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An engine is moving towards a wall with a velocity $50\, ms^{-1}$ emits a note of $1.2\, kHz$. The speed of sound in air is $350\, ms^{-1}$. The frequency of the note after reflection from the wall as heard by the driver of the engine is ..... $kHz$
$2.4$
$0.24$
$1.6$
$1.2$
Solution
The reflected sound appears to propagate in a direction opposite to that of direct sound of engine. Thus, the source and the observer can be presumed to approach, each other, with same velocity.
$\mathrm{v}^{\prime}=\frac{\mathrm{v}\left(\mathrm{u}+\mathrm{v}_{0}\right)}{\left(\mathrm{v}-\mathrm{v}_{\mathrm{s}}\right)}=\mathrm{v}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \quad\left[\because \mathrm{v}_{0}=\mathrm{v}_{\mathrm{s}}\right]$
or $\mathrm{v}^{\prime}=1.2\left(\frac{350+50}{350-50}\right)=\frac{1.2 \times 400}{300}=1.6 \mathrm{\,kHz}$