The Doppler formula, sign by sign

By the end you can derive the Doppler formula, choose the correct sign for an approaching or receding source, and rearrange it to find a source speed.

The main symbols used in this lesson:

Symbol	Meaning
$\lambda$	wavelength

The formula has four symbols, and mixing up two of them is the usual reason answers go wrong. Before writing anything down, name them:

$f_s$ is the frequency emitted by the source (s for source). It does not change.
$f_o$ is the frequency detected by the observer (o for observer). This is what changes.
$v$ is the speed of the wave through the air — about $340\ \text{m s}^{-1}$ for sound.
$v_s$ is the speed of the source. It is a speed of an object, never a speed of sound.

Notice the two speeds are different animals. $v$ belongs to the wave, $v_s$ belongs to the ambulance.

Take a source moving at speed $v_s$ directly towards a stationary observer, and let the air be still.

Step 1 — one period of the source. In the time $T = 1/f_s$ between two wavefronts, the source moves a distance $v_s T$ towards the observer.

Step 2 — one wavelength. The first wavefront has travelled $vT$ from where it started. The second wavefront starts $v_s T$ closer to the observer, so the gap between the two wavefronts is

$\lambda_{\text{ahead}} = vT - v_sT = (v - v_s)T = \frac{v - v_s}{f_s}$

Step 3 — what the observer counts. The observer is stationary, so wavefronts pass at speed $v$ with this new spacing. The frequency detected is

$f_o = \frac{v}{\lambda_{\text{ahead}}} = \frac{v}{(v - v_s)/f_s} = \frac{f_s\, v}{v - v_s}$

For a source moving away, the second wavefront starts further from the observer, so the wavelength is $(v + v_s)/f_s$ and the plus sign appears in the denominator instead.

Notice what did not change: $v$ never changed, and $f_s$ never changed. Only the spacing of the wavefronts did.

For a source moving relative to a stationary observer:

$f_o = \frac{f_s\, v}{v \pm v_s}$

Use the minus sign when the source is moving towards the observer. A smaller denominator gives a higher detected frequency.
Use the plus sign when the source is moving away from the observer. A larger denominator gives a lower detected frequency.

$v$ is the speed of the wave in the air and $v_s$ is the speed of the source.

A car horn emits a note of frequency $500\ \text{Hz}$ . The car moves directly towards a stationary pedestrian at $25\ \text{m s}^{-1}$ . Take the speed of sound in air as $340\ \text{m s}^{-1}$ .

Find the frequency the pedestrian detects.

The same car has passed the pedestrian and is now moving directly away at $25\ \text{m s}^{-1}$ , still sounding $500\ \text{Hz}$ .

Find the new detected frequency and the change from the approaching case.

A source of sound moves towards a stationary observer. Which expression gives the frequency the observer detects?

A siren emits $800\ \text{Hz}$ and moves directly towards a stationary observer at $30\ \text{m s}^{-1}$ . Take the speed of sound as $340\ \text{m s}^{-1}$ .

What frequency does the observer detect, in Hz?

The same siren, emitting $800\ \text{Hz}$ , has passed the observer and now moves directly away at $30\ \text{m s}^{-1}$ . Take the speed of sound as $340\ \text{m s}^{-1}$ .

What frequency does the observer detect, in Hz?

The Doppler formula, sign by sign

Key to the symbols

Name the four quantities first

Where the formula comes from