Question #164220

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


1
Expert's answer
2021-03-08T08:33:04-0500

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:


f1=fs(vv+vs),f_{1}=f_s(\dfrac{v}{v+v_s}),f1=100 Hz(340 ms340 ms+10 ms)=97.1 Hz.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:


f2=fs(vvvs),f_{2}=f_s(\dfrac{v}{v-v_s}),f2=100 Hz(340 ms340 ms10 ms)=103 Hz.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:


fbeat=f2f1=103 Hz97.1 Hz=5.9 Hz.f_{beat}=|f_2-f_1|=|103\ Hz-97.1\ Hz|=5.9\ Hz.

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