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 921 Hz, is shifted by 56.1 Hz. Is this shift an increase or a decrease in the frequency? Increase Decrease Can't tell from the information provided How long has your friend been in the air when she emits the yell whose frequency shift you hear? Take 345 m/s for the speed of sound in air and 9.80 m/s^2 for the acceleration due to gravity. Number S

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- hence it increases) 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 = 921 Hz, f = 56.1 Hz and c = 345 m/s into this equation gives

(921 + 56.1)/921 = 345/(345 - Vs)

which may be rearranged to give the instantaneous velocity

Vs = 345*56.1/(921 + 56.1) = 19.808 m/s

Since the source was initially at rest, and is falling freely with a gravitational acceleration of 9.80 m/s², the time t required to reach this velocity is given by

Then, using kinematic equations,

Vs = V0+g.t , V0=0;

Vs = g.t

that is, in a time

t = Vs/g = 2.025 s.

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