Why does an ambulance siren drop in pitch the moment it passes you?

How to use this simulation

1. Start in the Source moves scene with the default speed of 150 m/s (Mach 0.44). The amber dot is the source and the rings are its wavefronts. Notice that they sit close together on the right and far apart on the left.
2. Read the two listeners. At 500 Hz the listener ahead hears 895 Hz and the listener behind hears 347 Hz. The source always sends 500 Hz; the change happens on the way.
3. Slide the speed to zero. The rings become perfect concentric circles and both listeners hear exactly 500 Hz. Then raise the speed slowly and watch the graph: the ahead curve climbs and the amber behind curve falls.
4. Push the speed toward 340 m/s. The rings ahead pile up into one line and the ahead curve shoots upward without limit. That wall of compressed air is what pilots called the sound barrier.
5. Go past 340 m/s. A red V appears, the Mach cone, and the readout shows its half-angle. The listener ahead hears nothing until the cone sweeps over them as a bang.
6. Switch to Listener moves. The source stands still and you walk the listener toward it or away from it. The wavelength stays the same, but the listener meets crests more often or less often, so the frequency changes.

The ambulance that changes its note

Stand on a footpath and let an ambulance come toward you with its siren on. The note is high and anxious. At the moment it passes, the note slides down, and as it drives away it stays low. Ask the driver, and he will tell you the siren sounded the same all along. He is right. What changed was what reached your ears.

That change in the frequency you receive, caused by motion between the source and the listener, is the Doppler effect. It was predicted by the Austrian physicist Christian Doppler in 1842 for light and sound, and it was first tested with trumpeters playing on a moving train.

This topic appears in HSC physics under waves and sound, and it is trimmed from some Indian syllabi, so check your own syllabus before you decide how deeply to prepare. Even if your board skips it, the idea is worth ten minutes, because it explains radar guns, ultrasound scans and the expansion of the universe.

What really happens to the waves

Sound travels through air at a fixed speed, about 340 m/s here, and that speed depends only on the air, not on how the source moves. Once a crest has left the source it travels outward on its own as a growing circle, whatever the source does next.

Now let the source move. Each new crest starts from a place slightly further along the road. In front of the source the crests start from places closer and closer to the previous ones, so they bunch up. Behind the source each crest starts further from the last, so they spread out.

That is the whole effect. Ahead of the source the wavelength is shorter, so a stationary listener, who receives waves at the fixed speed v, counts more crests every second. Behind the source the wavelength is longer, so fewer crests arrive. Frequency is just the number of crests per second.

The simulation uses the default of 150 m/s and 500 Hz. The wavelength is 0.680 m at rest, 0.380 m ahead and 0.980 m behind. You can read both on the panel and compare them with the picture.

Moving source, still listener

In one period T = 1/f the source emits one crest and then moves a distance vₛT toward the listener. The two crests are therefore not v T apart but (v − vₛ) T apart. The new wavelength is λ′ = (v − vₛ)/f.

The waves still travel at v, so the listener receives f′ = v/λ′ = f v/(v − vₛ). If the source is moving away, the gap grows by vₛT instead, and the denominator becomes v + vₛ.

Still source, moving listener

Now keep the source still. The wavelength is fixed at λ = v/f and the waves travel at v relative to the air. A listener walking toward the source meets the waves at the relative speed v + vₒ, so more crests per second pass them.

The frequency becomes f′ = (v + vₒ)/λ = f (v + vₒ)/v for a listener approaching, and f (v − vₒ)/v for one moving away. Here the wavelength does not change at all, which is the difference from the moving source case.

Take a 500 Hz source and a listener moving at 10 m/s. Going toward it, the listener hears 514.7 Hz. Going away, the listener hears 485.3 Hz. The two shifts are not exactly symmetric, because the formula divides by v and not by v ∓ vₒ.

One formula for everything

When both move, combine the two results. Take the direction from the listener toward the source as positive. Then f′ = f (v + vₒ)/(v − vₛ), where vₒ is positive when the listener moves toward the source and vₛ is positive when the source moves toward the listener. If either one moves the other way, put in a negative number.

The numerator comes from the listener (how fast waves are met) and the denominator from the source (how tightly waves are packed). Never swap them. If you are unsure, check the limit: a source speed close to v must make the answer blow up, and only the denominator can do that.

How the pitch changes with speed

The table fixes the source at 500 Hz and raises its speed. Look at the last column: it divides the pitch ahead by the pitch behind, which is the jump a listener hears at the instant the source passes.

The ahead column grows faster and faster, because v − vₛ is in the denominator and keeps shrinking. The behind column falls slowly and can never reach zero. That asymmetry is why the rise sounds sharper than the fall.

Solved problem 1: the ambulance

An ambulance siren emits 700 Hz. The ambulance moves at 30 m/s (108 km/h) along a straight road. Take the speed of sound as 340 m/s. Find the frequency heard by a still listener (a) as it approaches and (b) as it recedes, and the drop at the moment it passes.

Solved problem 2: a train and a walker

A train whistle emits 400 Hz. The train moves at 20 m/s (72 km/h) toward a station. (a) What does a person standing on the platform hear while the train approaches, and after it passes? (b) A cyclist rides toward the station at 10 m/s while a train whistles 600 Hz at 20 m/s, coming toward the cyclist. What does the cyclist hear?

Solved problem 3: find the speed from the pitches

A listener standing by a road hears a passing car horn at 540 Hz as it approaches and 480 Hz as it recedes. Find the speed of the car and the true frequency of the horn.

Past the speed of sound: the Mach cone

What happens when the source moves as fast as sound? The source keeps up with its own wavefronts. All the crests pile up on the same line in front of it, and the formula’s denominator v − vₛ becomes zero. A huge pressure ridge forms. This is the sound barrier.

Go faster still and the source overtakes its own waves. Each circle is left behind, and the circles are enveloped by a cone, called the Mach cone, with the source at its tip. Sound can only be heard once the cone sweeps across the listener, and it arrives all at once as a sharp bang, the sonic boom.

The Mach number is M = vₛ/v. The half-angle μ of the cone satisfies sin μ = v/vₛ = 1/M. The faster the source, the narrower the cone. At M = 2 the half-angle is exactly 30°.

At 340 m/s the speed of sound is 1,224 km/h in the air of this simulation, so a jet at Mach 1.5 is flying at 510 m/s, or 1,836 km/h, and its cone has a half-angle of 41.8°.

Mistakes that lose marks

Check these before you hand in an answer.

• Writing that the source frequency changes. It is the apparent frequency that changes.
• Using the same formula for a moving listener and a moving source. They are different, and the wavelength changes only in the second case.
• Mixing up the signs. Approaching means higher pitch, so the denominator shrinks or the numerator grows.
• Forgetting to convert km/h to m/s. Divide by 3.6.
• Applying the sound formulas to a source faster than sound. For vₛ ≥ v there is no frequency ahead; you must talk about the cone.
• Expecting a change when the source moves at right angles to the line of sight. At that instant the pitch is almost exactly the true one.

Where the Doppler effect works for us

Police radar guns send a radio beam at a car and measure how much the reflected beam is shifted; the shift gives the speed. Weather radar does the same with raindrops, which lets forecasters see rotation inside a storm.

In a medical Doppler scan, ultrasound bounces off moving blood cells, and the shift reveals the speed and direction of flow in an artery. The same trick lets a bat tell whether a moth is flying toward or away from it.

For light, the effect gives the red shift. Distant galaxies show their spectral lines moved toward red, which tells us that they are moving away. The light version has a slightly different formula because light needs no medium, but the idea of crests bunching or stretching is the same.

If you want to see how waves from two sources add up, try the wave interference animation next. For the vibrations that make the sound in the first place, open the standing waves animation.

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