Find pressure and displacement modes in an open tube

Name the quantity before naming a node

An air column can support standing sound waves. At an ideal open end, the pressure perturbation is zero because the opening communicates with the surrounding ambient pressure. This is a pressure node. It is not zero absolute pressure and it does not prevent air from moving.

In the ideal one-dimensional mode, that open end is a displacement antinode: longitudinal displacement amplitude is maximal there. A pressure node and a displacement node are different conditions. Compare the two diagrams at the same endpoint rather than calling it simply a “node” without specifying the quantity.

The curves are graphs of longitudinal displacement and pressure, not pictures of air moving vertically. Each is divided by its own positive amplitude. A value of +1 on the displacement graph means its own positive displacement extreme; +1 on the pressure graph means its own positive pressure extreme. Those heights cannot be compared as equal physical quantities, and neither is a predicted amplitude in metres or pascals.

Fit the fundamental between two open ends

For the lowest nonzero-frequency mode, pressure nodes occur at both ends with one pressure antinode between them. Equivalently, displacement antinodes occur at both ends with one displacement node in the middle. The column spans half a wavelength: L = λ/2, so λ = 2L.

At L = 1 m and the model’s sound speed c = 340 m/s, wavelength is 2 m and fundamental frequency is c/λ = 170 Hz. The displacement node is x = 0.5 m. The pressure nodes are x = 0 and 1 m. Inspect the endpoint ledger: nonzero normalized displacement coexists with zero normalized pressure perturbation.

At this chosen phase the two displacement endpoints have opposite signs. That does not mean the air travels from one end to the other. It describes opposite local displacements within a standing oscillation. Over a cycle these signs reverse while the node positions remain fixed.

Add allowed half-wavelength segments

A higher mode fits an integer number n of half-wavelengths between open ends: L = nλ/2. Thus λn = 2L/n and fn = nc/(2L). Here n is a positive integer and equals the harmonic number.

At length 1 m, the first five frequencies are 170, 340, 510, 680 and 850 Hz. Mode 2 has wavelength 1 m, pressure nodes at 0, 0.5 and 1 m, and displacement nodes at 0.25 and 0.75 m. Check these positions in the plots and ledger. More oscillations fit spatially, while the sound speed remains fixed.

Halving length at a fixed mode halves wavelength and doubles frequency. The plots retain a fixed 0–1.5 m horizontal scale, so a shorter tube visibly occupies less of the graph. Blank space beyond the chosen right endpoint is outside the modeled column, not an extension of its curve.

Connect the two profiles

The selected open-open displacement shape is proportional to cos(nπx/L). Pressure is proportional to the negative spatial derivative of displacement, so its normalized shape is sin(nπx/L) with the displayed phase convention. A displacement extremum has zero gradient and corresponds to a pressure node; a displacement node corresponds to a pressure antinode for this single sinusoidal mode.

This standing-wave relation must not be confused with the single traveling-wave rule Δp = ρcu. A standing pattern contains oppositely traveling components; a simple pressure-to-velocity ratio for just one direction no longer describes the entire field. The browser shows mode profiles at one chosen phase, not an animation or a driven response.

Where the ideal endpoint approximation breaks down

Real openings radiate sound and are not perfect pressure-release boundaries at a sharply defined plane. The effective acoustic length can exceed the geometric length; this is described by an end correction. The present calculation deliberately omits it. Tube diameter, opening geometry, losses and source coupling are not controls, so the model cannot supply a precise correction or predict the measured resonance amplitude.

Compare the closed-open column lesson to see how a different left boundary changes the allowed frequencies, even with the same length and sound speed. Do not describe the two sets as interchangeable instruments.

Transfer to a different column

An ideal open-open column is 0.85 m long with c = 340 m/s. Find its fundamental and third harmonic, then locate the fundamental displacement node.

Check the open-column transfer

The fundamental is 340/(2 × 0.85) = 200 Hz. The third harmonic is 600 Hz. The fundamental displacement node is at the midpoint, x = 0.425 m; its pressure nodes are at the open endpoints. These predictions omit end correction and losses.

OpenStax normal modes of a standing sound wave describes the ideal air-column boundary conditions. The normalized graphs here use an explicit original phase convention.