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Derivation

Standing Waves and Boundary Quantization

D-066 Home PU-103 Threads waves · symmetry Depends on The Wave Equation as a Continuum Limit, d'Alembert's General Solution
Statement

For the transverse displacement \( u(x,t) \) of a uniform string of length \( L \) held fixed at both ends, the wave equation \( \partial_t^2 u = c^2\,\partial_x^2 u \) together with the boundary conditions \( u(0,t)=u(L,t)=0 \) admits nontrivial time-harmonic solutions only for a discrete set of wavenumbers \( k_n = n\pi/L \) with \( n = 1,2,3,\dots \); the corresponding angular frequencies form the spectrum \( \omega_n = n\pi c/L \), and the associated spatial profiles (normal modes) are \( \sin(k_n x) \).

Why it matters

This is the archetype of boundary-value quantization: a continuous field equation, once confined to a finite region with homogeneous boundary conditions, no longer supports arbitrary oscillation frequencies but only a countable ladder of them. The same logic reappears verbatim in acoustic pipes, electromagnetic cavities, and — with a different operator — the energy levels of a particle in a box.

Physically it explains why a plucked string of fixed length sounds a definite pitch and its integer harmonics rather than a continuum of tones: the geometry, not the excitation, sets the allowed frequencies.

Assumptions
The wave equation \( \partial_t^2 u = c^2\partial_x^2 u \) holds with constant \( c \).If tension or linear mass density varies along the string, \( c=c(x) \) and separation still works but the spatial modes are no longer sines; the spectrum shifts and need not be an integer ladder.
Both ends are rigidly fixed (Dirichlet conditions \( u(0,t)=u(L,t)=0 \)).Replacing a fixed end by a free end (Neumann, \( \partial_x u=0 \)) changes the selection rule to half-integer wavenumbers and removes the fundamental at \( n=1 \) in favour of \( k_n=(2n-1)\pi/2L \).
Small-amplitude (linear) motion, so distinct modes superpose without interacting.At large amplitude the tension itself depends on \( \partial_x u \), the equation becomes nonlinear, and modes exchange energy rather than persist independently.
The string is lossless and the ends are perfectly reflecting.With damping or partial transmission at the supports the eigenvalues \( k_n \) acquire imaginary parts; the "modes" become decaying quasi-normal modes and the real spectrum is only approximate.
Derivation
1
\[ u(x,t) = X(x)\,T(t) \]
Seek separable solutions; legitimate because the equation is linear with coefficients independent of \( x \) and \( t \), and the full solution is later recovered by superposition. A
2
\[ X(x)\,\ddot{T}(t) = c^2\,X''(x)\,T(t) \]
Substitute the product form into \( \partial_t^2 u = c^2\partial_x^2 u \); dots denote \( d/dt \), primes denote \( d/dx \). A
3
\[ \frac{\ddot{T}(t)}{c^2\,T(t)} = \frac{X''(x)}{X(x)} = -k^2 \]
Divide by \( c^2 X T \). The left side depends only on \( t \), the right only on \( x \); equality for all \( x,t \) forces both to a common constant, named \( -k^2 \) (sign chosen anticipating oscillatory, bounded solutions). B
4
\[ X''(x) + k^2 X(x) = 0 \quad\Longrightarrow\quad X(x) = A\sin(kx) + B\cos(kx) \]
Solve the spatial ordinary differential equation; this is the general solution of a linear second-order equation with constant coefficients and real \( k \). A
5
\[ X(0) = 0 \;\Rightarrow\; B = 0 \]
Apply the fixed left end \( u(0,t)=X(0)\,T(t)=0 \) for all \( t \); since \( T \) is not identically zero, \( X(0)=0 \), and \( \cos(0)=1 \) kills \( B \). A
6
\[ X(L) = A\sin(kL) = 0 \]
Apply the fixed right end \( u(L,t)=X(L)\,T(t)=0 \). Discarding the trivial \( A=0 \) (which gives \( u\equiv 0 \)), a nontrivial mode requires \( \sin(kL)=0 \). B
7
\[ \sin(kL)=0 \;\Longrightarrow\; kL = n\pi,\quad n = 1,2,3,\dots \]
The eigenvalue condition. Zeros of the sine occur at integer multiples of \( \pi \); \( n=0 \) yields \( X\equiv 0 \), and negative \( n \) merely flip the sign of \( \sin(kx) \), reproducing the same mode. This is precisely a Sturm–Liouville eigenvalue problem with a discrete real spectrum. C
8
\[ k_n = \frac{n\pi}{L} \]
Solve the quantization condition for the allowed wavenumbers. A
9
\[ \ddot{T}(t) + c^2 k^2\,T(t) = 0 \;\Rightarrow\; \omega = c k \]
Return to the temporal equation from Step 3; it is simple harmonic with angular frequency \( \omega = ck \) — the dispersion relation of the non-dispersive string. A
10
\[ \omega_n = c\,k_n = \frac{n\pi c}{L} \]
Insert the quantized \( k_n \) into \( \omega = ck \) to obtain the frequency spectrum. A
Result
\[ \boxed{\,k_n = \frac{n\pi}{L}, \qquad \omega_n = \frac{n\pi c}{L}, \qquad u_n(x,t)=\sin\!\left(\frac{n\pi x}{L}\right)\bigl[a_n\cos\omega_n t + b_n\sin\omega_n t\bigr],\quad n=1,2,3,\dots\,} \]

Reading. Only wavelengths \( \lambda_n = 2\pi/k_n = 2L/n \) fit between the two fixed ends — an integer number of half-wavelengths spans \( L \). Each such spatial pattern oscillates in time at its own frequency \( \omega_n \), an integer multiple of the fundamental \( \omega_1 = \pi c/L \). The general motion is a superposition \( u=\sum_n u_n \), a Fourier sine series whose coefficients are fixed by the initial shape and velocity.

Units check. \( c \) has units \( \mathrm{m\,s^{-1}} \) and \( L \) has units \( \mathrm{m} \), so \( n\pi c/L \) has units \( \mathrm{s^{-1}} \), i.e. \( \mathrm{rad\,s^{-1}} \), as required of an angular frequency. Likewise \( k_n = n\pi/L \) has units \( \mathrm{m^{-1}} \), correct for a wavenumber.

Limiting cases
  • Fundamental \( n=1 \): a single half-wave arch, \( \omega_1 = \pi c/L \); the lowest pitch the string can sound.
  • \( L\to\infty \): the spacing \( \Delta\omega = \pi c/L \to 0 \), the ladder collapses to a continuum and quantization disappears — the free string supports any frequency.
  • Large \( n \): mode spacing stays uniform, \( \omega_{n+1}-\omega_n = \pi c/L \), the hallmark of a non-dispersive (linear) medium.
  • \( c\to\infty \) (very stiff, light string): all frequencies scale up together; the spectrum stretches but keeps its integer ratios.
Breaks when
  • Dispersive stiffness. A real string has bending stiffness, adding a \( +\alpha\,\partial_x^4 u \) term. Then \( \omega_n \) grows faster than linearly in \( n \), overtones are stretched sharp, and the harmonics are no longer exact integer multiples — the reason piano tuning uses inharmonicity-corrected "stretched" octaves.
  • Non-rigid or lossy ends. If the supports move, radiate, or absorb energy, the pure Dirichlet condition fails; eigenvalues become complex, modes decay, and each sharp line broadens into a resonance of finite width.
  • Large amplitude. When \( \partial_x u \sim O(1) \) the tension varies with displacement, the equation turns nonlinear, and modes couple — energy leaks between harmonics and \( \omega_n \) becomes amplitude-dependent.
Failure modes
  • Allowing \( n=0 \) as a "mode" from \( kL=n\pi \) — it gives \( X\equiv 0 \), no physical vibration.
  • Confusing wavenumber quantization \( k_n=n\pi/L \) with wavelength: forgetting the factor of two, \( \lambda_n = 2L/n \), not \( L/n \).
  • Using \( f_n \) (in Hz) and \( \omega_n \) (in rad/s) interchangeably; they differ by \( 2\pi \), so \( f_n = nc/2L \).
  • Applying free-end logic (\( \partial_x u=0 \)) while keeping the sine that satisfies fixed ends — mixing incompatible boundary conditions.
  • Assuming the fundamental is the "average" of the motion; it is just one term of the Fourier sum, whose weight depends on the initial condition.
  • Treating \( c \) as the speed of the visible pattern; the standing wave does not travel — \( c \) is the speed of the two counter-propagating waves that build it.
Discussion

A standing wave is not a fundamentally new object but the interference of two identical waves travelling in opposite directions: \( \sin(k_n x)\cos(\omega_n t) = \tfrac12\left[\sin(k_n x - \omega_n t) + \sin(k_n x + \omega_n t)\right] \). The boundary conditions are what force the reflected wave to exist and to carry exactly the right phase, so the true content of the derivation is that reflection plus confinement quantizes. The nodes at \( x = mL/n \) never move, and between them the whole string oscillates in phase.

The mathematical structure is a Sturm–Liouville eigenvalue problem: the operator \( -\,d^2/dx^2 \) on the interval \( [0,L] \) with Dirichlet conditions is self-adjoint, which guarantees real eigenvalues \( k_n^2 \), a discrete spectrum, and eigenfunctions \( \sin(k_n x) \) that are mutually orthogonal, \( \int_0^L \sin(k_m x)\sin(k_n x)\,dx = \tfrac{L}{2}\delta_{mn} \). This orthogonality is exactly what makes the Fourier sine series work and lets arbitrary initial data be projected uniquely onto the modes.

The integer ladder \( \omega_n = n\omega_1 \) is special to the one-dimensional, non-dispersive, uniform, fixed–fixed string. It is the reason bowed and plucked strings produce a clear musical pitch with a harmonic overtone series, whereas a drumhead (two-dimensional, with Bessel-function modes) yields inharmonic overtones and no definite pitch. Geometry writes the spectrum.

Seen from the modern viewpoint, this problem is the classical skeleton of quantization. Replace \( -\partial_x^2 \to \hat{H} \) and the eigenvalue equation \( X'' + k^2 X = 0 \) with Dirichlet conditions becomes the time-independent Schrödinger equation for an infinite square well, \( -\tfrac{\hbar^2}{2m}\psi'' = E\psi \), whose energies \( E_n = \hbar^2 n^2\pi^2/2mL^2 \) come from the identical boundary quantization. The lesson generalizes: confining a wave to a finite domain with homogeneous boundary conditions replaces a continuous parameter by a discrete index — the mechanism behind atomic spectra, cavity photon modes, and lattice band structure alike.

Common misconceptions. The standing wave does not propagate — its envelope is fixed in space; only the amplitude at each point oscillates. And the discreteness is not caused by the driving or plucking: an idealized fixed–fixed string has this discrete spectrum whether or not it is excited. The excitation only decides which modes carry energy and how much.

Worked examples
1
Guitar-type string: \( L = 0.648\ \mathrm{m} \), tension \( \mathcal{T} = 84\ \mathrm{N} \), linear density \( \mu = 6.3\times10^{-4}\ \mathrm{kg\,m^{-1}} \). Find the fundamental frequency and the third harmonic.
Set up. First get the wave speed \( c=\sqrt{\mathcal{T}/\mu} \), then apply \( \omega_n=n\pi c/L \) and \( f_n=\omega_n/2\pi = nc/2L \). A
2
\[ c = \sqrt{\frac{\mathcal{T}}{\mu}} = \sqrt{\frac{84}{6.3\times10^{-4}}} = \sqrt{1.333\times10^{5}} = 365.2\ \mathrm{m\,s^{-1}} \]
Wave speed from tension and mass density (a prior result for the loaded string). A
3
\[ f_1 = \frac{c}{2L} = \frac{365.2}{2(0.648)} = 281.8\ \mathrm{Hz}, \qquad f_3 = 3f_1 = 845.5\ \mathrm{Hz} \]
Fundamental from \( n=1 \), third harmonic from \( n=3 \). A
\[ f_1 \approx 282\ \mathrm{Hz}, \qquad f_3 \approx 846\ \mathrm{Hz} \]

Reading. These parameters place the fundamental in the neighbourhood of the note D♯₄; the third harmonic sits an octave-plus-a-fifth above it. Units check. \( \sqrt{\mathrm{N}/(\mathrm{kg\,m^{-1}})} = \sqrt{\mathrm{kg\,m\,s^{-2}\cdot kg^{-1}\,m}} = \mathrm{m\,s^{-1}} \); then \( c/L \) gives \( \mathrm{s^{-1}} \).

1
A string with \( c = 200\ \mathrm{m\,s^{-1}} \) and \( L = 0.50\ \mathrm{m} \), fixed at both ends, is released from rest in the shape \( u(x,0) = 0.004\,\sin(3\pi x/L)\ \mathrm{m} \). Describe the subsequent motion and give its frequency.
Set up. The initial shape is exactly the \( n=3 \) eigenfunction, so only that mode is excited; its time dependence is \( \cos\omega_3 t \) because the release is from rest. B
2
\[ \omega_3 = \frac{3\pi c}{L} = \frac{3\pi(200)}{0.50} = 3.77\times10^{3}\ \mathrm{rad\,s^{-1}} \]
Apply the spectrum with \( n=3 \). A
3
\[ u(x,t) = 0.004\,\sin\!\left(\frac{3\pi x}{L}\right)\cos(\omega_3 t)\ \mathrm{m}, \qquad f_3 = \frac{\omega_3}{2\pi} = 600\ \mathrm{Hz} \]
Zero initial velocity kills the \( \sin\omega_3 t \) part, leaving a pure cosine oscillation of the single mode. B
\[ f_3 = 600\ \mathrm{Hz}, \qquad \text{interior nodes at } x = \tfrac{L}{3},\ \tfrac{2L}{3} \]

Reading. Because the initial data is a single eigenmode, the string vibrates purely at \( 600\ \mathrm{Hz} \) forever (in the idealized lossless limit), with fixed nodes at one-third and two-thirds of its length. Units check. \( (\mathrm{rad\,s^{-1}})/(2\pi) = \mathrm{Hz} \); the amplitude stays in metres.

Problems
  1. A string of length \( L = 1.2\ \mathrm{m} \) has wave speed \( c = 90\ \mathrm{m\,s^{-1}} \) and is fixed at both ends. Find the fundamental frequency \( f_1 \) in Hz.
    Solution \( f_1 = c/2L = 90/(2\times1.2) = 90/2.4 = 37.5\ \mathrm{Hz}. \)
  2. For the string in Problem 1, what is the frequency spacing \( \Delta f = f_{n+1}-f_n \) between consecutive harmonics, and which harmonic \( n \) lies closest to \( 500\ \mathrm{Hz} \)?
    Solution \( \Delta f = c/2L = 37.5\ \mathrm{Hz} \) (uniform, since the medium is non-dispersive). Then \( n \approx 500/37.5 = 13.3 \), so \( n=13 \) at \( 487.5\ \mathrm{Hz} \) is closest (\( n=14 \) gives \( 525\ \mathrm{Hz} \), which is farther).
  3. A wire is fixed at both ends with \( L = 0.80\ \mathrm{m} \), \( \mu = 5.0\times10^{-3}\ \mathrm{kg\,m^{-1}} \), and tension \( \mathcal{T} = 320\ \mathrm{N} \). Find the frequency of the second harmonic \( f_2 \).
    Solution \( c=\sqrt{\mathcal{T}/\mu}=\sqrt{320/(5.0\times10^{-3})}=\sqrt{6.4\times10^{4}}=253\ \mathrm{m\,s^{-1}} \). Then \( f_2 = 2c/2L = c/L = 253/0.80 = 316\ \mathrm{Hz}. \)
  4. A string is released from rest with initial shape \( u(x,0)=A\sin(\pi x/L)+\tfrac{1}{2}A\sin(4\pi x/L) \). Write \( u(x,t) \) and state which frequencies are present. Take \( c \) and \( L \) as given.
    Solution Each sine is an eigenmode; released from rest each carries a \( \cos\omega_n t \). Thus \( u(x,t)=A\sin(\pi x/L)\cos(\omega_1 t)+\tfrac12 A\sin(4\pi x/L)\cos(\omega_4 t) \) with \( \omega_1=\pi c/L \) and \( \omega_4=4\pi c/L \). Only the \( n=1 \) and \( n=4 \) frequencies appear — the fundamental and its fourth harmonic; the third harmonic is absent because it was not present in the initial shape.
  5. Show that if the right end is free instead of fixed, \( \partial_x u(L,t)=0 \), the allowed wavenumbers become \( k_n=(2n-1)\pi/2L \). Then give \( \omega_1 \) for \( L=0.60\ \mathrm{m} \), \( c=150\ \mathrm{m\,s^{-1}} \).
    Solution With \( X=A\sin(kx) \) (from the fixed left end), \( X'(x)=Ak\cos(kx) \); the free right end requires \( \cos(kL)=0 \), i.e. \( kL=(2n-1)\pi/2 \), giving \( k_n=(2n-1)\pi/2L \). The fundamental is \( n=1 \): \( \omega_1 = ck_1 = c\pi/2L = 150\pi/(2\times0.60) = 150\pi/1.2 = 392.7\ \mathrm{rad\,s^{-1}} \) (\( f_1 = 62.5\ \mathrm{Hz} \)). Note the fundamental is now a quarter-wave, exactly half the frequency of the fixed–fixed string of the same length.