Q141 / 180 · COMEDK 2017 · Physics

The apparent frequency of a note is $200 \mathrm{~Hz}$. When a listener is moving with a velocity of $40 \mathrm{~ms}^{-1}$ towards a stationary source. When he moves away from the same source with the same speed, the apparent frequency of the same note is $160 \mathrm{~Hz}$. The velocity of sound in air (in $\mathrm{m} / \mathrm{s}$ ) is

  • A
    340
  • B
    330
  • Correct answer
    C
    360
  • D
    320

Solution

Let $f_{s}$ be the frequency of source. The apparent frequency heard when listener moves towards source is $f^{\prime}=f_{s}\left(\frac{v_{s}+v_{0}}{v_{s}}\right)$ $\Rightarrow \quad 200=f_{s}\left(\frac{v_{s}+40}{v_{s}}\right)...(i)$ The apparent frequency heard, when listener moves away from source is $\begin{aligned} f^{\prime \prime} &=f_{s}\left(\frac{v_{s}-v_{o}}{v_{s}}\right) \\ \Rightarrow \quad 160 &=f_{s}\left(\frac{v_{s}-40}{v_{s}}\right)...(ii) \end{aligned}$ Dividing Eq. (i) by Eq. (ii), we get $\begin{aligned} & \frac{200}{160} &=\frac{v_{s}+40}{v_{s}-40} \\ \Rightarrow & & \frac{5}{4} &=\frac{v_{s}+40}{v_{s}-40} \\ \Rightarrow & 5 v_{s}-200 &=4 v_{s}+160 \\ \Rightarrow & v_{s} &=360 \mathrm{~m} / \mathrm{s} \end{aligned}$

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

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