Question

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

Answer 1

Given, the frequency of the whistle is 400 Hz, the speed of the train is 10 m/s and the speed of sound in air is 340 m/s .

i)

a)

The train is approaching the platform, so the relative velocity of sound is the difference in the velocity of sound in air and the speed of the train.

The formula to calculate the frequency of the whistle is,

f 1 = v v− v 1 ( f )

Here, the speed of sound in air is v, the speed of the train is v 1 and the frequency of the whistle is f.

Substituting the values in the above equation, we get:

f 1 = 340 340−10 ( 400 ) = 340( 400 ) 330 =412 Hz

Thus, the frequency of the whistle for the platform observer when the train approaches the platform is 412 Hz.

b)

The train recedes from the platform, so the relative velocity of sound is the sum of the velocity of sound in air and the speed of the train.

The formula to calculate the frequency of the whistle is,

f 2 = v v+ v 1 ( f )

Here, the speed of sound in air is v, the speed of the train is v 1 and the frequency of the whistle is f.

Substituting the values in the above equation, we get:

f 1 = 340 340+10 ( 400 ) = 340( 400 ) 350 =389 Hz

Thus, the frequency of the whistle for the platform observer when the train recedes from the platform is 389 Hz.

ii)

The speed of sound is the same in both the cases because the speed of sound is not relative to the motion of the bodies. It depends on the medium in which the sound waves travel, which is the same in both the cases, which is air. The speed of sound in air in both the cases is 340 m/s .