The difference between the apparent frequency of a source of sound as perceived by the observer during its approach and recession is 2% of the frequency of the source. If the speed of the sound in air is $300m{{s}^{-1}}$, the velocity of the source is:
A. $1.5m{{s}^{-1}}$
B. $12m{{s}^{-1}}$
C. $6m{{s}^{-1}}$
D. $3m{{s}^{-1}}$
Answer
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Hint: The difference in the apparent frequency and emitting frequency from the source is due to relative motion between source and observer. In this question the difference in frequency is noticed while source is approaching and receding. Therefore we apply the formula for change in frequency due to approach and due to recession and calculate the velocity of source.
Formula Used:
According to Doppler:
Apparent frequency observed by stationary observer while source is approaching towards observer is given by:
${{n}_{1}}=n\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Apparent frequency observed by stationary observer while source is approaching towards observer is given by:
${{n}_{2}}=n\left( \dfrac{v}{v+{{v}_{s}}} \right)$
Where:
$n$ is emitting frequency or original frequency
$v$ is the velocity of sound
${{v}_{s}}$ is the velocity of source
Complete step by step answer:
Change in frequency observed when the source is approaching and receding the stationary observer is known as beat.
Therefore,
$\Delta n={{n}_{1}}-{{n}_{2}}$
Apparent frequency due to approaching ${{n}_{1}}=n\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Apparent frequency due to recession ${{n}_{2}}=n\left( \dfrac{v}{v+{{v}_{s}}} \right)$
$\begin{align}
& \Delta n=n\left( \dfrac{v}{v-{{v}_{s}}} \right)-n\left( \dfrac{v}{v+{{v}_{s}}} \right) \\
& \Delta n=nv\left[ \dfrac{v+{{v}_{s}}-v+{{v}_{s}}}{{{v}^{2}}-{{v}_{s}}^{2}} \right] \\
& \Delta n=\dfrac{2nv{{v}_{s}}}{{{v}^{2}}\left( 1-\dfrac{{{v}_{s}}^{2}}{{{v}^{2}}} \right)} \\
& v>>{{v}_{s}} \\
& \therefore \dfrac{{{v}_{s}}^{2}}{{{v}^{2}}}<<1 \\
& \Delta n=\dfrac{2n{{v}_{s}}}{v} \\
\end{align}$
It is given that $\Delta n=\dfrac{2}{100}n$
Therefore:
$\begin{align}
& \dfrac{2}{100}n=\dfrac{2n{{v}_{s}}}{v} \\
& {{v}_{s}}=\dfrac{v}{100} \\
& {{v}_{s}}=\dfrac{300}{100}=3m{{s}^{-1}} \\
\end{align}$
Hence, the velocity of source is $3m{{s}^{-1}}$
Therefore, option D. is the correct answer
Note: Condition when Doppler Effect is not observed:
When both observer and source are at rest.
Both moving in the same direction with the same velocity.
Both moving parallel to each other with the same velocity.
Both moving perpendicular to each other.
When the medium alone is moving.
There is a limitation when Doppler Effect is not applicable is velocity of source, observer and medium must be less than the velocity of sound.
Formula Used:
According to Doppler:
Apparent frequency observed by stationary observer while source is approaching towards observer is given by:
${{n}_{1}}=n\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Apparent frequency observed by stationary observer while source is approaching towards observer is given by:
${{n}_{2}}=n\left( \dfrac{v}{v+{{v}_{s}}} \right)$
Where:
$n$ is emitting frequency or original frequency
$v$ is the velocity of sound
${{v}_{s}}$ is the velocity of source
Complete step by step answer:
Change in frequency observed when the source is approaching and receding the stationary observer is known as beat.
Therefore,
$\Delta n={{n}_{1}}-{{n}_{2}}$
Apparent frequency due to approaching ${{n}_{1}}=n\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Apparent frequency due to recession ${{n}_{2}}=n\left( \dfrac{v}{v+{{v}_{s}}} \right)$
$\begin{align}
& \Delta n=n\left( \dfrac{v}{v-{{v}_{s}}} \right)-n\left( \dfrac{v}{v+{{v}_{s}}} \right) \\
& \Delta n=nv\left[ \dfrac{v+{{v}_{s}}-v+{{v}_{s}}}{{{v}^{2}}-{{v}_{s}}^{2}} \right] \\
& \Delta n=\dfrac{2nv{{v}_{s}}}{{{v}^{2}}\left( 1-\dfrac{{{v}_{s}}^{2}}{{{v}^{2}}} \right)} \\
& v>>{{v}_{s}} \\
& \therefore \dfrac{{{v}_{s}}^{2}}{{{v}^{2}}}<<1 \\
& \Delta n=\dfrac{2n{{v}_{s}}}{v} \\
\end{align}$
It is given that $\Delta n=\dfrac{2}{100}n$
Therefore:
$\begin{align}
& \dfrac{2}{100}n=\dfrac{2n{{v}_{s}}}{v} \\
& {{v}_{s}}=\dfrac{v}{100} \\
& {{v}_{s}}=\dfrac{300}{100}=3m{{s}^{-1}} \\
\end{align}$
Hence, the velocity of source is $3m{{s}^{-1}}$
Therefore, option D. is the correct answer
Note: Condition when Doppler Effect is not observed:
When both observer and source are at rest.
Both moving in the same direction with the same velocity.
Both moving parallel to each other with the same velocity.
Both moving perpendicular to each other.
When the medium alone is moving.
There is a limitation when Doppler Effect is not applicable is velocity of source, observer and medium must be less than the velocity of sound.
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