Φ21 Fall 2006 1
HW20 Solutions
Problem K20.53
(Some of the numbers in this problem are randomized.) One way to monitor global warming is to measure the average temperature of the ocean. Researchers are doing this by measuring the time it takes sound pulses to travel underwater over large distances. At a depth of 1000 m, where ocean temperatures hold steady near 4 ◦ C, the average sound speed is 1480 m/s. It's known from laboratory measurements that the sound speed increases 4.0 m/s for every 1.0 ◦ C increase in temperature. In one experiment, where sounds generated near California are detected in the South Paci c, the sound waves travel 7700 km. If the smallest time change that can be reliably detected is 1.0 s, what is the smallest change in average temperature that can be measured? Part A.
Solution:
The speed of sound in water could be written as v (T ) = 1480 + 4 (T − 4 ◦ C)
For the given distance, at 4◦ C, the travel time is ( ) t(4◦ C) = 7700 × 103 m / (1480 m/s) = 5202.7 s
At a temperature of 5◦ C, the travel time would be ( ) t (5◦ C) = 7700 × 103 m / (1484 m/s) = 5188.7 s
So near this temperature, one degree C corresponds to a ∆t of 14 s. So if we can measure a ∆t of 1.0 s, then we can see a temperature dierence of ∆T =
2
1s = 0.07 ◦ C 14 s
Problem K21.21
(Some numbers in this problem have been randomized.) Two loudspeakers emit 700 Hz sound waves along the x-axis. Take the speed of sound to be 343 m/s. If the speakers are in phase, what is the smallest distance between the speakers for which the interference of the sound waves is destructive? (The speakers are in line on the x axis.) Part A.
If the speakers are separated by 1/2 λ, then the sound from the rst will be out of phase with the sound from the second.
Solution:
d=
λ v/f (343 m/s) / (700 Hz) = = = 0.245 m 2 2 2
If the speakers are out of phase, what is the smallest distance between the speakers for which the interference of the sound waves is constructive? Part B.
Solution: This is the same as Part A. The sound a distance λ/2 from the rst speaker will be out of phase with the sound at that rst speaker. Since the second speaker is described as being out of phase with the rst, placing them λ/2 apart will make their sound constructively interfere.
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Problem K21.40
(Some numbers in this problem have been randomized.) A violinist places her nger so that the vibrating section of a 1.30 g/m string has a length of 40.0 cm, then she draws her bow across it. A listener nearby hears a note with a wavelength of 50.0 cm. Take the speed of sound in air to be 343 m/s. What is the tension in the string?
√
The velocity of the wave on the string isv = T /µ so the tension is T = µv2 . To nd v, we'll use v = f λ. The wavelength on the string is λ = 2L = 2 (0.4 m) = 0.8 m. The frequency is found from the sound speed and wavelength in air to be Solution:
f=
vair 343 m/s = 686 Hz = λair 0.5 m
Plugging this all in, we get (( 2
T = µv 2 = µ (f λstring ) = µ
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vair λair
)
)2 (( ) )2 343 m/s (2L) = (0.0013 kg/m) (2 · 0.4 m) = 391.5 N 0.5 m
OH Violin String
The D string of a violin is 0.327 m long and is adjusted for a fundamental frequency of 293 Hz. Part A.
What is the wave propagation speed for a transverse wave on the string?
Solution:
The fundamental wavelength is twice the length of the string. So the wave speed is v = f λ = f (2L) = (293 Hz) (0.327 m · 2) = 191.6 m/s
Part B.
What is the frequency of the second harmonic?
Solution: Part C.
The frequency of the second harmonic is twice the fundamental frequency. So f2 = 2f = 586 Hz.
What is the frequency of the third harmonic?
Solution:
The frequency of the third harmonic is thrice the fundamental frequency. So f3 = 3f = 879 Hz.
What is the frequency of the sound wave produced in air by the fundamental vibration of the D string? Take the speed of sound in air to be 345 m/s, and give your answer in Hz. Part D.
Solution:
The frequency in all media are the same. So fair = f = 293 Hz.
What is the wavelength of the sound wave produced in air by the fundamental vibration of the D string? Give your answer in meters. Part E.
Solution:
Since v = f λ, it follows that λ = v/f = (345 m/s) / (293 Hz) = 1.18 m.
Parts F and G. Solution:
0.59 m.
What are the frequency and wavelength in air of the second harmonic?
The frequency is f2 = 586 Hz (Part B above). The wavelength is λ = v/f = (345 m/s) / (586 Hz) =
2
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Problem K20.50
One cue your hearing system uses to localize a sound (i.e., to tell where a sound is coming from) is the slight dierence in the arrival times of the sound at your ears. Your ears are spaced approximately 20 cm apart. Consider a sound source 5.0 m from the center of your head along a line 45◦ to your right. What is the dierence in arrival times? Give your answer in microseconds. Solution:
The distances of the two ears from the source are √(
dl dr
)2 5 √ = + + 0.1 = 5.07 m 2 √( )2 ( )2 5 5 √ = + √ − 0.1 = 4.93 m 2 2 5 √ 2
)2
(
Since the dierence in distance is ∆d = 0.14 m, the dierence in time is ∆t = ∆d/v = (0.14 m) / (345 m/s) =
406 µs.
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