A stationary source emits sound of frequency $f_{0} = 492 H z .$ The sound is reflected by a large car approaching the source with a speed of $2 m s^{- 1} .$ The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is 330 $m s^{- 1}$ and the car reflects the sound at the frequency it has received.)

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Solution

Frequency received by the approaching car (as observer)
$f_{1} = (\frac{v + v_{0}}{v}) f$
Frequency of the reflected wave as received by the source
$f_{2} = (\frac{v}{v - v_{0}}) f_{1} = (\frac{v + v_{0}}{v - v_{0}}) f .$
$∴$ beat frequency $= f_{2} - f = (\frac{v + v_{0}}{v - v_{0}} - 1) f = (\frac{332}{328} - 1) 492 s^{- 1} = 6 s^{- 1}$

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Frequency received by the approaching car (as observer) f1=(v+v0v)f Frequency of the reflected wave as received by the source f2=(vv−v0)f1=(v+v0v−v0)f. ∴ beat frequency =f2−f=(v+v0v−v0−1)f=(332328−1)492 s−1=6 s−1