Question

When observer moves toward stationary source with $v_1$, apparent frequency is $F_1$. When moving away with $v_1$, it is $F_2$. If $F_1/F_2 = 2$, find $V/v_1$.

• A. 3
• B. 2
• C. 1
• D. 4

Hint:

$f_{app}$ increases when the gap between source and observer closes.

Answer 1

The Correct Option is A

Step 1: Formula
Doppler effect for moving observer: $f' = f(\frac{V \pm v_o}{V})$.

Step 2: Analysis
- $F_1 = f(\frac{V+v_1}{V})$
- $F_2 = f(\frac{V-v_1}{V})$
Ratio $\frac{F_1}{F_2} = \frac{V+v_1}{V-v_1} = 2$.

Step 3: Calculation
$V + v_1 = 2(V - v_1) \implies V + v_1 = 2V - 2v_1$
$3v_1 = V \implies V/v_1 = 3$.

Step 4: Conclusion
Hence, $V/v_1 = 3$. Final Answer: (A)