Locate standing-wave nodes and antinodes

Find what stays fixed through time

A standing wave has a spatial amplitude pattern that stays in place while the medium oscillates. In this investigation, two equal-amplitude, equal-frequency waves travel in opposite directions. Their sum has nodes, positions with zero resultant displacement throughout the cycle, and antinodes, positions with the largest oscillation amplitude.

At phase offset 0°, inspect times 0, 0.25 and 0.50 s. Nodes stay at x = 0, 1, 2, 3 and 4 m. Antinodes lie halfway between them at x = 0.5, 1.5, 2.5 and 3.5 m. The tracked point at x = 0.5 m oscillates between +0.100 and −0.100 m, while the node at x = 1 m stays at zero.

The open circles identify persistent nodes calculated from the model, not zeros found in one plotted snapshot. At 0.25 s the entire resultant displacement is zero, but only the marked node positions remain zero at every time. Most other positions have nonzero velocity at that instant.

Interpret the envelope

The dashed curves show the upper and lower amplitude envelope: the maximum positive and negative displacement attainable at each position over a cycle. They are limits, not additional waves or the medium’s current shape. The instantaneous resultant remains between them.

The largest amplitude is 0.100 m, twice the amplitude of either traveling component. At a node, the envelope is zero. Between a node and antinode, the local amplitude lies between those limits. A point’s amplitude does not change as time advances in one selected pattern, even though its instantaneous displacement changes continuously.

Adjacent nodes are separated by half the traveling wavelength. Here λ = 2 m, so node spacing is 1 m. A node and its nearest antinode are separated by λ/4 = 0.5 m. Adjacent antinodes oscillate in opposite phase: when one is at its positive extreme, its neighbor is at its negative extreme. They do not all move upward together.

Shift the phase and locate the new nodes

Set phase to 90°. The nodes in the window move to x = 0.75, 1.75, 2.75 and 3.75 m. At 180°, they are at 0.5, 1.5, 2.5 and 3.5 m. The tracked point x = 0.5 m is now a node and retains zero resultant displacement and velocity at every selected time.

This is a comparison of different phase settings, not motion of a node within one steady pattern. At any fixed phase, the node positions stay constant as time advances. Their spacing remains 1 m, even though their positions relative to the viewing window change.

Optional: factor the sum into space and time

With A = 0.050 m, k = π radians per metre, omega = 2π radians per second and phi the phase offset in radians, the sum is y = 2A sin(kx + phi/2) cos(omega t + phi/2). The local amplitude is 2A times the absolute value of sin(kx + phi/2). Nodes satisfy kx + phi/2 = nπ for integer n. Thus x = n − phi/(2π) metres for these fixed parameters. If phi is entered in degrees, the shift is phi/360 metres. This separated expression is equivalent to adding the two traveling components.

Separate this pattern from a resonating instrument

The browser prescribes two waves over a viewing window. It does not simulate a finite string, a reflected pulse, a driver, losses or resonance growth. At zero phase, the endpoints of the window happen to be nodes; other phase settings need not satisfy a hypothetical fixed-end constraint. Do not describe every setting as a valid mode of the same clamped string.

For an ideal string fixed at both ends, each end must be a node. A length L then accommodates a positive integer n of half-wavelength segments: L = nλ/2, so λ = 2L/n. This is a boundary condition added to the wave model, not a consequence of merely drawing a finite plot. Resonance also involves driving and energy exchange; a stationary envelope by itself does not demonstrate resonance buildup.

Transfer to a bounded string

An ideal 3 m string fixed at both ends has three equal loops, each between neighboring nodes. Determine the traveling wavelength. If its wave speed is 12 m/s, determine the corresponding frequency. Keep these conditions separate from the browser’s fixed 2 m/s example.

Check the node transfer

Three half-wavelength segments occupy 3 m, so each half-wavelength is 1 m and λ = 2 m. Frequency is speed divided by wavelength: 12/2 = 6 Hz. The boundary condition selects an allowed spatial pattern; the calculation alone does not predict its driven amplitude or damping.

OpenStax standing waves and resonance develops standing patterns and fixed-end modes. The node and antinode positions here follow the explicitly stated phase convention.