$A$ 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 $340\; m s^{-1}$.

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(D) For the stationary observer: Frequency $= 400\; Hz$,Wavelength $= 0.85\; m$,Speed of sound $= 350\; m s^{-1}$.
For the running observer: Frequency $\approx 411.76\; Hz$,Wavelength $= 0.85\; m$,Speed of sound $= 340\; m s^{-1}$.
$1$. For the stationary observer:
Since the source and observer are stationary relative to each other,the frequency remains $400\; Hz$. The wind blows towards the observer,so the effective speed of sound is $v_e = v + w = 340 + 10 = 350\; m s^{-1}$. The wavelength is $\lambda = v_e / f = 350 / 400 = 0.875\; m$.
$2$. For the running observer:
Here,the air is still,so the speed of sound is $340\; m s^{-1}$. The wavelength remains $\lambda = v / f = 340 / 400 = 0.85\; m$. The observer moves towards the source at $10\; m s^{-1}$,so the frequency heard is $f' = f(v + v_o) / v = 400(340 + 10) / 340 = 411.76\; Hz$.
Conclusion: The two situations are not identical because the wavelength and the effective speed of sound differ in both cases.

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