Physics 9702 · Paper 1

The Doppler Effect

When a source of sound moves, a stationary listener hears a different frequency: higher as it comes towards you, lower as it goes away.
You have heard it whenever an ambulance passes: the siren drops in pitch the moment it goes by. This is the Doppler effect.

Why the pitch changes

The source keeps making the same number of waves per second, but each new wave is sent from a slightly moved position.
Ahead of the source the wavefronts are closer together (shorter wavelength, higher frequency); behind it they are further apart (lower frequency).
Drag the source speed below to see the wavefronts move closer together ahead of the source.

source speed v_s

v = 340 m s⁻¹·f_s = 500 Hz·v_s = 100 m s⁻¹·f_o = 708.3 Hz

source approaching → pitch up

An ambulance driving to the right along a road past a stationary person on the pavement. — A moving source and a stationary listener: the pitch heard changes as the ambulance passes.

The Doppler equation

f_o = \dfrac{f_s\, v}{v \pm v_s}

Symbols

$f_o$ = frequency the observer hears (Hz)
$f_s$ = frequency the source emits (Hz)
$v$ = speed of sound in the medium (m s⁻¹)
$v_s$ = speed of the source (m s⁻¹)

Use the minus sign when the source moves towards the observer (smaller denominator, higher $f_o$ ).
Use the plus sign when it moves away (bigger denominator, lower $f_o$ ).
The speed of sound $v$ is fixed by the air, not by the source. A moving source does not change how fast its sound travels.

Worked example

A train whistle approaching

A train sounds a whistle of frequency 800 Hz and moves towards a stationary observer at 60 m s⁻¹. The speed of sound is 330 m s⁻¹. What frequency is heard?

Approaching, so use the minus sign.
$f_o = \dfrac{800\times 330}{330 - 60} = \dfrac{264000}{270}$ .
$f_o \approx 978\text{ Hz}$ .

Answer

f_o \approx 978\text{ Hz}

Common mistake

Choosing the wrong sign. Check the direction: approaching must give a higher frequency, receding a lower one. If your answer goes the wrong way, you picked the wrong sign.

Variations you will meet

Reverse questions give you $f_o$ and ask for the source speed: rearrange the same equation.
Rotating source (a buzzer swung in a circle on a string): the highest pitch is heard when it moves straight at you, the lowest when it moves straight away.
The syllabus only asks about a moving source with a stationary observer.

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q24)

An aircraft emits sound of frequency 30.0 Hz. A stationary observer hears 20.0 Hz as it recedes. The speed of sound is 330 m s⁻¹. Find the aircraft's speed. (9702/11/M/J/25 Q24)

Receding, so use the plus sign: $20.0 = \dfrac{30.0\times 330}{330 + v_s}$ .
Rearrange: $330 + v_s = \dfrac{30.0\times 330}{20.0} = 495$ .
$v_s = 495 - 330 = 165\text{ m s}^{-1}$ .

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