The Doppler effect

Key ideas

The Doppler effect is the change in the frequency you hear when a source of waves moves relative to you.
At Level 3 the standard covers a moving source and a stationary observer (for mechanical waves such as sound). A moving observer is not in the standard.
The physics: a moving source chases its own wavefronts.
- Ahead of the source, wavefronts are squashed together — shorter wavelength — so a stationary listener hears a higher frequency.
- Behind, they are stretched apart — longer wavelength — a lower frequency.
The observed frequency is:

$f' = f\,\frac{v_w}{v_w \pm v_s}$

Each variable, with sub-bullets:
- $f$ — the frequency the source actually emits (Hz)
- $f'$ — the frequency the stationary observer hears (Hz)
- $v_w$ — the speed of the waves (e.g. sound, $\approx 340$ m s⁻¹ in air)
- $v_s$ — the speed of the source (m s⁻¹)
Choosing the sign: the source approaching squeezes the waves, so the denominator must shrink — use minus. Receding stretches them — use plus. Check the answer moved the right way rather than memorising the sign.

A sound source moving to the right, with its circular wavefronts bunched close together ahead of it and spread far apart behind; a stationary listener ahead of the source hears a higher frequency, and one behind hears a lower frequency.

Worked ExampleAn ambulance drives past

An ambulance siren emits $700$ Hz. The ambulance travels at $25$ m s⁻¹ and the speed of sound is $340$ m s⁻¹. Find the frequency heard by a stationary pedestrian as it approaches, and after it passes.

Step 1 — Approaching: wavefronts squashed, so the denominator shrinks (minus)

$f' = f\,\frac{v_w}{v_w - v_s} = 700 \times \frac{340}{340 - 25} = 700 \times \frac{340}{315} = 756 \text{ Hz}$

Step 2 — Receding: wavefronts stretched (plus)

$f' = 700 \times \frac{340}{340 + 25} = 700 \times \frac{340}{365} = 652 \text{ Hz}$

Step 3 — Sanity check

Higher while approaching, lower after passing — the familiar nee-naw drop as it goes by.

Tips

Reason the sign, then check the direction of your answer. Approaching must give $f' > f$ ; receding must give $f' < f$ . If your numbers go the other way, the sign is wrong — fix it before moving on.
The change is not symmetric: the rise while approaching ( $+56$ Hz above) is bigger than the drop after passing ( $-48$ Hz), because $v_s$ sits in the denominator. Don't average them and expect $f$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A train sounding its $400$ Hz horn approaches a stationary observer at $20$ m s⁻¹. The speed of sound is $340$ m s⁻¹.

Show that the observer hears about $425$ Hz.

Merit

A car horn emits $500$ Hz. A stationary listener hears $470$ Hz. The speed of sound is $340$ m s⁻¹.

Calculate the car's speed and state whether it is approaching or receding.

Excellence

Explain, in terms of wavefronts, why a stationary observer ahead of a moving sound source hears a higher frequency — and why the frequency the source emits is unchanged.

The Doppler effect

Covers AS91523

Key ideas

The Doppler effect is the change in the frequency you hear when a source of waves moves relative to you.
At Level 3 the standard covers a moving source and a stationary observer (for mechanical waves such as sound). A moving observer is not in the standard.
The physics: a moving source chases its own wavefronts.
- Ahead of the source, wavefronts are squashed together — shorter wavelength — so a stationary listener hears a higher frequency.
- Behind, they are stretched apart — longer wavelength — a lower frequency.
The observed frequency is:

$f' = f\,\frac{v_w}{v_w \pm v_s}$

Each variable, with sub-bullets:
- $f$ — the frequency the source actually emits (Hz)
- $f'$ — the frequency the stationary observer hears (Hz)
- $v_w$ — the speed of the waves (e.g. sound, $\approx 340$ m s⁻¹ in air)
- $v_s$ — the speed of the source (m s⁻¹)
Choosing the sign: the source approaching squeezes the waves, so the denominator must shrink — use minus. Receding stretches them — use plus. Check the answer moved the right way rather than memorising the sign.

Worked ExampleAn ambulance drives past

Step 1 — Approaching: wavefronts squashed, so the denominator shrinks (minus)

$f' = f\,\frac{v_w}{v_w - v_s} = 700 \times \frac{340}{340 - 25} = 700 \times \frac{340}{315} = 756 \text{ Hz}$

Step 2 — Receding: wavefronts stretched (plus)

$f' = 700 \times \frac{340}{340 + 25} = 700 \times \frac{340}{365} = 652 \text{ Hz}$

Step 3 — Sanity check

Higher while approaching, lower after passing — the familiar nee-naw drop as it goes by.

Tips

Reason the sign, then check the direction of your answer. Approaching must give $f' > f$ ; receding must give $f' < f$ . If your numbers go the other way, the sign is wrong — fix it before moving on.
The change is not symmetric: the rise while approaching ( $+56$ Hz above) is bigger than the drop after passing ( $-48$ Hz), because $v_s$ sits in the denominator. Don't average them and expect $f$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A train sounding its $400$ Hz horn approaches a stationary observer at $20$ m s⁻¹. The speed of sound is $340$ m s⁻¹.

Show that the observer hears about $425$ Hz.

Merit

A car horn emits $500$ Hz. A stationary listener hears $470$ Hz. The speed of sound is $340$ m s⁻¹.

Calculate the car's speed and state whether it is approaching or receding.

Excellence

Explain, in terms of wavefronts, why a stationary observer ahead of a moving sound source hears a higher frequency — and why the frequency the source emits is unchanged.