Harmonics and overtones — on strings and in pipes: Visual walkthrough

Step 1 — What a wave even is, and the one equation we lean on

WHAT. A wave is a moving ripple. Two numbers describe its shape and speed:

The wavelength $λ$ — the distance from one crest to the next identical crest. Think "how long is one full wiggle."
The frequency $f$ — how many full wiggles pass a fixed point each second, measured in hertz (Hz).

WHY this equation. If one wiggle is $λ$ metres long, and $f$ of them pass every second, then the ripple advances $f \times λ$ metres per second. That is the speed $v$ :

$v = wiggles per second f \times metres per wiggle λ \Rightarrow f = \frac{v}{λ} .$

That last rearrangement is the whole strategy of this page: the medium fixes $v$ (see Speed of a wave on a string and Speed of sound in gases); the boundaries will fix $λ$ ; then we simply divide to get $f$ . We never need anything harder than this.

PICTURE. One wiggle laid on a ruler, its length labelled $λ$ , an arrow showing it slide right at speed $v$ .

Figure — Harmonics and overtones — on strings and in pipes

Step 2 — Why a trapped wave becomes a standing pattern

WHAT. Clamp the ends. Now the wave can't escape — it hits a boundary and reflects back on itself (this is exactly Reflection of waves at boundaries). The forward wave and its reflection overlap and add together — this adding is Standing waves and superposition.

WHY it matters. When you add a wave to its own reflection, some points never move and some points swing hardest. The result no longer travels — it just stands and breathes in place. We call it a standing wave.

Two named spots appear, and every later step depends on them:

A node — a point that stays perfectly still (the two waves always cancel there).
An antinode — a point that swings with maximum amplitude (the two waves always reinforce there).

PICTURE. Blue forward wave + yellow reflected wave; their sum in green shows still points (red dots = nodes) and big-swing points (green arrows = antinodes).

Step 3 — The two rulers hidden inside a standing wave

WHAT. Before we fit anything into a length $L$ , we need two distances off the standing-wave shape.

WHY. The boundaries will demand "node here, antinode there." To count how many patterns fit, we must know how far apart these features sit.

Read straight off the wave shape: a full wiggle is $λ$ ; it crosses zero (a node) twice per wiggle, so consecutive nodes are half a wiggle apart. The peak (antinode) sits exactly halfway between, a quarter-wiggle from each node.

PICTURE. One standing-wave loop with a $λ /2$ bracket between two red nodes and a $λ /4$ bracket from a node to the green antinode.

Step 4 — Case 1: string fixed at both ends (node–node)

WHAT. A guitar string is clamped at both ends, so both ends must be nodes (a clamped point cannot move — see Reflection of waves at boundaries).

WHY the shapes are quantised. Between two nodes you must fit a whole number of node-to-node gaps, each $λ /2$ long. You cannot fit one-and-a-half gaps — a half gap would leave a moving point (antinode) sitting on the clamp, which is forbidden. So:

$L = number of loops n \times one loop’s length \frac{λ _{n}}{2}, n = 1, 2, 3, \dots$

Solve for the allowed wavelengths, then feed each into $f = v / λ$ from Step 1:

$λ_{n} = \frac{2 L}{n} ⟹ f_{n} = \frac{v}{λ _{n}} = \frac{v}{2 L / n} = \frac{n v}{2 L} = n f_{1}, f_{1} = \frac{v}{2 L} .$

Term by term: $L$ is the fixed string length; $n$ counts the loops (also the harmonic number); $v$ is set by tension and mass — $v = T / μ$ , see Speed of a wave on a string; $f_{1}$ is the deepest note the string can make. Every integer $n$ works, so all harmonics exist.

PICTURE. Three stacked shapes for $n = 1, 2, 3$ , red nodes pinned at both walls, loops counted.

Step 5 — Case 2: pipe open at both ends (antinode–antinode)

WHAT. In an open-ended pipe, air rushes freely in and out at each open end, so both ends must be antinodes.

WHY the math is identical to the string. Look back at Step 3: antinode → next antinode is also $λ /2$ (it's just the pattern shifted by a quarter-wave; the spacing between same-type features is unchanged). So the fitting rule is the same equation with the same $λ /2$ brick:

$L = n \frac{λ _{n}}{2} ⟹ λ_{n} = \frac{2 L}{n} ⟹ f_{n} = \frac{n v}{2 L} = n f_{1} .$

Here $v$ is the Speed of sound in gases instead of a string speed — that's the only physical difference. All harmonics present, exactly like the string.

PICTURE. Open pipe drawn as a tube, green antinodes at both mouths, modes $n = 1, 2, 3$ inside.

Step 6 — Case 3: pipe closed at one end (node–antinode)

WHAT. Cap one end. Air can't move at the cap → that end is a node. The open end still swings freely → antinode. Now the two ends are different types.

WHY only quarter-wavelengths — and only odd ones. The shortest shape that puts a node at one end and an antinode at the other is a single node-to-antinode span, which Step 3 tells us is $λ /4$ . To get the next legal shape you must add another full node-to-antinode-to-node stretch ( $λ /2$ ) so the ends keep their required types. Adding $λ /2$ repeatedly to $λ /4$ gives $\frac{1}{4} λ, \frac{3}{4} λ, \frac{5}{4} λ, \dots$ — odd quarter-wavelengths:

$L = 1, 3, 5, \dots (2 n - 1) \frac{λ _{n}}{4}, n = 1, 2, 3, \dots$

Solve and convert with $f = v / λ$ :

$λ_{n} = \frac{4 L}{2 n - 1} ⟹ f_{n} = \frac{( 2 n - 1 ) v}{4 L} = (2 n - 1) f_{1}, f_{1} = \frac{v}{4 L} .$

The factor $(2 n - 1)$ generates only $1, 3, 5, \dots$ — even harmonics are forbidden. And because the fundamental fits just $λ /4$ (a longer $λ$ than the open pipe's $λ /2$ ), $f_{1}$ is halved: a closed pipe sounds an octave below an open pipe of the same length.

PICTURE. Closed pipe: red node at the capped end, green antinode at the open mouth, modes for $n = 1$ (3rd harmonic) and $n = 2$ (which is the 3rd harmonic), even shape crossed out.

Step 7 — The degenerate & edge cases (never leave the reader stranded)

WHAT could break? Let's sweep the corners so no scenario surprises you.

$n = 0$ ? That would mean $L = 0$ — zero length, no pipe, no string. Not a physical mode; the ladder starts at $n = 1$ . There is no "zeroth harmonic."
A closed pipe's 2nd harmonic ( $2 f_{1}$ )? It simply does not exist. Feeding $λ = 2 L /2 = L$ would demand a node and an antinode swap that violates one boundary. This is why "1st overtone = 2nd harmonic" fails for closed pipes — the 1st overtone jumps to $3 f_{1}$ .
Do an open and closed pipe of equal length ever share a pitch? Open allows $170, 340, 510, 680, 850, 1020 \dots$ ; closed allows $170, 510, 850, 1190 \dots$ (using $f_{1} = 170$ ). They share $170, 510, 850, \dots$ — exactly the odd open-pipe harmonics. Any even open harmonic (like $1020$ ) is missing from the closed list.
Very high $n$ ? Nothing new: $λ_{n} \to 0$ and $f_{n} \to \infty$ in principle, but real materials damp tiny wiggles, so only the first several matter for the sound you hear (their mix is the timbre).

PICTURE. Two harmonic ladders side by side (open vs closed), shared rungs highlighted green, forbidden even closed rungs greyed out and crossed.

The one-picture summary

Everything above collapses into a single diagram: boundary type → allowed feature spacing → wavelength → frequency ladder. The string and open pipe walk the same road ( $λ /2$ bricks, all $n$ ); the closed pipe takes the $λ /4$ turn and drops every even rung.

Recall Feynman retelling of the whole walkthrough

Grab a rope tied to a wall and shake it. The wall end can't move, so it's a dead spot (a node). To make a neat standing shape, you have to fit a whole number of "loops" between the two dead ends — one hump, two humps, three humps — never a hump and a half, because a half would put a wiggling spot right on the wall, which isn't allowed. Each neat shape needs a shorter and shorter wiggle-length, and shorter wiggles hum higher, so you climb a ladder of pitches: 1×, 2×, 3× the lowest note. A pipe open at both ends is the same story with air puffing freely at each mouth — same ladder. But cap one end of the pipe and you make it lopsided: one end pinned (node), one end free (antinode). The smallest fit is now only a quarter of a wiggle, so its lowest note is deeper — an octave below the open pipe. Worse, to keep the ends the right types you can only add half-wiggles, which lands you on odd multiples only: 1×, 3×, 5×. That skipping of every other pitch is why a stopped organ pipe sounds hollow and different from a flute. The whole derivation is just: decide node-or-antinode at each end, count how many $λ /2$ or $λ /4$ bricks fit, then divide $v$ by that length.

Connections

Standing waves and superposition — Step 2's adding of wave + reflection.
Reflection of waves at boundaries — why fixed/closed ends force nodes.
Speed of a wave on a string — the $v$ in Case 1.
Speed of sound in gases — the $v$ in Cases 2 & 3.
Beats and resonance — how a driven column locks onto these frequencies.
Timbre and Fourier synthesis — why the mix of these harmonics defines an instrument.