Search any question & find its solution
Question:
Answered & Verified by Expert
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 $10 \mathrm{~m} / \mathrm{s}$. 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 $10 \mathrm{~ms}^{-1}$ ? The speed of sound in still air can be taken as $340 \mathrm{~ms}^{-1}$
Solution:
1861 Upvotes
Verified Answer
$v=400 \mathrm{~Hz}, v_{\mathrm{m}}=10 \mathrm{~m} / \mathrm{s}, v=340 \mathrm{~m} / \mathrm{s}$
When the wind is blowing in the direction of sound
$\therefore$ Effective speed of sound $=v+v_{\mathrm{m}}=340+10=350 \mathrm{~m} / \mathrm{s}$
As the source and listener both are at rest, therefore frequency remains unchanged, i.e. $400 \mathrm{~Hz}$.
Wavelength of sound for the stationary observer.
$$
\lambda^{\prime}=\frac{v+v_{\mathrm{m}}}{v}=\frac{350}{400}=0.875 \mathrm{~m}
$$
In case the air is still but the observer is running towards the yard.
When the observer moves towards the stationary engine (source) in still air,
$$
\begin{aligned}
v_{\mathrm{o}} &=-10 \mathrm{~m} / \mathrm{s}, v_{\mathrm{s}}=0 \\
v^{\prime} &=\frac{v-v_o}{v-v_s} \times v^{\prime}=\frac{340+10}{340-0} \times 400 \\
&=\frac{350}{340} \times 400=411.8 \mathrm{~Hz} .
\end{aligned}
$$
As wavelength of sound waves is not affected by motion of the observer, it remains unchanged.
Speed of sound relative to the observer,
$$
v^{\prime \prime}=340+10=350 \mathrm{~ms}^{-1} \text {. }
$$
Thus the situations in both the cases are not exactly identical.
When the wind is blowing in the direction of sound
$\therefore$ Effective speed of sound $=v+v_{\mathrm{m}}=340+10=350 \mathrm{~m} / \mathrm{s}$
As the source and listener both are at rest, therefore frequency remains unchanged, i.e. $400 \mathrm{~Hz}$.
Wavelength of sound for the stationary observer.
$$
\lambda^{\prime}=\frac{v+v_{\mathrm{m}}}{v}=\frac{350}{400}=0.875 \mathrm{~m}
$$
In case the air is still but the observer is running towards the yard.
When the observer moves towards the stationary engine (source) in still air,
$$
\begin{aligned}
v_{\mathrm{o}} &=-10 \mathrm{~m} / \mathrm{s}, v_{\mathrm{s}}=0 \\
v^{\prime} &=\frac{v-v_o}{v-v_s} \times v^{\prime}=\frac{340+10}{340-0} \times 400 \\
&=\frac{350}{340} \times 400=411.8 \mathrm{~Hz} .
\end{aligned}
$$
As wavelength of sound waves is not affected by motion of the observer, it remains unchanged.
Speed of sound relative to the observer,
$$
v^{\prime \prime}=340+10=350 \mathrm{~ms}^{-1} \text {. }
$$
Thus the situations in both the cases are not exactly identical.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.