A train emits a sound at 600 Hz and moves at 30 m/s towards a stationary observer. If the speed of sound in air is 330 m/s, what observed frequency will the observer hear as the train approaches and then recedes?

Step-by-step solution

1. For approaching source, use Doppler formula $f^{\prime }=f\frac{v}{v-v_s}$ where $v$ is speed of sound and $v_s$ is source speed.
2. Insert $f=600\text{ Hz}$, $v=330\text{ m/s}$, $v_s=30\text{ m/s}$: $f^{\prime }=600\times \frac{330}{330-30}=600\times \frac{330}{300}=600\times 1.1=660\text{ Hz}$.
3. For receding source, use $f^{\prime }=f\frac{v}{v+v_s}$.
4. Compute: $f^{\prime }=600\times \frac{330}{330+30}=600\times \frac{330}{360}=600\times 0.9167=550\text{ Hz}$ (rounded).
5. Thus, the observer hears 660 Hz when the train approaches and 550 Hz when it moves away.

Concept tested: Doppler effect for sound

Published 4 October 2026 · How this solution was prepared and checked