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An observer and a source emitting sound of frequency $120 \mathrm{~Hz}$ are on the $X$-axis. The observer is stationary while the source of sound is in motion given by the equation $x=3 \sin \omega t$ ( $x$ is in metres and $t$ is in seconds). If the difference between the maximum and minimum frequencies of the sound observed by the observers is $22 \mathrm{~Hz}$, then the value of $\omega$ is ( ppeed of sound in air $=330 \mathrm{~ms}^{-1}$ )
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The correct answer is:
$10 \mathrm{rad} \mathrm{s}^{-1}$
Instantaneous speed of source is $v=\frac{d x}{d t}=3 \omega \sin \omega t$
Difference between maximum and minimum frequencies is $22 \mathrm{~Hz}$.
So, $f_{\max }-f_{\min }=f\left(\frac{v}{v-v_s}\right)-f\left(\frac{v}{v+v_s}\right)=22$
Now here, $f=120 \mathrm{~Hz}, v=330 \mathrm{~ms}^{-1}, v_s=3 \omega$
Substituting these values in Eq (i), we get
$$
120\left(\frac{330}{330-3 \omega}-\frac{330}{330+3 \omega}\right)=22 \Rightarrow \omega=10 \mathrm{~s}^{-1}
$$
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