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Question

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

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

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.

$\mathrm{v}^{\prime}=\frac{\mathrm{v}\left(\mathrm{u}+\mathrm{v}_{0}ight)}{\left(\mathrm{v}-\mathrm{v}_{\mathrm{s}}ight)}=\mathrm{v}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}ight) \quad\left[\because \mathrm{v}_{0}=\mathrm{v}_{\mathrm{s}}ight]$

or $\mathrm{v}^{\prime}=1.2\left(\frac{350+50}{350-50}ight)=\frac{1.2 	imes 400}{300}=1.6 \mathrm{\,kHz}$

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 ..... $kHz$

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