Harmonics in air columns

Blow across the mouth of a bottle, or into a flute, and you set up an invisible standing wave inside a column of air — the note you hear is fixed entirely by the pipe's length and by whether its ends are open or closed. A closed end forces the air to stay still (a node); an open end lets it swing freely (an antinode). That single boundary rule, applied to a pipe closed at one end versus a pipe open at both ends, explains why a closed pipe sounds an octave lower than an open pipe of the same length, and why a closed pipe can never produce even-numbered harmonics while an open pipe produces every one. This article builds the frequency formulas for both cases from that node/antinode picture, works two numerical problems, and flags the mix-ups that cost the most marks.

The core insight

A closed end is a displacement node and an open end is a displacement antinode, and fitting these boundary conditions into a pipe of length $l$ gives $f_1=v/4l$ with only odd harmonics for a closed pipe, and $f_1'=v/2l$ with all harmonics for an open pipe.

The core idea

Sound travelling in a pipe is a longitudinal stationary wave, built from a forward wave and its reflection overlapping inside the tube. What each end does to that reflection fixes the pattern:

Closed end. The rigid boundary stops air molecules from moving along the tube axis — a closed end is always a displacement node.
Open end. Air there connects to the open atmosphere and swings with maximum amplitude — an open end is (to good approximation) always a displacement antinode.

Closed pipe (closed at one end, open at the other, length $l$ ). One end is a node, the other an antinode, and the shortest stationary pattern satisfying both has a quarter wavelength inside the pipe:

l=\frac{\lambda_1}{4} \implies \lambda_1 = 4l, \qquad f_1=\frac{v}{4l}

where: $v$ is the speed of sound in air, $l$ is the pipe length, $\lambda_1$ is the fundamental wavelength, and $f_1$ is the fundamental frequency. Every higher allowed mode must still keep a node at the closed end and an antinode at the open end, which restricts the pipe to an odd number of quarter-wavelengths:

l=(2n-1)\frac{\lambda_n}{4}, \quad n=1,2,3,\dots \implies f_n=(2n-1)f_1

so a closed pipe supports only odd harmonics: $f_1,\ 3f_1,\ 5f_1,\dots$

Open pipe (open at both ends, length $l$ ). Both ends are antinodes, and the shortest pattern fitting this has a node in the middle with half a wavelength across the pipe:

l=\frac{\lambda_1}{2} \implies \lambda_1'=2l, \qquad f_1'=\frac{v}{2l}

The general condition $l=n\,\lambda_n/2$ gives $f_n=nf_1'$ , so an open pipe supports every harmonic, odd and even alike — just like a stretched string.

Comparing the two. For the same length $l$ and speed $v$ , the closed pipe's fundamental ( $v/4l$ ) is exactly half the open pipe's fundamental ( $v/2l$ ): a closed pipe sounds an octave lower than an open pipe of equal length. Real pipes also need a small "end correction," since the antinode bulges slightly beyond the physical opening — a practical refinement, not needed for basic numericals here.

Figure

Node and antinode patterns for the fundamental mode of a closed pipe and an open pipe of the same length

The fundamental frequency of a closed pipe

Derivation AFundamental frequency of a pipe closed at one end

The goal is to turn the geometric boundary condition of a closed pipe (node at the closed end, antinode at the open end) into an explicit formula for its fundamental frequency $f_1$ in terms of pipe length $l$ and sound speed $v$ , and then to show why the higher allowed modes skip every even harmonic.

Step 1. Fix the wave shape from the boundary conditions. A closed end must be a node; an open end must be an antinode. The simplest continuous stationary-wave pattern joining a node to an antinode is one quarter of a full wave cycle, so the pipe length must equal a quarter wavelength: $l=\lambda_1/4$ .

Step 2. Solve for the fundamental wavelength. Rearranging, $\lambda_1=4l$ — the fundamental wavelength is four times the physical pipe length, twice as long as the pipe itself would suggest at first glance, because only a quarter of the wave pattern fits inside.

Step 3. Convert wavelength to frequency using $v=f\lambda$ . Substituting $\lambda_1=4l$ gives $f_1=v/\lambda_1=v/(4l)$ , the fundamental (first harmonic) of the closed pipe.

Step 4. Extend to higher modes and show the odd-harmonic rule. Any allowed mode must still place a node at the closed end and an antinode at the open end; this is satisfied only when an odd number of quarter-wavelengths spans the pipe, $l=(2n-1)\lambda_n/4$ for $n=1,2,3,\dots$ , which on inversion gives $f_n=(2n-1)f_1$ . Even multiples of $f_1$ would require a node exactly where the open end's antinode must sit, which is impossible — hence only odd harmonics ever appear.

Worked example

A closed pipe has length $l=0.25\ \text{m}$ ; take the speed of sound as $v=340\ \text{m/s}$ . Find (a) the fundamental frequency, (b) the frequency of the first overtone, (c) the frequency of the second overtone, and (d) the fundamental frequency of an open pipe of the same length.

Step 1. Fundamental of the closed pipe: $f_1=\dfrac{v}{4l}=\dfrac{340}{4\times0.25}=\dfrac{340}{1}=340\ \text{Hz}$ .

Step 2. First overtone is the next allowed mode above the fundamental — for a closed pipe that is the third harmonic (the second harmonic does not exist): $f=3f_1=3\times340=1020\ \text{Hz}$ .

Step 3. Second overtone is the fifth harmonic: $f=5f_1=5\times340=1700\ \text{Hz}$ .

Step 4. For an open pipe of the same length, $f_1'=\dfrac{v}{2l}=\dfrac{340}{2\times0.25}=\dfrac{340}{0.5}=680\ \text{Hz}$ , exactly double the closed pipe's fundamental — confirming the closed pipe sounds an octave lower.

Answer: $f_1=340\ \text{Hz}$ , first overtone $=1020\ \text{Hz}$ , second overtone $=1700\ \text{Hz}$ , and the equal-length open pipe's fundamental is $680\ \text{Hz}$ .

Competitive stretch

If instead the open pipe's fundamental of $680\ \text{Hz}$ is the given data, the pipe length follows from $l=v/(2f_1')=340/(2\times680)=0.25\ \text{m}$ — the same length recovered without needing the closed pipe's number at all, a useful cross-check in exam problems that swap which pipe is "given."

Common traps

Thinking both pipe types produce all harmonics — only the open pipe does; the closed pipe produces only odd harmonics ( $f_1,3f_1,5f_1,\dots$ ) and skips every even one.
Placing a node at an open end or an antinode at a closed end — it is the reverse: closed end = node (air cannot move), open end = antinode (air moves freely).
Using $\lambda_1=2l$ (the open-pipe relation) for a closed pipe instead of the correct $\lambda_1=4l$ , which halves the calculated fundamental by mistake.
Mislabelling "first overtone" — for an open pipe it is the 2nd harmonic, but for a closed pipe it is the 3rd harmonic, since the closed pipe has no 2nd harmonic at all.

Ask the tutor: Want to see the second and third harmonic patterns sketched for both pipe types, or try another resonance-tube numerical?

Why can a closed pipe not produce even harmonics at all?

Every allowed standing wave in the closed pipe must keep a node exactly at the closed end and an antinode exactly at the open end, because those boundary conditions never change. Fitting that node-to-antinode pattern into the pipe only works when the pipe holds an odd number of quarter-wavelengths — one, three, five, and so on. An even multiple of the fundamental would force a node where the open end needs an antinode, which the physics simply will not allow, so those modes never exist.

Why does an open pipe sound higher than a closed pipe of the same length?

The open pipe fits only half a wavelength into its length ( $l=\lambda_1/2$ ), while the closed pipe fits just a quarter wavelength ( $l=\lambda_1/4$ ). A shorter fraction of the wavelength squeezed into the same physical length means the closed pipe's wavelength is effectively longer, and since $f=v/\lambda$ , a longer wavelength gives a lower frequency. The result is that the closed pipe's fundamental is exactly half the open pipe's — an octave lower — for identical lengths.

Does the resonance-tube method use this same node and antinode idea?

Yes. In a resonance tube, one end is effectively closed (by a water surface or a stopper) and a tuning fork of known frequency f is held at the open end. As the air-column length is varied, loud resonance occurs whenever the column's own natural length matches lambda/4, 3 lambda/4, 5 lambda/4, and so on, where lambda = v/f is the fork's wavelength. Measuring the resonant length lets you calculate the speed of sound v, which is exactly how school laboratory experiments determine v experimentally.