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  4. A sound source emitting sound of frequency 450 Hz is approaching a stationary observer with velocity 30 m s- 1 and another identical source is going away from the observer with the same velocity. If the velocity of sound is 330 m s- 1 , then the difference of frequencies heard by observer is

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Q. A sound source emitting sound of frequency $450 \, Hz$ is approaching a stationary observer with velocity $30 \, m \, s^{- 1}$ and another identical source is going away from the observer with the same velocity. If the velocity of sound is $330 \, m \, s^{- 1}$ , then the difference of frequencies heard by observer is

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Solution:

Solution
$f_{1}=\left(\frac{v}{v - v_{s}}\right)f_{0}=\frac{330}{330 - 30}\times 450$
$f_{1}=\frac{330}{300}\times 450=11\times 45$
$f_{1}=495 \, Hz$
$f_{2}=\frac{330}{360}\times 450$
$f_{2}=412.5 \, Hz$
$\Delta f=82.5 \, Hz$