Doppler Effect Formula: Signs and Worked Sound Examples

A horn sounds at one fixed frequency. As its vehicle approaches, you hear a higher pitch; after it passes, a lower one. The source did not retune the horn. Motion changed the rate at which the wavefronts reach you. The Doppler effect formula puts a number on that change, but its two signs are easy to swap if you memorize only a plus-over-minus pattern.

We will use sound traveling through still air, with the source and listener moving directly along the line between them. One table makes the sign choice explicit. Then the same emitted tone is worked twice, first by moving the source and then by moving the listener at the same speed. A final case moves both. This paired comparison is useful because sound has a medium: those first two moves are similar, but not mathematically identical.

Name the source, listener and medium first

Call the emitted frequency $f_s$ and the frequency heard by the listener $f_o$. Let $v$ be the speed of sound relative to the air, $v_s$ the source speed relative to that air, and $v_o$ the listener speed relative to the air. All three speeds must be in the same unit, normally metres per second. The examples use $v=340\ \mathrm{m/s}$ as a stated value for the problem, not a universal speed for all air conditions.

The distinction between source and listener is physical. A moving source emits the next crest from a new position. In front of it, crests are closer together, so the wavelength in the air is shorter. A moving listener does not change that wavelength; it crosses the existing crests more or less often. OpenStax derives both cases from these two observations.

Draw one arrow from source to listener before substituting. Mark which way each object moves along that arrow. A car moving across your field of view is not covered by simply putting its full speed into this one-dimensional formula. For an angled path, the relevant speeds are the components along the line of sight at that moment; for an introductory problem, use the geometry and convention your course specifies.

The sound formula and a sign-choice table

For motion along one line in still air, write the formula as

$$f_o=f_s\frac{v\pm v_o}{v\mp v_s}.$$

The paired symbols are shorthand, not an instruction to pick both top signs together. Choose the numerator from the listener's motion and the denominator from the source's motion independently:

Set the speed of any stationary object to zero. If both objects approach, both choices raise the result: $f_o=f_s(v+v_o)/(v-v_s)$. If they separate, both choices lower it: $f_o=f_s(v-v_o)/(v+v_s)$. A direction can reverse as an object passes the listener, so do not use the approaching sign after it has passed.

A quick limit check catches the worst sign errors: set $v_s=v_o=0$. The formula must return $f_o=f_s$. Approach must give a value above the emitted frequency, separation a value below it. If your answer violates either check, revisit the diagram before the arithmetic. For slower motion, the shifts are modest; a factor of ten usually means mixed units or a denominator entered with the wrong sign.

Example 1: the source moves toward a still listener

A horn emits $f_s=680\ \mathrm{Hz}$. A vehicle carries it toward a stationary listener at $v_s=34\ \mathrm{m/s}$. Use $v=340\ \mathrm{m/s}$. The listener is still, so $v_o=0$. The source approaches, so subtract its speed in the denominator:

$$f_o=680\frac{340}{340-34} =680\frac{340}{306} =755.6\ \mathrm{Hz}.$$

That is about $75.6\ \mathrm{Hz}$ above the emitted tone. Check the factor: $340/306=1.111\ldots$ is greater than one, as an approaching source requires. The units of speed cancel; hertz remains. The result is an ideal instantaneous frequency for the simple, steady-motion situation, not a prediction of loudness.

What if that same vehicle has passed and is moving away at the same speed? Only the denominator's sign changes:

$$f_o=680\frac{340}{340+34} =618.2\ \mathrm{Hz}.$$

The approaching and receding frequencies do not sit exactly the same distance above and below $680\ \mathrm{Hz}$. The denominators are $306$ and $374$, not equal reciprocal changes. Do not subtract an approach shift from the source frequency to guess the receding tone.

You can see the wavelength explanation numerically. The period between emitted crests is $1/680\ \mathrm{s}$. During that period the source advances $34/680=0.050\ \mathrm{m}$; a wavefront advances $340/680=0.500\ \mathrm{m}$ through the air. Ahead of the moving source, consecutive crests are $0.500-0.050=0.450\ \mathrm{m}$ apart. A still listener receives them at $340/0.450=755.6\ \mathrm{Hz}$, the same answer by a second route.

Example 2: move the listener instead

Reset the horn so it is stationary in the air and still emits $680\ \mathrm{Hz}$. Now the listener approaches it at $v_o=34\ \mathrm{m/s}$. The wavelength in the air stays $v/f_s=340/680=0.500\ \mathrm{m}$. The listener meets crests at a relative speed of $340+34=374\ \mathrm{m/s}$:

$$f_o=680\frac{340+34}{340} =748.0\ \mathrm{Hz}.$$

Compare $748.0\ \mathrm{Hz}$ with $755.6\ \mathrm{Hz}$ in Example 1. Both are higher than $680\ \mathrm{Hz}$, but the moving source produces a slightly larger shift at this speed. Replacing both objects with one generic "relative speed" would conceal that difference for sound in a medium. The source changes wavelength; the listener changes the rate of crossing it.

If the listener moves away instead, use $v-v_o$ in the numerator: $680(340-34)/340=612.0\ \mathrm{Hz}$. That is lower than the moving-source recession result, $618.2\ \mathrm{Hz}$. The same-speed comparison works on both sides of the pass.

This table is the original worked check for this guide. Each column has the same emitted tone, sound speed and moving-object speed, so only which object moves changes. It is a useful self-test after any Doppler calculation.

Example 3: both move toward one another

Use the same $680\ \mathrm{Hz}$ horn and $340\ \mathrm{m/s}$ sound speed. This time the source moves toward the listener at $20\ \mathrm{m/s}$, while the listener moves toward the source at $10\ \mathrm{m/s}$. Both choices must raise the observed frequency. The listener adds to the numerator; the source subtracts from the denominator:

$$f_o=680\frac{340+10}{340-20} =680\frac{350}{320} =743.75\ \mathrm{Hz}.$$

The factor $350/320=1.09375$ is above one. As another check, remove the listener's motion: $680(340/320)=722.5\ \mathrm{Hz}$. Adding an approaching listener raises that to $743.75\ \mathrm{Hz}$, exactly as the diagram predicts. If the listener moved away at $10\ \mathrm{m/s}$ while the source still approached, the two effects would compete: $680(330/320)=701.25\ \mathrm{Hz}$. You must choose the two signs separately instead of labeling the whole situation simply "approaching".

Boundaries that keep the result meaningful

These are classical sound examples. The speeds refer to a specified medium, here still air, and the straight-line formula uses the motion along the sound path. Wind changes velocities relative to the air; a sideways path needs components. We keep the source below sound speed. As an approaching source approaches $v_s=v$, the denominator tends to zero. That signals the failure of the separated-wavefront picture used in this formula; it does not mean a listener hears an infinite frequency. Sonic booms and shock fronts need another model.

Light also shows Doppler shifts, but it travels in vacuum without an air-like medium. Do not put the speed of light into this sound fraction and use independent source/listener speeds against imaginary stationary air. OpenStax's light chapter gives the relativistic relation in terms of relative motion. The physical idea of approach raising frequency carries over, while the calculation changes.

For a fresh sound problem, write $f_s$, $v$, $v_s$ and $v_o$ with units, sketch the motion, then use the sign table. Check the answer against approach or separation and the zero-speed limit. If setting up moving-object directions is the hard part, the kinematics guide gives a way to keep velocity signs consistent. The waves and sound topic in Physics Zen includes Doppler practice with fresh numbers, instant checking and worked solutions. One clear diagram and two checks will help you more than memorizing a row of signs.