Question

A train blowing the whistle moves with a constant velocity $V$ away from an observer standing on the platform. The ratio of the natural frequency of the whistle $n$ to the apparent frequency is $1.2:1$. If the train is at rest and the observer moves away from it at the same velocity $V$, the ratio of $n$ to the apparent frequency is

• A. $0.51:1$
• B. $1.25:1$
• C. $2.05:1$
• D. $1.52:1$

Hint:

Effect of motion of source and observer on frequency is different in Doppler effect.

Answer 1

The Correct Option is B

Step 1: Doppler effect when source moves away.
\[ n' = n\left(\frac{v}{v+V}\right) \] Given:
\[ \frac{n}{n'} = \frac{v+V}{v} = 1.2 \]
Step 2: Finding velocity ratio.
\[ \frac{v+V}{v} = 1.2 \Rightarrow \frac{V}{v} = 0.2 \]
Step 3: Observer moving away, source at rest.
\[ n' = n\left(\frac{v-V}{v}\right) \]
Step 4: Ratio of frequencies.
\[ \frac{n}{n'} = \frac{v}{v-V} = \frac{1}{1-0.2} = 1.25 \]
Step 5: Conclusion.
The required ratio is $1.25:1$.