S1 and S are two identical sound sources of frequency 656 Hz. The source S1 is located at O, and S₂ moves anticlockwise with a uniform speed 4 √ 2 m/s on a circular path around O, as shown in the figure. There are three points P, Q, and R on this path such that P and R are diametrically opposite, while Q is equidistant from them. A sound detector is placed at point P. The source S₁ can move along the direction OP.

When only S₂ is emitting sound and it is at Q, the frequency of sound measured by the detector in Hz is:

The correct answer is 648.

1. Understanding the Physical Setup:

From the given figure:

• The center of the circular path is located at point O.
• A detector is stationary at point P.
• P and R are diametrically opposite points, and Q is equidistant from P and R. Thus, the line OQ is perpendicular to the diameter PR.
• Source S₂ moves anticlockwise along the circle with a uniform speed $v_s=4\sqrt{2}\text{m/s}$.
• The speed of sound in air is standard, $v=324\text{m/s}$ (or $330\text{m/s}$ depending on standard question data; using standard JEE problem data, speed of sound $v=324\text{m/s}$).

2. Motion of Source S₂ when at Q:

When source S₂ is at point Q, it is moving anticlockwise along the circular arc.
The velocity vector of S₂ at point Q is directed tangentially upwards (parallel to PR, pointing towards R).

3. Determining the Line of Sight and Velocity Component:

The line joining the source S₂ (at Q) to the detector (at P) makes an angle with the tangent at Q.
Since OP and OQ are perpendicular radii of equal length (say R), triangle OPQ is an isosceles right-angled triangle.
Therefore, $\angle OPQ=\angle OQP=45°$.

The tangential velocity of S₂ at Q is perpendicular to radius OQ. Thus, the angle between the velocity vector of S₂ and the line of sight QP (pointing towards the detector at P) is $45°$ away from detector P.

The component of source velocity away from the detector along the line QP is:

$v_{\text{app}}=v_{s}cos(45°)=\left(4\sqrt{2}\right)\times \frac{1}{\sqrt{2}}=4\text{m/s}$

4. Apparent Frequency using Doppler's Effect:

Since the source S₂ is moving away from the stationary detector at point P along the line joining them with speed component $4\text{m/s}$, the apparent frequency $f\prime$ measured by the detector is given by Doppler's formula:

$f\prime =f\left(\frac{v}{v+v_{\text{app}}}\right)$

Substituting the given values $f=656\text{Hz}$, $v=324\text{m/s}$, and $v_{\text{app}}=4\text{m/s}$:

$f\prime =656\times \left(\frac{324}{324+4}\right)$

$f\prime =656\times \left(\frac{324}{328}\right)$

$f\prime =2\times 324=648\text{Hz}$

Thus, the frequency of sound measured by the detector when S₂ is at Q is 648 Hz.