by observer before and after crossing are $n_{1}$ and $n_{2}$ respectively, then
$n_{1}=\frac{300+2}{300-10} 150=156.20 \mathrm{Hz}$
$n_{2}=\frac{300-2}{300+10} 150=144.19 \mathrm{Hz}$
$n_{1}-n_{2}=12 H z(\text {approx})$
[Given: The speed of sound in air is $324 ms ^{-1}$ ]
($1$) When only $S_2$ is emitting sound and it is $Q$, the frequency of sound measured by the detector in $Hz$ is. . . . . .
($2$) Consider both sources emitting sound. When $S_2$ is at $R$ and $S_1$ approaches the detector with a speed $4 ms ^{-1}$, the beat frequency measured by the detector is $\qquad$ $Hz$.
