Question

If a source emitting waves of frequency \( F \) moves towards an observer with a velocity \( \frac{V}{3} \) and the observer moves away from the source with a velocity \( \frac{V}{4} \), the apparent frequency as heard by the observer will be ( \( V \) = velocity of sound)

• A. \( \frac{9}{8} F \)
• B. \( \frac{8}{9} F \)
• C. \( \frac{3}{4} F \)
• D. \( \frac{4}{3} F \)

Hint:

"Towards" increases frequency (denominator gets smaller), "Away" decreases frequency (numerator gets smaller).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a standard Doppler Effect problem where both source and observer are moving.
Step 2: Key Formula or Approach:
Apparent frequency \( f' = f \left( \frac{V \pm v_o}{V \mp v_s} \right) \).
Step 3: Detailed Explanation:
Source velocity \( v_s = \frac{V}{3} \) (towards observer \( \implies \) denominator is \( V - v_s \)).
Observer velocity \( v_o = \frac{V}{4} \) (away from source \( \implies \) numerator is \( V - v_o \)).
\[ f' = F \left( \frac{V - V/4}{V - V/3} \right) \] \[ f' = F \left( \frac{3V/4}{2V/3} \right) = F \left( \frac{3}{4} \times \frac{3}{2} \right) \] \[ f' = \frac{9}{8} F \] Step 4: Final Answer:
The apparent frequency is \( \frac{9}{8} F \).