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 915 Hz, is shifted by 54.3 Hz. Is this shift an increase or a decrease in the frequency?

b) How long has your friend been in the air when she emits the yell whose frequency shift you hear? Take 343 m/s for the speed of sound in air and 9.80 m/s2 for the acceleration due to gravity.

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Explanation / Answer

This is an example of the Doppler effect, a shift ?f in the apparent frequency detected by a receiver arising from the relative motion of the source. With the receiver stationary and the source moving directly towards it at a velocity Vs and emitting sound with a true frequency fo, the shift is positive (ie towards a higher frequency) and given by

(fo + ?f)/fo = c/(c - Vs)

where c is the velocity of sound in the intervening medium.

Substituting the given values fo = 915 Hz, ?f = 54.3 Hz and c = 343 m/s into this equation gives

(915 + 54.3)/915 = 343/(343 - Vs)

which may be rearranged to give the instantaneous velocity

Vs = 343*54.3/(915 + 54.3) = 19.21 m/s

Since the source was initially at rest, and is falling freely with a gravitational acceleration of 9.80 m/s

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