A bat emits short pulses of sound at a frequency of 1.60 times 10^5 Hz. As the b

Question

A bat emits short pulses of sound at a frequency of 1.60 times 10^5 Hz. As the bat swoops toward a flat wall at speed 30.0 m/s, this sound is reflected from the wall back to the bat. What is the frequency of sound incident on the wall? Express your answer to two significant figures and include the appropriate units. Consider the wall as a sound source at the frequency calculated in part (a). What frequency of sound does the bat hear coming from the wall? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Part A :Here the bat is approaching the wall so the frequency of sound incident on the wall can be find out by using

Doppler effect for source approaching .

fWall = fBat [ v / ( v - vBat ) ] --------Source approaching the observer.

here  fBat = frequency emitted by the bat =1.60 x 105 Hz

vBat = Speed of the bat approaching the wall = 30.0 m/s.

v =speed of the sound at room temperature = 343 m/s

On substituting values we get

fWall = (1.60 x 105 Hz ) [ (343 m/s ) / ( 343 m/s  - 30.0 m/s ) ]

fWall = 1.75 x 105 Hz.

Part B: Here the frequency is emitted by thewall is fWall , The frequency hear by the bat is

fBat = fWall  [ 1 + ( vBat / v ) ] ------------- Observer approaching the source

{Here observer is bat and the source is wall }

fBat = ( 1.75 x 105 Hz ) [ 1 + (30.0 m/s) / (343 m/s) ]

fBat = 1.90 x 105 Hz.

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