Ultrasounds?

Question

A bat locates a moth and emits an ultrasound pulse that takes 1s to return to the bat.
(a) Estimate the distance from the bat to the moth.
(b) Estimate the lower frequency of the pulse emitted by the bat.

Answer 1

a)Since the sound took 1 second for a round trip, the distance is the distance travelled by sound in 0.5 seconds. Velocity of sound in air is approximately 334 m/s and so distance will be 334/2 = 167 meters.

b) The information given in the problem does not allow us to estimate the frequency of the ultrasound. However for the sound to be reflected by the moth, the wave length has to be the of the same order of magnitude. Let us assume that the moth is 1 cm in size (could be quite wrong but for this purpose it will do), the wavelength of the sound will also be 1 cm. Hence we can write:

Frequency X Wave Length = Velocity

Frequency = Velocity / Wave length

Frequency = 334000/10 = 33400 = 33 kHz approx.

Our answer may not be correct but is not absurd since it is above 20 kHz, which is the threshold for ultrasound.