
A train at rest blows a whistle of frequency $600 Hz$ in still air. What is the frequency of whistle for a platform observer when the train (a) approaches the platform with a speed of $15 ms^{-1}$ (b) recedes from the platform with a speed of $15 ms^{-1}$? What is the speed of sound in each case? Speed of sound in the air is $345 ms^{-1}$.
Answer
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Hint: Doppler shift causes sound waves transmitted by stationary and migrating sources in a stable air mass. Each disturbance diffuses out spherically from the position at which the sound is released. If the source is motionless, then all of the spheres expressing the air compressions in the sound wave are focused on the same point, and the stationary spectators on either side listen to the same wavelength and frequency as released by the source.
Complete step by step solution:
Given: $f_{o} = 600 Hz$,
$V_{s} = 15 ms^{-1}$
Speed of sound in air is $V = 345 ms^{-1}$.
Here, train is source
a) Train approaching the platform, then apparent frequency will be:
$f = f_{o} \left( \dfrac{V-0}{V-V_{s}} \right)$
$\implies f = 600 \left( \dfrac{345-0}{345-15} \right)$
$\implies f = 627.27 Hz$
This apparent frequency will increase as train approaches the platform.
b) Train recedes from the platform, then apparent frequency will be:
$f = f_{o} \left( \dfrac{V-0}{V+V_{s}} \right)$
$\implies f = 600 \left( \dfrac{345-0}{345+15} \right)$
$\implies f = 575 Hz$
This apparent frequency will decrease as the train recedes from the platform.
The apparent variation in the frequency of sounds is produced by the relative movements of the source and the observer. These dependent motions produce no influence on the speed of sound. Therefore, the sound speed in the air in both cases keeps the same.
Note: If the source is traveling, the situation is different. Each air compression moves out in a sphere from the position at which it was released but the point of emission movements. This moving emission point creates the air compressions to be closer collectively on one side and farther separate on the other. Thus, the wavelength is lower in the direction the source progresses and longer in the reverse direction. Finally, if the observers travel, the frequency at which they take the compression modifications. The observer traveling toward the source gets them at a more significant frequency, and the person passing away from the source takes them at a lower frequency.
Complete step by step solution:
Given: $f_{o} = 600 Hz$,
$V_{s} = 15 ms^{-1}$
Speed of sound in air is $V = 345 ms^{-1}$.
Here, train is source
a) Train approaching the platform, then apparent frequency will be:
$f = f_{o} \left( \dfrac{V-0}{V-V_{s}} \right)$
$\implies f = 600 \left( \dfrac{345-0}{345-15} \right)$
$\implies f = 627.27 Hz$
This apparent frequency will increase as train approaches the platform.
b) Train recedes from the platform, then apparent frequency will be:
$f = f_{o} \left( \dfrac{V-0}{V+V_{s}} \right)$
$\implies f = 600 \left( \dfrac{345-0}{345+15} \right)$
$\implies f = 575 Hz$
This apparent frequency will decrease as the train recedes from the platform.
The apparent variation in the frequency of sounds is produced by the relative movements of the source and the observer. These dependent motions produce no influence on the speed of sound. Therefore, the sound speed in the air in both cases keeps the same.
Note: If the source is traveling, the situation is different. Each air compression moves out in a sphere from the position at which it was released but the point of emission movements. This moving emission point creates the air compressions to be closer collectively on one side and farther separate on the other. Thus, the wavelength is lower in the direction the source progresses and longer in the reverse direction. Finally, if the observers travel, the frequency at which they take the compression modifications. The observer traveling toward the source gets them at a more significant frequency, and the person passing away from the source takes them at a lower frequency.
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