A source emitting 500 Hz moves towards a stationary listener at 30 m/s. If sound travels at 330 m/s, what frequency does the listener hear?

1. The source approaches a stationary listener, so use $n' = \left(\dfrac{v}{v - v_s}\right)n$. 2. List the data: $v = 330$, $v_s = 30$, $n = 500$. 3. Substitute: $n' = \dfrac{330}{330 - 30} \times 500$. 4. Simplify the denominator: $\dfrac{330}{300} = 1.1$. 5. So $n' = 1.1 \times 500 = 550\,\text{Hz}$. 6. Trap A (455) uses $v + v_s$, the receding case; trap B (500) forgets the motion entirely. _Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 5 Acoustics "Acoustics", p.75_