Q. 275.0( 1 Vote )

A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

Answer :

Given,


Ultrasonic beep frequency emitted by the bat, f = 40kHz=40×103 Hz


Velocity of the bat, vb = 0.03v


where, v = velocity of sound in air


The apparent frequency of the sound striking the wall is given by


f’ = [v/(v-vb)]f


f’ = [v/(v-0.03v)]f


f’ = f/0.97


f’ = (40×103 Hz)/0.97


f’ = 41.237 × 103 Hz


f’ = 41.237 kHz


This frequency (fs) is reflected back from the stationary wall. Let f’’ be the frequency received by the bat.


f’’ = [(v+v0)/v]f’


f’’ = [(v+0.03v)/v]f’


f’’ = 1.03f’


f’’ = 1.03×41.237 kHz


f’’ = 42.474 kHz


NOTE: The Doppler effect is the phenomenon due to which there is a change in frequency or wavelength of a wave received by an observer who is moving relative to the wave source.


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