Question

A whistle '$\text{S}$' of frequency $f_0$ revolves in a circle of radius $R$ at a constant speed $v_s$. If the speed of sound in stationary medium is $v$, what is the ratio of frequency detected by a detector ${}^{\prime }{\text{D}}^{\prime }$ at rest in the two cases when source was on topmost point and when the source was at bottom most point?

• A. $\frac{\sqrt{5}v+2v_s}{\sqrt{5}v-2v_s}$
• B. $\frac{\sqrt{5}v-v_s}{\sqrt{5}v+v_s}$
• C. $\left(\frac{v+v_s}{v-v_s}\right)$
• D. $\left(\frac{v-v_s}{v+v_s}\right)$

Answer 1

The correct option is A $\frac{\sqrt{5}v+2v_s}{\sqrt{5}v-2v_s}$

Given that, the detector ${}^{\prime }{\text{D}}^{\prime }$ is at a finite distance from centre of the circular path.

But as the source moves along the circle, the relative position of source changes with the observer, this apparently changes the frequency of sound heard by the observer.



When the source is moving perpendicular to the line joining source and observer at points $C$ and $D$, there is no change in frequency of sound.

But when the source is moving from points $D$ to $C$ along $A$, it is moving towards the observer and when the source is moving from points $C$ to $D$ along $B$, it is moving away from the observer.

From Doppler effect, we know that,

$f^{\prime }=f_0\left(\frac{v\pm v_o}{v\mp v_s\cos \theta }\right)$

Since the observer is stationary,
$f^{\prime }=f_0\left(\frac{v}{v\mp v_s\cos \theta }\right)....\left(1\right)$

From the diagram, we can conclude that that maximum and minumum values of $v+v_s\cos \theta$ occurs at points $B$ and $A$ respectively.

From the figure $\cos \theta =\frac{2R}{\sqrt{5}R}=\frac{2}{\sqrt{5}}$

From $\left(1\right)$, When the source is point $A$
$f_{max}=f_0\left(\frac{v}{v-v_s\times \frac{2}{\sqrt{5}}}\right)....\left(2\right)$

From $\left(1\right)$, When the source is point $B$
$f_{min}=f_0\left(\frac{v}{v+v_s\times \frac{2}{\sqrt{5}}}\right)....\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$, we can conclude that,
$\frac{f_{max}}{f_{min}}=\frac{\sqrt{5}v+2v_s}{\sqrt{5}v-2v_s}$