Stationary waves

Pluck a guitar string and something odd happens: the string does not send a single pulse racing off and vanishing — it settles into a pattern that stays put, some points frozen while others swing wildly. That fixed pattern is a stationary (or standing) wave, and it appears whenever two identical waves travel through the same medium in opposite directions, most commonly because a wave reflects straight back on itself — on a plucked string, in an air column, on a vibrating membrane. Unlike a travelling wave, which carries a disturbance (and energy) from one place to another, a stationary wave carries nothing anywhere: it just breathes in place. This article builds the stationary-wave equation from first principles by adding two oppositely-travelling waves, locates the special points called nodes and antinodes, and contrasts stationary waves with the progressive waves you already know.

The core insight

Superposing two identical waves travelling in opposite directions gives $y=2y_m\sin(kx)\cos(\omega t)$ , a pattern whose amplitude $2y_m\sin(kx)$ is fixed in space — zero at nodes, maximum at antinodes — so the wave oscillates without travelling and transports no energy.

The core idea

Two waves, one medium, opposite directions. Take two waves of identical amplitude $y_m$ , angular frequency $\omega$ and wavenumber $k$ , one moving in $+x$ and one in $-x$ : $y_1=y_m\sin(kx-\omega t)$ and $y_2=y_m\sin(kx+\omega t)$ . This is exactly the situation created by a reflection — the incident and reflected waves are identical in everything except direction.

Adding them. By the principle of superposition the medium simply carries $y=y_1+y_2$ . Using $\sin(A-B)+\sin(A+B)=2\sin A\cos B$ :

y=2y_m\sin(kx)\cos(\omega t)

where: $y_m$ is the amplitude of each individual travelling wave, $k=2\pi/\lambda$ is the wavenumber, $\omega=2\pi f$ is the angular frequency, $x$ is position and $t$ is time. The crucial feature is that $x$ and $t$ sit in separate factors — not combined as $kx\pm\omega t$ the way a travelling wave demands.

Position-dependent amplitude. Group the result as $y=A(x)\cos(\omega t)$ with $A(x)=2y_m\sin(kx)$ . Every particle at a given $x$ still oscillates in simple harmonic motion in time, but each has its own fixed amplitude set by where it sits — the pattern itself never shifts left or right.

Nodes and antinodes. Nodes occur where $A(x)=0$ , i.e. $\sin(kx)=0$ , giving $x=0,\lambda/2,\lambda,\dots$ — points permanently at rest. Antinodes occur where $A(x)=\pm2y_m$ is maximum, i.e. $\sin(kx)=\pm1$ , giving $x=\lambda/4,3\lambda/4,\dots$ — points oscillating with double the amplitude of either travelling wave. Consecutive nodes (or consecutive antinodes) are $\lambda/2$ apart; a node and its neighbouring antinode are $\lambda/4$ apart. Between two nodes, all particles move in phase; across a node, particles on either side are exactly out of phase.

Figure

A snapshot of a stationary wave on a string showing fixed nodes at rest and antinodes swinging between two dashed envelope curves

Comparison with progressive waves

A progressive wave moves through the medium, every particle sharing the same amplitude while phase varies smoothly along $x$ , and it transports energy from source to receiver. A stationary wave stays fixed in space, its amplitude ranging from $0$ at nodes to $2y_m$ at antinodes, with particles between a pair of nodes in phase and particles across a node out of phase by $\pi$ ; no net energy is transported — it stays trapped, converting locally between kinetic and potential energy as the pattern breathes.

Feature	Progressive wave	Stationary wave
Pattern	Moves through the medium	Fixed in space
Amplitude	Same for every particle	Varies with position, $0$ to $2y_m$
Phase	Varies continuously along $x$	In phase between nodes; out of phase by $\pi$ across a node
Energy	Transported along the wave	Not transported — stored, oscillating locally

This table is worth memorising directly: exam questions often ask you to "distinguish" the two wave types, and these four rows cover essentially every mark on offer.

Derivation ASuperposing two opposite travelling waves into a stationary wave

The goal is to show, starting only from two ordinary travelling waves of equal amplitude, frequency and wavenumber moving in opposite directions, that their sum is not a travelling wave at all but an equation in which position and time appear in separate factors — the defining algebraic signature of a stationary wave — and then to read off where that separated amplitude factor vanishes (nodes) and where it peaks (antinodes).

Step 1. Write the two waves. A wave moving in $+x$ is $y_1=y_m\sin(kx-\omega t)$ ; an identical wave reflected back and moving in $-x$ is $y_2=y_m\sin(kx+\omega t)$ . Both share the same amplitude $y_m$ , wavenumber $k$ and angular frequency $\omega$ because reflection changes direction, not amplitude or frequency.

Step 2. Superpose. The principle of superposition says the medium's net displacement is simply $y=y_1+y_2=y_m\sin(kx-\omega t)+y_m\sin(kx+\omega t)$ .

Step 3. Apply the identity $\sin(A-B)+\sin(A+B)=2\sin A\cos B$ with $A=kx$ , $B=\omega t$ . This gives $y=2y_m\sin(kx)\cos(\omega t)$ . Notice $x$ now appears only inside $\sin(kx)$ and $t$ only inside $\cos(\omega t)$ — they never combine into a single $kx\pm\omega t$ argument, so the pattern cannot slide left or right the way a travelling wave does.

Step 4. Identify the position-dependent amplitude $A(x)=2y_m\sin(kx)$ and locate its zeros and extremes. Zeros: $\sin(kx)=0\Rightarrow kx=0,\pi,2\pi,\dots\Rightarrow x=0,\lambda/2,\lambda,\dots$ — these are the nodes. Extremes: $\sin(kx)=\pm1\Rightarrow kx=\pi/2,3\pi/2,\dots\Rightarrow x=\lambda/4,3\lambda/4,\dots$ — these are the antinodes, each oscillating with amplitude $2y_m$ , twice that of either original wave.

Worked example

Two identical waves, each of amplitude $y_m=0.02\ \text{m}$ and wavelength $\lambda=0.4\ \text{m}$ , travel in opposite directions along a string and set up a stationary wave. Find (a) the resultant amplitude at the antinodes, (b) the spacing between consecutive nodes, and (c) the resultant amplitude at a point $0.1\ \text{m}$ from a node (taking that node as $x=0$ ).

Step 1. Antinode amplitude is double the individual amplitude: $2y_m=2\times0.02=0.04\ \text{m}$ .

Step 2. Node spacing is half a wavelength: $\lambda/2=0.4/2=0.2\ \text{m}$ .

Step 3. Find the wavenumber, $k=2\pi/\lambda=2\pi/0.4=5\pi\ \text{rad/m}$ , then evaluate the amplitude function at $x=0.1\ \text{m}$ : $A(x)=2y_m\sin(kx)=0.04\sin(5\pi\times0.1)=0.04\sin(0.5\pi)=0.04\times1=0.04\ \text{m}$ .

Step 4. Interpret: since $x=0.1\ \text{m}=\lambda/4$ , this point is exactly an antinode, so getting the full $0.04\ \text{m}$ amplitude is expected and matches part (a) exactly, confirming the result. The antinode amplitude is $0.04\ \text{m}$ , consecutive nodes are $0.2\ \text{m}$ apart, and the point at $0.1\ \text{m}$ (an antinode) oscillates with amplitude $0.04\ \text{m}$ .

Common traps

Writing the stationary-wave equation as $y=A\sin(kx-\omega t)$ — that is a travelling-wave form. The defining feature of a stationary wave is $x$ and $t$ in separate factors, $[2y_m\sin(kx)]\cos(\omega t)$ .
Assuming every particle has the same amplitude — amplitude varies continuously with position from $0$ at nodes to $2y_m$ at antinodes; only $\omega$ (and hence period) is common to every particle.
Believing a stationary wave transports energy from end to end like a travelling wave — it does not; energy stays trapped, oscillating locally between kinetic and potential forms.
Mixing up which quantity is maximum where — nodes have zero displacement but maximum strain/pressure variation, while antinodes have maximum displacement but zero strain, a distinction that matters once air columns are introduced.

Ask the tutor: Want the trig-identity step in the derivation walked through even more slowly, or a worked problem that starts from the stationary-wave equation and asks you to find $\lambda$ , $f$ and node positions in reverse?

Why does adding two travelling waves ever produce something that does not travel at all?

Because the two waves' time-dependences cancel out of step with each other. Writing $y_1+y_2$ and applying the sum-to-product identity forces $x$ and $t$ apart into two separate factors, $\sin(kx)$ and $\cos(\omega t)$ , instead of the single combined argument $kx\pm\omega t$ that makes a wave travel. With $x$ and $t$ decoupled, no fixed shape moves across the medium — each point just oscillates in place with its own amplitude set by $x$ .

Do nodes actually have zero energy, or just zero displacement?

Nodes have zero displacement and stay at rest, but they are not zero-energy points — in fact for waves in a medium like an air column, nodes have maximum strain and pressure variation, which store energy. What is genuinely true is that no net energy flows past a node in either direction over a full cycle, since the pattern is stationary; energy simply sloshes back and forth locally between kinetic and potential forms across the whole pattern.

Can a stationary wave form from any two waves travelling in opposite directions?

No — you specifically need two waves of the same amplitude, frequency and wavelength travelling in exactly opposite directions through the same medium. If the amplitudes differ, the cancellation at the nodes is incomplete and you get a partially standing wave with imperfect nulls instead of true zero-displacement points; if the frequencies differ, the pattern itself drifts rather than staying fixed. The clean equation $y=2y_m\sin(kx)\cos(\omega t)$ assumes identical waves, which is exactly what a reflection naturally produces.

Stationary waves

The core idea

Comparison with progressive waves

Worked example

Common traps

Before this

Where this leads