A source of sound is moving along a circular orbit of radius 3 m with an angular velocity of $10{\mathrm{radss}}^{-1}$ sound detector located far away from the source is executing linear simple harmonic motion along the line BD with an amplitude $BC=CD=6\mathrm{m}$ The frequency of oscillation of the detector is $5/\pi \mathrm{Hz}$ The source is at point A when the detector is at point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and minimum frequencies recorded by the detector. The speed of sound in air $=340{\mathrm{ms}}^{-1}$ 

 

 

 

$r=3\mathrm{m},\omega =10{\mathrm{rads}}^{-1}$ The linear speed of the source along the circular orbit is

$u_{\mathrm{s}}=r\omega =3\times 10=30{\mathrm{ms}}^{-1}$

The detector is executing simple harmonic motion of amplitude $a=6$ m and frequency $\mathrm{v}=\frac{5}{\pi }\mathrm{Hz}$ The angular velocity of the detector is ${\omega }^{\mathrm{\prime }}=2\pi \nu =2\pi \times \frac{5}{\pi }$ $=10{\mathrm{rads}}^{-1}$ and the maximum linear velocity of the detector, is  $u_0=a{\omega }^{\mathrm{\prime }}=6\times 10=60{\mathrm{ms}}^{-1}$

Since the source and the detector have the same angular velocity $\left(\text{ i.e., }\omega ={\omega }^{\mathrm{\prime }}\right)$ they have the same time period T. At time $t=T/4$ 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 

$$\begin{gathered} v_{min}=v\left(\frac{v-u_0}{v+u_s}\right)=340\times \frac{(340-60)}{(340+30)} \\ =257.3\mathrm{Hz} \end{gathered}$$

$$\begin{gathered} v_{max}=\left(\frac{v+u_0}{v-u_s}\right)v=340\times \frac{(340+60)}{(340-30)} \\ =438.7\mathrm{Hz} \end{gathered}$$