A train approaching a railway platform with a speed of $20\text{ m}{{\text{s}}^{\text{-1}}}$ , starts blowing the whistle. Speed of sound in air is $340\text{ m}{{\text{s}}^{\text{-1}}}$ . If the frequency of the emitted sound from the whistle is $640\text{ Hz}$ , the frequency of sound to a person standing on the platform will appear to be :
A. $600\text{ Hz}$
B. $640\text{ Hz}$
C. $680\text{ Hz}$
D. $720\text{ Hz}$
Answer
638.4k+ views
Hint: As the train is approaching, the person (detector) standing on the platform, the frequency of sound detected by the person appears to be more than the original frequency. This frequency shift can be calculated using the Doppler effect equation.
Formula used:
According to Doppler Effect equation, the frequency perceived by a detector is given by:
$v'=v\left( \dfrac{u-{{u}_{d}}}{u-{{u}_{s}}} \right)$
Where speed of sound $u$ ,velocity of source x, is the velocity of detector and original frequency .
Complete answer:
Lets calculate and find the correct answer.
Given data:
Original frequency of sound $v=640\text{ Hz}$ ,
Speed of sound, $u=340\text{ m}{{\text{s}}^{\text{-1}}}$ ,and
Velocity of source (train), ${{u}_{s}}=20\text{ m}{{\text{s}}^{\text{-1}}}$ .
The apparent frequency of sound, observed by a stationary listener (\[{{u}_{d}}=0\]), when a source of sound is approaching to listener, is given by
$\begin{align}
& v'=v\left( \dfrac{u-{{u}_{d}}}{u-{{u}_{s}}} \right) \\
& \Rightarrow 640\text{ Hz}\left( \dfrac{340\text{ m}{{\text{s}}^{-1}}-0\text{ m}{{\text{s}}^{-1}}}{340\text{ m}{{\text{s}}^{-1}}-20\text{ m}{{\text{s}}^{-1}}} \right) \\
& \Rightarrow 640\text{ Hz}\left( \dfrac{340\text{ m}{{\text{s}}^{-1}}}{320\text{ m}{{\text{s}}^{-1}}} \right) \\
& \Rightarrow 640\text{ Hz}\times \dfrac{34}{32} \\
& \Rightarrow 20\text{ Hz}\times 34 \\
& \Rightarrow 680\text{ Hz} \\
\end{align}$
Therefore, the frequency of sound will be $680\text{ Hz}$ to a person standing on the platform.
Hence, option C is the correct answer.
Note:
According to Doppler Effect, for a source moving toward the detector and for a detector moving toward the source the detected frequency increases. Similarly, if the source moves away from the detector or if the detector moves away from the source, then detected frequency decreases.
Formula used:
According to Doppler Effect equation, the frequency perceived by a detector is given by:
$v'=v\left( \dfrac{u-{{u}_{d}}}{u-{{u}_{s}}} \right)$
Where speed of sound $u$ ,velocity of source x, is the velocity of detector and original frequency .
Complete answer:
Lets calculate and find the correct answer.
Given data:
Original frequency of sound $v=640\text{ Hz}$ ,
Speed of sound, $u=340\text{ m}{{\text{s}}^{\text{-1}}}$ ,and
Velocity of source (train), ${{u}_{s}}=20\text{ m}{{\text{s}}^{\text{-1}}}$ .
The apparent frequency of sound, observed by a stationary listener (\[{{u}_{d}}=0\]), when a source of sound is approaching to listener, is given by
$\begin{align}
& v'=v\left( \dfrac{u-{{u}_{d}}}{u-{{u}_{s}}} \right) \\
& \Rightarrow 640\text{ Hz}\left( \dfrac{340\text{ m}{{\text{s}}^{-1}}-0\text{ m}{{\text{s}}^{-1}}}{340\text{ m}{{\text{s}}^{-1}}-20\text{ m}{{\text{s}}^{-1}}} \right) \\
& \Rightarrow 640\text{ Hz}\left( \dfrac{340\text{ m}{{\text{s}}^{-1}}}{320\text{ m}{{\text{s}}^{-1}}} \right) \\
& \Rightarrow 640\text{ Hz}\times \dfrac{34}{32} \\
& \Rightarrow 20\text{ Hz}\times 34 \\
& \Rightarrow 680\text{ Hz} \\
\end{align}$
Therefore, the frequency of sound will be $680\text{ Hz}$ to a person standing on the platform.
Hence, option C is the correct answer.
Note:
According to Doppler Effect, for a source moving toward the detector and for a detector moving toward the source the detected frequency increases. Similarly, if the source moves away from the detector or if the detector moves away from the source, then detected frequency decreases.
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