Question

A bat flying at 4.80 m/s is chasing an insect flying in the same direction. The bat emits a 40.0-kHz chirp and receives back an echo at 40.4 kHz. (Take the speed of sound in air to be 340 m/s.)

What is speed of insect (m/s)

Answer 1

Solve this problem using Doppler's Effect -

The equation for Doppler's effect is given as -

f2 = (v+vo) / (v+ve) * f1

Where -

v is the speed of the propagating waves in the medium = 340 m/s

vo is the speed of the observer toward the emitter;

ve is the speed of the emitter away the observer;

f1 is the original frequency;

f2 is the frequency perceived by the emitter.

First we have to calculate the frequency received by the insect. From the insect's point of view, vo = -vi (the unknown speed of the insect away from the source);

and ve = -4.80 m/s (negative as the bat is going toward the observer).

So, we have

f' = (340-vi) / (340-4.8) * 40.0 k

Now, the wave reflected by the insect should have the same frequency f', and we can use the Doppler equation to find the frequency of the wave that reaches the bat.

Here, vo = 4.8 m/s (positive as the bat is going toward the source);

ve = +vi (positive as the emitter is going away from the observer).

So we have

40.4k = (340+ 4.8) / (340+vi) * f'

Putting both equations together, we get (with some algebra)

40.4k = (340+4.8) / (340+vi) * (340-vi) / (340-4.8) * 40k

=> (40.4/40)*(335.2/344.8) = (340-vi) / (340+vi)

=> 0.982 = (340-vi) / (340+vi)

=> 0.982*340 + 0.982*vi = 340 - vi

=> 1.982*vi = 6.12

=> vi = 6.12 / 1.982 = 3.09 m/s

Therefore, the speed of the insect is 3.09 m/s in the same direction as the bat is flying.