Question

A source of sound is moving along a circular path of radius $3m$ with angular velocity of $10rad/s$. A sound detector located far away from the source is executing linear $S.H.M.$ along the line $BD$ shown in the figure with an amplitude $BC=DC=6m$. The frequency of oscillation of the detector is $\frac{5}{\pi }Hz$. The source is at point $A$ when the detector is at point $B$. If the source emits a continuous sound wave frequency $340Hz$. Then,

• A. Minimum frequency recorded by detector will be $275Hz$.
• B. Minimum frequency recorded by detector will be $257Hz$.
• C. Maximum frequency recorded by detector will be $485.7Hz$.
• D. Maximum frequency recorded by detector will be $458.7Hz$.

Answer 1

Answer: (B), (C)

The correct options are
B Minimum frequency recorded by detector will be $257Hz$.
C Maximum frequency recorded by detector will be $485.7Hz$.

The speed of sound in air $=340m/s.$
$r=3m,\omega =10rad/s$. The linear speed of the source along the circular orbit is
$v_s=r\omega =3\times 10m/s$

The detector is executing $SHM$ of amplitude $a=6m$ and frequency $n=\frac{5}{\pi }Hz$. The angular velocity of the detector is
${\omega }^{{}^{\prime }}=2\pi n=2\pi \times \frac{5}{\pi }=10\text{rad/s}$
and the linear velocity for the detector is
$u_0=a{\omega }^{{}^{\prime }}=6\times 10=60m/s$
Since the source and the detector have the same angular velocity (i.e., $\omega ={\omega }^{{}^{\prime }}$), they have the same time period $T$. At time $t=\frac{T}{4}$, the source will be at point $P$ and the detector will be at point $C$.

Since they are receding from each other at the maximum relative velocity, the recorded frequency will be minimum which is given by
${\nu }_{\text{min}}=\nu \left(\frac{v-u_0}{v+u_s}\right)=340\times \left(\frac{340-60}{340+30}\right)$
${\nu }_{min}=257Hz$
At a time $t=\frac{3T}{4}$, the source will be at point $Q$ and the detector will be at point $C$ approaching the source with the maximum relative velocity. Hence the recorded frequency will be maximum which is given by
${\nu }_{\text{max}}=\nu \left(\frac{v+u_0}{v-u_0}\right)=340\times \left(\frac{340+60}{340-30}\right)$
${\nu }_{max}=485.7Hz$.