Question

A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what speed is the bat gaining on its prey? Take the speed of sound in air to be 340 m/s.

Answer 1

The speed the bat is gaining on its prey is 0.03m/s

Step-by-step explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

`F_0 = F_s[(v+v_0)/(v-v_s) ]\\\\(F_0)/(F_s) = [(v+v_0)/(v-v_s) ]\\\\(36.79)/(36) = (340+3.7)/(340-v_s)\\\\1.0219 = (343.7)/(340-v_s)\\\\ 340-v_s = (343.7)/(1.0219)\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s`

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s