 log10(Isignal/Inoise);

58 dB = 10 log10(Isignal/Inoise), which gives Isignal/Inoise = 6.3 105.

14. (a) Using the intensity from Table 122, we haveP = IA = I2?r2 = (3 106 W/m2)2?(0.50 m)2 = 5 106 W.

(b) We find the number of people required fromN= Ptotal/P = (100 W)/(5 106 W) = 2 107.

15. (a) We find the intensity of the sound from= 10 log10(I/I0);

50 dB = 10 log(I/1012 W/m2), which gives I= 1.0 107 W/m2.The rate at which energy is absorbed is the power of the sound wave:

P = IA = (1.0 107

W/m2

)(5.0 105

m2

) = 5.0 1012

W.(b) We find the time fromt = E/P = (1.0 J)/(5.0 1012 W) = 2.0 1011 s = 6.3 103 yr.


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16. (a) If we assume that one channel is connected to the speaker, the rating is the power in the sound, andthe sound spreads out uniformly, we have

I= P/4?r2, so we getI1 = (250 W)/4?(3.0 m)2 = 2.21 W/m2;I2 = (40 W)/4?(3.0 m)2 = 0.354 W/m2.

For the intensity levels we have1 = 10 log10(I1/I0) = 10 log10[(2.21 W/m2)/(1012 W/m2)] = 123 dB;2 = 10 log10(I2/I0) = 10 log10[(0.354 W/m

2)/(1012 W/m2)] = 115 dB.(b) The difference in intensity levels is 8 dB. A change of 10 dB corresponds to a doubling of the

loudness, so the expensive amp is almost twice as loud.

17. (a) We find the intensity of the sound from= 10 log10(I/I0);

130 dB = 10 log(I/1012 W/m2), which gives I= 10 W/m2.The power output of the speaker is

P = IA = I4?r2 = (10 W/m2)4?(2.5 m)2 = 7.9 102 W.(b) We find the intensity of the sound from

= 10 log10(I/I0);

90 dB = 10 log(I/1012 W/m2), which gives I= 1.0 103 W/m2.We find the distance from

P = IA ;7.9 102 W = (1.0 103 W/m2)4?r2 , which gives r = 2.5 102 m.

18. (a) The intensity of the sound wave is given byI= 2?2f2x0

2v.Because the frequency, density, and velocity are the same, for the ratio we have

I2/I1 = (x02/x01)2 = 32 = 9.

(b) We find the change in intensity level from= 10 log10(I2/I1) = 10 log10(9) = 9.5 dB.

19. (a) We find the intensity of the sound fromP = IA = I2?r2;5.0 105 W = I4?(30 m)2 , which gives I= 44.2 W/m2.

We find the intensity level from= 10 log10(I/I0) = 10 log10[(44.2 W/m

2)/(1012 W/m2)] = 136 dB.(b) We find the intensity of the sound without air absorption from

P = I1A1 = I12?r12;5.0 105 W = I14?(1000 m)2, which gives I1= 3.98 102 W/m2.

We find the intensity level from10 = 10 log10(I1/I0) = 10 log10[(3.98 10

2 W/m2)/(1012 W/m2)] = 106 dB.When we consider air absorption, we have

1 =10 (7.0 dB/km)r1 = 106 dB (7.0 dB/km)(1.0 km) = 99 dB.(c) We find the intensity of the sound without air absorption from

P = I2A2 = I22?r22;5.0 105 W = I24?(5000 m)2, which gives I2= 1.59 103 W/m2.

We find the intensity level from20 = 10 log10(I2/I0) = 10 log10[(1.59 10

3 W/m2)/(1012 W/m2)] = 92 dB.When we consider air absorption, we have


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2 =20 (7.0 dB/km)r2 = 92 dB (7.0 dB/km)(5.0 km) = 57 dB.


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20. The intensity of the sound wave is given byI= 2?2f2x0

2v.Because the amplitude, density, and velocity are the same, for the ratio we have

I2/I1 = (f2/f1)2 = 22 = 4.

21. The intensity of the sound wave is given byI = 2?2f2x0

2v

= 2?2(1.29 kg/m3)(260 Hz)2(1.3 103 m)2(343 m/s) = 1.0 103 W/m2.We find the intensity level from

= 10 log10(I/I0) = 10 log10[(1.0 103 W/m2)/(1012 W/m2)] = 150 dB.

22. We find the intensity of the sound from= 10 log10(I/I0);

120 dB = 10 log(I/1012 W/m2), which gives I= 1.00 W/m2.We find the maximum displacement from

I= 2?2

f2

x02

v1.00 W/m2 = 2?2(1.29 kg/m3)(131 Hz)2x02(343 m/s), which gives x0 = 8.17 10

5 m.

23. From Figure 127, we see that a 100-Hz tone with an intensity level of 50 dB has a loudness level of 20 phons.If we follow this loudness curve to 6000 Hz, we find that the intensity level must be 25 dB.

24. From Figure 127, we see that an intensity level of 30 dB intersects the loudness level of 0 phons at 150 Hz. Atthe high frequency end, the highest frequency is reached, so the range is 150 Hz to 20,000 Hz.

25. (a) From Figure 127, we see that at 100 Hz the lowest threshold is 5 109 W/m2 and the threshold ofpain is 1 W/m2. For the ratio we have

(1 W/m2)/(5 109 W/m2) = 2 108.(b) From Figure 127, we see that at 5000 Hz the lowest threshold is 5 1013 W/m2 and the threshold

of pain is 1 101 W/m2. For the ratio we have(1 101 W/m2)/(5 1013 W/m2) = 2 1011.

26. The wavelength of the fundamental frequency for a string is = 2L, so the speed of a wave on the string isv = f= 2(0.32 m)(440 Hz) = 282 m/s.

We find the tension fromv = [FT/(m/L)]1/2;

282 m/s = {FT/[(0.35 103

kg)/(0.32 m)]}1/2

, which gives FT = 87 N.

27. The wavelength of the fundamental frequency for a string is = 2L. Because the speed of a wave on the stringdoes not change, we have

v = 1f1 = 2f2;2(0.70 m)(330 Hz) = 2L2(440 Hz), which gives L2 = 0.525 m.

Thus the finger must be placed 0.70 m 0.525 m = 0.17 m from the end.


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28. We find the speed of sound at the temperature fromv = (331 + 0.60T) m/s = [331 + (0.60)(21C)] m/s = 344 m/s.

The fundamental wavelength for an open pipe has an antinode at each end, so the wavelength is = 2L. Wefind the length from

v = f;344 m/s = 2L(262 Hz), which gives L = 0.66 m.

29. (a) The empty soda bottle is approximately a closed pipe with a node at the bottom and an antinode atthe top. The wavelength of the fundamental frequency is = 4L. We find the frequency from

v = f;343 m/s = 4(0.15 m)f, which givesf= 570 Hz.

(b) The length of the pipe is now% of the original length. We find the frequency fromv = f;343 m/s = 4(%)(0.15 m)f, which givesf= 860 Hz.

30. For an open pipe the wavelength of the fundamental frequency is = 2L. We find the required lengths from

v = f = 2Lf;343 m/s = 2Llowest(20 Hz), which gives Llowest = 8.6 m;343 m/s = 2Lhighest(20,000 Hz), which gives Lhighest = 8.6 103 m = 8.6 mm.

Thus the range of lengths is 8.6 mm < L < 8.6 m.

31. (a) For a closed pipe the wavelength of the fundamental frequency is 1 = 4L. We find the fundamentalfrequency from

v = 1f1;343 m/s = 4(1.12 m)f1, which givesf1 = 76.6 Hz.

Only the odd harmonics are present, so we havef3 = 3f1 = (3)(76.6 Hz) = 230 Hz;

f5 = 5f1 = (5)(76.6 Hz) = 383 Hz;f7 = 7f1 = (7)(76.6 Hz) = 536 Hz.

(b) For an open pipe the wavelength of the fundamental frequency is 1 = 2L. We find the fundamentalfrequency from

v = 1f1;343 m/s = 2(1.12 m)f1, which givesf1 = 153 Hz.

All harmonics are present, so we havef2 = 2f1 = (2)(153 Hz) = 306 Hz;f3 = 3f1 = (3)(153 Hz) = 459 Hz;f4 = 4f1 = (4)(153 Hz) = 612 Hz.

32. The fundamental wavelength for a flute has an antinode at each end, so the wavelength is = 2L. Uncoveringa hole shortens the length. We find the new length from

v = f;343 m/s = 2L(294 Hz), which gives L = 0.583 m.

Thus the hole must be 0.655 m 0.583 m = 0.072 m = 7.2 cm from the end.


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33. Without the support the bridge is like a string, so the wavelength is 1 = 2L. Adding the support creates anode at the middle, so the wavelength is 2 = L. The wave speed in the bridge has not changed. We find thenew frequency from

v = 1f1 = 2f2;2L(4.0 Hz) = Lf2, which givesf2 = 8.0 Hz.

Yes, because this is higher than the expected earthquake frequencies, the modification did some good.

34. We assume that the length, and thus the wavelength, has not changed. The frequency change is due to thechange in the speed of sound. We find the speed of sound at 5C:

v = (331 + 0.60T) m/s = [331 + (0.60)(5C)] m/s = 334 m/s.For the percent change in frequency we have

(f/f)100 = (v/v)100 = [(334 m/s 343 m/s)/(343 m/s)]100 = 2.6%.

35. (a) We find the speed of sound at 5C:v = (331 + 0.60T) m/s = [331 + (0.60)(15C)] m/s = 340 m/s.

The wavelength of the fundamental frequency is 1 = 4L. We find the length from

v = 1f1 = 4Lf1;340 m/s = 4L(294 Hz), which gives L = 0.289 m.

(b) For helium we havev = 1f1 = 4Lf1;1005 m/s = 4(0.289 m)f1, which givesf1 = 869 Hz.

Note that we have no correction for the 5 C temperature change.

36. (a) For an open pipe all harmonics are present, the difference in frequencies is the fundamentalfrequency, and all frequencies will be integral multiples of the difference. For a closed pipe only oddharmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencieswill not be integral multiples of the difference but odd multiples of half the difference. For this pipe

we havef= 616 Hz 440 Hz = 440 Hz 264 Hz = 176 Hz.Because we have frequencies that are not integral multiples of this, the pipe is closed.

(b) The fundamental frequency isf1 = !f= !(176 Hz) = 88 Hz.

Note that the given frequencies are the third, fifth, and seventh harmonics.

37. For an open pipe all harmonics are present, the difference in frequencies is the fundamental frequency, and allfrequencies will be integral multiples of the difference. Thus we have

f1 = f= 330 Hz 275 Hz = 55 Hz.The wavelength of the fundamental frequency is 1 = 2L. We find the speed of sound from

v = 1f

1 = 2Lf1 = 2(1.80 m)(55 Hz) = 198 m/s.


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38. For an open pipe all harmonics are present, the difference in frequencies is the fundamentalfrequency, and all frequencies will be integral multiples of the difference. For a closed pipe only oddharmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencieswill not be integral multiples of the difference but odd multiples of half the difference. For this pipewe have

f= 280 Hz 240 Hz = 40 Hz.

Because we have frequencies that are integral multiples of this, the pipe is open, with a fundamentalfrequency of 40 Hz.The wavelength of the fundamental frequency is 1 = 2L. We find the length from

v = 1f1 = 2Lf1 ;343 m/s = 2L(40 Hz), which gives L = 4.3 m.

39. (a) The wavelength of the fundamental frequency is 1 = 2L. We find the fundamental frequency fromv = 1f1 = 2Lf1;343 m/s = 2(2.44 m)f1, which givesf1 = 70.3 Hz.

We find the highest harmonic fromfn = nf1;

20,000 Hz = n(70.3 Hz), which gives n = 284.5.Because all harmonics are present in an open pipe and the fundamental frequency is within the audiblerange, 284 harmonics are present, which means 283 overtones.

(b) The wavelength of the fundamental frequency is 1 = 4L. We find the fundamental frequency fromv = 1f1 = 4Lf1;343 m/s = 4(2.44 m)f1, which givesf1 = 35.1 Hz.

We find the highest harmonic fromfn = nf1;20,000 Hz = n(35.1 Hz), which gives n = 569.

Because only the odd harmonics are present in a closed pipe and the fundamental frequency is withinthe audible range, 569 harmonics are present, which means 284 overtones.

40. If we consider the ear canal as a closed pipe, the wavelength of the fundamental frequency is 1 = 4L. We takethe most sensitive frequency as the fundamental frequency to find the length of the ear canal:

v = 1f1 = 4Lf1;343 m/s = 4L(3500 Hz), which gives L = 2.5 102 m = 2.5 cm.

41. Because the intensity is proportional to the square of both the amplitude and the frequency, we haveI2/I1 = (A2/A1)

2(f2/f1)2 = (0.4)2(2)2 = 0.64;

I3/I1 = (A3/A1)2(f3/f1)

2 = (0.15)2(3)2 = 0.20.We find the relative intensity levels from

2 1 = 10 log10(I2/I1) = 10 log10(0.64) = 2 dB;

3 1 = 10 log10(I3/I1) = 10 log10(0.20) = 7 dB.

42. The beat frequency is the difference in frequencies, so we havef=fbeat = 1/(2.0 s) = 0.50 Hz.

Note that the frequency of the second string could be higher or lower.


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43. The beat frequency is the difference in frequencies, so we havefbeat = f= 277 Hz 262 Hz = 15 Hz.

Because the human ear can detect beats up to about 8 Hz, this will not be audible.If each frequency is reduced by a factor of 4, the beat frequency will be reduced by the same factor:

fbeat = (fbeat = ((15 Hz) = 3.8 Hz.This will be audible.

44. The beat frequency is the difference in frequencies, so we havefbeat = f=f2 f1; 5.0 kHz =f2 23.5 kHz, which givesf2 = 18.5 kHz, 28.5 kHz.

Because the second whistle cannot be heard by humans, its frequency is 28.5 kHz.

45. Because the beat frequency increases for the fork with the higher frequency, the string frequency must bebelow 350 Hz. Thus we have

f=f1 fbeat1 =f2 fbeat2 = 350 Hz 4 Hz = 355 Hz 9 Hz = 346 Hz.

46. (a) The beat frequency will be the difference in frequencies. Because the second frequency could behigher or lower, we have

fbeat = f=f2 f1; (3 beats)/(2.0 s) =f2 132 Hz, which givesf2 = 130.5 Hz, 133.5 Hz.

(b) We assume that the change in tension does not change the mass density, so the velocityvariation depends only on the tension. Because the wavelength does not change, we have

= v1/f1 = v2/f2, or FT2/FT1 = (f2/f1)2.For the fractional change we have

(FT2 FT1)/FT1 = (FT2/FT1) 1 = (f2/f1)2 1 = [(f2 f1)/f1]2;FT/FT = [( 1.5 Hz)/(132 Hz)]2, which gives FT/FT = 0.023 = 2.3%.

47. The beat frequency will be the difference in frequencies. We assume that the change in tension does notchange the mass density, so the velocity variation depends only on the tension. Because the wavelengthdoes not change, we have

= v1/f1 = v2/f2, or (FT2/FT1)1/2 = (f2/f1)2.If we write the changes as

FT2 = FT1 + FT, and f2 =f1 + f, we get[(FT1 + FT)/FT1]1/2 = [(f1 + f)/f1], or [1 + (FT/FT1)]1/2 = 1 + (f/f1).

Because (FT/FT1) 1, we can expand the left hand side to get1 + !(FT/FT1) = 1 + (f/f1), or

!(FT/FT1) = f/f1;

!( 0.015) = f/(294 Hz), which gives f = 2.2 Hz.

48. For destructive interference, the path difference is an odd multiple of half the wavelength.(a) Because the path difference is fixed, the lowest frequency corresponds to the longest

wavelength. We find this from L = 1/2, so we havef1 = v/1 = (343 m/s)/2(3.5 m 3.0 m) = 343 Hz.

(b) We find the wavelength for the next frequency from L = 32/2, so we havef2 = v/2 = (343 m/s)/[2(3.5 m 3.0 m)/3] = 1030 Hz.

We find the wavelength for the next frequency from L = 53/2, so we have


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f3 = v/3 = (343 m/s)/[2(3.5 m 3.0 m)/5] = 1715 Hz.


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49. (a) We find the two frequencies fromf1 = v/1 = (343 m/s)/(2.64 m) = 129.9 Hz;f2 = v/2 = (343 m/s)/(2.76 m) = 124.3 Hz.

The beat frequency isfbeat = f=f2 f1 = 129.9 Hz 124.3 Hz = 5.6 Hz.

(b) The intensity maxima will travel with the speed of sound, so the separation of regions of maximumintensity is the wavelength of the beats:

beat = v/fbeat = (343 m/s)/(5.6 Hz) = 61 m.

50. We see that there is no path difference for a listener on the bisector of thetwo speakers. The maximum path difference occurs for a listener on theline of the speakers. If this path difference is less than /2, there will beno location where destructive interference can occur. Thus we have

path difference = dB dA = d ? /2.

51. Because the police car is at rest, the wavelength traveling toward you is1 = v/f0 = (343 m/s)/(1800 Hz) = 0.191 m.

(a) This wavelength approaches you at a relative speed ofv + vL. You hear a frequencyf1 = (v + vL)/1 = (343 m/s + 30.0 m/s)/(0.191 m) = 1950 Hz.

(b) The wavelength approaches you at a relative speed ofv vL. You hear a frequencyf2 = (v vL)/1 = (343 m/s 30.0 m/s)/(0.191 m) = 1640 Hz.

52. Because the bat is at rest, the wavelength traveling toward the object is1 = v/f0 = (343 m/s)/(50,000 Hz) = 6.86 10

3 m.The wavelength approaches the object at a relative speed ofv vobject. The sound strikes and reflects from theobject with a frequency

f1 = (v vobject)/1 = (343 m/s 25.0 m/s)/(6.86 103 m) = 46,360 Hz.

This frequency can be considered emitted by the object, which is moving away from the bat. Because thewavelength behind a moving source increases, the wavelength approaching the bat is

2 = (v + vobject)/f1 = (343 m/s + 25.0 m/s)/(46,360 Hz) = 7.94 103 m.

This wavelength approaches the bat at a relative speed ofv, so the frequency received by the bat isf2 = v/2 = (343 m/s)/(7.94 103 m) = 43,200 Hz.

53. Because the wavelength in front of a moving source decreases, the wavelength from the approaching tuba is1 = (v v1)/f0 = (343 m/s 10.0 m/s)/(75 Hz) = 4.44 m.

This wavelength approaches the stationary listener at a relative speed ofv, so the frequency heard by thelistener is

f1 = v/1 = (343 m/s)/(4.44 m) = 77 Hz.Because the frequency from the stationary tube is unchanged, the beat frequency is

A

B

Listener

d

no path

difference


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fbeat = f=f1 f0 = 77 Hz 75 Hz = 2 Hz.


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54. Because the wavelength in front of a moving source decreases, the wavelength from the approachingautomobile is

1 = (v v1)/f0 = (343 m/s 15 m/s)/f0.This wavelength approaches the stationary listener at a relative speed ofv, so the frequency heard by thelistener is

f1 = v/1 = (343 m/s)/[(343 m/s 15 m/s)/f0] = (343 m/s)f0/(328 m/s).

Because the frequency from the stationary automobile is unchanged, the beat frequency isfbeat = f=f1 f0 ;5.5 Hz = [(343 m/s)f0/(328 m/s)] f0, which givesf0 = 120 Hz.

55. We convert the speed of the train: (40 km/h)/(3.6 ks/h) = 11.1 m/s.Because the wavelength behind a moving source increases, the wavelength from the receding train is

1 = (v + v1)/f0 = (343 m/s + 11.1 m/s)/(277 Hz) = 1.28 m.This wavelength approaches the stationary observer at a relative speed ofv, so the frequency heard by theobserver is

f1 = v/1 = (343 m/s)/(1.28 m) = 268 Hz.Because the frequency from the stationary train is unchanged, the beat frequency is

fbeat = |f| = |f1 f0| = |268 Hz 277 Hz| = 9 Hz.

56. Because the wavelength in front of a moving source decreases, the wavelength from the approaching source is1 = (v v1)/f0.


f1 = v/1 = v/[(v v1)/f0]= vf0/(v v1) = (343 m/s)(2000 Hz)/(343 m/s 15 m/s) = 2091 Hz.The wavelength from a stationary source is

2 = v/f0.This wavelength approaches the moving receiver at a relative speed ofv + v1, so the frequency heard is

f2 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 15 m/s)(2000 Hz)/(343 m/s) = 2087 Hz.The two frequencies are not exactly the same, but close.For the other speeds we have,at 150 m/s:

f3 = v/1 = v/[(v v1)/f0]= vf0/(v v1) = (343 m/s)(2000 Hz)/(343 m/s 150 m/s) = 3554 Hz;

f4 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 150 m/s)(2000 Hz)/(343 m/s) = 2875 Hz.at 300 m/s:

f3 = v/1 = v/[(v v1)/f0]= vf0/(v v1) = (343 m/s)(2000 Hz)/(343 m/s 1300 m/s) = 15950 Hz;f4 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 300 m/s)(2000 Hz)/(343 m/s) = 3750 Hz.

The Doppler formulas are not symmetric; the speed of the source creates a greater shift than the speed of theobserver.We can write the expression for the frequency from an approaching source as

f1 = v/1 = vf0/(v v1) =f0/[1 (v1/v)].

Ifv1v, we use 1/[1 (v1/v)] 1 + (v1/v), so we havef1f0[1 + (v1/v)],

which is the expressionf2 = (v + v1)f0/v for the frequency for a stationary source and a moving listener.


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57. Because the source is at rest, the wavelength traveling toward the blood is1 = v/f0.

This wavelength approaches the blood at a relative speed ofv vblood. The ultrasound strikes and reflectsfrom the blood with a frequency

f1 = (v vblood)/1 = (v vblood)/(v/f0) = (v vblood)f0/v.This frequency can be considered emitted by the blood, which is moving away from the source. Because thewavelength behind the moving blood increases, the wavelength approaching the source is

2 = (v + vblood)/f1 = (v + vblood)/[(v vblood)f0/v] = [1 + (vblood/v)]v/f0[1 (vblood/v)].This wavelength approaches the source at a relative speed ofv, so the frequency received by the source is

f2 = v/2 = v/{[1 + (vblood/v)]v/f0[1 (vblood/v)]} = [1 (vblood/v)]f0/[1 + (vblood/v)].

Because vbloodv, we use 1/[1 + (vblood/v)] 1 (vblood/v), so we havef2f0[1 (vblood/v)]

2f0[1 2(vblood/v)].For the beat frequency we have

fbeat =f0 f2 = 2(vblood/v)f0 = 2[(0.020 m/s)/(1540 m/s)](500 103 Hz) = 13 Hz.

58. Because the source is at rest, the wavelength traveling toward the heart is

1 = v/f0.If we assume that the heart is moving away, this wavelength approaches the heart at a relative speed ofv vheart. The ultrasound strikes and reflects from the heart with a frequency

f1 = (v vheart)/1 = (v vheart)/(v/f0) = (v vheart)f0/v.This frequency can be considered emitted by the heart, which is moving away from the source. Because thewavelength behind the moving heart increases, the wavelength approaching the source is

2 = (v + vheart)/f1 = (v + vheart)/[(v vheart)f0/v] = [1 + (vheart/v)]v/f0[1 (vheart/v)].This wavelength approaches the source at a relative speed ofv, so the frequency received by the source is

f2 = v/2 = v/{[1 + (vheart/v)]v/f0[1 (vheart/v)]} = [1 (vheart/v)]f0/[1 + (vheart/v)].

Because vheartv, we use 1/[1 + (vheart/v)] 1 (vheart/v), so we havef2f0[1 (vheart/v)]

2f0[1 2(vheart/v)].

The maximum beat frequency occurs for the maximum heart velocity, so we havefbeat =f0 f2 = 2(vheart/v)f0 ;

600 Hz = 2[vheart/(1.54 103 m/s)](2.25 106 Hz), which gives vheart = 0.205 m/s.

59. In Problem 58, we assumed that the heart is moving away from the source. Because the heart velocity is muchless than the wave speed, the same beat frequency will occur when the heart is moving toward the source.Thus the maximum beat frequency will occur twice during each beat of the heart, so the heartbeat rate is

!(180 maxima/min) = 90 beats/min.

60. From the diagram we see that

sin = (vwavet)/(vboatt) = vwave/vboat;sin 20 = (2.0 m/s)/vboat, which gives vboat = 5.8 m/s.

61. (a) From the definition of the Mach number, we havev = (Mach number)vsound = (0.33)(343 m/s) = 1.1 102 m/s.

(b) We find the speed of sound fromv = (Mach number)vsound;(3000 km/h)/(3.6 ks/h) = (3.2)vsound, which gives vsound = 2.6 102 m/s.

vboatt

vwavet


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62. In a time t the shock wave moves a distance vsoundt perpendicular to thewavefront. In the same time the object moves vobjectt. We see from thediagram that

sin = (vsoundt)/(vobjectt) = vsound/vobject.

63. (a) We find the angle of the shock wave fromsin = vsound/vobject = vsound/(Mach number)vsound

= 1/(Mach number) = 1/2.3 = 0.435, so = 26.(b) From the diagram we see that, when the shock wave hits

the ground, we havetan = h/vairplanet = h/(Mach number)vsoundt;tan 26 = (7100 m)/(2.1)(310 m/s)t, which gives t = 23 s.

64. (a) From the definition of the Mach number, we havev = (Mach number)vsound;

(15,000 km/h)/(3.6 ks/h) = (Mach number)(35 m/s), which gives Mach number = 120.(b) We find the angle of the shock wave from

sin = vsound/vobject = (35 m/s)/[(15,000 km/h)/(3.6 ks/h)] = 0.0084, so = 0.48.Thus the apex angle is 2= 0.96.

65. (a) We find the angle of the shock wave in the air fromsin = vsound/vobject = (343 m/s)/(8000 m/s) = 0.0429, so = 2.46.

(b) We find the angle of the shock wave in the ocean fromsin = vsound/vobject = (1540 m/s)/(8000 m/s) = 0.193, so = 11.1.

66. (a) From the diagram we see that, when the shock wave hitsthe ground, we havetan = h/dplane = (1.5 km)/(2.0 km) = 0.75, so = 37.

(b) We find the speed of the plane fromsin = vsound/vobject = vsound/(Mach number)vsound = 1/(Mach number);sin 37 = 1/(Mach number), which gives Mach number = 1.7.

67. Because the frequency doubles for each octave, we havefhigh/flow = 2x;20,000 Hz)/(20 Hz) = 2x, orlog10(1000) = x log10(2). which gives x 10.

vobjectt

vsoundt

vairplanet

vsoundt

h

dplane = vplanet

vsoundt

h


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68. We choose a coordinate system with origin at the top of the cliff, down positive, and t = 0 when the stone isdropped. If we call t1 the time of fall for the stone, we have

y =y01+ v01t1 + !gt12;

h = 0 + 0 + !gt12, or t1 = (2h/g)

1/2.For the time t2 for the sound to reach the top of the cliff, we have

t2 = h/vsound.Thus for the total time we have

t = t1 + t2 = (2h/g)1/2 + h/vsound;

3.5 s = [2h/(9.80 m/s2)]1/2 + [h/(343 m/s)].This is a quadratic equation for h1/2, which has the positive result h1/2 = 7.39 m1/2,so the height of the cliff is 55 m.

69. The intensity for a 0 dB sound is I0 = 1012 W/m2. For 1000 mosquitoes, we find the intensity level from

= 10 log10(I1/I0) = 10 log10(1000I0/I0) = 30 dB.

70. Because the wavelength in front of a moving source decreases, the wavelength from the approaching car is1 = (v vcar)/f0.


f1 = v/1 = v/[(v vcar)/f0]= vf0/(v vcar).Because the wavelength behind a moving source increases, the wavelength from the receding car is

2 = (v + vcar)/f0.This wavelength approaches the stationary listener at a relative speed ofv, so the frequency heardby the listener is

f2 = v/2 = v/[(v + vcar)/f0]= vf0/(v + vcar).If the frequency drops by one octave, we have

f1/f2 = [vf0/(v vcar)]/[vf0/(v + vcar)] = (v + vcar)/(v vcar) = 2, or2(v vcar) = v + vcar, which gives vcar =@v =@(343 m/s) = 114 m/s = 410 km/h (257 mi/h).

71. The wavelength of the fundamental frequency for a string is = 2L. Because the speed of a wave on the stringdoes not change, we have

v = 1f1 = 1f1;2L[(600 Hz)/3] = 2(0.60L)f1, which givesf1 = 333 Hz.

72. The tension and mass density of the string determines the velocity:v = (FT/)1/2.

Because the strings have the same length, the wavelengths are the same, so for the ratio of frequencies wehavefn+1/fn = vn+1/vn = (n/n+1)1/2 = 1.5, or n+1/n = (1/1.5)2 = 0.444.

With respect to the lowest string, we haven+1/1 = (0.444)n.

If we call the mass density of the lowest string 1, we have1, 0.444, 0.198, 0.0878, 0.0389.


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73. (a) The wavelength of the fundamental frequency for a string is = 2L, so the speed of a wave on thestring is

v = f= 2(0.32 m)(440 Hz) = 2.8 102 m/s.We find the tension from

v = (FT/)1/2;282 m/s = [FT/(5.5 104 kg/m)]1/2 , which gives FT = 44 N.

(b) For a closed pipe the wavelength of the fundamental frequency is = 4L. We find the requiredlength from

v = f = 4Lf;343 m/s = 4L(440 Hz), which gives L = 0.195 m = 19.5 cm.

(c) All harmonics are present in the string, so the first overtone is the second harmonic:f2 = 2f1 = 2(440 Hz) = 880 Hz.

Only the odd harmonics are present in a closed pipe, so the first overtone is the third harmonic:f3 = 3f1 = 3(440 Hz) = 1320 Hz.

74. We find the ratio of intensities from= 10 log10(I2/I1);

10 dB = 10 log10(I2/I1), which gives I2/I1 = 0.10.If we assume uniform spreading of the sounds, the intensity is proportional to the power output, so we have

P2/P1 = I2/I1;P2/(100 W) = 0.10, which gives P2 = 10 W.

75. At each resonant position the top of the water column is a node.Thus the distance between the two readings corresponds to thedistance between adjacent nodes, or half a wavelength:

L = /2, or = 2 L.For the frequency we have

f= v/= (343 m/s)/2(0.395 m 0.125 m) = 635 Hz.

76. We find the gain from= 10 log10(P2/P1) = 10 log10[(100 W)/(1 10

3 W)] = 50 dB.

L1

L2


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77. (a) Because both sources are moving toward the observer at the same speed, they will have thesame Doppler shift, so the beat frequency will be 0.

(b) Because the wavelength in front of a moving source decreases, the wavelength from theapproaching loudspeaker is

1 = (v vcar)/f0 = (343 m/s 10.0 m/s)/(200 Hz) = 1.665 m.This wavelength approaches the stationary listener at a relative speed ofv, so the frequency heard by

the listener isf1 = v/1 = (343 m/s)/(1.665 m) = 206 Hz.

Because the wavelength behind a moving source decreases, the wavelength from the recedingloudspeaker is

2 = (v + vcar)/f0 = (343 m/s + 10.0 m/s)/(200 Hz) = 1.765 m.This wavelength approaches the stationary listener at a relative speed ofv, so the frequency heard bythe listener is

f2 = v/2 = (343 m/s)/(1.765 m) = 194 Hz.The beat frequency is

fbeat = f= 206 Hz 194 Hz = 12 Hz.Note that this frequency may be too high to be audible.

(c) Because both sources are moving away from the observer at the same speed, they will have thesame Doppler shift, so the beat frequency will be 0.

78. Because the wavelength in front of a moving source decreases, the wavelength from the approaching trainwhistle is

1 = (v vtrain)/f0.Because you are stationary, this wavelength approaches you at a relative speed ofv, so the frequency heard byyou is

f1 = v/1 = v/[(v vtrain)/f0] = vf0/(v vtrain).Because the wavelength behind a moving source increases, the wavelength from the receding trainwhistle is

2 = (v + vtrain)/f0.

This wavelength approaches you at a relative speed ofv, so the frequency heard by you isf2 = v/2 = v/[(v + vtrain)/f0] = vf0/(v + vtrain).

For the ratio of frequencies, we getf2/f1 = (v vtrain)/(v + vtrain);(486 Hz)/(522 Hz) = (343 m/s vtrain)/(343 m/s + vtrain), which gives vtrain = 12.3 m/s.

79. For a closed pipe the wavelength of the fundamental frequency is 1pipe = 4L. Thus the frequency of the thirdharmonic is

f3pipe = 3f1pipe = 3v/4L = 3(343 m/s)/4(0.75 m) = 343 Hz.This is the fundamental frequency for the string. The wavelength of the fundamental frequency for the string

is 1string = 2L, so we havevstring =f1string/2L = [FT/(m/L)]1/2;(343 Hz)/2(0.75 m) = {FT/[(0.00180 kg)/(0.75 m)]}1/2, which gives FT = 6.4 102 N.


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82. Because the wind velocity is a movement of the medium, it adds or subtracts from the speed of sound in themedium.(a) Because the wind is blowing away from the observer, the effective speed of sound is v vwind.

Therefor the wavelength traveling toward the observer isa = (v vwind)/f0.

This wavelength approaches the observer at a relative speed ofv vwind. The observer will hear a

frequencyfa = (v vwind)/a = (v vwind)/[(v vwind)/f0] =f0 = 645 Hz.

(b) Because the wind is blowing toward the observer, the effective speed of sound is v + vwind.From the analysis in part (a), we see that there will be no change in the frequency: 645 Hz.

(c) Because the wind is blowing perpendicular to the line toward the observer, the effective speed of soundis v. Because there is no relative motion of the whistle and the observer, there will be no change inthe frequency: 645 Hz.

(d) Because the wind is blowing perpendicular to the line toward the observer, the effective speed of soundis v. Because there is no relative motion of the whistle and the observer, there will be no change

in the frequency: 645 Hz.(e) Because the wind is blowing toward the cyclist, the effective speed of sound is v + vwind.

Therefore the wavelength traveling toward the cyclist is

e = (v + vwind)/f0.This wavelength approaches the cyclist at a relative speed ofv + vwind + vcycle. The cyclist will heara frequency

fe = (v + vwind + vcycle)/e = (v + vwind + vcycle)/[(v + vwind)/f0]= (v + vwind + vcycle)f0/(v + vwind)= (343 m/s + 9.0 m/s + 13.0 m/s)(645 Hz)/(343 m/s + 9.0 m/s) = 669 Hz.

(f) Because the wind is blowing perpendicular to the line toward the cyclist, the effective speed of soundis v. Therefore the wavelength traveling toward the cyclist is

f= v/f0.This wavelength approaches the cyclist at a relative speed ofv + vcycle. The cyclist will heara frequency

ff = (v + vcycle)/

f= (v + vcycle)/(v/f0)= (v + vcycle)f0/v = (343 m/s + 13.0 m/s)(645 Hz)/(343 m/s) = 669 Hz.

83. When the person is equidistant from the sources, there is constructive interference. To move to a point wherethere is destructive interference, the path difference from the sources must be L = n/2. If we assume that n= 1, we find the frequency from

f= v/= v/2 L = (343 m/s)/2(0.22 m) = 780 Hz.Because larger values ofn would give a frequency out of the given range, this is the result.

84. (a) We find the wavelength from= v/f = (1560 m/s)/(200,000 Hz) = 7.8 103 m.

(b) We find the transit time fromt = 2d/v = 2(100 m)/(1560 m/s) = 0.13 s.

85. IfTis the period of the pulses and t is the duration of a pulse, and the moth must recieve the entire reflectedpulse, the maximum time for the signal to travel to the moth and return is tmax = T t.We find the distance from

2dmax = vtmax = (343 m/s)(70.0 ms 3.0 ms)(103 s/ms), which gives dmax = 11.5 m.


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86. Because the wavelength in front of a moving source decreases, the wavelength approaching the moth is1 = (v vbat)/f0.

The wavelength approaches the moth at a relative speed ofv + vmoth. The sound strikes and reflects from theobject with a frequency

f1 = (v + vmoth)/1 = (v + vmoth)/[(v vbat)/f0] = (v + vmoth)f0/(v vbat).This frequency can be considered emitted by the moth, which is moving toward the bat. Because thewavelength in front of a moving source decreases, the wavelength approaching the bat is

2 = (v vmoth)/f1 = (v vmoth)/[(v + vmoth)f0/(v vbat)] = (v vmoth)(v vbat)/(v + vmoth)f0/(v vbat)].This wavelength approaches the bat at a relative speed ofv + vbat, so the frequency received by the bat is

f2 = (v + vbat)/2 = (v + vbat)/[(v vmoth)/f1] = (v + vbat)(v + vmoth)f0/(v vmoth)(v vbat)= (343.0 m/s + 8.0 m/s)(343.0 m/s + 5.0 m/s)(51.53 kHz)/(343.0 m/s 5.0 m/s)(343.0 m/s 8.0 m/s)= 55.39 kHz.

Ch 12 Physics Answers

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