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?

Question

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?

Philoid · Accepted Answer

Given,

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

Velocity of the bat, v_b = 0.03v

where, v = velocity of sound in air

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

f’ = [v/(v-v_b)]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.