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: Doppler Effect Formula
$F' = F \left( \frac{V \pm v_o}{V \mp v_s} \right)$
Step 2: Assigning Signs
Observer moves away: $-v_o$. Source moves towards: $-v_s$.
$v_o = V/4$, $v_s = V/3$.
Step 3: Calculation
$F' = F \left( \frac{V - V/4}{V - V/3} \right) = F \left( \frac{3V/4}{2V/3} \right)$
$F' = F \left( \frac{3}{4} \times \frac{3}{2} \right) = \frac{9}{8}F$
Step 4: Conclusion
The apparent frequency is $\frac{9}{8}F$.
Final Answer:(A)