KinexiaLEARN PHYSICS

Doppler Effect — Approaching Train Horn (Worked Example)

Medium
Problem

A train horn sounds at 400 Hz while the train approaches at 30 m/s. What frequency does a stationary observer hear? (v_sound = 343 m/s)

Given

  • f = 400 Hz

  • vs = 30 m/s (approaching)

  • v = 343 m/s

Find

The observed frequency f'.

Principle

Approaching source compresses wavefronts: f' = f·v/(v − vs) — higher pitch.

Formula

Solution

  1. Write the formula: f′=f⋅v/(v−vs)f' = f \cdot v/(v - v_s)

  2. Substitute: f′=400×343/(343−30)f' = 400 \times 343/(343 - 30)

  3. Denominator: 313313; f′=400×343/313f' = 400 \times 343/313

  4. Calculate: f′=438 Hzf' = 438\ \mathrm{Hz}

438 Hz

Unit check

Hz × (m/s ÷ m/s) = Hz — correct.

What this means physically

438 Hz — nearly a semitone sharper than the true 400 Hz. As the train passes, the pitch suddenly drops below 400 Hz — the classic Doppler swing.

Common mistake

Using (v + vs) for approaching. Approaching shrinks the denominator (v − vs) — plus is for receding.