An observer travelling with a constant velocity of $20 m s^{- 1}$ passes close to a stationary source of sound and notices that there is a change of frequency of $50 H_{Z}$ as he passes the source. What is the frequency of the source? Speed of sound in air = 340 m/s.

450 Hz

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425 Hz

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400 Hz

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500 Hz

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Solution

The correct option is B 425 Hz
Speed of sound air = $340 m s^{- 1}$
When the observer approaches the source:
$f^{'} = \frac{f (v + v_{0})}{v}$
$f^{'} = \frac{f (340 + 20)}{340}$
$f^{'} = \frac{18 f}{17} \to (1)$

When the observer moves away from the source:
$f "= \frac{f (v - v_{0})}{v}$
$f^{''} = \frac{f (340 - 20)}{340}$
$f "= \frac{16 f}{17} \to (2)$

Change in frequency,
$f^{'} - f "= 50$
$∴ \frac{18 f}{17} - \frac{16 f}{17} = 50$
$f = 425 H z$

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An observer travelling with a constant velocity of 20 ms−1 passes close to a stationary source of sound and notices that there is a change of frequency of 50 HZ as he passes the source. What is the frequency of the source? Speed of sound in air = 340 m/s.

An observer travelling with a constant velocity of $20 m s^{- 1}$ passes close to a stationary source of sound and notices that there is a change of frequency of $50 H_{Z}$ as he passes the source. What is the frequency of the source? Speed of sound in air = 340 m/s.