A source of sound is moving with a velocity of \( 50 \mathrm{~ms}^{-1} \) towards a stationary observer. The observer measures the frequency of sound as \( 500 \mathrm{~Hz} \). The apparent frequency of sound as heard by the observer when source is moving away from him with the same speed is (Speed of sound at room temperature \( 350 \mathrm{~ms}^{-1} \))

Given, velocity of source of sound \( =50 \mathrm{~ms}^{-1} \); frequency measured by observer \( =500 \mathrm{~Hz} \)
Now, frequency of sound when source is moving towards stationary observer is
\(f^{\prime}=f\left(\frac{v}{v-v_{s}}\right) \rightarrow(1)\)
Frequency of sound when source is moving away from the stationary observer is
\(f^{\prime \prime}=f\left(\frac{v}{v+v_{s}}\right) \rightarrow(2)\)
Using Eqs. (1) and (2), we have
\(\frac{f^{\prime}}{f^{\prime \prime}}=\frac{f\left(\frac{v}{v-v_{s}}\right)}{f\left(\frac{v}{v+v_{s}}\right)} \)
\(\Rightarrow \frac{f^{\prime}}{f^{\prime \prime}}=\frac{\left(v+v_{s}\right)}{\left(v-v_{s}\right)}\)
Substitute \( v=350 \mathrm{~ms}^{-1} ; v_{s}=50 \mathrm{~ms}^{-} \)
\(f^{\prime}=500 H z \)
\(\frac{500}{f^{\prime \prime}}=\frac{350+50}{350-50} \)
\(\Rightarrow \frac{500}{f^{\prime \prime}}=\frac{400}{300} \)
\(\Rightarrow f^{\prime \prime}=\frac{300 \times 500}{400}\)
Therefore, \( f^{\prime \prime}=375 \mathrm{~Hz} \)
Thus, apparent frequency of sound as heard by the observer when source is moving away from him is \( 375 \mathrm{~Hz} \).