Question

An observer and a source emitting sound of frequency $120Hz$ 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 $22Hz$, then the value of $\omega$ is $($ speed of sound in air $=330m{s}^{-1}$ )

• A. $33rad{s}^{-1}$
• B. $36rad{s}^{-1}$
• C. $20rad{s}^{-1}$
• D. $10rad{s}^{-1}$

Answer 1

Answer: (D)

Instantaneous speed of source is $v=\frac{dx}{dt}=3\omega \sin \omega t$
Difference between maximum and minimum frequencies is $22Hz$.
So, $f_{\max }-f_{\min }=f\left(\frac{v}{v-v_s}\right)-f\left(\frac{v}{v+v_s}\right)=22$
$\Rightarrow f\left\{\frac{v}{v-v_s}-\frac{v}{v+v_s}\right\}=22$ ...(i)
Now here, $f=120Hz,v=330m{s}^{-1},v_s=3\omega$
Substituting these values in Eq (i), we get
$120\left(\frac{330}{330-3\omega }-\frac{330}{330+3\omega }\right)=22$
$\Rightarrow \omega =10s^{-1}$