Interference and Superposition

Beginner

When two waves meet, they do not bounce off each other like balls. They simply add up: at each point the height is what the first wave would make plus what the second would make. Afterwards each goes on its way unchanged, just as two conversations cross a room without spoiling each other. This adding-up is called superposition.

The animation shows two waves (dashed) on the same string and their sum (solid). On the top string the waves are in step: crest meets crest and trough meets trough, so the string swings twice as high. This is constructive interference.

On the lower string one wave is shifted by half a wavelength, so every crest meets a trough. One wave pushes up exactly as much as the other pulls down, and the string barely moves: destructive interference. Noise-cancelling headphones use this trick, playing an upside-down copy of the noise into your ear.

Where does the energy go at a quiet spot? It is not destroyed. With two loudspeakers playing the same steady note, you can walk across a room and find quiet places right next to loud ones: the energy is only moved around. Perfect silence is rare, because walls reflect sound too.

When two notes are almost, but not quite, the same, the waves drift slowly in and out of step, and the sound swells and fades: wah, wah, wah. These are beats, and there are as many per second as the difference between the two frequencies. A guitarist tunes two strings by turning a peg until the beats slow down and stop.

Check yourself: tuning forks of 440 Hz and 443 Hz sound together. How many beats per second do you hear? Answer: three.

Intermediate

Waves add, point by point

Noise-cancelling headphones and soap-bubble colours rest on one rule. Where two waves pass through the same place, each point of the medium moves by the sum of the displacements that each wave alone would give it:
$y=y_1+y_2$

This is the principle of superposition. Displacements carry signs, so a crest of 1 cm meeting a trough of 1 cm gives zero. It holds in a linear medium, one whose restoring force is proportional to the displacement: strings, air and water at small amplitudes, and light in vacuum.

Each wave carries on as if the other were not there: two pulses running towards each other along a rope add while they overlap, then separate unchanged. Note what adds: displacements, not intensities.

Phase difference and path difference

Whether two waves of the same frequency help or cancel each other depends on their phase difference $\phi$, how far one is ahead of the other in its cycle ($2\pi$ for a whole cycle). For sources that vibrate in step, it comes from the path difference $\Delta =r_2-r_1$, the extra distance one wave travels to the point; each wavelength of extra path is one cycle of delay:
$\phi =\frac{2\pi \Delta }{\lambda }$

So $\Delta =0,\lambda ,2\lambda ,\dots$ brings the waves in step, and $\Delta =\lambda /2,3\lambda /2,\dots$ brings crest onto trough. Only $\Delta$ in wavelengths matters: 3.2 wavelengths acts like 0.2.

A steady pattern needs coherent sources, whose phase difference stays fixed, like two loudspeakers fed by one signal. If the frequencies differ slightly, the phase difference drifts, and the loudness swells and fades as many times per second as the frequencies differ:
$f_{\mathrm{beat}}=|f_1-f_2|$
Strings at 196 Hz and 198 Hz give two beats per second; the animation's expert layer shows them on a third string.

The conditions for loud and quiet

For sources in step and waves of amplitudes $A_1$ and $A_2$ at the point, the two extreme cases are

• $\Delta =m\lambda \to A=A_1+A_2\text{(constructive)}$
• $\Delta =\left(m+\frac{1}{2}\right)\lambda \to A=|A_1-A_2|\text{(destructive)}$

Here $m=0,1,2,\dots$, and $A$ is the amplitude of the combined oscillation. Complete silence needs equal amplitudes.

For two equal amplitudes $a$, any path difference in between gives
$A=2a\left|\cos \frac{\pi \Delta }{\lambda }\right|$
which falls smoothly from $2a$ at $\Delta =0$ to zero at $\Delta =\lambda /2$.

What we hear or see is the intensity $I$, the power through each square metre, which grows as the square of the amplitude. Two equal sources that each give $I_0$ alone therefore produce
$I=4I_0{\cos }^2\left(\frac{\pi \Delta }{\lambda }\right)$
Bright places get $4I_0$, dark places nothing, and the average is $2I_0$, as without interference: energy is moved from quiet places to loud ones, not destroyed.

Worked example: the strings in the animation

Each dashed wave has amplitude $a=1\mathrm{cm}$, wavelength $\lambda =10\mathrm{cm}$ (the ruler above the top string) and frequency 0.5 Hz, so each point repeats its motion every 2 s and the waves move at $v=f\lambda =5cm/s$.

On the top string the gold wave B lies exactly on the blue wave A, so $\Delta =0$ and the amplitudes add:
$A=A_1+A_2=1+1=2\mathrm{cm}$
The readout at P swings between $-2.0$ and $+2.0\mathrm{cm}$, and the red sum is twice as tall as either dashed wave, with $2^2=4$ times the intensity of one.

On the second string wave B is shifted by half a wavelength, $\Delta =\lambda /2=5\mathrm{cm}$, so every crest of A meets a trough of B:
$A=|A_1-A_2|=|1-1|=0$
The red sum lies flat, and the readout stays at 0.0 cm.

In between, a quarter-wavelength shift of 2.5 cm would give $A=2\cos (\pi /4)\approx 1.41\mathrm{cm}$ and exactly $2I_0$; a shift of $\lambda /3$ gives $A=2\cos (\pi /3)=1\mathrm{cm}$, no more than one wave alone.

Worked example: two loudspeakers

Two small loudspeakers are driven in phase by the same 686 Hz signal. With the speed of sound at 343 m/s, the wavelength is $\lambda =v/f=343/686=0.50\mathrm{m}$. You stand 3.00 m from one speaker and 3.75 m from the other, so
$\Delta =3.75-3.00=0.75\mathrm{m}=1.5\lambda$
That is $(m+\frac{1}{2})\lambda$ with $m=1$: you are at a quiet spot.

How quiet? The amplitude of a small source falls as $1/r$, so the far speaker's amplitude is $3.00/3.75=0.80$ of the near one's. What is left is $1-0.80=0.20$, an intensity of $0.20^2=0.04$ of the near speaker alone, while a nearby loud spot gets $1.80^2=3.24$: 81 times more, about 19 dB.

The spot also depends on the note: 0.75 m is an odd number of half wavelengths at 229, 686 and 1143 Hz, which are quiet there, but a whole number of wavelengths at 457 and 915 Hz, which are loud. So music is never silenced at one place.

Adding waves by eye

To sketch a superposition, add the signed heights of the two waves at a few points and join the results: where one wave is at zero the sum equals the other, and where the two are equal and opposite it vanishes. In the animation the dashed blue and gold curves are the waves, the solid red one their sum, and the dashed line through P marks where the readouts measure $y$.

Two waves of the same frequency travelling the same way add up to a wave of that frequency and speed: the red crests on the top string keep pace with the dashed ones at 5 cm/s. The flat second string, with no wave and no energy left, is best read as what arrives at one point by two paths; with separate sources, cancellation in one place always comes with reinforcement elsewhere.

Common mistakes

• Adding intensities instead of amplitudes: two equal waves in step give $2^2=4$ times the intensity of one, not 2; only the average over the pattern is twice one source.
• Believing that destructive interference destroys energy: the quiet spot's 0.04 comes with loud spots at 3.24, whose average, 1.64, is exactly $1+0.80^2$, the sum of the two speakers' separate intensities.
• Expecting silence whenever waves are out of step: that needs equal amplitudes, and waves of 4 cm and 1 cm still leave 3 cm.
• Thinking that the distance itself decides: only the difference counts, so 3.00 m and 3.75 m, or 10.00 m and 10.75 m, both give quiet spots at 686 Hz.
• Expecting a pattern from any two sources: two lamps, or two violinists on one note, have wandering phases, so only the summed intensity, $2I_0$, remains.

Where you meet it

Noise-cancelling headphones play an inverted copy of incoming noise and work best on steady low hums, such as an aircraft cabin's. Soap bubbles get their colours from light reflected by the two surfaces of a thin film, and the anti-reflection coating of a camera lens is about 100 nm thick so that two reflections of green light cancel.

Wi-Fi at 2.4 GHz (wavelength 12.5 cm) reaches a laptop by several paths, so moving it a few centimetres can turn a weak spot into a strong one, and gravitational-wave detectors use interference to sense length changes far smaller than a proton. Standing Waves adds waves running in opposite directions, and Diffraction and the Double Slit applies these conditions to light.

Advanced

Why waves simply add

Everything in this lesson rests on one fact: where two waves overlap, their displacements add. This section shows why that is so, and the next sections draw the consequences.

Superposition is not an extra law of nature. It is a property of the equation the wave obeys. Take a stretched string and let $y(x,t)$ be the small sideways displacement, at time $t$, of the piece of string at position $x$. Newton's second law applied to each short piece gives, for small displacements, the wave equation, in which $v$ is the speed of waves on the string:
$\frac{{\partial }^{2}y}{\partial t^2}=v^2\frac{{\partial }^{2}y}{\partial x^2}$

This equation is linear: each term contains $y$ once and only to the first power, with no $y^2$ and no product of $y$ with its own derivatives. Suppose $y_1$ and $y_2$ are both solutions, and try $y=c_1{y}_1+c_2{y}_2$ with any constants $c_1$ and $c_2$. The derivative of a sum is the sum of the derivatives, and constants come out in front, so each side of the equation splits into a $y_1$ part and a $y_2$ part. Each part balances on its own, so the combination is a solution too. That is the principle of superposition: two waves pass through each other, and at every point the displacement is simply $y_1+y_2$.

The general solution of the wave equation, $y=F(x-vt)+G(x+vt)$, shows it at work: $F$ is a pulse of any shape moving right, $G$ one moving left, and the two cross with their shapes intact. Small pressure changes in air, light in vacuum, whose field equations are exactly linear, and the wavefunction of quantum mechanics all obey linear equations of this kind, so everything below applies to them as well.

Adding two oscillations with phasors

At any one point, each wave makes the medium oscillate. To find the combined motion we must add two oscillations of the same frequency but with different amplitudes and phases. Adding cosines directly is clumsy; drawing each oscillation as a turning arrow turns the sum into simple geometry.

Legend

$A_1$	Amplitude of the first oscillation: the length of the blue arrow.	$\mathrm{m}$
$A_2$	Amplitude of the second oscillation: the length of the gold arrow.	$\mathrm{m}$
$A$	Amplitude of the sum: the red arrow, which closes the triangle.	$\mathrm{m}$
$\omega t+{\phi }_1$	Phase of the first oscillation: the angle of the blue arrow from the axis, growing at the rate $\omega$.	$\mathrm{rad}$
$\phi$	Phase difference ${\phi }_1-{\phi }_2$: the angle between the two arrows, fixed while both turn.	$\mathrm{rad}$
$y_1$	Displacement due to the first oscillation: the projection of the blue arrow on the axis.	$\mathrm{m}$
$y_2$	Displacement due to the second oscillation: the projection of the gold arrow on the axis.	$\mathrm{m}$
$y$	Displacement of the sum, $y_1+y_2$: the projection of the red arrow.	$\mathrm{m}$

The two oscillations are $y_1=A_1\cos (\omega t+{\phi }_1)$ and $y_2=A_2\cos (\omega t+{\phi }_2)$, with the same angular frequency $\omega$. Figure 1 draws each one as an arrow, called a phasor. The length of the arrow is the amplitude: $A_1$ for the blue arrow, $A_2$ for the gold one. Its angle from the horizontal axis is the phase: $\omega t+{\phi }_1$ for the blue arrow, marked at its foot, and $\omega t+{\phi }_2$ for the gold one. The oscillation itself is the projection of the arrow on the axis, its shadow: an arrow of length $A_1$ at the angle $\omega t+{\phi }_1$ has the projection $A_1\cos (\omega t+{\phi }_1)$, which is exactly $y_1$. As time passes the arrow turns anticlockwise at the rate $\omega$, and its shadow moves back and forth.

Now look at three things in the figure. First, the gold arrow starts at the tip of the blue one, head to tail, and the red arrow $A$ runs from the foot of the blue arrow to the tip of the gold one: $A$ is the vector sum of the two. Second, both arrows turn at the same rate $\omega$, so the angle between them, $\phi ={\phi }_1-{\phi }_2$, marked between the gold arrow and the dashed extension of the blue one, never changes. The shaded triangle therefore keeps its shape and turns as one rigid piece, and $A$ turns at the rate $\omega$ too. Third, along the axis the shadows $y_1$ and $y_2$ lie one after the other and together span $y$, the shadow of $A$, because the projection of a sum of vectors is the sum of their projections. So $y=y_1+y_2$ is itself an oscillation at the same angular frequency $\omega$, and its amplitude is the length $A$.

The length $A$ follows from the triangle. The angle inside the triangle where the two arrows meet is $\pi -\phi$, because $\phi$ is measured from the extension of the blue arrow. The law of cosines, together with $\cos (\pi -\phi )=-\cos \phi$, gives
$A^2=A_1^2+A_2^2+2A_1{A}_2\cos \phi$
The amplitude is largest, $A_1+A_2$, when $\phi =0$ and the arrows point the same way, and smallest, $|A_1-A_2|$, when $\phi =\pi$ and they point opposite ways.

In complex numbers the same sum reads as follows. Each oscillation is the real part of $A_j{e}^{i(\omega t+{\phi }_j)}$, the common factor $e^{i\omega t}$ comes out, and $A^2=|A_1{e}^{i{\phi }_1}+A_2{e}^{i{\phi }_2}{|}^2$; the argument of that complex sum is the phase of the resultant. Multiplying out the squared modulus gives the law of cosines again.

Where does $\phi$ come from? Take two sources of the same angular frequency $\omega$, with their own phases ${\alpha }_1$ and ${\alpha }_2$, and a point at distances $r_1$ and $r_2$ from them. A wave travelling at speed $v$ needs the time $r_j/v$ to arrive, so what reaches the point at time $t$ left source $j$ at the time $t-r_j/v$. Its phase is therefore $\omega (t-r_j/v)+{\alpha }_j=\omega t+{\alpha }_j-k{r}_j$, where $k=\omega /v=2\pi /\lambda$ is the wavenumber and $\lambda$ the wavelength. So ${\phi }_j={\alpha }_j-k{r}_j$, and in the difference the common $\omega t$ drops out:
$\phi ={\phi }_1-{\phi }_2=\frac{2\pi \Delta }{\lambda }+({\alpha }_1-{\alpha }_2),\Delta =r_2-r_1$
Here $\Delta$ is the path difference, the extra distance travelled by the wave from the second source. Each wavelength of extra path adds $2\pi$ to the phase difference.

For two equal amplitudes $a$ from sources in step (${\alpha }_1={\alpha }_2$), the law of cosines becomes $A^2=2a^2(1+\cos \phi )$. The identity $1+\cos \phi =2{\cos }^2(\phi /2)$ turns this into $A^2=4a^2{\cos }^2(\phi /2)$, and since $\phi /2=\pi \Delta /\lambda$,
$A=2a\left|\cos \frac{\pi \Delta }{\lambda }\right|$
Special cases can be read straight off the triangle. At $\Delta =\lambda /3$ the phase difference is $2\pi /3$, or 120°, so the angle inside the triangle is 60°; two equal sides around a 60° angle make an equilateral triangle, and $A=a$. At $\Delta =\lambda /4$ the arrows are perpendicular, $A$ is the hypotenuse of a right isosceles triangle, and $A=\sqrt{2}a$.

Intensity and where the energy goes

An ear or an eye does not register the amplitude but the intensity $I$, the power carried through each square metre, which is proportional to the square of the amplitude. Turning the amplitude law into an intensity law also shows that interference only moves energy around.

Multiply the law of cosines by the constant that turns $A^2$ into intensity. The terms $A_1^2$ and $A_2^2$ become $I_1$ and $I_2$, the intensities each source gives alone, and $A_1{A}_2$ becomes $\sqrt{I_1{I}_2}$:
$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \phi$
For two equal sources in step that each give $I_0$ alone, $\phi =2\pi \Delta /\lambda$ and
$I=4I_0{\cos }^2\left(\frac{\pi \Delta }{\lambda }\right)$
The last term of the general law, $2\sqrt{I_1{I}_2}\cos \phi$, is called the cross term, and it is the whole of interference: without it the intensities would simply add. Across a screen or a room, $\phi$ runs through many whole cycles, and the average of $\cos \phi$ over a whole cycle is zero. What remains on average is $I_1+I_2=2I_0$: the bright fringes, at $4I_0$, gain exactly what the dark ones lose.

That energy balance assumes the sources are many wavelengths apart. Close together they are coupled: each source pushes on a medium that the wave of the other is also moving. If that wave arrives in step with the push, the source works against a higher pressure, or field, and radiates more; if it arrives out of step, the source radiates less. Two small sources a distance $d$ apart, driven with equal strength and a phase difference $\alpha$, radiate in total
$P=2P_0\left(1+\cos \alpha \frac{\sin kd}{kd}\right)$
where $P_0$ is the power one source radiates alone. Two checks: when $kd$ is large, $\sin kd/(kd)$ is small and $P\approx 2P_0$, the sum of the separate powers; when $d\to 0$ the factor tends to 1, so two sources in step act as one source of double strength, $P=4P_0$, while two in opposite phase cancel.

For example, two loudspeakers 1.5 m apart playing 50 Hz (wavelength 6.86 m in air, so $kd=1.37$) radiate $3.43P_0$ in step, but only $0.57P_0$ with one wired backwards, which reverses its phase. That is a factor of 6, almost 8 dB of bass lost. Energy is still conserved: the air loads each speaker differently, so the amplifier simply delivers a different power.

Path difference far away, and steering a beam

Antennas, loudspeakers and slits are usually observed from far away, at many times their separation. Then the path difference depends only on the direction, and the interference pattern becomes a pattern of directions. Figure 2 finds that path difference.

Legend

$S_1$	First source.
$S_2$	Second source.
$d$	Distance between the two sources.	$\mathrm{m}$
$\theta$	Direction of the distant point, measured from the perpendicular bisector of $S_1{S}_2$.	$\mathrm{rad}$
$\Delta$	Path difference: how much farther the wave from $S_2$ travels than the wave from $S_1$.	$\mathrm{m}$

In Figure 2 the sources $S_1$ and $S_2$ are a distance $d$ apart, and we look towards a distant point in the direction at the angle $\theta$ from their perpendicular bisector, the horizontal dashed line. Because the point is so far away, the two rays to it, blue from $S_1$ and gold from $S_2$, are practically parallel. The dashed line from $S_1$ to the gold ray meets that ray at a right angle. Beyond this line both paths are equally long, so the only extra length of the path from $S_2$ is the purple piece between $S_2$ and the foot of the dashed line. That piece is $\Delta$.

The purple piece is one side of a right triangle whose hypotenuse is $S_1{S}_2$, of length $d$. The angle of that triangle at $S_1$ equals $\theta$: the line $S_1{S}_2$ is at right angles to the bisector, the dashed line is at right angles to the rays, and turning two lines by the same right angle leaves the angle between them unchanged. The purple side lies opposite that angle, so
$\Delta =d\sin \theta$
For sources in step, the waves therefore arrive in step, and form a strong beam, in every direction where $d\sin \theta$ is a whole number of wavelengths.

The phases of the sources can also be set on purpose. Let $S_1$ lag behind $S_2$ by the phase $\alpha$, so that ${\alpha }_1-{\alpha }_2=-\alpha$. Then $\phi =2\pi d\sin \theta /\lambda -\alpha$, and the strongest direction, where $\phi =0$, moves to
$\sin \theta =\frac{\alpha \lambda }{2\pi d}$
The beam swings towards the side of the delayed source, $S_1$: its shorter path makes up exactly for its late start. A 10 GHz radar ($\lambda =3.0\mathrm{cm}$) with elements 1.5 cm apart, each a quarter cycle (90°) behind its neighbour, aims where $\sin \theta =0.5$, at 30°. Changing the phases electronically steers the beam with no moving parts: this is how phased-array antennas work.

Beats: interference in time

So far both waves had the same frequency, so the phase difference at a point stayed fixed. If the frequencies differ slightly, the phase difference grows steadily with time, and the combined loudness at one place rises and falls. These swells are beats, which musicians use to tune their instruments.

Legend

$a$	Amplitude of each oscillation: the length of the blue arrow and of the gold arrow.	$\mathrm{m}$
$A$	Amplitude of the sum: the length of the red arrow, plotted as the red curve.	$\mathrm{m}$
$2a$	Largest value of $A$, reached when the two arrows point the same way.	$\mathrm{m}$
$\Delta \omega t$	Angle by which the gold arrow has dropped back from the blue one; it grows in proportion to the time $t$.	$\mathrm{rad}$
$\pi$	Half a turn: the arrows point opposite ways and $A=0$.	$\mathrm{rad}$
$2\pi$	A whole turn: the arrows line up again, and one beat is complete.	$\mathrm{rad}$

Figure 3 shows beats with the same arrows. Both oscillations have the amplitude $a$, and their angular frequencies ${\omega }_1$ and ${\omega }_2$ differ by a small amount $\Delta \omega ={\omega }_1-{\omega }_2$. Look at the arrows from the blue one, as if you turned with it at ${\omega }_1$, so that it stays still along the axis. The gold arrow, turning at the slower ${\omega }_2$, falls steadily behind: after a time $t$ it has dropped back by the angle $\Delta \omega t$. Its foot stays at the tip of the blue arrow, so its own tip runs round the dashed circle of radius $a$ centred there. The red sum $A$ therefore stretches and shrinks: it is $2a$ when the two arrows line up and $0$ when they point opposite ways, at $\Delta \omega t=\pi$.

The two equal arrows and their sum form an isosceles triangle, which gives the length of the sum at any moment, $A=2a|\cos (\Delta \omega t/2)|$. The graph on the right plots it against $\Delta \omega t$, and its dot marks the moment drawn on the left. The curve repeats every $2\pi$ of $\Delta \omega t$, that is, every $2\pi /\Delta \omega$ in time. That is one beat: the time the faster arrow needs to gain a whole turn on the slower one.

Algebra gives the same result. At a fixed point take $y=a\sin {\omega }_{1}t+a\sin {\omega }_{2}t$. The identity $\sin p+\sin q=2\cos \frac{p-q}{2}\sin \frac{p+q}{2}$ turns this into
$y=2a\cos \left(\frac{\Delta \omega t}{2}\right)\sin (\bar{\omega }t)$
where $\bar{\omega }=({\omega }_1+{\omega }_2)/2$ is the mean angular frequency. The second factor is a fast oscillation at the mean frequency; the first is a slow envelope that sets its size.

The third string of the animation, shown at this level, adds two waves of amplitude $a=1\mathrm{cm}$ with the frequencies $f_1=1.5\mathrm{Hz}$ and $f_2=1.25\mathrm{Hz}$ (in hertz, $f=\omega /(2\pi )$). The fast oscillation runs at their mean, 1.375 Hz, and the envelope's cosine at half their difference, 0.125 Hz: one period of the cosine per 8 s loop. Loudness follows the size of the envelope, not its sign, and that size peaks twice in each period of the cosine, once where the cosine is $+1$ and once where it is $-1$. So the beats come at the full difference,
$f_{\mathrm{beat}}=|f_1-f_2|$
here 0.25 Hz, one beat every 4 s.

On the graph of $y(t)$ at the point P, the purple trace swells to almost 2 cm near $t=0.4\mathrm{s}$ and $4.4\mathrm{s}$ and pinches to zero near 2.4 s and 6.4 s, always inside the dashed envelope $\pm 2a$.

Each beat holds $1.375\cdot 4=5.5$ fast oscillations, because in 4 s wave 1 makes 6 cycles and wave 2 makes 5. The extra half cycle means that the fast motion restarts upside down after every pinch, which is why the envelope's cosine needs two beats to repeat. In Figure 3 this is the faster arrow overtaking the slower one once per beat. With unequal amplitudes $A_1$ and $A_2$, the law of cosines with $\phi =\Delta \omega t$ gives the intensity
$I\propto A_1^2+A_2^2+2A_1{A}_2\cos (\Delta \omega t)$
which swings between $(A_1+A_2{)}^2$ and $(A_1-A_2{)}^2$. The beats are still there, but the sound never falls silent.

Modulation: beats read backwards

Beats show that two close frequencies added together look like one frequency with a slowly changing amplitude. Read the other way round, a wave whose amplitude is made to change contains extra frequencies. This is how AM radio carries sound.

Let a carrier of angular frequency ${\omega }_c$ have its amplitude follow a tone of angular frequency ${\omega }_m$, with the modulation depth $b$: the signal is $(1+b\cos {\omega }_{m}t)\cos {\omega }_{c}t$. The product rule $\cos p\cos q=\frac{1}{2}[\cos (p+q)+\cos (p-q)]$ splits it into the carrier itself plus two side bands of amplitude $b/2$, at ${\omega }_c+{\omega }_m$ and ${\omega }_c-{\omega }_m$. An AM station at 1.2 MHz carrying sound up to 4.5 kHz therefore fills the band from 1195.5 to 1204.5 kHz, 9 kHz wide, which is why AM channels in Europe are spaced 9 kHz apart.

Many sources: a polygon of arrows

Gratings and antenna arrays combine many sources, not two. The phasor picture handles any number of them: the arrows are simply laid head to tail in a longer chain.

Legend

$N$	Number of equal sources, here 5.
$a$	Amplitude of one source: the length of each blue arrow.	$\mathrm{m}$
$A$	Amplitude of the sum: the red arrow from the start of the chain to its end.	$\mathrm{m}$
$\phi$	Phase step between neighbouring sources: the angle each arrow is turned from the one before.	$\mathrm{rad}$
$N\phi$	Angle that the whole chain, from the first arrow to the last, spans at the centre of the dashed circle.	$\mathrm{rad}$

Figure 4 takes $N$ equal sources in a row, each a phase $\phi$ ahead of the one before; on the left $N=5$. All $N$ arrows have the same length $a$, and each is turned by the same angle $\phi$ from the one before it. A chain with equal sides and the same turn at every corner is part of a regular polygon, so its corners lie on one circle, the dashed one; let $R$ be its radius. Seen from the centre, each side spans the angle $\phi$, and the whole chain, from the foot of the first arrow to the tip of the last, spans $N\phi$. The sum $A$, the red arrow, closes the chain.

A chord that spans the angle $\beta$ at the centre of a circle of radius $R$ has the length $2R\sin (\beta /2)$: the line from the centre to its midpoint cuts the isosceles triangle into two right triangles. So one side is $a=2R\sin (\phi /2)$, and the closing chord is $A=2R|\sin (N\phi /2)|$. Dividing the second by the first removes the unknown radius $R$:
$A=a\left|\frac{\sin (N\phi /2)}{\sin (\phi /2)}\right|$
For $N=2$, the identity $\sin \phi =2\sin (\phi /2)\cos (\phi /2)$ brings back $A=2a|\cos (\phi /2)|$.

Two cases stand out. When $\phi$ is a multiple of $2\pi$, all the arrows point the same way, so $A=Na$ and the intensity is $I=N^2{I}_0$, where $I_0$ is the intensity of one source alone. As $\phi$ grows from zero the chain curls up, and the first zero comes at $\phi =2\pi /N$: the $N$ turns of $\phi$ then add up to a whole turn and the polygon closes on itself, as on the right of Figure 4.

With ten sources the peaks are 100 times as intense as one source, and their first zeros lie at $\phi =2\pi /10$ instead of $\pi$ for two sources, so the peaks are five times narrower. The average intensity is still $10I_0$, because the cross terms average to zero: the energy is gathered into narrow, bright peaks. This is the principle of diffraction gratings and radio-telescope arrays.

Standing waves and wave packets

Superposition also explains two patterns met in later lessons: waves that stand still, and waves confined to a short packet.

Two equal waves running in opposite directions add, by the identity used for beats, to
$a\sin (kx-\omega t)+a\sin (kx+\omega t)=2a\sin kx\cos \omega t$
Every point oscillates in step with every other, with an amplitude $2a|\sin kx|$ fixed by its position. Where $\sin kx=0$, every $\lambda /2$, there are nodes that never move: this is a standing wave.

Adding many neighbouring frequencies, spread over a band of width $\Delta f$, makes a single wave packet. The components agree in phase at one moment and drift out of step within a time of about $1/\Delta f$, so the shorter the packet, the wider the band it needs: roughly $\Delta f\Delta t\approx 1$.

The packet's envelope moves at the group velocity $d\omega /dk$; for two waves this is the speed $\Delta \omega /\Delta k$ of the beat pattern. On the animation's string every frequency travels at the same 5 cm/s, so $\omega =vk$ and the group velocity equals the wave speed: the beat envelope, which repeats every $v/\Delta f=20\mathrm{cm}$ along the string, keeps pace with the waves. In deep water, where long waves run faster than short ones, it would not: there the envelope moves at only half the speed of the crests.

Worked problem: a detuned loudspeaker

This problem brings together the amplitude law, the contrast of a pattern and beats.

Take the two loudspeakers of the intermediate level, in step at 686 Hz, so that $\lambda =0.50\mathrm{m}$ in air at 343 m/s. A listener sits 3.00 m from one and 3.75 m from the other. The amplitude of each speaker falls off as $1/r$, so at the listener the far one is weaker by the ratio $3.00/3.75=0.80$. Find the contrast of the pattern, then let the far speaker drift to 686.25 Hz.

Before the drift, the path difference is $\Delta =0.75\mathrm{m}=1.5\lambda$, so $\phi =2\pi \cdot 1.5=3\pi$ and $\cos \phi =-1$: the listener sits at a quiet spot. Measure intensities in units of the near speaker alone, so that $A_1=1$ and $A_2=0.8$. The quietest and loudest values the pattern can take, at $\cos \phi =-1$ and $\cos \phi =+1$, are
$I_{\min }=(1-0.8{)}^2=0.04,I_{\max }=(1+0.8{)}^2=3.24$
The visibility, or contrast, of the pattern compares the two. Since $I_{\max }-I_{\min }=4A_1{A}_2$ and $I_{\max }+I_{\min }=2(A_1^2+A_2^2)$,
$V=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{2A_1{A}_2}{A_1^2+A_2^2}=\frac{1.6}{1.64}=0.98$
A 20 % mismatch in amplitude still leaves a dip of $10{\log }_{10}(3.24/0.04)\approx 19\mathrm{dB}$ below the loudest value: interference is forgiving.

Now detune the far speaker by 0.25 Hz. Its phase gains a whole cycle on the other every 4 s, so $\phi$ runs through all values, and the listener hears beats at 0.25 Hz, the rhythm of the animation's third string: 0.04 at $t=0$, 3.24 two seconds later, 0.04 again at 4 s.

The same drift moves the pattern in space. On the line between the speakers, a step $\delta$ towards one speaker shortens one path by $\delta$ and lengthens the other by $\delta$, so it changes $\Delta$ by $2\delta$; the quiet spots, one wavelength of $\Delta$ apart, are therefore $\lambda /2=0.25\mathrm{m}$ apart. As the phase drifts, the whole pattern slides by one spacing per beat, 0.25 m in 4 s or about 6 cm/s, towards the speaker with the lower frequency. Beats in time and fringes in space are one phenomenon: a pattern sweeping past a fixed ear.

Averaged over a beat, the cross term vanishes and leaves $1+0.8^2=1.64$, the sum of the two separate intensities. If the detuning wandered randomly by a few hertz, every seat would receive that same 1.64 on average: independent sources add intensities, not amplitudes.

Limits of validity

Superposition is exact only while the medium responds linearly, that is, while doubling the disturbance doubles the response. Strong waves break this rule, and coherence sets a second limit.

A sound wave cannot have a pressure amplitude larger than atmospheric pressure, because the pressure in a rarefaction cannot fall below zero; for a pure tone that limit is about 191 dB. Long before it is reached, the compressions of a loud wave travel faster than its rarefactions, and the wave steepens into a shock. A string swung too far stretches, so its tension and its pitch rise, and a water wave breaks when its height reaches about a seventh of its wavelength.

Light in vacuum obeys linear equations: the scattering of light by light that quantum theory predicts has been seen only in particle colliders. In some crystals, though, intense light drives the electrons beyond their linear response and creates new frequencies: a green laser pointer turns 1064 nm infrared light into 532 nm green light, at twice the frequency.

Steady fringes also need coherence: the two waves must keep a fixed phase difference. A real wave train with a spread of frequencies $\Delta f$ keeps a steady phase only over its coherence length, $L_c\approx c/\Delta f$, where $c$ is the speed of light. For sunlight, which spans the whole visible band, that is about 1 µm, which is why soap films show colours only while they are no thicker than about a micrometre. For a laser with $\Delta f\approx 1\mathrm{MHz}$ it is some 300 m. Detectors average too: eyes and photodiodes record only the mean intensity, and the ear follows beats only up to some 15 per second.

Where this leads

Two equal waves running in opposite directions give the fixed nodes of Standing Waves. Many coherent sources along a slit or a grating make the fringes of Diffraction and the Double Slit, where again $\Delta =d\sin \theta$. Light is a travelling pattern of electric and magnetic fields, as Electromagnetic Waves explains, and two equal perpendicular components a quarter cycle apart give the circular states of Polarization of Light.

Incoherent sources add intensities, which is how decibels combine in Loudness, Intensity and Decibels; two coherent sources in step add 6 dB at a bright spot instead of 3 dB, because the intensity there is four times, not twice, that of one source. Electrons and single photons build up fringes dot by dot: amplitudes add, and probabilities are their squares. That idea, the heart of Wave–Particle Duality and of Schrödinger's Equation and Tunnelling, is superposition at its deepest.