When an engine passes near to a stationary observer then its apparent frequencies occurs in the ratio $5/3$. If the velocity sound is $340 m/s$, than the velocity of engine is .... $m/s$
Medium
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(c) When engine approaches towards observer $n' = n\,\left( {\frac{v}{{v - {v_s}}}} \right)$
when engine going away from observer $n'' = \left( {\frac{v}{{v + {v_S}}}} \right)n$
$\therefore $ $\frac{{n'}}{{n''}} = \frac{{v + {v_s}}}{{v - {v_s}}}$
$\Rightarrow \frac{5}{3} = \frac{{340 + {v_s}}}{{340 - {v_s}}}$
$\Rightarrow {v_s} = 85$$m/s$.
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