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 (a) 4 (b) 8 (c) 2 (d) zero

Answer: (a). Moving towards Q: fQ = f v+u/v; moving away from P: fP = f v-u/v. Beats = fQ - fP = 2fu/v = (2 × 660 × 1)/(330) So the answer is (a) 4 beats per second.

Two sources P and Q produce notes of frequency 660 Hz. each. A… | NEET 2013

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Doppler Effect

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

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Answer: (a)

Moving towards Q: $f_Q = f\,\dfrac{v+u}{v}$ ; moving away from P: $f_P = f\,\dfrac{v-u}{v}$ .
Beats $= f_Q - f_P = \dfrac{2fu}{v} = \dfrac{2 \times 660 \times 1}{330}$
So the answer is (a) 4 beats per second.

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