Question

When the observer moves towards a stationary source with velocity '\(V_1\)', the apparent frequency of the emitted note is '\(F_1\)'. When the observer moves away from the source with velocity '\(V_1\)', the apparent frequency is '\(F_2\)'. If 'V' is the speed of sound in air and \(F_1/F_2 = 2\), then \(V/V_1 =\)

• A. \(2\)
• B. \(3\)
• C. \(6\)
• D. \(8\)

Hint:

Write apparent frequencies for approach and recession, then take the ratio.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a stationary source and a moving observer, apparent frequency is \(f' = f\,\dfrac{V \pm V_o}{V}\), with a plus sign for approach and a minus sign for recession.

Step 2: Write F1 and F2:
\(F_1 = f\,\dfrac{V + V_1}{V}\) and \(F_2 = f\,\dfrac{V - V_1}{V}\).

Step 3: Ratio:
\[ \frac{F_1}{F_2} = \frac{V + V_1}{V - V_1} = 2 \Rightarrow V + V_1 = 2V - 2V_1 \Rightarrow V = 3V_1 \]

Step 4: Result:
\(\dfrac V{V_1} = 3\), option (B).

Final Answer:
(V + V1)/(V - V1) = 2 gives V = 3 V1. \[ \boxed{\text{(B) }3} \]