An engine moving towards a wall with a velocity 50 ~m / s emits a...

The reflected sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity. v^{\prime}=\frac{v\left(v+v_o\right)}{\left(v-v_s\right)} $=v\left(\frac{v+v_s}{v-v_s}\right) \quad\left(\because v_o=v_s\right)$\Rightarrow \quad v^{\prime}=1.2\left(\frac{350+50}{350-50}\right)$=\frac{1.2 \times 400}{300}$ $=1.6 \mathrm{kHz}$