Question 25

A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m s$$^{-1}$$. The frequency observed by the observer will be (given speed of sound = 330 m s$$^{-1}$$) _____ Hz.

Correct Answer: 400

Solution :

For a source approaching a stationary observer, apparent frequency is given by Doppler’s formula :

$$f' = \frac{v}{v-v_s}f$$

Given :

$$f = 320\text{ Hz}$$

$$v = 330\text{ m s}^{-1}$$

$$v_s = 66\text{ m s}^{-1}$$

Substituting :

$$f' = \frac{330}{330-66}\times 320$$

$$= \frac{330}{264}\times 320$$

$$= 1.25 \times 320$$

$$= 400\text{ Hz}$$

Final Answer :

$$400$$

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