Starting from rest, your friend dives from a high cliff into a deep lake below,

Question

Starting from rest, your friend dives from a high cliff into a deep lake below, yelling in excitement at the thrill of free-fall on her way down. You watch her, as you stand on the lake shore, and at a certain instant your keen hearing recognizes that the usual frequency of her yell, which is 909 Hz, is shifted by 52.9 Hz. Is this shift an increase or a decrease in the frequency?

How long has your friend been in the air when she emits the yell whose frequency shift you hear? Take 341 m/s for the speed of sound in air and 9.80 m/s2 for the acceleration due to gravity. Answer in s thank you!

Explanation / Answer

Since your friend is going away from you, the apparent frequency will appear reduced.   Thus you will notice a decrease in the frequency

From the Doppler Effect we can find the velocity at the time producing the frequency shift

f' = f(v/v + vs)

909 - 52.9 = (909)(341/(341 + vs)

vs = 21.1 m/s

Then, using kinematic equations,

vf = vo + at

(21.1) = (0) + (9.8)(t)

t = 2.15 sec

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