Question

When the observer moves towards a stationary source with velocity \(\mathrm{V}_1\), the apparent frequency of emitted note is \(F_1\). When observer moves away from the source with velocity \(\mathrm{V}_1\), the apparent frequency is \(\mathrm{F}_2\). If V is the velocity of sound in air and \(F_1 / F_2=2\) then \(V / V_1\) is equal to

• A. 3
• B. 2
• C. 1
• D. 4

Hint:

Write the apparent frequency for an observer approaching and receding, then divide the two.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a moving observer and stationary source, \(F=f\dfrac{V\pm V_{1}}{V}\).

Step 2: Write both frequencies
Towards the source: \(F_{1}=f\dfrac{V+V_{1}}{V}\).
Away from the source: \(F_{2}=f\dfrac{V-V_{1}}{V}\).

Step 3: Use the ratio
\(\dfrac{F_{1}}{F_{2}}=\dfrac{V+V_{1}}{V-V_{1}}=2\), so \(V+V_{1}=2V-2V_{1}\).

Step 4: Result
\(V=3V_{1}\), hence \(\dfrac{V}{V_{1}}=3\).

Final Answer:
The ratio V/V1 equals 3. \[ \boxed{\text{(A) }3} \]