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 939 Hz, is shifted by 53.9 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/s^2 for the acceleration due to gravity.

Explanation / Answer

Since she is coming toward you, the shift is an increase in frequency.

By the Doppler Effect...

f' = f(v/(v - vs))

992.9 = 939(341/(341 - vs)

vs = 18.5 m/s

Then apply vf = vo + at

18.5 = 0 + 9.8(t)

t = 1.89 s

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