The Doppler Effect Explained
Everyone has heard it. An ambulance approaches with a high, urgent siren, passes, and the pitch drops in an instant even though the siren itself never changed. The Doppler effect is one of the few physics topics where the everyday experience is completely reliable, which makes it a good place to build confidence - as long as you can get the signs right under exam pressure.
That is where most marks are lost. The physics is a single, short formula, but it carries two plus-or-minus signs, and guessing them is a coin flip. This guide explains what is physically happening, sets out one sign convention and sticks to it, and works through numbers you can check yourself.
What is really changing
Sound travels through air at a speed set by the air itself, close to 343 m/s at 20 °C, and that speed does not care how fast the source is moving. What the source's motion changes is where each new wavefront starts. A stationary whistle emits crests from the same point, so they spread out as evenly spaced spheres. Move the whistle forward and each new crest is emitted a little further along, so the crests ahead of it are squeezed closer together and the crests behind are stretched further apart.
Since the wave speed is fixed, squeezed wavelengths must arrive more often, and stretched wavelengths less often. That is the entire effect: a change in the received frequency caused by a change in spacing, not by any change in what the source is doing. Recognising this also tells you what does not happen - the emitted frequency, and the speed of the sound in air, both stay exactly the same.
The formula and one sign convention
The general expression covers every case at once: f' = f·(v ± v_o)/(v ∓ v_s), where f is the emitted frequency, f' is the frequency heard, v is the speed of sound in the medium, v_o is the observer's speed, and v_s is the source's speed. The signs are not free - they encode the direction of motion.
Use this convention throughout and you will not get lost: the upper signs apply to motion that closes the gap. If the observer moves toward the source, take +v_o in the numerator; if the observer moves away, take −v_o. If the source moves toward the observer, take −v_s in the denominator; if the source moves away, take +v_s. Anything at rest contributes a zero and drops out.
Before touching a calculator, decide from physics alone whether the answer should be higher or lower than f. Approaching always raises the pitch, receding always lowers it. If your arithmetic disagrees with that prediction, you picked a sign wrongly, and you have caught the error for free.
- Numerator handles the observer; denominator handles the source.
- Closing the gap: +v_o on top, −v_s on the bottom, and f' rises.
- Opening the gap: −v_o on top, +v_s on the bottom, and f' falls.
- Only the component of velocity along the line joining source and observer counts.
Worked examples with real numbers
Take a siren emitting f = 800 Hz in air at 20 °C, so v = 343 m/s, and a vehicle travelling at 30 m/s past a stationary listener.
- Approaching: the source moves toward a stationary observer, so f' = 800 × 343/(343 − 30) = 800 × 343/313 ≈ 877 Hz.
- Receding: the source now moves away, so f' = 800 × 343/(343 + 30) = 800 × 343/373 ≈ 736 Hz.
- The audible drop as it passes is roughly 877 − 736 ≈ 141 Hz, close to a whole tone and a half - big enough that nobody misses it.
- Now swap the roles: a listener drives at 30 m/s toward a stationary 800 Hz siren, giving f' = 800 × (343 + 30)/343 = 800 × 373/343 ≈ 870 Hz.
- Compare the two approach cases: 877 Hz versus 870 Hz. Same relative speed, different answers.
Why the two cases are not symmetric
That last comparison bothers people, and it should - until you look at the wavelengths. When the source moves, it physically alters the wave pattern in the air. The wavelength ahead of a source approaching at 30 m/s is (343 − 30)/800 = 0.391 m instead of the stationary 343/800 = 0.429 m, and behind it the wavelength is stretched to (343 + 30)/800 = 0.466 m. The medium now carries a genuinely distorted wave.
When only the observer moves, the wave in the air is untouched - the crests are still 0.429 m apart everywhere. The listener simply runs into them at a different rate, sweeping up crests faster than a stationary ear would. Different mechanisms, so different formulas, and the difference grows as speeds approach the wave speed. This is a real asymmetry, and it exists because the air provides a preferred frame of reference for sound.
Mistakes that cost marks
Most Doppler errors are procedural rather than conceptual, and they repeat. Reading through this list before an exam is worth more than re-deriving the formula.
- Putting the source's speed in the numerator. The observer is always on top.
- Assuming the pitch falls continuously as the vehicle approaches. It stays constant and high while approaching, then switches to constant and low once it has passed - the change happens at closest approach.
- Using a relative speed of source and observer as a single number. Each has its own term, and they do not combine into one velocity.
- Forgetting that only the line-of-sight component matters. A car passing at a distance has almost no radial velocity at the closest point, so the shift is momentarily near zero.
- Changing the speed of sound because the source is fast. It depends on the air, mainly its temperature, and not on the source at all.
- Using the sound formula for light. It does not apply.
Faster than the wave, and the case of light
Push the source speed up to the wave speed and the denominator heads for zero, which is the mathematics warning you that the wavefronts pile up on top of each other. Beyond that speed the source outruns its own sound and the crests form a cone behind it, whose half-angle satisfies sin θ = 1/M, where the Mach number M is the source speed divided by the speed of sound. The bang you hear is that cone sweeping past - a continuous phenomenon, not a one-off event at the moment of breaking the sound barrier.
Light is a different story, because there is no medium and no preferred frame, so the sound formula genuinely does not apply. For motion directly along the line of sight, relativity gives f' = f·√((1 − β)/(1 + β)) for a source and observer separating, with β = v/c. When v is far smaller than c this reduces to the handy approximation Δλ/λ ≈ v/c, where a positive value means recession and a redshift. That approximation is how astronomers measure the motion of stars and galaxies, and it is the same physics as the siren wearing very different mathematics.
If you want to compare the general, approaching, and receding forms side by side with their variables defined, the waves category on PhysRef lists them together along with the beat-frequency and Mach-number relations that often appear in the same questions.
Frequently asked questions
Does the siren's pitch drop gradually as the ambulance comes toward me?
No. While it is approaching along your line of sight the received frequency is steady and high, and once it is receding it is steady and low. The audible slide happens over the short interval around closest approach, when the line-of-sight component of the velocity swings from positive to negative.
Why are the moving-source and moving-observer formulas different?
A moving source physically changes the wavelength in the air, bunching crests ahead of it. A moving observer leaves the wave untouched and merely encounters the existing crests at a different rate. Because air sets a preferred frame for sound, those two situations are not equivalent.
How do I remember which sign goes where?
Motion that closes the gap uses the upper signs: +v_o on top, −v_s on the bottom, which makes the fraction bigger and raises the pitch. Then check the result against your physical expectation - approach means higher, recede means lower - before writing it down.
Does the speed of sound change when the source moves fast?
No. Sound travels at a speed fixed by the medium, about 343 m/s in air at 20 °C, rising with temperature. A fast source changes where wavefronts are emitted, not how quickly they travel.
Can I use the same formula for light from a distant galaxy?
Not the sound formula. Light needs the relativistic expression, which for motion along the line of sight is f' = f·√((1 − β)/(1 + β)) with β = v/c. For speeds well below c it simplifies to Δλ/λ ≈ v/c, with a positive shift meaning the source is receding.