Standing waves on a string: why only the harmonics fit

Regime 3 — an algebra-only topic.

Regime 3 · algebra-only Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A string of length $L = 1$ m, fixed at both ends, carries transverse waves at speed $v = 120$ m/s. It cannot vibrate at just any frequency — only at a discrete ladder of standing waves, the harmonics $f_n = nv/2L$ . The $n$ -th harmonic has $n$ half-wavelengths between the walls and $n-1$ interior nodes that never move. Drag the mode number: the shape changes, the nodes stay pinned, and the frequency climbs in exact integer steps. Why only integers? Because the ends are pinned — and that boundary condition is what quantizes the modes.

Fundamental frequency

f_1 = v/2L

60 Hz

Fundamental wavelength

\lambda_1 = 2L

2 m

Frequency of harmonic

n=3

f_n = nv/2L

180 Hz

Wavelength of harmonic

n=3

\lambda_n = 2L/n

0.6667 m

The string obeys the wave equation; its standing solutions separate

\partial_{tt} y = v^2\,\partial_{xx} y \ \Longrightarrow\ y(x,t) = A\sin(kx)\cos(\omega t),\quad \omega = vk

Every small piece of the string obeys the wave equation. Looking for solutions that oscillate in place — a fixed shape times a time wobble — gives $y = A\sin(kx)\cos(\omega t)$ , which solves it whenever $\omega = vk$ . So far $k$ could be anything.

check $\partial_{tt}y = v^2\partial_{xx}y$ ; fixed ends $\sin(kL)=0$ ; standing = two travelling waves; $f_n = nv/2L$ ; $\lambda_n = 2L/n$

✓ The mode solves the wave equation $\partial_{tt} y = v^2\,\partial_{xx} y$ — back-substituting $y = A\sin(kx)\cos(\omega t)$ leaves zero, with $\omega = vk$ . [structural]
✓ The string is pinned at both ends: $y(0)=0$ automatically, and $y(L)=0$ forces $\sin(kL)=\sin(n\pi)=0$ — the boundary condition that quantizes the modes to integer $n$ . [structural]
✓ A standing wave is two counter-propagating travelling waves superposed: $\tfrac12 A[\sin(kx-\omega t)+\sin(kx+\omega t)] = A\sin(kx)\cos(\omega t)$ . [simplify]
✓ The harmonic frequencies fall out: $f_n = \omega_n/2\pi = nv/(2L)$ . [structural]
✓ The wavelengths are $\lambda_n = 2\pi/k_n = 2L/n$ — an integer number of half-wavelengths fits the length. [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A string can be made to vibrate as a standing wave at any frequency you choose.”

It cannot. The fixed ends force a node at each end, so only shapes with a whole number of half-wavelengths between them survive — the harmonics $f_n = nv/2L$ . Drive the string at some other frequency and the reflections off the ends no longer line up: the wave interferes with itself and dies out instead of building a steady standing pattern. This is the same quantization that, one domain over, restricts an electron in a box to discrete energies — boundary conditions select a discrete ladder.

Modeling assumptions — author-asserted, disclosed not discharged

An ideal string fixed rigidly at both ends, so there is a displacement node at each end at all times.
The wave speed $v$ is constant (a uniform string under constant tension); damping is neglected, so the modes are sharp.

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):