A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s-1 (b) recedes from the platform with a speed of 10 m s-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s-1.


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Given: vs = 10 m/s, v = 340 m/s, n0 = 400 Hz

Apparent frequency (n), velocity of sound (vs) in each case

Formulae:

i. n = n0 \((\frac{v}{v-v_s})\)

ii. n = n0 \(\frac{v}{v+v_s}\)

Calculation:

a. As the train approaches the platform, using formula (i),

n = 400 \((\frac{340}{340-10})\) = 421.12 Hz

b. As the train recedes from the platform, using formula (ii),

n = 400 \((\frac{340}{340+10})\) = 388.57 Hz

ii. The relative motion of source and observer results in the apparent change in the frequency but has no effect on the speed of sound. 

Hence, the speed of sound remains unchanged in both the cases.

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