MCQ 11 Mark
A train whistling at constant frequency 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 expected curve:
- A

- B

- ✓

- D

Answer
View full question & answer→Correct option: C. 

When qbserver is at rest and source of sound id moving towards observer then observed frequency n'.
$\text{n}'=\Big(\frac{\text{v}}{\text{V}-\text{v}_\text{s}}\Big)\text{n}_0$
Where $n_0$ original frequency of source of sound
v = speed of sound in medium
$\therefore\text{n}'>\text{n}_0\ \ \text{v}_\text{s}=$ speed of source
When source is moving away from observer
$\text{n}'=\frac{\text{v}}{(\text{v}+\text{v}_\text{s})}\text{n}_0\ \text{n}''<\text{n}_0$
Hence, the frequencies in both cases are same and $\text{n}'>\text{n}''.$ so graph (c) verifies the answer.
$\text{n}'=\Big(\frac{\text{v}}{\text{V}-\text{v}_\text{s}}\Big)\text{n}_0$
Where $n_0$ original frequency of source of sound
v = speed of sound in medium
$\therefore\text{n}'>\text{n}_0\ \ \text{v}_\text{s}=$ speed of source
When source is moving away from observer
$\text{n}'=\frac{\text{v}}{(\text{v}+\text{v}_\text{s})}\text{n}_0\ \text{n}''<\text{n}_0$
Hence, the frequencies in both cases are same and $\text{n}'>\text{n}''.$ so graph (c) verifies the answer.
