Two sources P and Q produce notes of frequency 660 Hz each. A listener moves from P to Q with a speed of 1 ms–1. If the speed of sound is 330 m/s, then the number of beats heard by the listener per second will be [NEET Kar. 2013]a)zerob)4c)8d)2Correct answer is option 'B'. Can you explain this answer?

Question

Two sources P and Q produce notes of frequency 660 Hz each. A listener moves from P to Q with a speed of 1 ms&ndash;1. If the speed of sound is 330 m/s, then the number of beats heard by the listener per second will be [NEET Kar. 2013]a)zerob)4c)8d)2Correct answer is option 'B'. Can you explain this answer?

Answer

-1. What is the beat frequency heard by the listener?

The beat frequency is given by the difference in frequency between the two sources, which is 0 Hz in this case since they both produce notes of 660 Hz. However, because the listener is moving towards one source and away from the other, the frequency of the sound waves reaching the listener will be Doppler shifted.

The Doppler shift in frequency is given by:

f' = f(v + vl) / (v + vs)

where f is the frequency of the source, v is the speed of sound, vl is the speed of the listener towards the source, and vs is the speed of the source away from the listener.

At P, the listener is stationary and hears a note of frequency 660 Hz. At Q, the listener is moving away from source Q and towards source P with a speed of 1 ms-1. Therefore, the frequency of the note heard at Q from source Q is:

fQ' = f(v + vl) / (v - vs)
= 660(343 + 1) / (343 - 1)
= 662.9 Hz

And the frequency of the note heard at Q from source P is:

fP' = f(v - vl) / (v + vs)
= 660(343 - 1) / (343 + 1)
= 657.1 Hz

The beat frequency heard by the listener is therefore the difference between these two frequencies:

fbeat = fQ' - fP'
= 662.9 - 657.1
= 5.8 Hz

So the listener hears a beat frequency of 5.8 Hz as they move from source P to Q.

Answer

Answer

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