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 $$340 m s^{–1}$$

Asked by Pragya Singh | 1 year ago |  149

#### 1 Answer

##### Solution :-

Frequency of the whistle = 400 Hz

Speed of wind, vw= 10 m/s

Speed of sound in still air,v= 340 m/s

Effective speed of the sound for an observer standing on the platform

v’ = v + v= 340 + 10 = 350 m/s

There is no relative motion between the source and the observer, therefore the frequency of the sound heard by the observer will be the same.

Therefore, f = 400 Hz

Wavelength of the sound heard by the observer

$$\dfrac{speed \;of\; wave}{frequency}$$

$$\dfrac{350}{400}$$ = 0.875 m

When the air is still and the observer runs towards the yard at a speed of 10 ms–1 then there is a relative motion between the observer the source with respect to the medium.

The medium is at rest. Therefore

v’ = v = 340 m/s

The change in frequency, f ‘

=$$\dfrac{(v + v_0)}{v}\times f$$

$$\dfrac{(340 + 10)}{400}$$ = 411.76 Hz

Wavelength = $$\dfrac{340}{411.76}$$ =  0.826  m

Obviously, the situations in the two cases are entirely different.

Answered by Abhisek | 1 year ago

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