Q121 / 180 · COMEDK 2013 · Physics

The difference between the apparent frequency of a source of sound as perceived by the observer during its approach and recession is $2 \%$ of the frequency of the source. If the speed of sound in air is $300 \mathrm{~ms}^{-1}$, then the velocity of the source is

  • A
    $1.5 \mathrm{~ms}^{-1}$
  • B
    $12 \mathrm{~ms}^{-1}$
  • C
    $6 \mathrm{~ms}^{-1}$
  • Correct answer
    D
    $3 \mathrm{~ms}^{-1}$

Solution

If frequency of source be $f_{s}$, then $f_{\text {approach }}-f_{\text {recession }}=\frac{2}{100} \times f_{o}$ $\Rightarrow \quad f_{s}\left(\frac{v+v_{s}}{v}\right)-f_{s}\left(\frac{v-v_{s}}{v}\right)=\frac{f_{s}}{50}$ $\Rightarrow \quad \frac{2 v_{s}}{v}=\frac{1}{50} \Rightarrow \frac{2 v_{s}}{300}=\frac{1}{50}$ $\Rightarrow \quad v_{s}=\frac{300}{2 \times 50}=3 \mathrm{~ms}^{-1}$

Hence, the correct answer is option: 'D'.

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