
Two vehicles, each moving with speed $u$ on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity $w$. One of these vehicles blows a whistle of frequency $f_{1}$. An observer in the other vehicle hears the frequency of the whistle to be $f_{2}$. The speed of the sound in still air is $V$. The correct statements are: (This question has multiple correct options)
A. if the wind blows from the observer to the source, $f_{2}>f_{1}$
B. if the wind blows from the source to the observer, $f_{2}>f_{1}$
C. if the wind blows from the observer to the source, $f_{1} < f_{2}$
D. if the wind blows from the source to the observer, $f_{1} < f_{2}$
Answer
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Hint: Doppler Effect is the change in frequency when the position of the observer changes with respect to the source. However, here we are assuming that the velocity of the wave is constant during the interaction. Also, the wave is either approaching the observer or moving away from the observer, only.
Formula:
$f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$, where, $f$ is the apparent frequency of the sound, $f_{0}$ is the actual or real frequency of the sound, $c$ is the speed of the sound wave and $v_{s},v_{o}$ is the speed of the moving source and observer respectively.
Complete answer:
Given that the velocity of the cars A and B are $u$, velocity of wind is $w$ and the speed of the sound is $V$. Let the frequency of the whistle from A be $f_{1}$ and the whistle from B be $f_{2}$.
We also know that the Doppler effect is given as $f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$.
We if the wind blows from the source to observer, as $f_{2}=f_{1}\left(\dfrac{V+w+u}{V+w-u}\right)$ clearly, $f_{2}>f_{1}$. Thus option B is correct.
Also, if the wind blows from the observer to source $f_{2}=f_{1}\left(\dfrac{V-w+u}{V-w-u}\right)$ clearly, $f_{2}>f_{1}$. Thus option A is correct.
Hence the correct answer is A. if the wind blows from the observer to the source, $f_{2}>f_{1}$ and, B. if the wind blows from the source to the observer, $f_{2}>f_{1}$
Note:
Doppler law is used only when the speed of the source and the observer are both less than the speed of the sound wave in the medium. Then clearly the relative frequency of a moving source or observer or both, is lesser than the frequency of the sound wave. This law can be extended to any waves.
Formula:
$f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$, where, $f$ is the apparent frequency of the sound, $f_{0}$ is the actual or real frequency of the sound, $c$ is the speed of the sound wave and $v_{s},v_{o}$ is the speed of the moving source and observer respectively.
Complete answer:
Given that the velocity of the cars A and B are $u$, velocity of wind is $w$ and the speed of the sound is $V$. Let the frequency of the whistle from A be $f_{1}$ and the whistle from B be $f_{2}$.
We also know that the Doppler effect is given as $f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$.
We if the wind blows from the source to observer, as $f_{2}=f_{1}\left(\dfrac{V+w+u}{V+w-u}\right)$ clearly, $f_{2}>f_{1}$. Thus option B is correct.
Also, if the wind blows from the observer to source $f_{2}=f_{1}\left(\dfrac{V-w+u}{V-w-u}\right)$ clearly, $f_{2}>f_{1}$. Thus option A is correct.
Hence the correct answer is A. if the wind blows from the observer to the source, $f_{2}>f_{1}$ and, B. if the wind blows from the source to the observer, $f_{2}>f_{1}$
Note:
Doppler law is used only when the speed of the source and the observer are both less than the speed of the sound wave in the medium. Then clearly the relative frequency of a moving source or observer or both, is lesser than the frequency of the sound wave. This law can be extended to any waves.
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