PROBLEM 11 & 12 April 9 2010

PROBLEM 11 & 12 April 9 2010 - as measured relative...

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SOLVING PROBLEM 11 April 9, 2010 (like 13.26): How can a dolphin in sea water where the velocity of sound is 1,533 m/s retrieve a quarter from the bottom his pool? Assuming he uses sonar to find things under water, what must be the frequency of Flipper’s squeak when he can just detect a 0.025 m diameter object? a) 6 kHz; b) 60 kHz; c) 600 kHz; d) 6 MHz; e) 60 Hz. What do you know? The resolution limit is about 1 wavelength, so 0.025 m. The frequency is f = v = 1533/0.025 = 60 kHz. Problem 12 SOLVING PROBLEM 12 (like 13.39): The sound of a 1 kHz fog horn is carried down wind to a person trying to get to sleep across the bay. If the wind speed is 10 m/s and the speed of sound is 343 m/s, what frequency does the person hear? a) 343 Hz; b) 1000 Hz; c) 10000 Hz; d) 100 Hz; e) 1343 Hz. What do you know? For sound waves, the Doppler shifts are thought of
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Unformatted text preview: as measured relative to the stationary medium, i.e. the air. In a measurement frame in which the air is stationary, the horn and the sleeper are both traveling at the same velocity so the sleeper hears no Doppler shift and the perceived frequency is 1000 Hz. What frequency does a seagull blown along with the wind hear? a) 179 Hz; b) 990 Hz; c) 1010 Hz; d) 1029 Hz; e) 971 Hz. What do you know? The bird is stationary in the moving air and thinks the fog horn is traveling away from him. This produces a lowering of the frequency he hears because the wavelength in the air is longer than it would be if the air and the horn were at rest relative to one another. Thus Hz Hz v v v v f f S O 971 10 343 343 1000 '...
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