The Doppler Effect

An ambulance comes down the street with its siren going, passes you, and drives off. The pitch you hear is high while it approaches, and low once it has gone by. The driver hears neither of those. Inside the cab the siren sounds exactly the same the whole time, because the siren never changes what it is doing.

So the change is not in the source. It is in the trip the sound takes from the source to your ear. In the previous lesson you counted crests arriving from two fixed sources and watched them beat against each other. Here you count crests again, but now the counting itself is what motion interferes with. Every result below comes from one honest question: in one second, how many crests actually reach your ear?

The air sets the speed, and nothing else does

Sound is a pressure wave in air. Once a crest has left the source it travels outward at the speed the air allows, about $v = 343$ m/s at room temperature, and it has no memory of how the source was moving when it was born. A crest from a parked truck and a crest from a truck doing 30 m/s spread out as circles growing at the same rate.

That single fact is the whole lesson. The source cannot speed its sound up. All it can do is move to a new place before it emits the next crest. And the observer cannot slow the sound down. All they can do is run into the crests, or run away from them.

Two mechanisms, then, and they are not the same mechanism. Take them one at a time.

A moving source squeezes the wavelength

Let the source emit crests at a steady rate $f$, one every period $T = 1/f$, and let it move toward you at speed $v_s$.

Emit a crest. It starts spreading. Wait one period. In that time the crest has run a distance $vT$ toward you, and the source has moved $v_s T$ toward you as well. Now the source emits the next crest, from that new closer spot. The gap between the two crests (the wavelength in the air ahead of the source) is what is left over:

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

Behind the source the source has been running away from its own old crest, so the gap opens up instead: $\lambda_{\text{behind}} = (v + v_s)T$.

This is a real, physical change in the air. If you could photograph the pressure pattern you would see the crests crowded on one side and stretched on the other. It is not an illusion of hearing.

You stand still, so those crests reach you at the ordinary speed $v$. The rate they arrive is speed divided by spacing:

$f' = \frac{v}{\lambda_{\text{ahead}}} = \frac{v}{(v - v_s)T} = f\,\frac{v}{v - v_s}$

Read that. The denominator is smaller than $v$, so $f'$ is bigger than $f$: approaching means higher pitch, as it should. If the source recedes, put $-v_s$ in for $v_s$ and the denominator grows, so the pitch drops.

A moving observer runs into the crests

Now park the source and move yourself. The source is at rest, so it lays down an ordinary wavelength $\lambda = v/f$ in the air, evenly spaced, unchanged. Nothing is squeezed.

What changes is how fast you sweep through that pattern. Sit still and crests pass you at $v$. Walk toward the source at $v_o$ and, in your frame, the crests come at you at $v + v_o$. Same spacing, faster closing speed, so you meet more of them per second:

$f' = \frac{v + v_o}{\lambda} = f\,\frac{v + v_o}{v}$

Move away instead and you are partly outrunning them: $f' = f(v - v_o)/v$. If you could run at $v_o = v$ you would keep pace with a crest and hear nothing at all.

Put the two together and you have the general result, with the source speed in the denominator and the observer speed in the numerator:

$f' = f\,\frac{v \pm v_o}{v \mp v_s}$

The signs are not a table to learn. Ask one question before you write anything: does the motion bring them closer together? If yes, the frequency must go up, so choose the signs that make the numerator bigger or the denominator smaller. If no, choose the other ones. Then substitute numbers.

Worked example 1: the ambulance, both ways

A siren emits $f = 800$ Hz. The ambulance drives toward a listener standing still at $v_s = 25.0$ m/s, with $v = 343$ m/s.

Source moving, observer still, closing. So the denominator shrinks:

$f' = 800 \times \frac{343}{343 - 25.0} = 800 \times \frac{343}{318} = 800 \times 1.07862 = 862.9 \text{ Hz}$

After it passes, the same ambulance recedes at 25.0 m/s. Now the denominator grows:

$f' = 800 \times \frac{343}{343 + 25.0} = 800 \times \frac{343}{368} = 800 \times 0.93207 = 745.7 \text{ Hz}$

A drop of 117 Hz across the pass, about two semitones, which is exactly the swoop you hear.

Notice the shift up (62.9 Hz) is bigger than the shift down (54.3 Hz). It is not symmetric, because $v/(v-v_s)$ and $v/(v+v_s)$ are not symmetric about 1. That asymmetry is what lets you work backwards from the two heard pitches to the speed, which you will do in example 3.

Worked example 2: both of them moving

A truck carries a 500 Hz siren and drives at $v_s = 25.0$ m/s toward a motorcyclist, who rides at $v_o = 10.0$ m/s toward the truck. Take $v = 343$ m/s.

Before any algebra, decide the direction of the answer: they are closing, so the pitch must rise. Both effects push the same way. The observer's motion enlarges the numerator, the source's motion shrinks the denominator:

$f' = 500 \times \frac{343 + 10.0}{343 - 25.0} = 500 \times \frac{353}{318}$

Do the division: $353/318 = 1.11006$. So $f' = 555.0$ Hz.

The same recipe handles a chase. A van's 400 Hz horn moves away from a car at 20.0 m/s while the car chases it at 15.0 m/s. The van running away enlarges the denominator; the car giving chase enlarges the numerator:

$f' = 400 \times \frac{343 + 15.0}{343 + 20.0} = 400 \times \frac{358}{363} = 400 \times 0.98623 = 394.5 \text{ Hz}$

Still below 400 Hz, because the van is winning the race. If the car chased faster than the van fled, the pitch would come out above 400 Hz. The formula gets the crossover right on its own; you do not have to decide in advance.

Worked example 3: reading the speed off the pitch drop

Here is one that needs an idea rather than a substitution. A bystander hears a passing siren drop from 870 Hz on the way in to 800 Hz on the way out. Neither the true frequency $f$ nor the ambulance speed $u$ is given. Two unknowns, so you need two equations, and you have them:

$f_{\text{in}} = f\,\frac{v}{v - u}, \qquad f_{\text{out}} = f\,\frac{v}{v + u}$

The trick is that $f$ appears in both. Divide one by the other and it cancels, leaving only the thing you want:

$\frac{f_{\text{in}}}{f_{\text{out}}} = \frac{v + u}{v - u} = \frac{870}{800} = 1.0875$

Cross-multiply: $v + u = 1.0875(v - u)$, so $v + u = 1.0875v - 1.0875u$, so $u(1 + 1.0875) = 0.0875v$, giving

$u = \frac{0.0875 \times 343}{2.0875} = \frac{30.01}{2.0875} = 14.38 \text{ m/s}$

About 32 mph. Now put that back into either equation for the true pitch:

$f = f_{\text{in}}\,\frac{v - u}{v} = 870 \times \frac{343 - 14.38}{343} = 870 \times \frac{328.62}{343} = 833.5 \text{ Hz}$

Worth noticing: 833.5 Hz is not the average of 870 and 800, which would be 835. The true pitch sits slightly below the midpoint, because the upward shift is the larger of the two.

Worked example 4: echoes, done in two steps

A bat flies at 5.00 m/s straight at a cave wall, calling steadily at 50.0 kHz. What beat frequency does it hear between its own call and the echo?

Never try to do a reflection in one formula. Do it in two, because there genuinely are two Doppler shifts and one relay station in between.

Step one, wall as observer. The wall is a still listener; the bat is a source closing at 5.00 m/s. So the frequency arriving at the wall is

$f_1 = 50000 \times \frac{343}{343 - 5.00} = 50000 \times \frac{343}{338} = 50739.6 \text{ Hz}$

Step two, wall as source. The wall re-radiates whatever it received, at that same $f_1$, and it is not moving, so it is a stationary source. The bat is now the observer, closing at 5.00 m/s, so the numerator grows:

$f_2 = f_1 \times \frac{343 + 5.00}{343} = 50739.6 \times \frac{348}{343} = 51479.3 \text{ Hz}$

The two factors of 343 cancel if you chain them, which is the tidy way to write it:

$f_2 = f\,\frac{v + v_{\text{bat}}}{v - v_{\text{bat}}} = 50000 \times \frac{348}{338} = 51479 \text{ Hz}$

The beat frequency is the difference between the two tones present at the bat's ear, the call and the echo:

$f_{\text{beat}} = 51479.3 - 50000 = 1479 \text{ Hz}$

That is a beat the bat can hear directly, and it is how a bat measures closing speed: the same trick as police radar, done with sound.

If the reflector moves instead, run the same two steps with the roles swapped: a 2000 Hz stationary speaker aimed at a panel approaching at 5.00 m/s gives $f_1 = 2000 \times 348/343$ at the panel (moving observer), then $f_2 = f_1 \times 343/338$ back at the microphone (moving source), which is $2000 \times 348/338 = 2059$ Hz.

Wind, light, and going supersonic

Three extensions cover everything else that gets asked.

Wind. Air that is itself moving changes only $v$, and only for the leg of the trip between source and observer. Add the wind component pointing from source to observer: a 10.0 m/s tailwind makes the effective sound speed $v = 353$ m/s, a headwind makes it 333 m/s. The source and observer speeds stay measured against the ground, unchanged. So a 500 Hz source closing at 25.0 m/s on a still listener, with a 10.0 m/s tailwind, gives $500 \times 353/(353 - 25.0) = 538.1$ Hz. Do it with the headwind and you get 540.6 Hz: slightly more, not less, because shrinking $v$ makes the ratio $v/(v - v_s)$ more lopsided. That surprises people.

Light. Light needs no medium, so there is no rest frame to ask about and only the relative speed can matter. For speeds well below $c$ the fractional change in wavelength is just the speed ratio:

$\frac{\Delta\lambda}{\lambda} \approx \frac{v}{c}$

A hydrogen line emitted at 656.3 nm seen at 658.0 nm has shifted by 1.7 nm, so $v = c\,\Delta\lambda/\lambda = 3.00\times10^8 \times 1.7/656.3 = 7.771\times10^5$ m/s, or 777 km/s. Longer wavelength means receding, and that is a redshift.

Supersonic. As $v_s \to v$ the denominator $v - v_s$ goes to zero and the formula blows up. That is not a real infinity. It is the algebra telling you the crests ahead have piled onto each other into a single shock front. Past $v_s > v$ the source outruns its own sound and the fronts wrap into a cone whose half-angle $\theta$ obeys

$\sin\theta = \frac{v}{v_s} = \frac{1}{M}$

where $M = v_s/v$ is the Mach number. At Mach 2.00, $\sin\theta = 0.500$ so $\theta = 30.0^\circ$. Backwards: a measured cone of $25.0^\circ$ gives $M = 1/\sin 25.0^\circ = 1/0.4226 = 2.366$, so $v_s = 2.366 \times 343 = 811.6$ m/s.

The timing question follows from geometry. A plane at height $h = 8000$ m flying at Mach 1.50 passes overhead. The cone trails behind it at half-angle $\theta = \arcsin(1/1.50) = 41.81^\circ$. The boom reaches the ground observer when the cone's edge sweeps over her, and by then the plane has flown a horizontal distance $h/\tan\theta$ past overhead. With $\tan 41.81^\circ = 0.8944$ that distance is $8000/0.8944 = 8944$ m, covered at $v_s = 1.50 \times 343 = 514.5$ m/s, so

$t = \frac{8944}{514.5} = 17.38 \text{ s}$

She sees it overhead in silence and hears it seventeen seconds later.

Recognising the technique

Reach for Doppler when a problem gives you a frequency plus a velocity, and asks for a different frequency. That pairing is the tell, and it is almost the only one you need. Then sort out which case you are in:

• One speed, one thing moving. Decide whether the mover is the source or the observer, decide whether the gap is closing, and write the single ratio.
• Two speeds. Both numerator and denominator are in play. Observer speed on top, source speed on the bottom, every time.
• The words "echo", "reflect", "radar", "beat frequency", or any wall or panel. Two shifts, done in order, with the reflector acting first as an observer and then as a source at the frequency it received.
• Two frequencies given and a speed asked for. Divide the two equations to kill the unknown true frequency, as in example 3.
• "Wind" or "moving air". Adjust $v$ only, by the wind component along the source-to-observer line.
• A Mach number, a cone, a boom, or a speed above 343 m/s. Not the frequency formula at all. Use $\sin\theta = 1/M$ and then plain trigonometry.
• A wavelength shift in a spectrum. Light, so use $\Delta\lambda/\lambda = v/c$ and no medium.

One habit will save you more marks than any of this: before substituting, say out loud whether the answer should be higher or lower than the emitted frequency. Then check the number you get against that. A sign slip is the most common Doppler mistake by a wide margin, and this catches every one of them.

Sound crossing from one material into another bends and partly bounces back, and that is what happens next in reflection and refraction.