𝑆1 and 𝑆2 are two identical sound sources of frequency 656 Hz. The source 𝑆1 is located at 𝑂 and 𝑆2 moves anti-clockwise with a uniform speed 4√2 m s −1 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 direction 𝑂𝑃. [Given: The speed of sound in air is 324 m s −1 ] When only 𝑆2 is emitting sound and it is at 𝑄, the frequency of sound measured by the detector in Hz is _________.

 To determine the frequency of sound measured by the detector at point 

$P$ when only $S_2$ is emitting sound and it is at point $Q$, we need to consider the Doppler effect. Here's a step-by-step breakdown of the solution:

1. Understanding the Setup

• Sound Sources: $S_1$ and $S_2$ are identical sound sources with a frequency of $656$ Hz.

• Movement of $S_2$: $S_2$ moves anti-clockwise around $O$ with a uniform speed of $4\sqrt{2}$ m/s on a circular path.

• Points on the Path: Points $P$, $Q$, and $R$ lie on this circular path. $P$ and $R$ are diametrically opposite, and $Q$ is equidistant from $P$ and $R$.

• Detector Placement: A sound detector is placed at point $P$.

• Speed of Sound: Given as $324$ m/s.

2. Visualizing the Scenario

Since $P$ and $R$ are diametrically opposite, the distance between $P$ and $R$ is the diameter of the circular path. Point $Q$ is equidistant from $P$ and $R$, meaning $Q$ is located at a point that forms a right angle with $P$ and $R$ on the circle.

3. Determining the Position of $S_2$ at Point $Q$

When $S_2$ is at point $Q$, it is moving along the circular path. The velocity of $S_2$ at $Q$ can be broken down into components relative to the detector at $P$.

4. Applying the Doppler Effect

The Doppler effect describes the change in frequency of a wave in relation to an observer moving relative to the wave source. The formula for the observed frequency $f^{\mathrm{\prime }}$ when the source is moving towards the observer is:

$$f^{\mathrm{\prime }}=\left(\frac{v+v_o}{v+v_s}\right)f$$

Where:

• $v$ = speed of sound in air ($324$ m/s)

• $v_o$ = speed of the observer relative to the medium (since the detector is stationary, $v_o=0$)

• $v_s$ = speed of the source relative to the medium ($4\sqrt{2}$ m/s)

• $f$ = emitted frequency ($656$ Hz)

However, since $S_2$ is moving on a circular path, we need to consider the component of its velocity that is directly towards or away from the detector at $P$.

5. Calculating the Radial Component of $S_2$'s Velocity

At point $Q$, $S_2$ is moving tangentially to the circle. The velocity vector at $Q$ is perpendicular to the line $OQ$. Since $Q$ is equidistant from $P$ and $R$, and $P$ and $R$ are diametrically opposite, the angle between $OQ$ and $OP$ is $90^{\circ }$.

Therefore, the component of $S_2$'s velocity towards $P$ is:

$$v_{s,\text{radial}}=v_s\cos (90^{\circ })=4\sqrt{2}\times 0=0\text{ m/s}$$

This means that at point $Q$, $S_2$ is not moving towards or away from $P$ radially; it's moving tangentially.

6. Re-evaluating the Doppler Effect

Since the radial component of $S_2$'s velocity relative to $P$ is zero, there is no relative motion along the line connecting $S_2$ and $P$ at the instant $S_2$ is at $Q$. Therefore, the Doppler effect does not cause a change in frequency in this specific position.

7. Conclusion

Given that there is no relative radial motion between $S_2$ and the detector at $P$ when $S_2$ is at $Q$, the frequency detected at $P$ remains the same as the emitted frequency.

Final Answer: The frequency of sound measured by the detector at point $P$ when $S_2$ is at $Q$ is $656$ Hz.