A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to

(a) 666 Hz

(b) 753 Hz

(c) 500 Hz

(d) 333 Hz

Solution

Frequency of the sound produced by the open flute

f=2(v/2l) = (2 x 330)/(2 x 0.5) = 660 Hz

Velocity of observer, v0 = 10 x (5/18) = (25/9) m/s

As the source is moving towards the observer, according to the Doppler effect.

Frequency detected by observer

f’ = {(v + v0)/v}f = {((25/9) + 330)/330}660

= 2 ((25/9) + 330)

f’ = 665.55 ≈ 666 Hz

Answer: (a) 666 Hz

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