IB Physics HLTopic 3 — Oscillations & WavesPaper 1 & 2f′ = f v/(v ± us)~17 min read
Doppler Equations for Sound
For light we got away with an approximation, because nothing goes anywhere near the speed of light. Sound is different. A motorbike at 30 m s−1 is already doing nearly a tenth of the speed of sound, and the neat Δλ/λ ≈ v/c shortcut falls apart. Sound needs exact equations — and, surprisingly, a moving source and a moving observer need different ones.
Moving observer, stationary source: f′ = f (v ± uo)/v
In terms of wavelength, for a moving source: λ′ = λ(1 ± us/v)
v is the wave speed — about 340 m s−1 for sound in air
The ± is decided by one question: should f′ come out bigger or smaller?
Source towards → denominator v − us. Observer towards → numerator v + uo
The two situations are not equivalent, even at the same relative speed
A moving source: where the equation comes from
The source emits one wavefront, waits one period T, then emits the next. In that time the wavefront has run ahead a distance vT — but the source has also crept forward by usT. The gap left between them is the wavelength the observer in front actually receives.
The source runs after its own wave and eats into the gap. Divide the wave speed by that shortened wavelength and you get the frequency heard.
Since f′ = v/λ′ and T = 1/f, that single picture gives you the whole equation:
Moving source, stationary observerf′ = f(v / (v ± us) )towards the observer → v − us • away → v + us
Where f′ is the observed frequency (Hz), f the source frequency (Hz), v the wave speed and us the source speed (both m s−1). The same picture written in wavelengths:
Now hold the source still and run towards it. Nothing squashes the waves — the wavefronts sit in the air exactly λ apart, undisturbed. But you are sweeping through them, so you meet them more often. The waves approach you at v + uo.
Compare with the moving source, where λ itself was squashed. Here λ is untouched — it is the rate of arrival that changes.
Moving observer, stationary sourcef′ = f( (v ± uo) / v)towards the source → v + uo • away → v − uo
Notice the signs sit in different places. For a moving source the speed goes on the bottom; for a moving observer it goes on the top. Don’t memorise four rules. Memorise one: closing the gap means a higher frequency. Then look at your equation and ask “does this make f′ bigger?” If not, flip the sign. That takes five seconds and never fails.
Who moves
Direction
Equation
f′
Source
Towards observer
f′ = f v/(v − us)
Higher
Source
Away from observer
f′ = f v/(v + us)
Lower
Observer
Towards source
f′ = f(v + uo)/v
Higher
Observer
Away from source
f′ = f(v − uo)/v
Lower
Why the two cases differ
At small speeds the two give nearly the same answer. Push the speed up and they part company dramatically: a moving observer’s shift grows in a straight line, while a moving source’s shift blows up as us approaches the wave speed.
Both start at 1 and agree for slow speeds. But drive the source at the speed of sound and λ′ hits zero — the equation predicts infinite frequency. That is the sound barrier.
🚓 Getting the ± right, every time
Label the source and the observer on the question. Literally write S and O on the paper.
Which one is moving? That picks the equation — us on the bottom, uo on the top.
Are they closing or separating? Closing → f′ must be bigger than f.
Choose the sign that does that. Smaller denominator, or bigger numerator.
Check your answer against step 3. Higher when approaching, lower when receding.
WE 1
A train sounds its horn at 320 Hz and travels at 30 m s⁻¹ along a straight track. The speed of sound is 340 m s⁻¹. Calculate the frequency heard by a stationary observer as the train (a) approaches and (b) recedes.
Step 1 — the source is moving, so us goes on the bottomf′ = f v / (v ± us)(a) approaching — f′ must be bigger, so use v − usf′ = 320 × 340 / (340 − 30) = 320 × 340 / 310f′ = 351 Hz(b) receding — f′ must be smaller, so use v + usf′ = 320 × 340 / 370f′ = 294 HzAs it passes, the pitch falls 351 → 294 Hz. Notice the rise on approach (+31 Hz) is bigger than the drop on recession (−26 Hz). The shift is never symmetric — the denominator is doing different work each time.
WE 2
A bell hangs still and rings at 600 Hz. A cyclist rides directly towards it at 8.0 m s⁻¹. Taking the speed of sound as 340 m s⁻¹, calculate the frequency the cyclist hears, and state what happens to the wavelength of the sound in the air.
Step 1 — the observer is moving, so uo goes on the topf′ = f (v ± uo) / vStep 2 — riding towards, so f′ must be bigger: use v + uof′ = 600 × (340 + 8.0) / 340 = 600 × 348/340f′ = 614 HzStep 3 — the wavelength in the airunchanged
The source is still, so nothing squashes the wavefronts. The cyclist simply meets them more often.
This is the whole difference between the two cases. Moving source → λ changes. Moving observer → λ doesn’t.
WE 3
A car horn emits a note of 500 Hz. A stationary pedestrian hears it at 470 Hz. Taking the speed of sound as 340 m s⁻¹, determine the speed of the car and state its direction of travel.
Step 1 — which way is it going?
The heard frequency is lower, so the car is receding. Use v + us.
Step 2 — write the equation and rearrangef′ = f v / (v + us) → v + us = f v / f′us = v (f/f′ − 1)Step 3 — substituteus = 340 × (500/470 − 1) = 340 × 0.0638us = 22 m s⁻¹, driving awaySanity check: 22 m s⁻¹ is about 78 km/h. Plausible for a car. If you had got 220 m s⁻¹, you would know a sign had gone astray.
💡 Top tips
Write S and O on the question paper. Half the sign errors vanish immediately.
Source speed lives on the bottom. Observer speed lives on the top. Remember the position, not four formulas.
Closing the gap → higherf′. Pick whichever sign delivers that.
Moving source changes λ. Moving observer does not. Worth a mark on its own.
Use v = 340 m s−1 for sound in air unless told otherwise.
Finish by checking your answer moved the right way. It costs seconds.
⚠ Common mistakes
Putting us on the top, or uo on the bottom — the two equations are not interchangeable
Using + for an approaching source. Approaching means a smaller denominator, so it is −
Thinking the wavelength changes when only the observer moves
Assuming the rise in pitch on approach equals the fall on recession — they are not symmetric
Using the light equation Δλ/λ ≈ v/c for sound — a source speed is rarely small compared with the wave speed, so the approximation fails
Forgetting that f is the source frequency and f′ the observed one, then inverting the fraction
Quick recap: For a moving source, f′ = f v/(v ± us) and λ′ = λ(1 ± us/v) — the wavelength itself is squashed or stretched. For a moving observer, f′ = f(v ± uo)/v — the wavelength is unchanged, but the wavefronts arrive more often. In both, choose the sign so that closing the gap raises the frequency.
And that finishes the Doppler Effect. Trace what you’ve built: wavefronts bunching up in front of an ambulance, the same idea stretched across the sky into red-shifted galaxies and an expanding universe, and now the exact arithmetic for a train horn. One piece of physics — a source chasing its own waves — running all the way from a passing bicycle bell to the age of the cosmos.
Struggling to pick the right sign?
Book a free meeting and we’ll drill moving-source and moving-observer questions until the ± is automatic.
