MCQ
$A$ and $B$ are two sources generating sound waves. A listener is situated at $C$ . The frequency of the source at $A$ is $500\,Hz$ . $A,$ now, moves towards $C$ with a speed $4\,m/s.$ The number of beats heard at $C$ is $6.$ When $A$ moves away from $C$ with speed $4\,m/s,$ the number of beats heard at $C$ is $18.$ The speed of sound is $340\,m/s.$ The frequency of the source at $B$ is ..... $Hz$
  • A
    $500$
  • B
    $506$
  • $512$
  • D
    $494$

Answer

Correct option: C.
$512$
c
$Case\, 1:$ When source is moving towards stationary listener

apparent frequency $\eta^{\prime}=\eta\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$

$=500\left(\frac{340}{336}\right)=506 \mathrm{Hz}$

$Case\, 2:$ When source is moving away from the stationary listener

$\eta^{\prime \prime}=\eta\left(\frac{v}{v+v_{s}}\right)=500\left(\frac{340}{344}\right)=494 \mathrm{Hz}$

In case $1$ rumber of beats heard is $6$ and in

$case\, 2$ number of beats heard is $18$ therefore frequency of the source at $\mathrm{B}=512 \mathrm{Hz}$

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