- A$450$
- B$540$
- C$270$
- ✓$648$
$\mathrm{f}^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{0}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ from Doppler's effect
where $v_{0}=v_{s}=30 \mathrm{m} / \mathrm{s},$ velocity of observer and source
Speed of sound $v=330 \mathrm{m} / \mathrm{s}$
$\because$ Frequency of whistle $(\mathrm{f})=540 \mathrm{Hz}$
$\therefore \mathrm{f}^{\prime}=\frac{330+30}{330-30} \times 540=648 \mathrm{Hz}$
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[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 $\qquad$ $Hz$.