tarting from rest, your friend dives from a high cliff into a deep lake below, y
ID: 1569662 • Letter: T
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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