Superposition — what happens when waves meet — Physics

Before this

This leads to

Two things in the same place at the same time

Two friends talk to you at once. The sound from both of them travels through the same air, at the same moment, to the same ear. Yet you can still pick out each voice. Neither sound has been damaged by meeting the other.

Try that with anything solid. Two balls rolled at each other collide, bounce, and change direction. Two cars in the same place at the same time is an accident. So how can two sounds share the same air without crashing?

Now a stranger fact. Noise-cancelling headphones make silence by adding more sound. A small speaker inside each cup plays a second sound on top of the noise, and the two together are quieter than the noise alone. Adding a sound to a sound can give less sound. This lecture explains both facts with one simple rule.

The rule: displacements add

The lecture on wave properties showed that a wave is a travelling pattern. Each particle of the medium only moves back and forth about its rest position. Its distance from that position, at any moment, is its displacement.

Now let two waves arrive at the same particle at the same time. Each wave, on its own, would give the particle a displacement. The particle cannot be in two places at once. So where does it go?

The answer is the simplest one possible. The particle's displacement is the sum of the two displacements, each with its sign. If one wave would lift a point of a rope by 3 cm and the other by 2 cm, the point rises 5 cm. If one would lift it by 3 cm and the other would pull it down by 3 cm, it stays at zero.

The principle of superposition

When two or more waves meet at a point, the resultant displacement at that point is the sum of the displacements that each wave would cause on its own.

Displacements are added with their signs: up is positive, down is negative.

Some books call this a "vector sum", because displacement has a direction. For a rope or a water surface, where every displacement is up or down along the same line, a vector sum is simply adding with signs.

There is a second half to the principle, and it answers the hook. Each wave carries on as if the other were not there. After they have passed through each other, both waves continue with their own shape, speed and direction. A wave is a pattern of motion, not a lump of matter, so two patterns can pass through the same place. That is why you can still hear both friends.

See it happen on a rope. Two pulses run towards each other. Watch what the rope does when they overlap, and what happens afterwards.

Simulation — two pulses meet on a rope

Choose the size and direction of each pulse, then press Play. Watch the solid line (the real rope) as the pulses overlap. Use Step to stop exactly at the meeting moment.

Pulse 1 (moving right): 2 cm

Pulse 2 (moving left): 2 cm

Presets:

A crest 2 cm high moves right and a trough 2 cm deep moves left along the same rope. Dashed lines: each pulse on its own. Solid line: the rope.

A crest (+2 cm) moving right meets a trough (−2 cm) moving left. Solid line: the rope. At the meeting moment the rope is flat, and the black lines show how fast each point is moving: down on the left, up on the right.

Before they meet, the rope shows both pulses. At the moment they overlap exactly, every displacement adds to zero, and the rope is flat. Afterwards, both pulses carry on, unchanged. With two crests of 2 cm, the rope would instead rise to 4 cm at the meeting moment.

What should you have seen? With two crests of 2 cm, the rope rose to 4 cm for an instant. With a crest and a trough of the same size, the rope went completely flat for an instant. In both cases the pulses came out on the other side unchanged. A big crest and a small trough gave a smaller crest, 3 − 1 = 2 cm, at the meeting moment.

Constructive and destructive interference

When waves add to give a bigger displacement, we call it constructive interference: crest meets crest, or trough meets trough. When they add to give a smaller displacement, we call it destructive interference: crest meets trough. If the crest and trough are the same size, the cancellation is complete.

Stop and think

At the moment the crest and the equal trough overlap, the rope is perfectly flat. Each pulse carried energy. Where is that energy at that instant?

Show the answer

In the motion of the rope. The rope is flat, but it is not still. On one side of the middle, both pulses are moving the rope downwards. On the other side, both are moving it upwards. So the rope passes through the flat shape at its fastest, and all the energy is kinetic energy for that instant. A moment later the rope bends again, and the pulses re-form. Destructive interference never destroys energy. It only means that, at that place and time, the displacements cancel.

Waves that keep on meeting

Pulses meet once. Continuous waves from two sources meet all the time. Suppose two dippers tap the surface of a ripple tank at the same rate, side by side. Circular waves spread from each. Everywhere the two sets of waves overlap, they add.

Use the words from the lecture on wave properties. If, at some point, the two waves always arrive in phase, crest with crest and trough with trough, the water there moves with twice the amplitude of one wave. If they always arrive in antiphase, crest with trough, the water there stays almost still. Most places are in between.

Waves spreading from two sources, S₁ and S₂, 6.0 cm apart, with wavelength 2.0 cm. Solid blue circles are crests, dashed grey circles are troughs. At P, a crest from each source arrives together: P is 10 cm from S₁ and 12 cm from S₂. At Q, a crest from S₁ meets a trough from S₂: Q is 9 cm from S₁ and 10 cm from S₂.

What decides which kind of place a point is? The two waves start in step at the sources. They travel different distances to reach the point. The difference in distance is called the path difference.

path difference = ∣ S_{1} P - S_{2} P ∣

This is the hard part of the lecture. Spend time here. Every extra wavelength of path puts one wave a whole cycle behind the other. A whole cycle behind is exactly in step again. So a path difference of 1λ, 2λ or 3λ makes no difference at all to how the waves meet. But an extra half wavelength puts a crest where a trough would be. This is the phase-difference rule from the wave properties lecture, applied to paths:

phase difference = \frac{path difference}{λ} \times 36 0^{\circ}

Path difference, for two sources in phase

If the path difference is a whole number of wavelengths, $nλ$ , the waves arrive in phase: constructive interference.

If it is a whole number plus a half, $(n + \frac{1}{2}) λ$ , they arrive in antiphase: destructive interference.

Check it on the figure. At P the path difference is 12 − 10 = 2 cm, which is exactly 1λ, so P is a place of strong waves. At Q it is 10 − 9 = 1 cm, which is ½λ, so Q is a still place.

Notice the words "for two sources in phase". If the sources themselves start in antiphase, one wave is already half a cycle behind before it sets off. Then every rule swaps: a whole number of wavelengths gives cancellation, and a half number gives reinforcement. Try both in the explorer.

Explore — loud and quiet places near two sources

Two sources, 6.0 cm apart, send out waves with wavelength 2.0 cm (an invented ripple tank). Move the point P up and down a line 10 cm away. Read the two distances, the path difference, and what the water does at P. Then switch the sources to antiphase.

Position of P along the line: 0 cm

Two sources 6.0 cm apart, in phase, wavelength 2.0 cm. The point P is on a line 10 cm away, at a height y above the middle. For example, at y ≈ 3.7 cm, S₁P ≈ 10.02 cm and S₂P ≈ 12.02 cm.

y (cm)	Path difference (cm)	In wavelengths	Water at P
0	0	0	strong
≈ 1.8	1.0	0.5	still
≈ 3.7	2.0	1	strong
≈ 6.0	3.0	1.5	still
≈ 9.2	4.0	2	strong

Where the path difference is a whole number of wavelengths, the waves arrive in phase and the water moves strongly. Where it is a whole number plus a half, they arrive in antiphase and the water is still. If the sources are in antiphase, every strong place becomes still and every still place becomes strong.

That is it. That is the whole idea. Superposition says the displacements add. Path difference tells you whether, at a given point, they add up or cancel out.

Why the sources must be coherent

A pattern of strong and still places stays fixed only if the two sources keep the same relationship to each other. They must have the same frequency, and a constant phase difference between them. Such sources are called coherent.

If the frequencies were different, the two waves would drift in and out of step, and every still place would soon become a strong place. If the phase difference kept changing at random, the pattern would move around faster than any eye or ear could follow, and you would see or hear only an average. This is why two torches shone at a wall never show light and dark bands. Each one gives out light in short, random bursts. Two loudspeakers driven by the same signal generator are coherent, and you can walk through their pattern of loud and quiet places. A laser gives coherent light, which is why lasers are used to show interference of light.

Worked examples

Worked Example 1 — Adding displacements

At one instant, two waves pass the same point P. Find the resultant displacement at P when the separate displacements are (a) +3 cm and +2 cm, (b) +3 cm and −3 cm, (c) +3 cm and −1 cm.

(a) $+ 3 + 2 = + 5$ cm. Both waves lift P, so the displacements reinforce: constructive.

(b) $+ 3 + (- 3) = 0$ cm. One lifts P and the other pulls it down by the same amount: complete cancellation.

Check: in every case the answer lies between the difference and the sum of the two sizes. ✓

Worked Example 2 — Strong or still?

Two coherent sources in phase send out waves of wavelength 4 cm. Point A is 12 cm from one source and 20 cm from the other. Point B is 10 cm and 14 cm away. Is each point a place of constructive or destructive interference? Give two path differences that would give destructive interference.

A: path difference $= 20 - 12 = 8$ cm. Divide by the wavelength: $8 \div 4 = 2$ . A whole number, so A is constructive.

B: path difference $= 14 - 10 = 4$ cm. $4 \div 4 = 1$ . Also a whole number, so B is constructive too.

Destructive needs $(n + \frac{1}{2}) λ$ : $0.5 \times 4 = 2$ cm, $1.5 \times 4 = 6$ cm, $2.5 \times 4 = 10$ cm, and so on.

Worked Example 3 — Divide by the wavelength first

Two coherent sources in phase have wavelength 3 cm. A point P is 15 cm from one and 10.5 cm from the other. Constructive or destructive?

Path difference $= 15 - 10.5 = 4.5$ cm.

Many students stop here: 4.5 is "not a whole number", so they are unsure. But the question is not whether the path difference is a whole number of centimetres. It is whether it is a whole number of wavelengths. So divide: $4.5 \div 3 = 1.5$ .

1.5 wavelengths is $(1 + \frac{1}{2}) λ$ , so the waves arrive in antiphase: destructive interference.

Check with the phase rule: $1.5 \times 36 0^{\circ} = 54 0^{\circ}$ , which is $36 0^{\circ} + 18 0^{\circ}$ . One whole cycle (no effect) plus half a cycle. ✓

Worked Example 4 — Two loudspeakers (invented)

Two loudspeakers, connected to the same signal generator, play a note of 686 Hz in phase. The speed of sound is 343 m/s. Mei stands 3.00 m from one speaker and 3.75 m from the other. Does she hear a loud or a quiet note? What if the speakers were wired in antiphase?

Wavelength: $λ = \frac{v}{f} = \frac{343}{686} = 0.500$ m.

Path difference $= 3.75 - 3.00 = 0.75$ m. In wavelengths: $0.75 \div 0.500 = 1.5$ . So the waves arrive in antiphase, and Mei is at a quiet place.

In antiphase, the speakers already start half a cycle apart. The extra 1.5 wavelengths of path adds another half cycle (after removing the whole one), so the waves arrive in phase, and the note is loud.

In a real room the quiet place is never perfectly silent. Sound also reaches her by reflecting from the walls, along other paths.

Three classic mistakes

Trap 1 — Judging the path difference in centimetres

A path difference of 4.5 cm is neither "whole" nor "half" until you compare it with the wavelength. Always divide by λ first. With λ = 3 cm it is 1.5λ (destructive); with λ = 1.5 cm it is 3λ (constructive).

Trap 2 — Forgetting that the sources must start in phase

"nλ is constructive" is true only for sources in phase. For sources in antiphase, the rules swap. Read the question to find out how the sources start.

Trap 3 — "Destructive interference destroys the waves"

The waves pass through each other unchanged, and their energy is not destroyed. At a still place the energy is simply not there: it arrives at the strong places instead. At a strong place, twice the amplitude means four times the energy of one wave, because energy goes with amplitude squared.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Two pulses on a rope overlap. At three points, the pulses on their own would give these displacements: X: +4 cm and +1 cm; Y: +2.5 cm and −2.5 cm; Z: −3 cm and +1.5 cm. Find the rope's displacement at each point, and say which kind of interference it is.

Hint 1
Add the two displacements, keeping the signs.

Hint 2
Same signs reinforce. Opposite signs cancel, completely or partly.

Full answer
X: $+ 4 + 1 = + 5$ cm, constructive. Y: $+ 2.5 + (- 2.5) = 0$ cm, complete destructive interference. Z: $- 3 + 1.5 = - 1.5$ cm, partial destructive interference: still a trough, but a smaller one. Check: each answer lies between the difference and the sum of the two sizes. ✓
Two coherent sources in phase have wavelength 2.0 cm. For each point, find the path difference in wavelengths, and decide: constructive, destructive, or in between. P: 13.0 cm and 9.0 cm. Q: 11.0 cm and 8.0 cm. R: 10.0 cm and 9.5 cm.

Hint 1
Subtract the distances, then divide by 2.0 cm.

Hint 2
A quarter of a wavelength is neither in phase nor in antiphase.

Full answer
P: $4.0 \div 2.0 = 2$ wavelengths: constructive. Q: $3.0 \div 2.0 = 1.5$ wavelengths: destructive. R: $0.5 \div 2.0 = 0.25$ wavelength, a phase difference of $0.25 \times 36 0^{\circ} = 9 0^{\circ}$ : in between, neither the largest nor zero. Check P: $2 \times 36 0^{\circ} = 72 0^{\circ}$ , two whole cycles, which is the same as in phase. ✓
Two loudspeakers in phase play a note of 1000 Hz. Sound travels at 343 m/s. A listener walks from a point where the paths are equal to the nearest quiet point. What is the path difference there? Then explain why the pattern would disappear if each speaker played music from a different phone.

Hint 1
Find the wavelength first, with $λ = v \div f$ .

Hint 2
The nearest quiet point has a path difference of half a wavelength. For the second part, think about coherence.

Full answer
$λ = \frac{343}{1000} = 0.343$ m. The first quiet point is at a path difference of $\frac{1}{2} λ = 0.1715 \approx 0.17$ m, about 17 cm. With two different phones, the two sounds have different and changing frequencies, so their phase difference changes all the time. They are not coherent, so no place stays quiet or loud. Check: $0.1715 \div 0.343 = 0.5$ . ✓
Two loudspeakers in phase stand 0.686 m apart on a line. A microphone is placed on the same line, beyond both speakers, so that one speaker is exactly 0.686 m further away than the other. The speed of sound is 343 m/s. (a) Find the three lowest frequencies that give a quiet signal at the microphone. (b) Explain why moving the microphone further along the same line does not change your answer. (c) Which of your frequencies would noise-cancelling headphones find easiest to cancel, and why?

Hint 1
The path difference is always 0.686 m. Quiet needs this to be $(n + \frac{1}{2}) λ$ .

Hint 2
Solve $0.686 = 0.5 λ$ , $1.5 λ$ , $2.5 λ$ , then use $f = v \div λ$ . For (c), compare the wavelengths with small errors in position.

Full answer
(a) $λ = \frac{0.686}{0.5} = 1.372$ m gives $f = \frac{343}{1.372} = 250$ Hz. $λ = \frac{0.686}{1.5} \approx 0.457$ m gives 750 Hz. $λ = \frac{0.686}{2.5} = 0.2744$ m gives 1250 Hz. (b) Beyond both speakers, moving the microphone adds the same distance to both paths, so the path difference stays 0.686 m. (c) 250 Hz. Its wavelength, 1.372 m, is long, so a small error in timing or position changes the phase only a little, and the cancellation still works. At short wavelengths, a shift of a few centimetres is already a large part of a cycle. This is one reason why noise-cancelling headphones work best on low, steady sounds such as engine hum. Check: $1.5 \times 0.457 \approx 0.686$ . ✓

The bigger idea

Superposition is how waves reveal themselves. Particles collide; waves pass through each other and add. So whenever something produces a pattern of strong and still places, that is evidence that it is a wave. The colours of a soap bubble are one example: light reflected from the front and back of the thin film travels slightly different distances, so some colours reinforce and others cancel, and the colours change as the film gets thinner. Noise-cancelling headphones are another: a microphone measures the noise, and a speaker adds a copy in antiphase, so the two cancel at your ear. They work best on low, steady sounds, below about 1000 Hz. The next lecture uses the same rule on light passing through two narrow slits. That experiment settled one of the great arguments in the history of physics.

Key concepts in this lesson

The principle of superposition says that where waves meet, the resultant displacement is the sum of the separate displacements, added with their signs.
Waves pass through each other and carry on unchanged, because a wave is a travelling pattern of motion, not a lump of matter.
Crest meeting crest gives constructive interference, and crest meeting trough gives destructive interference, which cancels completely when the two are the same size.
For two sources in phase, a path difference of $nλ$ gives constructive interference and $(n + \frac{1}{2}) λ$ gives destructive interference, and for sources in antiphase the rules swap.
A fixed interference pattern needs coherent sources, with the same frequency and a constant phase difference, and destructive interference moves energy to other places rather than destroying it.