Stationary waves in an open pipe
A pipe open at both ends has a displacement antinode at each end, so L = n(λ/2), giving λₙ = 2L/n and fₙ = nv/2L = n f₁ with all harmonics present. The fundamental has one node at the centre and λ = 2L.
λₙ = 2L/n · fₙ = nv/(2L) = n f₁ (n = 1, 2, 3, …)What this actually means
The boundary condition is the only thing that differs from the string, and it inverts: antinodes at the ends rather than nodes. The mathematics comes out identical, so an open pipe and a string of the same length have the same harmonic series.
The physical reason for the antinode is that air molecules at an open end are free to move, and the pressure there is fixed at atmospheric. Free to move means large displacement, so a displacement antinode and a pressure node.
The reflection is easy to forget. Even though nothing solid is there, a sound wave partially reflects at an open end because of the sudden change in the medium's effective cross-section, and it is that reflected wave which forms the stationary wave.
Sketching in displacement is fine and expected, but add a note that sound is longitudinal and the transverse-looking graph only shows the amplitude of the longitudinal vibration.
Count the loops carefully. An open pipe in its third harmonic holds three half-wavelengths, so it has three loops, four antinodes (two of them at the open ends) and three nodes inside.
Putting nodes at the open ends by copying the string diagram, which halves every wavelength you calculate.
Prove it — watch it be true
- Select the 3D open pipe and set the mode selector to n = 1.
- Confirm antinodes are drawn at both open ends with one node at the centre.
- Step up through n = 2 to n = 5 and confirm every integer mode is available, unlike the closed pipe.