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

Question

tarting 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 933 Hz, is shifted by 53.3 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 344 m/s for the speed of sound in air and 9.80 m/s2 for the acceleration due to gravity.

Explanation / Answer

V = 344 m/s

fo = 933+53.3 Hz , f= 933 Hz

when source approching observer, shift will increase

from doppler effect, source moving towards observer.

fo = f*v/(v-vs)

(933+53.3) = 933*344/(344-vs)

source velocity vs = 18.59 m/s

u =o

from kinematic equation

v = u+at

18.59 = 0+9.8*t

t = 1.9 s

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