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Home » A train, standing in a station-yard, blows a whistle of frequency in still air. The wind starts blowing in the direction from the yard to the station with a speed of . 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 ? The speed of sound in still air can be taken as
A train, standing in a station-yard, blows a whistle of frequency 400 \mathrm{~Hz} in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 \mathrm{~m} \mathrm{~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 \mathrm{~m} \mathrm{~s}^{-1} ? The speed of sound in still air can be taken as 340 \mathrm{~m} \mathrm{~s}^{-1}
CBSE | Class 11 | NCERT | Physics | Waves
A train, standing in a station-yard, blows a whistle of frequency 400 \mathrm{~Hz} in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 \mathrm{~m} \mathrm{~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 \mathrm{~m} \mathrm{~s}^{-1} ? The speed of sound in still air can be taken as 340 \mathrm{~m} \mathrm{~s}^{-1}

Frequency of the whistle is given as 400 \mathrm{~Hz}

Speed of wind is given as \mathrm{v}_{\mathrm{w}}=10 \mathrm{~m} / \mathrm{s}

Speed of sound in still air is given as v= 340 \mathrm{~m} / \mathrm{s}

Effective speed of the sound for an observer standing on the platform can be calculated as,

v^{\prime}=v+v_{w}=340+10=350 \mathrm{~m} / \mathrm{s}

Because the source and the observer have no relative motion, the frequency of the sound received by the observer will be the same.

So, f=400 \mathrm{~Hz}

Wavelength of the sound heard by the observer = speed of wave/frequency =350 / 400=0.875 \mathrm{~m}

There is a relative motion between the observer and the source with regard to the medium when the air is motionless and the observer sprints towards the yard at a speed of 10 ms^{-1}.

The medium is at rest. So we have

v^{\prime}=v=340 \mathrm{~m} / \mathrm{s}

The change in frequency, f^{\prime}=\left[\left(v+v_{0}\right) / v\right] \times f

=[(340+10) / 340] \times 400=411.76 \mathrm{~Hz}

Wavelength =340 / 411.76=0.826 \mathrm{~m}

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

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