Key Idea: The Doppler effect is one idea seen in two forms: when a wave source moves toward or away from you, the wave you receive is shifted. For sound the pitch changes — higher coming, lower going. For light the wavelength changes — blueshift (shorter) approaching, redshift (longer) receding. The source itself never changes; only what the observer measures does. It is examined on both papers. Paper 1A tends to be quick: which way the pitch shifts, the shape of the heard-frequency graph as a source passes, or whether a galaxy is approaching or receding. Paper 2 is the structured work — a sound calculation with f' = f·v/(v ± vₛ), or an astronomical one with Δλ/λ = v/c (speed of a star, a galaxy, or a rotating star's edge), plus an explain of why the shift happens.
📋 Key formulas
Both equations carry the data-booklet badge (look for it) — you are given them, so the skill is choosing the right one and the right sign, not memorising them.
- observed frequency — the pitch you hear (Hz)
- source frequency — the pitch actually emitted (Hz)
- speed of sound in the air (m s⁻¹)
- speed of the moving source (m s⁻¹)
- change in wavelength, observed − lab (m) — written Δλ
- the source's true (laboratory) wavelength (m)
- change in frequency, observed − lab (Hz) — written Δf
- the source's true (laboratory) frequency (Hz)
- speed of the source toward or away from us (m s⁻¹)
- the speed of light, 3.0 × 10⁸ m s⁻¹
Both come from the same physics: a moving source bunches the wavefronts ahead of it and stretches them behind. - Sound is slow enough that the source speed vₛ appears directly in f' = f·v/(v ± vₛ). - Light travels at c, so for ordinary speeds the shift is a tiny fraction v/c — hence Δλ/λ ≈ v/c.
⚖️ Sound vs light Doppler
🔵🔴 Approaching vs receding — read off the direction
Red = Receding (away, longer λ); blue = approaching (toward, shorter λ). For the sound sign, check the denominator: minus shrinks it → bigger f' (higher); plus grows it → smaller f' (lower). Always sanity-check the direction against your answer.
One edge of a spinning star (or the Sun) turns toward us → that edge is blueshifted (shorter λ). The opposite edge turns away → redshifted (longer λ). The edge speed v is the surface rotation speed, found from the shift of one edge — not the gap between the two.
✏️ Worked exam-style questions
IB-style question — pitch of an approaching siren (sound)
A fire engine sounds a steady 384 Hz siren as it drives directly toward a stationary observer at 28 m s⁻¹. Take the speed of sound in the air as 340 m s⁻¹. Find the frequency the observer hears.
Solution:
Approaching, so use the minus sign in the given sound equation:
Put in the numbers (f = 384, v = 340, vₛ = 28):
Tidy the denominator:
Work it out — keep the unit:
f' ≈ 418 Hz — higher than 384 Hz, as expected for an approaching source. If your answer came out below 384 Hz you used the wrong sign.
IB-style question — speed of a source from its pitch (sound)
A train sounds a steady 480 Hz whistle as it travels directly toward a stationary observer, who measures the pitch as 510 Hz. The speed of sound in the air is 340 m s⁻¹. Find the speed of the train.
Solution:
Approaching, so start from the minus form of the given equation:
Rearrange to isolate the source speed vₛ:
Put in the numbers (v = 340, f = 480, f' = 510):
Work it out — keep the unit:
vₛ = 20 m s⁻¹. When the unknown is the source speed, rearrange the given formula first — isolate (v ± vₛ), then subtract v.
IB-style question — recession speed of a galaxy (light)
A spectral line with a laboratory wavelength of 500.0 nm is observed at 508.0 nm in the light from a distant galaxy. Find the galaxy's speed relative to Earth, and state its direction of motion. (c = 3.0 × 10⁸ m s⁻¹.)
Solution:
Start with the given light-Doppler relation:
Rearrange for the speed v:
Find the shift Δλ = observed − lab:
Put in the numbers (λ and Δλ in the same units cancel):
Work it out — keep the unit:
v ≈ 4.8 × 10⁶ m s⁻¹. The wavelength is longer (a redshift), so the galaxy is moving away — consistent with an expanding Universe. Use the change Δλ on top, not the whole observed wavelength.
IB-style question — speed of a rotating star's edge (light)
A line that is 589.00 nm in the laboratory is observed at 588.80 nm in light from one edge of a rotating star. State whether that edge is approaching or receding, and find its speed. (c = 3.0 × 10⁸ m s⁻¹.)
Solution:
The observed value (588.80 nm) is shorter than the lab value (589.00 nm) — a blueshift — so this edge is approaching.
Start with the given relation, rearranged for v:
Use the size of the shift Δλ = 589.00 − 588.80 = 0.20 nm:
Work it out — keep the unit:
The edge is approaching (blueshift, shorter λ); its speed is v ≈ 1.0 × 10⁵ m s⁻¹. The opposite edge gives an equal redshift — that is the star's surface rotation speed.
🧠 Quick self-check
Tap each card to reveal the answer.
🎯 Exam tips
Exam Tips
- Sound: approaching → minus sign → higher f'; receding → plus sign → lower f'. Always sanity-check the direction against your number (approaching must give MORE than f).
- The heard-frequency graph of a passing source is high-flat, a sharp step down, then low-flat — never a smooth gradual slope. The step is at closest approach.
- Light: rearrange Δλ/λ = v/c to v = (Δλ ÷ λ) × c. Use the change Δλ (observed − lab) on top, NOT the whole observed wavelength.
- Keep λ and Δλ in the same units (e.g. both nm) — they cancel in Δλ/λ; only c must be in m s⁻¹.
- Red = Receding (away, longer λ); blue = approaching (toward, shorter λ). A rotating star shows BOTH at once, one edge each.
- Δλ/λ = v/c is valid only when v is much smaller than c — that is why everyday sources give an undetectable light shift (v/c ~ 10⁻⁷) but fast galaxies give a measurable one.
- In every case the source emits one fixed frequency/wavelength; it is the relative motion that changes what the observer measures.