A bat ing at a speed of 4.5 m/s pursues an insect flying in the same direction.

Question

A bat ing at a speed of 4.5 m/s pursues an insect flying in the same direction. The bat emits a 48000-Hz sonar pulse and hears the pulse reflected back from the insect at a frequency of 48000 + 590) Hz. Take the speed of sound to be 343 m/s. What is the speed of the insect, in meters per second, relative to the air? v_i = 4.13 This is a two-step process. First, the bat is the source, moving toward the insect, which is the receding observer. Then the insect is the source and is moving away from the bat, which is the approaching observer. Apply the Doppler-effect formula that is appropriate for each step. Be careful with the signs of the velocities in each step.

Explanation / Answer

This is a Doppler effect problem. The generic equation is:

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

Where v is the speed of the propagating waves in the medium

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 = -4.5 m/s (negative as the bat is going toward the observer).

So, we have

f' = (343 - vi) / (343 - 4.5) * 48000

Let's leave it like this for the moment. 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 = 4.5 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

48590 = (343 + 4.5) / (343 + vi) * f'

Putting both equations together, we get

48590 = (343 + 4.5) / (343 + vi) * (343 - vi) / (343 - 4.5) * 48000

(48590/48000)*(338.5/347.5) = (343 - vi) / (343 + vi)

0.986 = (343 - vi) / (343 + vi)

0.986*343 + 0.986*vi = 343 - vi

1.986 vi = 11

vi = 5.539 m/s

Since we've defined vi as the speed of thw insect away from the source, the insect is flying away from the bat as a

speed of 5.539 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.