Question

A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s¯¹. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s¯¹? 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 in still air, the speed of wind in the direction from the yard to the station is 10 m/s , the speed of the observer is 10 m/s and the speed of sound in still air is 340 m/s .

There is no relative motion between the observer and the source, so the frequency of sound for the observer is 400 Hz.

Let v a and v 0 be the speed of sound in still air and speed of wind respectively.

The effective speed of sound is,

v= v a + v 0

Substituting the values in the above equation, we get:

v=340 m/s +10 m/s =350 m/s

Thus, the speed of the sound wave for an observer is 350 m/s .

The wavelength of the sound wave is given by,

λ= v f

Substituting the values in the above equation, we get:

λ= 350 m/s 400 Hz =0.875 m

Hence, the frequency, wavelength and speed of the sound for an observer standing on the station’s platform is 400 Hz, 0.875 m and 350 m/s respectively.

The speed of the sound wave when the observer is running is given by the equation,

f'=( v+ v 0 v )f

Substituting the values in the above equation, we get:

v'=( 340 m/s +10 m/s 340 m/s )×400 Hz =412 Hz

Since the air is not moving, so the effective speed of the sound wave remains the same, that is 340 m/s . Also, the wavelength of the sound does not depend on the observer, so the wavelength of the sound also remains the same that is 0.875 m. But, from the above calculation, it is observed that the frequency of the sound wave when the observer is running is 412 Hz. So, the two conditions are not exactly identical.