# Horseshoe bats use the Doppler effect to determine their location. A horseshoe bat flies toward a wall at a speed of 19 m/s while emitting a sound of frequency 37 kHz. What is the beat frequency between the emission frequency and the echo?

Dec 8, 2015

$\Delta f = 4.09 \times {10}^{3} \text{Hz}$

#### Explanation:

The bat emits a sound which is reflected off the wall. This then interferes with the incoming signal to produce beats.

The incoming signal is "Doppler Shifted" twice because it has been reflected off a wall to which the bat is moving.

To get the beat frequency you find the difference between the two frequencies.

I will assume that at 20 degC the speed of sound $v$ is $343.7 \text{m/s}$

The beat frequency between a transmitted signal and a reflected signal off a moving object is:

f_("reflected")-f_("transmitted")=Deltaf=(2v_("target"))/(v).f

In relative terms we can consider the wall to be moving towards the bat.

$\therefore \Delta f = \frac{2 \times 19}{343.7} \times 37 \times {10}^{3}$

$\Delta f = 4.09 \times {10}^{3} \text{Hz}$

Police radar guns measure vehicle speeds like this.