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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 $10 \mathrm{~m} /$ s. Given that the speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$. Then
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The correct answers are:
the frequency of sound as heard by an observer standing on the platform is $400 \mathrm{~Hz}$
,
the speed of sound for the observer standing on the platform is $350 \mathrm{~m} / \mathrm{s}$
the frequency of sound as heard by an observer standing on the platform is $400 \mathrm{~Hz}$
,
the speed of sound for the observer standing on the platform is $350 \mathrm{~m} / \mathrm{s}$
As given that The frequency of source of sound is, $v_0=400 \mathrm{~Hz}$
Speed of sound in still air, $v_s=340 \mathrm{~m} / \mathrm{s}$
Velocity of wind, $v_w=10 \mathrm{~m} / \mathrm{s}$
(a) As both source and observer are stationary, hence frequency observed will be same as natural frequency $v_0=400 \mathrm{~Hz}$
(b) As the listener standing on platform, so the velocity of sound (with respect to listener),
$$
\begin{aligned}
v &=v_s+v_w \\
&=340+10=350 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
(c) and (d)
There is no relative motion between source and observer, Hence there will be no effect on frequency as heard by listener. Hence, (c), (d) are incorrect.
Speed of sound in still air, $v_s=340 \mathrm{~m} / \mathrm{s}$
Velocity of wind, $v_w=10 \mathrm{~m} / \mathrm{s}$
(a) As both source and observer are stationary, hence frequency observed will be same as natural frequency $v_0=400 \mathrm{~Hz}$
(b) As the listener standing on platform, so the velocity of sound (with respect to listener),
$$
\begin{aligned}
v &=v_s+v_w \\
&=340+10=350 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
(c) and (d)
There is no relative motion between source and observer, Hence there will be no effect on frequency as heard by listener. Hence, (c), (d) are incorrect.
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