Answer to Question #164220 in Optics for Gady

Question

A stationary observer hears beats when a source of sound of frequency 100Hz moves directly away from the observer with a speed of 10.0ms^-1 towards a vertical wall. Explain why beats are produced and calculate the beat frequency

Answer 1

The beats produced due to the interference between the sound waves coming from the source and the sound waves reflected from the wall. Let's first calculate the frequency of the sound waves coming directly from the source:

$$f_{1}=f_s(\dfrac{v}{v+v_s}),$$ $$f_{1}=100\ Hz\cdot(\dfrac{340\ \dfrac{m}{s}}{340\ \dfrac{m}{s}+10\ \dfrac{m}{s}})=97.1\ Hz.$$

Let's calculate the frequency of the sound waves reflected from the wall:

$$f_{2}=f_s(\dfrac{v}{v-v_s}),$$ $$f_{2}=100\ Hz\cdot(\dfrac{340\ \dfrac{m}{s}}{340\ \dfrac{m}{s}-10\ \dfrac{m}{s}})=103\ Hz.$$

Finally, we can calculate the beat frequency:

$$f_{beat}=|f_2-f_1|=|103\ Hz-97.1\ Hz|=5.9\ Hz.$$