KinexiaLEARN PHYSICS

Doppler Effect — Race Car (Worked Example)

Medium
Problem

A race car's engine note is 300 Hz at the car. It approaches a spectator at 60 m/s. What frequency does the spectator hear? (v = 343 m/s)

Given

  • f = 300 Hz

  • vs = 60 m/s (approaching)

  • v = 343 m/s

Find

The observed frequency f'.

Principle

Fast approach, big shift: f' = f·v/(v − vs).

Formula

Solution

  1. Substitute: f′=300×343/(343−60)f' = 300 \times 343/(343 - 60)

  2. Denominator: 283283

  3. Calculate: f′=300×343/283=364 Hzf' = 300 \times 343/283 = 364\ \mathrm{Hz}

364 Hz

Unit check

Hz × (m/s ÷ m/s) = Hz. The units confirm an observed frequency — correct.

What this means physically

364 Hz — over 20% sharper, more than three semitones. Race fans hear this dramatic rise-and-fall every lap as cars stream past the stands.

Common mistake

Forgetting vs cannot exceed v in this formula. At vs → v the formula blows up — that is the sonic boom regime, needing shock-wave physics instead.