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.

4 views

1 Answers

Blowing of wind changes the velocity of sound. As the wind is blowing in the direction of sound, effective speed of sound ve = v + vw

= 340 + 10 = 350 m/s

As the source and listener both are at rest,

frequency is unchanged, i.e., n = 400 Hz.

∴ wavelength, λ = \(\frac{v_e}{n}\) = \(\frac{350}{400}\) = 0.875 m

For still air, vw = 0 and ve = v = 340 m/s

Also, as observer runs towards the stationary train vL = 10 m/s and vs = 0

The frequency now heard by the observer,

n = n0 \((\frac{v+v_L}{v})\) = 400 \((\frac{340+10}{340})\)

= 411.76 Hz

As the source is at rest, wavelength does not change i.e., λ’ = λ = 0.875 m

Comparing the answers, it can be stated that, the situations in two cases are different.

4 views