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: Use the wavelength method:
Approaching: wavelength $\lambda_1=\dfrac{V-V_s}{n}$, so $n_1=\dfrac{V}{\lambda_1}$. Receding: $\lambda_2=\dfrac{V+V_s}{n}$.

Step 2: Compute the difference:
$n_1-n_2=nV\left[\dfrac{1}{V-V_s}-\dfrac1{V+V_s}\right]=nV\cdot\dfrac{(V+V_s)-(V-V_s)}{(V-V_s)(V+V_s)}$.

Step 3: Simplify:
$=\dfrac{2nVV_s}{V^2-V_s^2}$. Option A.

Final Answer:
Subtracting nV/(V + Vs) from nV/(V - Vs) gives 2nVVs/(V^2 - Vs^2). \[ \boxed{\text{(A) }\dfrac{2nVV_s}{V^2-V_s^2}} \]