Harmonics and overtones — on strings and in pipes

WHAT are harmonics and overtones?

The crucial subtlety: "harmonic" counts integer multiples of $f_1$; "overtone" counts the allowed modes above the fundamental. They only line up the same way when every harmonic is allowed (string, open pipe). In a closed pipe, even harmonics are forbidden, so the numbering shifts.

HOW the boundaries pick the frequencies (derivation from scratch)

A wave on a string travels at speed $v=\sqrt{T/\mu}$ ($T$ tension, $\mu$ mass per length). In a pipe, $v$ is the speed of sound. For any medium, frequency, wavelength, speed obey: $v = f\lambda \quad\Rightarrow\quad f = \frac{v}{\lambda}.$

So the whole game is: find which $\lambda$ the boundary allows, then convert to $f$.

Rule of the boundary

• A fixed end (string clamp) or closed pipe end must be a node (no displacement there).
• A free end or open pipe end must be an antinode (maximum displacement).

The distance from a node to the next node is $\lambda/2$; node to nearest antinode is $\lambda/4$.

Case 1 — String fixed at both ends (node–node)

We need a node at each end. The shortest fit is one loop: length $L = \lambda/2$. The next adds a full loop each time, so: $L = n\,\frac{\lambda_n}{2}, \quad n=1,2,3,\dots \;\Rightarrow\; \lambda_n=\frac{2L}{n}.$ Convert with $f=v/\lambda$: $\boxed{\,f_n = \frac{n v}{2L} = n f_1,\quad f_1=\frac{v}{2L}\,}$ ALL integer harmonics exist. Here overtone number = harmonic number $-1$.

Case 2 — Pipe open at both ends (antinode–antinode)

Same spacing as the string! Antinode-to-antinode is also $\lambda/2$, so the math is identical: $\lambda_n=\frac{2L}{n},\qquad \boxed{f_n=\frac{nv}{2L}=nf_1.}$ All harmonics present, just like the string.

Case 3 — Pipe closed at one end (node–antinode)

Closed end = node, open end = antinode. Shortest fit is a quarter wavelength: $L=\lambda/4$. Each next mode adds half a wavelength (node→antinode→node→antinode...), giving odd quarters: $L = (2n-1)\frac{\lambda_n}{4}, \quad n=1,2,3,\dots \;\Rightarrow\; \lambda_n=\frac{4L}{2n-1}.$ $\boxed{\,f_n=\frac{(2n-1)v}{4L}=(2n-1)f_1,\quad f_1=\frac{v}{4L}\,}$ Only odd harmonics ($1,3,5,\dots$) exist. The fundamental is half that of an open pipe of the same length — that's why a stopped organ pipe sounds an octave lower.

Worked examples

Steel-man your mistakes

Active recall

Connections

• Standing waves and superposition — the mechanism that creates nodes/antinodes.
• Reflection of waves at boundaries — why fixed/free ends invert or preserve phase.
• Speed of a wave on a string — origin of $v=\sqrt{T/\mu}$.
• Speed of sound in gases — sets $v$ for pipes.
• Beats and resonance — how driven columns lock onto these frequencies.
• Timbre and Fourier synthesis — why the mix of overtones defines an instrument's sound.

Concept Map

Hinglish (regional understanding)