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Physics 23
Assignment 2 Solutions
T. Tao^1 (^1) Department of Physics, University of California, Santa Barbara, CA 93106 (Dated: October 18, 2006)
E19-
a) In miles per second, the speed of sound is 02.06 mi/s. In five seconds, the sound of the lightning travels approximately 1.03 miles, which is about 3 percent too large. b) Simply count in seconds and divide by 3.
E19-
Let one person speak with an intensity I 1. N people would have an intensity I 2 = N I 1. The sound levels and intensities are related by
SL 1 − SL 2 = 10Log
(I 1
I 2
= 10Log
N
Using SL 1 = 65dB and SL 2 = 80dB, we obtain N ∼= 32.
E19-
a) The path length difference (the extra distance that a wave from the upper speaker has to travel compared to the lower one in order to reach the listenser) between the waves from the speakers is ∆L = [(2. 12 m)^2 + (3. 75 m)^2 ]^1 /^2 − 3. 75 m ∼= 0. 56 m. (2)
In order to have a minimum when reaching the listener, upon traveling the path length difference a wave from the upper speaker must be completely out of phase with the wave from the lower speaker. (we assume that the speakers emit waves at the same time with the same wavelength and frequency) Therefore ∆L must be a half integer number of the wavelengths so that ∆L =
n − (^12)
λ ∼= 0. 56 m, (3)
where n is an integer. Since f = c λ s, (4)
where cs is the speed of sound in air, the listener will hear a minimum at the frequencies
f ∼=
n − (^12)
615 Hz. (5)
b) Now ∆L must be an integer number of the wavelengths so that the waves arrive at the listener in phase. We therefore have
∆L = nλ ∼= 0. 56 m. (6)
The frequencies for which the listener hears a maximum are thus
f ∼= n(615Hz). (7)
E19-
Since the tunnel is open on both ends, the air pressure perturbation at the ends must be zero, therefore the resonance condition for sound waves inside the tunnel is exactly the same as for that of standing waves on a string,
f = nc 2 L s. (8)
Here n is an integer, L is the length of the tunnel and cs is the speed of sound in air. Assuming that 135Hz and 138Hz are two consecutive resonances (or overtones) of the tunnel, the conditions on the shortest possible L are
L = 2 ncfns (9)
and L = (n 2 + 1)fn+1cs , (10)
where fn = 135Hz and fn+1 = 138Hz. Solving the two equations, we obtain L ∼= 57. 2 m.
P19-
a) To obtain a minima at D, the wave traveling through SBD must destructively interfere with that through SAD. Conversely, position D will see a maxima only if waves from the
where vs is the speed of the source, f is the original frequency and f+ is the up shifted frequency due to the source moving closer, corresponding to the minus sign in the formula. f− gives the reverse. The observer hears the beat frequency of the up and down shifted waves: ∆f = f+ − f− ∼= 81Hz. (17)
b) Now the sources are stationary while the observer moves, resulting in Doppler shifted frequencies of f± = f cs^ ± cs^ vo. (18)
Here vo is the speed of the observer. The plus sign in the equation gives the up shifted frequency f+ with the observer moving towards the source. The minus sign indicates the opposite. The observer hears a beat frequency of approximately 81Hz.
P19-
The submarine moves at 75.2km/h wrt to the ocean floor but only 75. 2 km/h −
- 5 km/h ∼= 12. 4 m/s wrt to the medium, namely the ocean current. Since the Doppler shift equations are derived in the reference frame of the medium, we effectively have a sta- tionary observer listening to a moving source, since the observing ship is stationary wrt to the current. The resulting shifted frequency is therefore
f ′^ = f (^) cs c−s vs^ ∼= 997Hz. (19)
Note that the speed of sound in water is approximately 1482m/s, much greater than in air.
