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⁻¹. If the speed of sound is 330 m/s, then the number of beats heard by the listener per second will be(a) zero(b) 4(c) 8(d) 2

Answer 1

the correct answer is b that is 4 please mark as brainlest

Answer 2

Answer:

B) 4 beats

Explanation:

Frequency of each source = v = 660Hz

Speed of listener, v1 = 1 ms⁻¹

Speed of sound, v =330 ms⁻¹

Apparent frequency due to P = v' = (v-vl)/v

Apparent frequency due to Q = v'' = (v+vl)/v

Number of beats heard by the listener per second is -

v''-v' = (v+vl)/v - (v-vl)/v

= 2vvl/v

= 2 × 660 × 1/330

= 4

Thus, the number of beats heard by the listener per second will be 4 beats