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 ] Consider both sources emitting sound. When 𝑆2 is at 𝑅 and 𝑆1 approaches the detector with a speed 4 m s −1 , the beat frequency measured by the detector is _______Hz.

 To solve this problem, we need to calculate the beat frequency measured by the detector at point 

P when both sources S1 and S2 are emitting sound. Specifically, S2 is at point R, and S1 is moving towards the detector at P with a speed of 4m/s.

Step 1: Understand the Setup

  • SourcesS1 and S2 emit sound at a frequency of 656Hz.

  • Positions:

    • S1 is at O and moves towards P at 4m/s.

    • S2 is at R, which is diametrically opposite to P on the circular path.

  • Detector: Placed at P.

  • Speed of Soundv=324m/s.

Step 2: Calculate the Frequency Shift Due to S1

Since S1 is moving towards the detector at P, the frequency observed due to S1 will be higher than the emitted frequency due to the Doppler effect.

The Doppler effect formula when the source is moving towards the observer is:

f=(vvvs)f

Where:

  • v=324m/s (speed of sound)

  • vs=4m/s (speed of S1 towards P)

  • f=656Hz (emitted frequency)

Plugging in the values:

f=(3243244)×656=(324320)×656f=1.0125×656664.1Hz

Step 3: Calculate the Frequency Shift Due to S2

Since S2 is at R, which is diametrically opposite to P, the distance between S2 and P is the diameter of the circular path. However, S2 is stationary relative to P at this instant because it is moving along the circular path and its velocity is tangential to the circle. Therefore, there is no radial component of velocity towards or away from P.

Thus, the frequency observed due to S2 remains the same as the emitted frequency:

f=656Hz

Step 4: Calculate the Beat Frequency

The beat frequency is the absolute difference between the frequencies observed from S1 and S2:

Beat Frequency=ff=664.1656=8.1Hz

Final Answer

The beat frequency measured by the detector is 8.1 Hz.

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