Question

A train, standing in a station-yard, blows a whistle of frequency 400 Hz in st air. The wind starts blowing in the direction from the yard to the station with speed of 10 m s⁻¹. Is the situation exactly identical the case when the air is still and the observer runs towards the yard at a speed 10 m s⁻¹? The speed of sound in still air can be taken as 340 m s⁻¹.

Answer 1

$\textbf{For the stationary observer}$:
Frequency of the sound produced by the whistle, f = 400 Hz
Speed of sound, $v_s$ = 340 m/s
Velocity of the wind, $v_w$ = 10 m/s


As you can see that , there is no relative motion between the source and the observer, so, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400 Hz. The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units,
So, effective speed of the sound, $v_e$ = 340 + 10 = 350 m/s
The wavelength (λ) of the sound heard by the observer is given by
λ = $\frac{v_e}{f}$
⇒ λ = (350 m/s)/(400 Hz)
⇒ λ = 0.875 m

$\textbf{For the running observer}$:

Velocity of the observer, $v_0$ = 10 m/s
The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (f') which is given by
f’ = $\frac{v+v_0}{v}\times f$ [ from Doppler effect ]
⇒ f’ = [(340+10)/340] × 400
⇒ f’ = 411.76 Hz

Since the air is still, the effective speed of sound will be the same, i.e., 340 m/s. The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m. Hence, the given two situations are not exactly identical.