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
\[
\begin{array}{r}
\frac{f^{\prime}}{f^{\prime \prime}}=\frac{f\left(\frac{v}{v-v_{s}}\right)}{f\left(\frac{v}{v+v_{s}}\right)} \\
\Rightarrow \frac{f^{\prime}}{f^{\prime \prime}}=\frac{\left(v+v_{s}\right)}{\left(v-v_{s}\right)}
\end{array}
\]
Substitute \( v=350 \mathrm{~ms}^{-1} ; v_{s}=50 \mathrm{~ms}^{-} \)
\[
\begin{array}{l}
f^{\prime}=500 H z \\
\frac{500}{f^{\prime \prime}}=\frac{350+50}{350-50} \\
\Rightarrow \frac{500}{f^{\prime \prime}}=\frac{400}{300} \\
\Rightarrow f^{\prime \prime}=\frac{300 \times 500}{400}
\end{array}
\]
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 \( 375 \mathrm{~Hz} \).