Question

A small source of sound moves on a circle as shown in the figure and an observer is standing on O. Let ${n_1},{n_2}\& {n_3}$ be the frequency heard when the source is at A, B and C respectively. Then
(A) ${n_1} > {n_2} > {n_3}$
(B) ${n_2} > {n_3} > {n_1}$
(C) ${n_1} = {n_2} > {n_3}$
(D) ${n_2} > {n_1} > {n_3}$

Answer 1

Here, you are supposed to find the frequency heard by the observer standing at O as the sound source moves on the circle. Doppler effect is applied when there is change in frequency due to relative motion of the source and the observer.

Complete step by step solution:
The change in wave frequency due to relative motion between the wave source and the observer is given by Doppler Effect.
The observed frequency$n$ is given by$n = \left( {\dfrac{{u + \left( { \pm {v_O}} \right)}}{{u + \left( { \pm {v_S}} \right)}}} \right){n_0}$, where ${n_0}$ is the natural frequency emitted by the source,
$u$ is the speed of sound wave in the medium, ${v_O}$ is the speed of observer relative to medium,
${v_S}$ is the speed of the source relative to medium.

To find ${v_O}$ and ${v_S}$, draw an imaginary line connecting the source and the observer. Take the component of velocities of both along this imaginary line, these will be your ${v_O}$ and ${v_S}$. The sign convention will go as follows: If the observer is approaching or moving towards the source, then positive ${v_O}$ will be taken, else if receding or moving away from the source,then negative ${v_O}$ will be taken.

For source, if the source is approaching or moving towards the observer, then negative ${v_S}$ will be taken, else if receding or moving away from the observer, then positive ${v_S}$ will be taken.

Now, consider the above diagram, let the natural frequency be ${n_0}$ and here ${v_O} = 0$ as the observer is standing.

For point A, you can see that the source is going away from the observer, hence frequency observed
by the observer will be ${n_A} = \left( {\dfrac{u}{{u + {v_S}}}} \right){n_0} = {n_1}$.
For point B, you can see that the source is moving towards the observer, hence frequency observed
by the observer will be ${n_B} = \left( {\dfrac{u}{{u - {v_S}}}} \right){n_0} = {n_2}$.
For point C, you can see that the source is neither approaching nor receding the observer as there is no velocity component of the source in the direction of line joining the source and observer, hence ${v_S} = 0$ and frequency observed by the observer will be ${n_C} = \left( {\dfrac{u}{u}} \right){n_0} = {n_0} = {n_2}$.
From the above calculations, one can say that ${n_2} > {n_3} > {n_1}$
Hence, ${n_2} > {n_3} > {n_1}$

Option (B) is correct.

Note: Always keep in mind the sign convention. For an observer approaching the source, the speed of the observer is taken positive, else negative. For source approaching the observer, the speed of source is taken negative, else positive. Also remember that the speed of sound wave is taken as in the
medium.