Question

A train moves towards a stationary observer with speed $34m{{s}^{-1}}$. The train sounds a whistle and its frequency registered by observer is ${{f}_{1}}$. If the speed of the train is reduced to $17m{{s}^{-1}}$, the frequency registered is ${{f}_{2}}$. If the speed of the sound is $340m{{s}^{- 1}}$, then the ratio $\dfrac{{{f}_{1}}}{{{f}_{2}}}$ is
A. $\dfrac{18}{19}$
B. $\dfrac{1}{2}$
C. 2
D. $\dfrac{19}{18}$

Answer 1

Hint:When the source of sound and the observer are in motion, the frequency of the sound heard by the observer is different from the frequency of the actual frequency of the sound.

Use the formula for the apparent frequency when the source is moving towards the stationary observer.

Formula used:
$f'={{f}_{0}}\left( \dfrac{v}{v-{{v}_{s}}} \right)$

Complete step by step answer:
When the source of sound and the listener (observer) are in motion, the frequency of the sound heard by the observer is different from the frequency of the actual frequency of the sound, i.e. the frequency that the observer hears when both of them are stationary.

Suppose the frequency of sound heard by a stationary observer, emitted by a stationary source is
${{f}_{0}}$. If any one or both of them move, the frequency heard by the observer will not be equal to ${{f}_{0}}$. This different frequency is called apparent frequency.

When the source of sound moves towards a stationary observer, the apparent frequency is equal to $f'={{f}_{0}}\left( \dfrac{v}{v-{{v}_{s}}} \right)$,

where v is the speed of the sound and ${{v}_{s}}$ is the speed of the source.

In the first case, $v=340m{{s}^{-1}}$, ${{v}_{s}}=34m{{s}^{-1}}$ and $f'={{f}_{1}}$.
Then,
$\Rightarrow {{f}_{1}}={{f}_{0}}\left( \dfrac{340}{340-34} \right)={{f}_{0}}\left( \dfrac{340}{306}
\right)$ …. (i)

In the second case, $v=340m{{s}^{-1}}$, ${{v}_{s}}=17m{{s}^{-1}}$ and $f'={{f}_{2}}$.
Then,
$\Rightarrow {{f}_{2}}={{f}_{0}}\left( \dfrac{340}{340-17} \right)={{f}_{0}}\left( \dfrac{340}{323}
\right)$ …. (ii).

Now, divide (i) by (ii).
$\Rightarrow \dfrac{{{f}_{1}}}{{{f}_{2}}}=\dfrac{{{f}_{0}}\left( \dfrac{340}{306} \right)}{{{f}_{0}}\left(
\dfrac{340}{323} \right)}$
$\Rightarrow \dfrac{{{f}_{1}}}{{{f}_{2}}}=\dfrac{323}{306}=\dfrac{19}{18}$

Hence, the correct option is D.

Note:The above cases were when the observer is stationary and the source is moving to the observer. If the source is moving away from the stationary observer than the apparent frequency is equal to $f'={{f}_{0}}\left( \dfrac{v}{v+{{v}_{s}}} \right)$.

From the two formulae of apparent frequency, we can conclude that if the source moves towards the stationary observer, then the frequency heard by the observer is more than the actual frequency.

If the source moves away from the stationary observer, then the frequency heard by the observer is less than the actual frequency.