Standing Waves & Stationary-Wave Formation

Every musical instrument you have heard is a standing wave made audible. A guitar or violin string, clamped at both ends, can only hold standing waves with a node at each end, which is why it rings at a definite pitch and its overtones rather than a smear of frequencies — the string selects the wavelengths that fit. A flute or organ pipe does the same with a standing wave of air; the length of the pipe sets which wavelengths have nodes and antinodes in the right places, and that is the note you hear. The principle scales far beyond music. A microwave oven sets up a standing wave of microwaves between its walls, and the antinodes — the points of fiercest oscillation — are where the food heats fastest, which is exactly why the turntable exists, to drag cold spots at the nodes through the hot antinodes. (You can even estimate the speed of light by melting a chocolate bar in a microwave with the turntable removed and measuring the spacing of the melted patches: it is half a wavelength.) Laser cavities, radio antennas cut to a quarter or half wavelength, the resonant sway of bridges and skyscrapers in the wind — all are engineered around where the nodes and antinodes of a standing wave will fall.

A standing wave is not a new kind of wave; it is two ordinary travelling waves caught in the act of overlapping. Send a wave along a string and reflect it off a boundary, and the reflected wave runs back through the incoming one. Wherever they overlap — which, on a bounded string, is everywhere — superposition adds them, and the sum has a startling property: the places of maximum and minimum motion stop moving. Take the two waves to have the same amplitude $A$, wave number $k$ and angular frequency $\omega$, one going in $+x$ and one in $-x$. Adding them (the first derivation does the trig) gives

where: $2A\sin(kx)$ — a fixed spatial shape that depends only on position $x$; $\cos(\omega t)$ — a single clock that depends only on time. This separation is the whole phenomenon. Unlike a travelling wave, whose argument $(kx - \omega t)$ ties space and time together so a crest moves, a standing wave's space and time are divorced: the shape $2A\sin(kx)$ is stamped permanently along the string, and $\cos(\omega t)$ simply scales the whole thing up and down in unison. At one instant the string is a full sine curve of height $2A$; a quarter-period later $\cos(\omega t) = 0$ and the string is instantaneously flat; then it swings to the mirror-image curve. The pattern never slides left or right. Because the shape is fixed, the points where $2A\sin(kx) = 0$ are displaced by nothing at every instant — they are permanently at rest. These are the nodes:

where: the nodes fall at $x = 0, \lambda/2, \lambda, \dots$, so consecutive nodes are half a wavelength apart. Halfway between each pair, where $\lvert\sin(kx)\rvert = 1$, the swing is largest — amplitude $2A$. These are the antinodes:

where: the antinodes sit at $x = \lambda/4, 3\lambda/4, \dots$, each a quarter-wavelength from its neighbouring nodes. The distance from a node to the next antinode is therefore $\lambda/4$, and from node to node is $\lambda/2$ — the single most-tested measurement on this topic.

The motion of an individual particle follows from (1) by fixing its position $x$: it does simple harmonic motion in time, $y = [2A\sin(kx)]\cos(\omega t)$, with an amplitude

where: the amplitude depends on where the particle sits — zero at a node, a full $2A$ at an antinode, and something in between elsewhere. This is the sharpest break from a travelling wave, in which every particle has the same amplitude $A$. Two more readings follow from the single clock $\cos(\omega t)$: because every particle shares that one clock, all the particles in one loop (between two nodes) reach their peaks at the same instant — they move in phase. And because $\sin(kx)$ flips sign as you cross a node, particles in adjacent loops move in exact antiphase: as one loop swings up, its neighbours swing down.

A standing wave is built from two travelling waves that carry equal energy in opposite directions, so their energy flows cancel: there is no net transport of energy along a standing wave. This does not mean there is no energy — there is plenty. It sloshes back and forth within each loop, twice a period: at the instant the string is fully stretched into its sine shape it is momentarily at rest, and all the energy is potential (stored in the stretch); a quarter-period later the string is flat but every particle is moving fastest, and all the energy is kinetic. Energy trades between potential and kinetic inside each loop but never crosses a node — the nodes are perfect energy fences. A travelling wave ferries energy down the line; a standing wave locks it in place.

The comparison every mark scheme wants — progressive (travelling) wave versus standing (stationary) wave — is really a table of limits, the two extremes of what superposition can produce.

1 · Board level — locating the nodes, antinodes and amplitudes (CBSE/ICSE pattern, 3 marks)

Two identical waves, each of amplitude $4.0\ \mathrm{mm}$ and frequency $100\ \mathrm{Hz}$, travel in opposite directions on a string at $40\ \mathrm{m/s}$ and form a standing wave. Find (a) the wavelength and wave number, (b) the spacing between adjacent nodes and the antinode amplitude, (c) the amplitude of the particle at $x = 0.05\ \mathrm{m}$ and at $x = 0.20\ \mathrm{m}$ (measuring $x$ from a node).

Step 1. Wavelength and wave number: $\lambda = v/f = 40/100 = \boxed{0.40\ \mathrm{m}}$, and $k = 2\pi/\lambda = 2\pi/0.40 = \boxed{15.708\ \mathrm{rad/m}}$.

Step 2. Node spacing is half a wavelength: $\lambda/2 = \boxed{0.20\ \mathrm{m}}$; node to antinode is $\lambda/4 = 0.10\ \mathrm{m}$. The antinodes swing with amplitude $2A = 2(4.0) = \boxed{8.0\ \mathrm{mm}}$.

Step 3. Particle amplitude at $x = 0.05\ \mathrm{m}$, from equation (4). The phase is $kx = 15.708 \times 0.05 = 0.7854\ \mathrm{rad} = 45°$, so $A_{\text{particle}} = 2A\sin 45° = 8.0 \times 0.7071 = \boxed{5.657\ \mathrm{mm}}$.

Step 4. At $x = 0.20\ \mathrm{m}$: $kx = 15.708 \times 0.20 = \pi\ \mathrm{rad} = 180°$, so $\sin kx = 0$ and the amplitude is $\boxed{0}$ — this point is a node and never moves. Sensible: $x = 0.20\ \mathrm{m}$ is exactly one node spacing from the origin.

2 · Board level — the antinode's motion over one period (CBSE/NEET pattern, 3 marks)

For the same standing wave, take a point at an antinode. Using $y = 2A\cos(\omega t)$ there, find its displacement at $t = 0$, $T/8$, $T/4$, $3T/8$ and $T/2$, and state its angular frequency.

Step 1. Angular frequency: $\omega = 2\pi f = 2\pi \times 100 = \boxed{628.3\ \mathrm{rad/s}}$, and the period is $T = 1/f = 0.010\ \mathrm{s}$. At an antinode $\sin kx = 1$, so the displacement is simply $y = 2A\cos(\omega t) = 8.0\cos(\omega t)\ \mathrm{mm}$.

Step 2. At $t = 0$: $\cos 0 = 1$, so $y = \boxed{8.0\ \mathrm{mm}}$ — fully stretched. At $t = T/8$: $\omega t = \pi/4$, so $y = 8.0\cos 45° = \boxed{5.657\ \mathrm{mm}}$.

Step 3. At $t = T/4$: $\omega t = \pi/2$, so $y = 8.0\cos 90° = \boxed{0}$ — the whole string is momentarily flat, yet every particle is moving at top speed. At $t = 3T/8$: $y = 8.0\cos 135° = \boxed{-5.657\ \mathrm{mm}}$.

Step 4. At $t = T/2$: $y = 8.0\cos 180° = \boxed{-8.0\ \mathrm{mm}}$ — the mirror image of the start. The antinode has swung from $+8.0$ through $0$ to $-8.0\ \mathrm{mm}$ in half a period, doing SHM exactly like a particle on a spring — the same value $5.657\ \mathrm{mm}$ we found in space now appearing in time.

3 · Competitive level — where the amplitude takes a chosen value (JEE-Main pattern)

For the same standing wave ($\lambda = 0.40\ \mathrm{m}$, antinode amplitude $8.0\ \mathrm{mm}$), find the positions in the first loop (between $x = 0$ and $x = 0.20\ \mathrm{m}$) where a particle oscillates with amplitude exactly $4.0\ \mathrm{mm}$ — that is, half the antinode value.

Step 1. Set equation (4) to the target amplitude: $2A\lvert\sin kx\rvert = 4.0\ \mathrm{mm}$, so $8.0\lvert\sin kx\rvert = 4.0$, giving $\lvert\sin kx\rvert = 0.5$.

Step 2. Solve for $kx$ within the first loop ($0 < kx < \pi$): $\sin kx = 0.5$ gives $kx = 30° = \pi/6$ or $kx = 150° = 5\pi/6$.

Step 3. Convert to positions with $k = 15.708\ \mathrm{rad/m}$: $x = (\pi/6)/k = 0.5236/15.708 = \boxed{0.0333\ \mathrm{m}}$ (which is $\lambda/12$), and $x = (5\pi/6)/k = 2.618/15.708 = \boxed{0.1667\ \mathrm{m}}$ (which is $5\lambda/12$).

Step 4. Read the geometry: the two points sit symmetrically about the antinode at $x = 0.10\ \mathrm{m}$, each $\lambda/12$ from it. Amplitude is not linear in position — it follows $\sin kx$ — so "half the amplitude" occurs at a phase of $30°$, not at the halfway point of the loop. Timing note: about two minutes. The trap is to place the half-amplitude point at the midpoint between node and antinode; it is not, because amplitude goes as $\sin kx$, and $\sin$ reaches $0.5$ at $30°$, not at $45°$. Set up $\sin kx = 0.5$ immediately and solve for both solutions in the loop — examiners reward finding both the $\pi/6$ and the $5\pi/6$ position.

Saying the node spacing is a wavelength. It is half a wavelength. A full $\lambda$ contains two loops — node, antinode, node, antinode, node. If a problem gives node spacing $d$, the wavelength is $2d$, not $d$. This single slip halves or doubles every downstream answer.

Thinking every particle has the same amplitude. That is a travelling wave. In a standing wave the amplitude is $2A\lvert\sin kx\rvert$ — zero at nodes, $2A$ at antinodes, and position-dependent everywhere. Quoting "amplitude $A$" or "amplitude $2A$" for the whole string is wrong; only the antinodes reach $2A$.

Claiming a standing wave travels, or has a speed. The pattern is fixed. The constituent travelling waves have speed $v = \omega/k$, but their sum stands still. There is no such thing as "the speed of the standing wave."

Saying there is no energy in a standing wave. There is — it just does not go anywhere. The energy oscillates between kinetic and potential inside each loop, trapped between the nodes. What is zero is the net transport of energy, not the energy.

Confusing "in phase" with "same amplitude." Between two nodes, particles move in phase (peaking together) but with different amplitudes. Across a node they are in antiphase. Phase and amplitude are separate questions; a standing wave has uniform phase within a loop and non-uniform amplitude.

Board scope. Standing waves are examined by CBSE, ICSE, the JEE/NEET band, AP Physics 2, IB (SL and HL) and A-Level. They are not on IGCSE 0625, which stops at the travelling-wave model — if that is your board, this is enrichment, not syllabus.

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.7 (standing waves) · H. C. Verma, Concepts of Physics Vol. 1, "Waves on a String" — standing waves and normal modes · Halliday, Resnick and Walker, "Waves — I" · Cambridge A-Level 9702 topic 8.3 (stationary waves).

Provenance. Every number is recomputed in `verify_hs2-standing-waves.py`: $\lambda = 0.40$ m, $k = 15.708$ rad/m, $\omega = 628.3$ rad/s; node spacing $0.20$ m, antinode amplitude $8.0$ mm; the particle amplitude $5.657$ mm at $x = 0.05$ m and $0$ at $x = 0.20$ m; the antinode's displacements $8.0, 5.657, 0, -5.657, -8.0$ mm at $t = 0, T/8, T/4, 3T/8, T/2$; and the half-amplitude positions $0.0333$ m ($\lambda/12$) and $0.1667$ m ($5\lambda/12$).

Prerequisites. Superposition & Interference of Waves — the standing wave is the superposition of two opposite-going waves · Reflection of Waves & Phase Change at Boundaries — where the reflected wave comes from, and why a fixed end forces a node. Leads to. The modes and harmonics of a string fixed at both ends — the next part of this chapter — build directly on these nodes and antinodes. Go deeper. The university treatment, in which the allowed standing waves are the normal modes of the string and any motion is a sum of them, is Normal Modes and Standing Waves on a String — enrichment, not a dependency.