A bat, moving at 5.50 m/s, is chasing a flying insect. The bat emits a 39.90 -kH

Question

A bat, moving at 5.50 m/s, is chasing a flying insect. The bat emits a 39.90-kHz chirp and receives back an echo at 40.50 kHz. (Take the speed of sound in air to be v = 343 m/s.)

A bat, moving at 5.50 m/s, is chasing a flying insect. The bat emits a 39.90-kHz chirp and receives back an echo at 40.50 kHz. (Take the speed of sound in air to be v = 343 m/s.) What is the speed of the insect? Will the bat be able to catch the insect? No, because the bat is flying more slowly than the insect. No, because the bat is flying at the same speed as the insect. Yes, because the bat is flying faster than the insect.

Explanation / Answer

This is a double Doppler problem because first the signal from the bat is shifted when it strikes the insect (Here the velocity of both the source(bat) vs and the listener (insect) v are negative). Then this shifted signal is reflected back to the original source. Here the velocity of the listener (bat)
and the source (insect) are positive

So let f' be the frequency the insect "hears" and f" the frequency the bat finally receives

Now f' = 39.90x10^3*(343 -v)/(343 - 5.5)

and f" =f'*(343 +5.5)/(343 +v)

So f" = (39.90x10^3*(343 -v)/(343 - 5.5))*(343 +5.5)/(343 +v)

Now 40.50x10^3 = 39.90x10^3*(343 -v)/(343 - 5.5))*(343 +5.5)/(343 +v)

0.98296 = (343 - v)/(343 + v)

0.98296*(343 + v) = 343 - v

or 1.98296*v = 343 - 0.98296*343

v = 2.95m/s

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