Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe"that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed v_bal emits sound of frequency f_bat; the sound it hears reflected from an insect flying toward it has a higher frequency f_ref I If the bat emits a sound at a frequency of 80.8 kHz and hears it reflected at a frequency of 84.0 kHz while traveling at a speed of 3.8 m/s, calculate the speed of the insect. Use 344 m/s for the speed of sound in air. Express your answer using two significant figures.

Question

Physics

Donovan Rodriguez

7 July, 16:05

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe"that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed v_bal emits sound of frequency f_bat; the sound it hears reflected from an insect flying toward it has a higher frequency f_ref I If the bat emits a sound at a frequency of 80.8 kHz and hears it reflected at a frequency of 84.0 kHz while traveling at a speed of 3.8 m/s, calculate the speed of the insect. Use 344 m/s for the speed of sound in air. Express your answer using two significant figures.

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Answers (1)

Know the Answer?

Holden Barrett · Answer 1

2.88 m/s

Explanation:

From the Doppler's effect formula, we have

frequency, f = [ (v + vr) / (v + vs) ]*fo

where vr = velocity of the receiver, vs = velocity of the source

Given:

v = 344 m/s, vs = - 3.8 m/s (negative as observer is away from source)

fo = 80.8 KHz (frequency of source)

Hence, from bat to insect

f1 = [ (344 + v) / (344 + (-3.8)) ]*80.8

f1 = 0.2375 (344 + v)

= 0.2375 (344 + 3.8) = 82.61 Hz

from insect to bat

f2 = [ (344 + v) / (344 - v) ]*f1

where f2 = 84 KHz

hence 84 = [ (344 + v) / (344 - v) ]*82.61

Hence, 349.81 - 1.017v = 344 + v

solving, v = 2.88 m/s