Question

A source of sound is moving towards a stationary observer with velocity '\(V_S\)' and then moves away with velocity '\(V_S\)'. Assume that medium through which the sound waves travel is at rest. If 'V' is the velocity of sound and 'n' is the frequency emitted by the source then the difference between the apparent frequencies heard by the observer is

• A. \(\frac{2nVV_s}{V^2-V_s^2}\)
• B. \(\frac{nVV_s}{V^2-V_s^2}\)
• C. \(\frac{nVV_s}{V_s^2+V^2}\)
• D. \(\frac{2nVV_s}{V_s^2+V^2}\)

Hint:

Use f = nV/(V - Vs) for approach and f = nV/(V + Vs) for recession.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a source moves toward a stationary observer, the observed frequency rises. When it moves away, the frequency falls.

Step 2: Write the two frequencies:
Approaching: \(n_1=\dfrac{nV}{V-V_s}\). Receding: \(n_2=\dfrac{nV}{V+V_s}\).

Step 3: Subtract:
\[ n_1-n_2=nV\left[\dfrac{1}{V-V_s}-\dfrac{1}{V+V_s}\right]=nV\cdot\dfrac{2V_s}{V^2-V_s^2} \]
So the difference is \(\dfrac{2nVV_s}{V^2-V_s^2}\). Option A.

Step 4: Why the other options are wrong.
Option B misses the factor 2. Options C and D have \(V^2+V_s^2\) in the denominator, but the product \((V-V_s)(V+V_s)\) equals \(V^2-V_s^2\).

Final Answer:
The difference is 2nVVs / (V^2 - Vs^2). \[ \boxed{\text{(A) }\dfrac{2nVV_s}{V^2-V_s^2}} \]