A and B are two sources generating sound waves. A listener is situated at C. The frequency of the source at A is 500 Hz. A, now, moves towards C with a speed 4 m/s. The number of beats heard at C is 6. When A moves away from C with speed 4 m/s, the number of beats heard at C is 18. The speed of sound is 340 m/s. The frequency of the source at B is :

The apparent frequency ($$f$$) heard by a stationary listener when a source moves is given by $$f' = f \left( \frac{v}{v \mp v_s} \right)$$

Case 1: A moves towards C

$$f'_{A1} = 500 \left( \frac{340}{340 - 4} \right) = 500 \left( \frac{340}{336} \right) \approx \mathbf{506\text{ Hz}}$$

Case 2: A moves away from C

$$f'_{A2} = 500 \left( \frac{340}{340 + 4} \right) = 500 \left( \frac{340}{344} \right) \approx \mathbf{494\text{ Hz}}$$

Beat frequency is the absolute difference between two frequencies: $$f_{\text{beat}} = |f_A' - f_B|$$

From Case 1 (6 beats): $$|506 - f_B| = 6 \implies f_B = 506 + 6 = \mathbf{512\text{ Hz}} \text{ or } 506 - 6 = 500\text{ Hz}$$

From Case 2 (18 beats): $$|494 - f_B| = 18 \implies f_B = 494 + 18 = \mathbf{512\text{ Hz}} \text{ or } 494 - 18 = 476\text{ Hz}$$

Comparing both scenarios, the only consistent value for the frequency of source B is $$512\text{ Hz}$$.