Question

Some bat species eat up to 1,000 mosquitoes in an hour. A bat, flying with a speed of 4.36 m/s, is in pursuit of a mosquito. While in motion, the bat produces a 39.70 kHz sound and hears the sound which is reflected off the mosquito at a frequency of 40.36 kHz. (Assume the speed of sound in air is v = 343 m/s.) How fast is the mosquito flying (in m/s)?

Answer 1

f' = ( v + vo ) / ( v - vo ) x f

where:

f' = observed frequency

f = original frequency at source

v = speed of sound = 343 m/s

vo = speed of observer

vs = speed of source of sound

vm = speed of observer = mosquito speed

f = original frequency at source = 39700 Hz

vs = speed of source of sound = speed of bat = 4.36 m/s;

∴ f' = ( 343 - vm ) / ( 343 - 4.36 ) x39700 = 117.2336( 343 - vm )

Using the doppler equation again for the bat as now being the receiver of the sound,

vo = speed of observer = 4.36 m/s,

f = original frequency at source = 40591 Hz

v = speed of sound = 343 m/s

vs = speed of source of sound = vm

∴ f' = (343 + 4.36) / (343 + vm) x 117.2336( 343 - vm ) = 40360 Hz
(343 - vm)/(343 + vm) = 0.9911
vm = 1.5324 m/s