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?

Answer 1

Here,
Original frequency emitted by bat ( fo) = 40 kHz
speed of the bat ( Vb) = 0.03V
Where , V is speed of the sound .
Apparent frequency recieved at the wall = fo{ V/(V - Vo)} [ A/C to Doppler's effect . ]
= 40 × { V/(V - 0.03V)}
= 40 × V/0.97V
= 40 × 1/0.97 kHz

This frequency will be reflected by the stationary wall towards the bat .

Bat receives apparent of the reflected frequency as f'
f' = f{( V + Vb)/V}
= 40/0.97 { (V + 0.03V)/V}
= 1.03 × 40/0.97
= 42.67 kHz

Answer 2

ultrasonic frequency emitted by the bat I=40 kHZ

velocity of the bat,vb=0.03 v

where v=velocity of sound in air

the Apparent frequency of the sound striking
the wall is given as:

v1=(v/v-vb)

=(v/v-0.03v)×40

40/0.97 kHz

this frequency is reflected by the stationery wall (vs=0)

The frequency (v^n)of received sound is given by the relation

v^n=(v+vb/v)v^+

=(v+0.03v/v)×40/0.97

=1.03×40/0.97=42.47 kHz.




hope its helps