Q27 of 31 Page 388

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?

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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