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A train whistling at constant frequency $n$ is moving towards a station at a constant speed $v$. The train goes past a stationary observer on the station. The frequency $n$ of the sound as heard by the observer is plotted as a function of time $t$. Identify the correct curve.
Options:
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
Since, the observer is stationary, so when train is approaching towards the observer or moving towards the observer, apparent frequency is given as,
$n^{\prime}=n\left(\frac{v}{v-v_{s}}\right)$
where, $v_{s}$ is the speed of sound.
It is clear from the above expression that $n^{\prime}>n$. However, when the train is going away from the observer, then apparent frequency,
$\begin{array}{ll}
& n^{\prime}=n\left(\frac{v}{v+v_{s}}\right) \\
\therefore \quad & n^{\prime} < n
\end{array}$
Hence, graph represented in option (d) is correct.
$n^{\prime}=n\left(\frac{v}{v-v_{s}}\right)$
where, $v_{s}$ is the speed of sound.
It is clear from the above expression that $n^{\prime}>n$. However, when the train is going away from the observer, then apparent frequency,
$\begin{array}{ll}
& n^{\prime}=n\left(\frac{v}{v+v_{s}}\right) \\
\therefore \quad & n^{\prime} < n
\end{array}$
Hence, graph represented in option (d) is correct.
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