A source of 90 Hz approaches a stationary listener at one-tenth the speed of sound. What frequency does the listener hear?

1. Source approaching stationary listener: $n' = \left(\dfrac{v}{v - v_s}\right)n$. 2. With $v_s = v/10$, this becomes $\dfrac{v}{v - v/10} = \dfrac{10}{9}$. 3. $n' = \dfrac{10}{9} \times 90 = 100\,\text{Hz}$. 4. Trap A uses the receding factor 9/10; trap B ignores motion. _Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 5 Acoustics "Acoustics", p.74_