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?
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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