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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 have L = !vt = !(343 m/s)(1.5 s) = 2.6 × 10 2 m.

2. Because the sound travels down and back, we have L = !vt = !(1560 m/s)(2.0 s) = 1.6 × 10 3 m = 1.6 km.

3. (a) We find the extreme wavelengths from λ 1 = v/f 1 = (343 m/s)/(20 Hz) = 17 m; λ 2 = v/f 2 = (343 m/s)/(20,000 Hz) = 1.7 × 10 –2 m = 1.7 cm.

The range of wavelengths is 1.7 cm ² λ ² 17 m. (b) We find the wavelength from

λ = v/f = (343 m/s)/(10 × 10 6 Hz) = 3.4 × 10 –5 m.

4. For the travel times we have t wire = d/v wire = (8.4 m)/(5100 m/s) = 0.0016 s; t air = (d/v air ) = (8.4 m)/(343 m/s) = 0.024 s (≈ 15× t wire ).

5. The speed in the concrete is determined by the elastic modulus: v concrete = (E/ρ) 1/2 = [(20 × 10 9 N/m 2 )/(2.3 × 10 3 kg/m 3 )] 1/2 = 2.95 × 10 3 m/s.

For the time interval we have Æt = (d/v air ) – (d/v concrete ); 1.4 s = d[1/(343 m/s)] – [1/(2.95 × 10 3 m/s)], which gives d = 5.4 × 10 2 m.

6. (a) For the time interval in sea water we have Æt = d/v water = (1.0 × 10 3 m)/(1560 m/s) = 0.64 s.

(b) For the time interval in the air we have Æt = d/v air = (1.0 × 10 3 m)/(343 m/s) = 2.9 s.

7. If we let L 1 represent the thickness of the top layer, the total transit time is t = (L 1 /v 1 ) + [(L – L 1 )/v 2 )]; 4.5 s = [L 1 /(331 m/s)] + [(1500 m – L 1 )/(343 m/s)], which gives L 1 = 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 is Æd = t Æv; so the percentage change is (Æd/d)100 = (Æv/v)100 = (343 – 331 m/s)(100)/(343 m/s) = 3.5%.


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9. We find the intensity of the sound from β 1 = 10 log 10 (I 1 /I 0 ); 120 dB = 10 log(I 1 /10 –12 W/m 2 ), which gives I 1 = 1.0 W/m 2 .

For a whisper we have β 2 = 10 log 10 (I 2 /I 0 ); 20 dB = 10 log(I 1 /10 –12 W/m 2 ), which gives I 2 = 1.0 × 10 –10 W/m 2 .

10. We find the intensity level from β = 10 log 10 (I 1 /I 0 ) = 10 log 10 [(2.0 × 10 –6 W/m 2 )/(10 –12 W/m 2 )] = 63 dB.

11. We find the ratio of intensities of the sounds from β = 10 log 10 (I 2 /I 1 ); 2.0 dB = 10 log 10 (I 2 /I 1 ), which gives I 2 /I 1 = 1.58.

Because the intensity is proportional to the square of the amplitude, we have I 2 /I 1 = (A 2 /A 1 ) 2 ; 1.58 = (A 2 /A 1 ) 2 , which gives A 2 /A 1 = 1.3.

12. When three of the four engines are shut down, the intensity is reduced by a factor of 4, so we have β 2 – β 1 = 10 log 10 (I 2 /I 0 ) – 10 log 10 (I 1 /I 0 ) = 10 log 10 (I 2 /I 1 ); β 2 – 120 dB = 10 log((I 1 /I 1 ), which gives β 2 = 114 dB.

13. We find the ratio of intensities from β = 10 log 10 (I signal /I noise ); 58 dB = 10 log 10 (I signal /I noise ), which gives I signal /I noise = 6.3 × 10 5 .

14. (a) Using the intensity from Table 12–2, we have P = IA = I2¹r 2 = (3 × 10 –6 W/m 2 )2¹(0.50 m) 2 = 5 × 10 –6 W.

(b) We find the number of people required from N = P total /P = (100 W)/(5 × 10 –6 W) = 2 × 10 7 .

15. (a) We find the intensity of the sound from β = 10 log 10 (I/I 0 ); 50 dB = 10 log(I/10 –12 W/m 2 ), which gives I = 1.0 × 10 –7 W/m 2 .

The rate at which energy is absorbed is the power of the sound wave: P = IA = (1.0 × 10 –7 W/m 2 )(5.0 × 10 –5 m 2 ) = 5.0 × 10 –12 W.

(b) We find the time from t = E/P = (1.0 J)/(5.0 × 10 –12 W) = 2.0 × 10 11 s = 6.3 × 10 3 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, and the sound spreads out uniformly, we have

I = P/4¹r 2 , so we get I 1 = (250 W)/4¹(3.0 m) 2 = 2.21 W/m 2 ; I 2 = (40 W)/4¹(3.0 m) 2 = 0.354 W/m 2 .

For the intensity levels we have β 1 = 10 log 10 (I 1 /I 0 ) = 10 log 10 [(2.21 W/m 2 )/(10 –12 W/m 2 )] = 123 dB; β 2 = 10 log 10 (I 2 /I 0 ) = 10 log 10 [(0.354 W/m 2 )/(10 –12 W/m 2 )] = 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 β = 10 log 10 (I/I 0 ); 130 dB = 10 log(I/10 –12 W/m 2 ), which gives I = 10 W/m 2 .

The power output of the speaker is P = IA = I4¹r 2 = (10 W/m 2 )4¹(2.5 m) 2 = 7.9 × 10 2 W.

(b) We find the intensity of the sound from β = 10 log 10 (I/I 0 ); 90 dB = 10 log(I/10 –12 W/m 2 ), which gives I = 1.0 × 10 –3 W/m 2 .

We find the distance from P = IA ; 7.9 × 10 2 W = (1.0 × 10 –3 W/m 2 )4¹r 2 , which gives r = 2.5 × 10 2 m.

18. (a) The intensity of the sound wave is given by I = 2¹ 2 ρf 2 x 0

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

I 2 /I 1 = (x 02 /x 01 ) 2 = 3 2 = 9. (b) We find the change in intensity level from

Æβ = 10 log 10 (I 2 /I 1 ) = 10 log 10 (9) = 9.5 dB.

19. (a) We find the intensity of the sound from P = IA = I2¹r 2 ; 5.0 × 10 5 W = I4¹(30 m) 2 , which gives I = 44.2 W/m 2 .

We find the intensity level from β = 10 log 10 (I/I 0 ) = 10 log 10 [(44.2 W/m 2 )/(10 –12 W/m 2 )] = 136 dB.

(b) We find the intensity of the sound without air absorption from P = I 1 A 1 = I 1 2¹r 1

2 ; 5.0 × 10 5 W = I 1 4¹(1000 m) 2 , which gives I 1 = 3.98 × 10 –2 W/m 2 .

We find the intensity level from β 10 = 10 log 10 (I 1 /I 0 ) = 10 log 10 [(3.98 × 10 –2 W/m 2 )/(10 –12 W/m 2 )] = 106 dB.

When we consider air absorption, we have β 1 = β 10 – (7.0 dB/km)r 1 = 106 dB – (7.0 dB/km)(1.0 km) = 99 dB.

(c) We find the intensity of the sound without air absorption from P = I 2 A 2 = I 2 2¹r 2

2 ; 5.0 × 10 5 W = I 2 4¹(5000 m) 2 , which gives I 2 = 1.59 × 10 –3 W/m 2 .

We find the intensity level from β 20 = 10 log 10 (I 2 /I 0 ) = 10 log 10 [(1.59 × 10 –3 W/m 2 )/(10 –12 W/m 2 )] = 92 dB.

When we consider air absorption, we have


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β 2 = β 20 – (7.0 dB/km)r 2 = 92 dB – (7.0 dB/km)(5.0 km) = 57 dB.


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20. The intensity of the sound wave is given by I = 2¹ 2 ρf 2 x 0

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

I 2 /I 1 = (f 2 /f 1 ) 2 = 2 2 = 4.

21. The intensity of the sound wave is given by I = 2¹ 2 ρf 2 x 0

2 v = 2¹ 2 (1.29 kg/m 3 )(260 Hz) 2 (1.3 × 10 –3 m) 2 (343 m/s) = 1.0 × 10 3 W/m 2 .

We find the intensity level from β = 10 log 10 (I/I 0 ) = 10 log 10 [(1.0 × 10 3 W/m 2 )/(10 –12 W/m 2 )] = 150 dB.

22. We find the intensity of the sound from β = 10 log 10 (I/I 0 ); 120 dB = 10 log(I/10 –12 W/m 2 ), which gives I = 1.00 W/m 2 .

We find the maximum displacement from I = 2¹ 2 ρf 2 x 0

2 v 1.00 W/m 2 = 2¹ 2 (1.29 kg/m 3 )(131 Hz) 2 x 0

2 (343 m/s), which gives x 0 = 8.17 × 10 –5 m.

23. From Figure 12–7, we see that a 100Hz 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 12–7, we see that an intensity level of 30 dB intersects the loudness level of 0 phons at 150 Hz. At the high frequency end, the highest frequency is reached, so the range is 150 Hz to 20,000 Hz.

25. (a) From Figure 12–7, we see that at 100 Hz the lowest threshold is 5 × 10 –9 W/m 2 and the threshold of pain is 1 W/m 2 . For the ratio we have

(1 W/m 2 )/(5 × 10 –9 W/m 2 ) = 2 × 10 8 . (b) From Figure 12–7, we see that at 5000 Hz the lowest threshold is 5 × 10 –13 W/m 2 and the threshold

of pain is 1 × 10 –1 W/m 2 . For the ratio we have (1 × 10 –1 W/m 2 )/(5 × 10 –13 W/m 2 ) = 2 × 10 11 .

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

We find the tension from v = [F T /(m/L)] 1/2 ; 282 m/s = F T /[(0.35 × 10 –3 kg)/(0.32 m)] 1/2 , which gives F T = 87 N.

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

v = λ 1 f 1 = λ 2 f 2 ; 2(0.70 m)(330 Hz) = 2L 2 (440 Hz), which gives L 2 = 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 from v = (331 + 0.60T) m/s = [331 + (0.60)(21°C)] m/s = 344 m/s.

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

v = λf ; 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 at the top. The wavelength of the fundamental frequency is λ = 4L. We find the frequency from

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

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

30. For an open pipe the wavelength of the fundamental frequency is λ = 2L. We find the required lengths from v = λf = 2Lf; 343 m/s = 2L lowest (20 Hz), which gives L lowest = 8.6 m; 343 m/s = 2L highest (20,000 Hz), which gives L highest = 8.6 × 10 –3 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 λ 1 = 4L. We find the fundamental frequency from

v = λ 1 f 1 ; 343 m/s = 4(1.12 m)f 1 , which gives f 1 = 76.6 Hz.

Only the odd harmonics are present, so we have f 3 = 3f 1 = (3)(76.6 Hz) = 230 Hz; f 5 = 5f 1 = (5)(76.6 Hz) = 383 Hz; f 7 = 7f 1 = (7)(76.6 Hz) = 536 Hz.

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

v = λ 1 f 1 ; 343 m/s = 2(1.12 m)f 1 , which gives f 1 = 153 Hz.

All harmonics are present, so we have f 2 = 2f 1 = (2)(153 Hz) = 306 Hz; f 3 = 3f 1 = (3)(153 Hz) = 459 Hz; f 4 = 4f 1 = (4)(153 Hz) = 612 Hz.

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

v = λ′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 λ 1 = 2L. Adding the support creates a node at the middle, so the wavelength is λ 2 = L. The wave speed in the bridge has not changed. We find the new frequency from

v = λ 1 f 1 = λ 2 f 2 ; 2L(4.0 Hz) = Lf 2 , which gives f 2 = 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 the change in the speed of sound. We find the speed of sound at 5°C:

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

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

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

The wavelength of the fundamental frequency is λ 1 = 4L. We find the length from v = λ 1 f 1 = 4Lf 1 ; 340 m/s = 4L(294 Hz), which gives L = 0.289 m.

(b) For helium we have v = λ 1 f 1 = 4Lf 1 ; 1005 m/s = 4(0.289 m)f 1 , which gives f 1 = 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 fundamental frequency, and all frequencies will be integral multiples of the difference. For a closed pipe only odd harmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencies will not be integral multiples of the difference but odd multiples of half the difference. For this pipe we have

Æf = 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 is f 1 = !Æf = !(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 all frequencies will be integral multiples of the difference. Thus we have

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

v = λ 1 f 1 = 2Lf 1 = 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 fundamental frequency, and all frequencies will be integral multiples of the difference. For a closed pipe only odd harmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencies will not be integral multiples of the difference but odd multiples of half the difference. For this pipe we have

Æf = 280 Hz – 240 Hz = 40 Hz. Because we have frequencies that are integral multiples of this, the pipe is open, with a fundamental frequency of 40 Hz. The wavelength of the fundamental frequency is λ 1 = 2L. We find the length from

v = λ 1 f 1 = 2Lf 1 ; 343 m/s = 2L(40 Hz), which gives L = 4.3 m.

39. (a) The wavelength of the fundamental frequency is λ 1 = 2L. We find the fundamental frequency from v = λ 1 f 1 = 2Lf 1 ; 343 m/s = 2(2.44 m)f 1 , which gives f 1 = 70.3 Hz.

We find the highest harmonic from f n = nf 1 ; 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 audible range, 284 harmonics are present, which means 283 overtones.

(b) The wavelength of the fundamental frequency is λ 1 = 4L. We find the fundamental frequency from v = λ 1 f 1 = 4Lf 1 ; 343 m/s = 4(2.44 m)f 1 , which gives f 1 = 35.1 Hz.

We find the highest harmonic from f n = nf 1 ; 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 within the 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 λ 1 = 4L. We take the most sensitive frequency as the fundamental frequency to find the length of the ear canal:

v = λ 1 f 1 = 4Lf 1 ; 343 m/s = 4L(3500 Hz), which gives L = 2.5 × 10 –2 m = 2.5 cm.

41. Because the intensity is proportional to the square of both the amplitude and the frequency, we have I 2 /I 1 = (A 2 /A 1 ) 2 (f 2 /f 1 ) 2 = (0.4) 2 (2) 2 = 0.64; I 3 /I 1 = (A 3 /A 1 ) 2 (f 3 /f 1 ) 2 = (0.15) 2 (3) 2 = 0.20.

We find the relative intensity levels from β 2 – β 1 = 10 log 10 (I 2 /I 1 ) = 10 log 10 (0.64) = – 2 dB; β 3 – β 1 = 10 log 10 (I 3 /I 1 ) = 10 log 10 (0.20) = – 7 dB.

42. The beat frequency is the difference in frequencies, so we have Æf = f beat = 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 have f beat = Æf = 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:

f beat ′ = (f beat = ((15 Hz) = 3.8 Hz. This will be audible.

44. The beat frequency is the difference in frequencies, so we have f beat = Æf = f 2 – f 1 ; ± 5.0 kHz = f 2 – 23.5 kHz, which gives f 2 = 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 be below 350 Hz. Thus we have

f = f 1 – f beat1 = f 2 – f beat2 = 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 be higher or lower, we have

f beat = Æf = f 2 – f 1 ; ± (3 beats)/(2.0 s) = f 2 – 132 Hz, which gives f 2 = 130.5 Hz, 133.5 Hz.

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

λ = v 1 /f 1 = v 2 /f 2 , or F T2 /F T1 = (f 2 /f 1 ) 2 . For the fractional change we have

(F T2 – F T1 )/F T1 = (F T2 /F T1 ) – 1 = (f 2 /f 1 ) 2 – 1 = [(f 2 – f 1 )/f 1 ] 2 ; ÆF T /F T = [(± 1.5 Hz)/(132 Hz)] 2 , which gives ÆF T /F T = ± 0.023 = ± 2.3%.

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

λ = v 1 /f 1 = v 2 /f 2 , or (F T2 /F T1 ) 1/2 = (f 2 /f 1 ) 2 . If we write the changes as

F T2 = F T1 + ÆF T , and f 2 = f 1 + Æf , we get [(F T1 + ÆF T )/F T1 ] 1/2 = [(f 1 + Æf )/f 1 ], or [1 + (ÆF T /F T1 )] 1/2 = 1 + (Æf/f 1 ).

Because (ÆF T /F T1 ) « 1, we can expand the left hand side to get 1 + !(ÆF T /F T1 ) = 1 + (Æf/f 1 ), or !(ÆF T /F T1 ) = Æf/f 1 ; !(– 0.015) = Æf/(294 Hz), which gives Æf = 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 = λ 1 /2, so we have f 1 = v/λ 1 = (343 m/s)/2(3.5 m – 3.0 m) = 343 Hz.

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

We find the wavelength for the next frequency from ∆L = 5λ 3 /2, so we have


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f 3 = v/λ 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 from f 1 = v/λ 1 = (343 m/s)/(2.64 m) = 129.9 Hz; f 2 = v/λ 2 = (343 m/s)/(2.76 m) = 124.3 Hz.

The beat frequency is f beat = Æf = f 2 – f 1 = 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 maximum intensity is the “wavelength” of the beats:

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

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

path difference = d B – d A = d ³ λ/2.

51. Because the police car is at rest, the wavelength traveling toward you is λ 1 = v/f 0 = (343 m/s)/(1800 Hz) = 0.191 m.

(a) This wavelength approaches you at a relative speed of v + v L . You hear a frequency f 1 = (v + v L )/λ 1 = (343 m/s + 30.0 m/s)/(0.191 m) = 1950 Hz.

(b) The wavelength approaches you at a relative speed of v – v L . You hear a frequency f 2 = (v – v L )/λ 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 is λ 1 = v/f 0 = (343 m/s)/(50,000 Hz) = 6.86 × 10 –3 m.

The wavelength approaches the object at a relative speed of v – v object . The sound strikes and reflects from the object with a frequency

f 1 = (v – v object )/λ 1 = (343 m/s – 25.0 m/s)/(6.86 × 10 –3 m) = 46,360 Hz. This frequency can be considered emitted by the object, which is moving away from the bat. Because the wavelength behind a moving source increases, the wavelength approaching the bat is

λ 2 = (v + v object )/f 1 = (343 m/s + 25.0 m/s)/(46,360 Hz) = 7.94 × 10 –3 m. This wavelength approaches the bat at a relative speed of v, so the frequency received by the bat is

f 2 = v/λ 2 = (343 m/s)/(7.94 × 10 –3 m) = 43,200 Hz.

53. Because the wavelength in front of a moving source decreases, the wavelength from the approaching tuba is λ 1 = (v – v 1 )/f 0 = (343 m/s – 10.0 m/s)/(75 Hz) = 4.44 m.

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

f 1 = v/λ 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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f beat = Æf = f 1 – f 0 = 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 approaching automobile is

λ 1 = (v – v 1 )/f 0 = (343 m/s – 15 m/s)/f 0 . This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is

f 1 = v/λ 1 = (343 m/s)/[(343 m/s – 15 m/s)/f 0 ] = (343 m/s)f 0 /(328 m/s). Because the frequency from the stationary automobile is unchanged, the beat frequency is

f beat = Æf = f 1 – f 0 ; 5.5 Hz = [(343 m/s)f 0 /(328 m/s)] – f 0 , which gives f 0 = 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

λ 1 = (v + v 1 )/f 0 = (343 m/s + 11.1 m/s)/(277 Hz) = 1.28 m. This wavelength approaches the stationary observer at a relative speed of v, so the frequency heard by the observer is

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

f beat = |Æf| = |f 1 – f 0 | = |268 Hz – 277 Hz| = 9 Hz.

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

λ 1 = (v – v 1 )/f 0 . This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is

f 1 = v/λ 1 = v/[(v – v 1 )/f 0 ]= vf 0 /(v – v 1 ) = (343 m/s)(2000 Hz)/(343 m/s – 15 m/s) = 2091 Hz. The wavelength from a stationary source is

λ 2 = v/f 0 . This wavelength approaches the moving receiver at a relative speed of v + v 1 , so the frequency heard is

f 2 = (v + v 1 )/λ 2 = (v + v 1 )/(v/f 0 ) = (v + v 1 )f 0 /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:

f 3 = v/λ 1 = v/[(v – v 1 )/f 0 ]= vf 0 /(v – v 1 ) = (343 m/s)(2000 Hz)/(343 m/s – 150 m/s) = 3554 Hz; f 4 = (v + v 1 )/λ 2 = (v + v 1 )/(v/f 0 ) = (v + v 1 )f 0 /v = (343 m/s + 150 m/s)(2000 Hz)/(343 m/s) = 2875 Hz.

at 300 m/s: f 3 = v/λ 1 = v/[(v – v 1 )/f 0 ]= vf 0 /(v – v 1 ) = (343 m/s)(2000 Hz)/(343 m/s – 1300 m/s) = 15950 Hz; f 4 = (v + v 1 )/λ 2 = (v + v 1 )/(v/f 0 ) = (v + v 1 )f 0 /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 the observer. We can write the expression for the frequency from an approaching source as

f 1 = v/λ 1 = vf 0 /(v – v 1 ) = f 0 /[1 – (v 1 /v)]. If v 1 « v, we use 1/[1 – (v 1 /v)] ≈ 1 + (v 1 /v), so we have

f 1 ≈ f 0 [1 + (v 1 /v)], which is the expression f 2 = (v + v 1 )f 0 /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 is λ 1 = v/f 0 .

This wavelength approaches the blood at a relative speed of v – v blood . The ultrasound strikes and reflects from the blood with a frequency

f 1 = (v – v blood )/λ 1 = (v – v blood )/(v/f 0 ) = (v – v blood )f 0 /v. This frequency can be considered emitted by the blood, which is moving away from the source. Because the wavelength behind the moving blood increases, the wavelength approaching the source is

λ 2 = (v + v blood )/f 1 = (v + v blood )/[(v – v blood )f 0 /v] = [1 + (v blood /v)]v/f 0 [1 – (v blood /v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is

f 2 = v/λ 2 = v/[1 + (v blood /v)]v/f 0 [1 – (v blood /v)] = [1 – (v blood /v)]f 0 /[1 + (v blood /v)]. Because v blood « v, we use 1/[1 + (v blood /v)] ≈ 1 – (v blood /v), so we have

f 2 ≈ f 0 [1 – (v blood /v)] 2 ≈ f 0 [1 – 2(v blood /v)]. For the beat frequency we have

f beat = f 0 – f 2 = 2(v blood /v)f 0 = 2[(0.020 m/s)/(1540 m/s)](500 × 10 3 Hz) = 13 Hz.

58. Because the source is at rest, the wavelength traveling toward the heart is λ 1 = v/f 0 .

If we assume that the heart is moving away, this wavelength approaches the heart at a relative speed of v – v heart . The ultrasound strikes and reflects from the heart with a frequency

f 1 = (v – v heart )/λ 1 = (v – v heart )/(v/f 0 ) = (v – v heart )f 0 /v. This frequency can be considered emitted by the heart, which is moving away from the source. Because the wavelength behind the moving heart increases, the wavelength approaching the source is

λ 2 = (v + v heart )/f 1 = (v + v heart )/[(v – v heart )f 0 /v] = [1 + (v heart /v)]v/f 0 [1 – (v heart /v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is

f 2 = v/λ 2 = v/[1 + (v heart /v)]v/f 0 [1 – (v heart /v)] = [1 – (v heart /v)]f 0 /[1 + (v heart /v)]. Because v heart « v, we use 1/[1 + (v heart /v)] ≈ 1 – (v heart /v), so we have

f 2 ≈ f 0 [1 – (v heart /v)] 2 ≈ f 0 [1 – 2(v heart /v)]. The maximum beat frequency occurs for the maximum heart velocity, so we have

f beat = f 0 – f 2 = 2(v heart /v)f 0 ; 600 Hz = 2[v heart /(1.54 × 10 3 m/s)](2.25 × 10 6 Hz), which gives v heart = 0.205 m/s.

59. In Problem 58, we assumed that the heart is moving away from the source. Because the heart velocity is much less 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 θ = (v wave t)/(v boat t) = v wave /v boat ; sin 20° = (2.0 m/s)/v boat , which gives v boat = 5.8 m/s.

61. (a) From the definition of the Mach number, we have v = (Mach number)v sound = (0.33)(343 m/s) = 1.1 × 10 2 m/s.

(b) We find the speed of sound from v = (Mach number)v sound ; (3000 km/h)/(3.6 ks/h) = (3.2)v sound , which gives v sound = 2.6 × 10 2 m/s.

v boat t θ

v wave t


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62. In a time t the shock wave moves a distance v sound t perpendicular to the wavefront. In the same time the object moves v object t. We see from the diagram that

sin θ = (v sound t)/(v object t) = v sound /v object .

63. (a) We find the angle of the shock wave from sin θ = v sound /v object = v sound /(Mach number)v sound

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

the ground, we have tan θ = h/v airplane t = h/(Mach number)v sound t; 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 have v = (Mach number)v sound ; (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 θ = v sound /v object = (35 m/s)/[(15,000 km/h)/(3.6 ks/h)] = 0.0084, so θ = 0.48°.

Thus the apex angle is 2θ = 0.96°.

65. (a) We find the angle of the shock wave in the air from sin θ = v sound /v object = (343 m/s)/(8000 m/s) = 0.0429, so θ = 2.46°.

(b) We find the angle of the shock wave in the ocean from sin θ = v sound /v object = (1540 m/s)/(8000 m/s) = 0.193, so θ = 11.1°.

66. (a) From the diagram we see that, when the shock wave hits the ground, we have

tan θ = h/d plane = (1.5 km)/(2.0 km) = 0.75, so θ = 37°. (b) We find the speed of the plane from

sin θ = v sound /v object = v sound /(Mach number)v sound = 1/(Mach number); sin 37° = 1/(Mach number), which gives Mach number = 1.7.

67. Because the frequency doubles for each octave, we have f high /f low = 2 x ; 20,000 Hz)/(20 Hz) = 2 x , or log 10 (1000) = x log 10 (2). which gives x ≈ 10.

v object t

θ v sound t

v airp lane t

θ v sound t

h

d plane = v plane t

θ v sound t 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 is dropped. If we call t 1 the time of fall for the stone, we have

y = y 01 + v 01 t 1 + !gt 1 2 ;

h = 0 + 0 + !gt 1 2 , or t 1 = (2h/g) 1/2 .

For the time t 2 for the sound to reach the top of the cliff, we have t 2 = h/v sound .

Thus for the total time we have t = t 1 + t 2 = (2h/g) 1/2 + h/v sound ; 3.5 s = [2h/(9.80 m/s 2 )] 1/2 + [h/(343 m/s)].

This is a quadratic equation for h 1/2 , which has the positive result h 1/2 = 7.39 m 1/2 , so the height of the cliff is 55 m.

69. The intensity for a 0 dB sound is I 0 = 10 –12 W/m 2 . For 1000 mosquitoes, we find the intensity level from β = 10 log 10 (I 1 /I 0 ) = 10 log 10 (1000I 0 /I 0 ) = 30 dB.

70. Because the wavelength in front of a moving source decreases, the wavelength from the approaching car is λ 1 = (v – v car )/f 0 .

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

f 1 = v/λ 1 = v/[(v – v car )/f 0 ]= vf 0 /(v – v car ). Because the wavelength behind a moving source increases, the wavelength from the receding car is

λ 2 = (v + v car )/f 0 . This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is

f 2 = v/λ 2 = v/[(v + v car )/f 0 ]= vf 0 /(v + v car ). If the frequency drops by one octave, we have

f 1 /f 2 = [vf 0 /(v – v car )]/[vf 0 /(v + v car )] = (v + v car )/(v – v car ) = 2, or 2(v – v car ) = v + v car , which gives v car = @v = @(343 m/s) = 114 m/s = 410 km/h (257 mi/h).

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

v = λ 1 f 1 = λ 1 ′f 1 ′; 2L[(600 Hz)/3] = 2(0.60L)f 1 ′, which gives f 1 ′ = 333 Hz.

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

Because the strings have the same length, the wavelengths are the same, so for the ratio of frequencies we have

f n+1 /f n = v n+1 /v n = (µ n /µ n+1 ) 1/2 = 1.5, or µ n+1 /µ n = (1/1.5) 2 = 0.444. With respect to the lowest string, we have

µ n+1 /µ 1 = (0.444) n . If we call the mass density of the lowest string 1, we have

1, 0.444, 0.198, 0.0878, 0.0389.


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

v = λf = 2(0.32 m)(440 Hz) = 2.8 × 10 2 m/s. We find the tension from

v = (F T /µ) 1/2 ; 282 m/s = [F T /(5.5 × 10 –4 kg/m)] 1/2 , which gives F T = 44 N.

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

v = λf = 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: f 2 = 2f 1 = 2(440 Hz) = 880 Hz.

Only the odd harmonics are present in a closed pipe, so the first overtone is the third harmonic: f 3 = 3f 1 = 3(440 Hz) = 1320 Hz.

74. We find the ratio of intensities from Æβ = 10 log 10 (I 2 /I 1 ); – 10 dB = 10 log 10 (I 2 /I 1 ), which gives I 2 /I 1 = 0.10.

If we assume uniform spreading of the sounds, the intensity is proportional to the power output, so we have P 2 /P 1 = I 2 /I 1 ; P 2 /(100 W) = 0.10, which gives P 2 = 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 the distance between adjacent nodes, or half a wavelength:

ÆL = λ/2, or λ = 2 ÆL. For the frequency we have

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

76. We find the gain from β = 10 log 10 (P 2 /P 1 ) = 10 log 10 [(100 W)/(1 × 10 –3 W)] = 50 dB.

L 1

L 2


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

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

λ 1 = (v – v car )/f 0 = (343 m/s – 10.0 m/s)/(200 Hz) = 1.665 m. This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is

f 1 = v/λ 1 = (343 m/s)/(1.665 m) = 206 Hz. Because the wavelength behind a moving source decreases, the wavelength from the receding loudspeaker is

λ 2 = (v + v car )/f 0 = (343 m/s + 10.0 m/s)/(200 Hz) = 1.765 m. This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is

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

f beat = Æf = 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 the same 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 train whistle is

λ 1 = (v – v train )/f 0 . Because you are stationary, this wavelength approaches you at a relative speed of v, so the frequency heard by you is

f 1 = v/λ 1 = v/[(v – v train )/f 0 ] = vf 0 /(v – v train ). Because the wavelength behind a moving source increases, the wavelength from the receding train whistle is

λ 2 = (v + v train )/f 0 . This wavelength approaches you at a relative speed of v, so the frequency heard by you is

f 2 = v/λ 2 = v/[(v + v train )/f 0 ] = vf 0 /(v + v train ). For the ratio of frequencies, we get

f 2 /f 1 = (v – v train )/(v + v train ); (486 Hz)/(522 Hz) = (343 m/s – v train )/(343 m/s + v train ), which gives v train = 12.3 m/s.

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

f 3pipe = 3f 1pipe = 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 λ 1string = 2L, so we have

v string = f 1string /2L = [F T /(m/L)] 1/2 ; (343 Hz)/2(0.75 m) = F T /[(0.00180 kg)/(0.75 m)] 1/2 , which gives F T = 6.4 × 10 2 N.


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80. Because the source is at rest, the wavelength traveling toward the blood is λ 1 = v/f 0 .

This wavelength approaches the blood at a relative speed of v – v blood . The ultrasound strikes and reflects from the blood with a frequency

f 1 = (v – v blood )/λ 1 = (v – v blood )/(v/f 0 ) = (v – v blood )f 0 /v. This frequency can be considered emitted by the blood, which is moving away from the source. Because the wavelength behind the moving blood increases, the wavelength approaching the source is

λ 2 = (v + v blood )/f 1 = (v + v blood )/[(v – v blood )f 0 /v] = [1 + (v blood /v)]v/f 0 [1 – (v blood /v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is

f 2 = v/λ 2 = v/[1 + (v blood /v)]v/f 0 [1 – (v blood /v)] = [1 – (v blood /v)]f 0 /[1 + (v blood /v)]. Because v blood « v, we use 1/[1 + (v blood /v)] ≈ 1 – (v blood /v), so we have

f 2 ≈ f 0 [1 – (v blood /v)] 2 ≈ f 0 [1 – 2(v blood /v)]. For the beat frequency we have

f beat = f 0 – f 2 = 2(v blood /v)f 0 = 2[(0.32 m/s)/(1.54 × 10 3 m/s)](5.50 × 10 6 Hz) = 2.29 × 10 3 Hz = 2.29 kHz.

81. The initial path difference is ÆD = 2[(¬/2) 2 + d 2 ] 1/2 – ¬.

When the obstacle is moved Æd, the new path difference is ÆD′ = 2[(¬/2) 2 + (d + Æd) 2 ] 1/2 – ¬.

Because Æd « d, we get ÆD′ = 2[(¬/2) 2 + d 2 + Æd(2d + Æd)] 1/2 – ¬

= 2[(¬/2) 2 + d 2 + 2d Æd] 1/2 – ¬ = 2[(¬/2) 2 + d 2 ] 1/2 (1 + 2d Æd/[(¬/2) 2 + d 2 ]) 1/2 – ¬. = 2[(¬/2) 2 + d 2 ] 1/2 (1 + !2d Æd/[(¬/2) 2 + d 2 ]) – ¬. = 2[(¬/2) 2 + d 2 ] 1/2 (1 + d Æd/[(¬/2) 2 + d 2 ]) – ¬.

To create destructive interference, the change in path difference must be λ/2, so we have

ÆD′ – ÆD = λ/2; 2[(¬/2) 2 + d 2 ] 1/2 (1 + d Æd/[(¬/2) 2 + d 2 ]) – ¬ – 2[(¬/2) 2 + d 2 ] 1/2 – ¬ = λ/2; 2[(¬/2) 2 + d 2 ] 1/2 d Æd/[(¬/2) 2 + d 2 ] = λ/2, which gives Æd = (λ/4d)[(¬/2) 2 + d 2 ] 1/2 .

d

¬

S

Detector


Page 12 – 21

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

Therefor the wavelength traveling toward the observer is λ a = (v – v wind )/f 0 .

This wavelength approaches the observer at a relative speed of v – v wind . The observer will hear a frequency

f a = (v – v wind )/λ a = (v – v wind )/[(v – v wind )/f 0 ] = f 0 = 645 Hz. (b) Because the wind is blowing toward the observer, the effective speed of sound is v + v wind .

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 sound

is v. Because there is no relative motion of the whistle and the observer, there will be no change in the frequency: 645 Hz.

(d) Because the wind is blowing perpendicular to the line toward the observer, the effective speed of sound is 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 + v wind . Therefore the wavelength traveling toward the cyclist is

λ e = (v + v wind )/f 0 . This wavelength approaches the cyclist at a relative speed of v + v wind + v cycle . The cyclist will hear a frequency

f e = (v + v wind + v cycle )/λ e = (v + v wind + v cycle )/[(v + v wind )/f 0 ] = (v + v wind + v cycle )f 0 /(v + v wind ) = (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 sound is v. Therefore the wavelength traveling toward the cyclist is

λ f = v/f 0 . This wavelength approaches the cyclist at a relative speed of v + v cycle . The cyclist will hear a frequency

f f = (v + v cycle )/λ f = (v + v cycle )/(v/f 0 ) = (v + v cycle )f 0 /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 where there is destructive interference, the path difference from the sources must be ÆL = nλ/2. If we assume that n = 1, we find the frequency from

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

84. (a) We find the wavelength from λ = v/f = (1560 m/s)/(200,000 Hz) = 7.8 × 10 –3 m.

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

85. If T is the period of the pulses and Æt is the duration of a pulse, and the moth must recieve the entire reflected pulse, the maximum time for the signal to travel to the moth and return is t max = T – Æt. We find the distance from

2d max = vt max = (343 m/s)(70.0 ms – 3.0 ms)(10 –3 s/ms), which gives d max = 11.5 m.


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86. Because the wavelength in front of a moving source decreases, the wavelength approaching the moth is λ 1 = (v – v bat )/f 0 .

The wavelength approaches the moth at a relative speed of v + v moth . The sound strikes and reflects from the object with a frequency

f 1 = (v + v moth )/λ 1 = (v + v moth )/[(v – v bat )/f 0 ] = (v + v moth )f 0 /(v – v bat ). This frequency can be considered emitted by the moth, which is moving toward the bat. Because the wavelength in front of a moving source decreases, the wavelength approaching the bat is

λ 2 = (v – v moth )/f 1 = (v – v moth )/[(v + v moth )f 0 /(v – v bat )] = (v – v moth )(v – v bat )/(v + v moth )f 0 /(v – v bat )]. This wavelength approaches the bat at a relative speed of v + v bat , so the frequency received by the bat is

f 2 = (v + v bat )/λ 2 = (v + v bat )/[(v – v moth )/f 1 ] = (v + v bat )(v + v moth )f 0 /(v – v moth )(v – v bat ) = (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.

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