Question

A bat moving with speed of 3.25 m/s and emitting sound of 35.0 khz approaches a moth at rest on a tree trunk. (a) what frequency is heard by the moth? (b) if the speed of the bat is increased, is the frequency heard by the moth higher or lower? (c) calculate the frequency heard by the moth when the speed of the bat is 4.25 m/s

Answer 1

Use subscript 1 for the source (the bat), and 2 for the receiver (the moth).

Given:
v₁ = -3.25 m/s, velocity of the source (the bat moves toward the moth).
f₁ = 35.0 kHz, the frequency of the source
v₂ = 0, the velocity of the receiver (the moth)
c = 3 43 m/s, the speed of sound in air.

Let f₂ = the frequency received by the moth
Due to Doppler shift,

`f_(2) = f_(1) ( (c+v_(2))/(c+v_(1)))`

That is,
f₂ = (35.0 kHz)*(343/(343 - 3.25)
= 35.34 kHz

When v₁ = 4.25 m/s
f₂ = (35.0 kHz)*(343/(343 - 4.25))
= 35.44 kHz

Answers:
(a) 35.34 kHz
(b) As the speed of the bat increases, the frequency heard by the moth is higher.
(c) 35.44 kHz