Question

When an observer moves towards a stationary source with velocity '\(V_1\)', the apparent frequency of emitted note is '\(F_1\)'. When observer moves away from stationary source with velocity '\(V_1\)' the apparent frequency is '\(F_2\)'. If '\(v\)' is velocity of sound in air and \(\frac{F_1}{F_2} = 2\), then \(\frac{v}{V_1}\) is equal to

• A. 6
• B. 5
• C. 3
• D. 4

Hint:

Shortcut: If $F_1/F_2 = n$, then $v/V_1 = (n+1)/(n-1)$. Here $(2+1)/(2-1) = 3$.

Answer 1

The Correct Option is C

Step 1: Concept
Use the Doppler effect formula for a moving observer and stationary source: $f_{app} = f_0 \left( \frac{v \pm v_o}{v} \right)$.

Step 2: Meaning
$F_1 = f_0 \left( \frac{v + V_1}{v} \right)$ (towards) and $F_2 = f_0 \left( \frac{v - V_1}{v} \right)$ (away).

Step 3: Analysis
$\frac{F_1}{F_2} = \frac{v + V_1}{v - V_1} = 2$
$v + V_1 = 2v - 2V_1$
$3V_1 = v \implies \frac{v}{V_1} = 3$.

Step 4: Conclusion
The ratio $\frac{v}{V_1}$ is 3. Final Answer: (C)