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A bungee jumper jumps from rest and screams with a frequency of 517 Hz. The air temperature is 20 degrees C..?
A bungee jumper jumps from rest and screams with a frequency of 517 Hz. The air temperature is 20 degrees C. What is the frequency heard by the people on the ground below when she has fallen a distance of 9.89 m? Assume that the bungee cord has not yet taken effect, so she is in free-fall.
This is a doppler effect question so we want to find the difference in frequency of the sound, df.
df= -v_s/lambda
v_s is the velocity of the transmitter relative to the receiver in meters per second: negative when moving towards one another, positive when moving away and lambda the wavelength of the sound.
lambda = v/f = 343.14/517 = 0.6637m
(v is the speed of sound at 20 deg C)
v_s can be found by using a suvat equation assuming that she is starting from a velocity of 0m/s.
v^2=u^2+2as
u=0, a=9.81, s=9.98
Thus she is travelling at a velocity v_s=13.93m/s after falling 9.989m.
Subbing all this into the doppler equation,
df=-21Hz
Thus the sound heard by the observers is at a pitch of f=517-21=496Hz
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