
A train is approaching towards a platform with a speed of 10$m{{s}^{-1}}$ while blowing a whistle of frequency 340 Hz. What is the frequency of whistles heard by a stationary observer on the platform?
Given speed of sound = 340$m{{s}^{-1}}$.
A. 330 Hz
B. 350 Hz
C. 340 Hz
D. 360 Hz
Answer
586.5k+ views
Hint: When the source of the sound waves is in motion, the frequency heard by a stationary observer is not equal to the actual frequency of sound emitted by the source. This frequency is called apparent frequency (f’) given by $f'=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$, where f is the actual frequency, v is the speed of the sound in that medium and ${{v}_{s}}$ is the speed of the source.
Formula used:
$f'=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Complete step-by-step answer:
Suppose there is a source of sound at a distance from a stationary observer. Even the source of sound is at rest. Let the frequency of sound that the source is emitting be f. and let the speed of the sound be v.
We know that the observer will hear the sound of frequency equal to the frequency of the sound waves emitted by the source.
However, if we set the source in a uniform motion (constant speed), something different will happen. When the sound waves are emitted by a moving source, the frequency that the stationary observer will hear is not equal to the frequency f.
The new frequency is called apparent frequency. Let the apparent be f’.
Apparent frequency f’ is given as $f'=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$.
Where, ${{v}_{s}}$ is the speed of the source.
The direction in which the sound wave travels is taken as a positive direction.
Let us calculate the apparent frequency in the given case.
Here, f = 340 Hz, v = 340$m{{s}^{-1}}$ and ${{v}_{s}}$ = 10 $m{{s}^{-1}}$.
Therefore, $f'=340\left( \dfrac{340}{340-10} \right)=340\left( \dfrac{340}{330} \right)=350.3Hz\approx 350Hz$
Hence, the correct option is B.
Note: The apparent frequency heard by a stationary observer can be more than or less than the actual frequency of the sound waves emitted by the source. It depends on the direction of the speed of the source.
If the source is moving towards the observer, the frequency heard by the observer will be more than the actual frequency.
If the source is moving away from the observer, the frequency heard by the observer will be less than the actual frequency.
Formula used:
$f'=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Complete step-by-step answer:
Suppose there is a source of sound at a distance from a stationary observer. Even the source of sound is at rest. Let the frequency of sound that the source is emitting be f. and let the speed of the sound be v.
We know that the observer will hear the sound of frequency equal to the frequency of the sound waves emitted by the source.
However, if we set the source in a uniform motion (constant speed), something different will happen. When the sound waves are emitted by a moving source, the frequency that the stationary observer will hear is not equal to the frequency f.
The new frequency is called apparent frequency. Let the apparent be f’.
Apparent frequency f’ is given as $f'=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$.
Where, ${{v}_{s}}$ is the speed of the source.
The direction in which the sound wave travels is taken as a positive direction.
Let us calculate the apparent frequency in the given case.
Here, f = 340 Hz, v = 340$m{{s}^{-1}}$ and ${{v}_{s}}$ = 10 $m{{s}^{-1}}$.
Therefore, $f'=340\left( \dfrac{340}{340-10} \right)=340\left( \dfrac{340}{330} \right)=350.3Hz\approx 350Hz$
Hence, the correct option is B.
Note: The apparent frequency heard by a stationary observer can be more than or less than the actual frequency of the sound waves emitted by the source. It depends on the direction of the speed of the source.
If the source is moving towards the observer, the frequency heard by the observer will be more than the actual frequency.
If the source is moving away from the observer, the frequency heard by the observer will be less than the actual frequency.
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