Fourier Decomposition into Normal Modes

Derivation

Fourier Decomposition into Normal Modes

D-067 Home PU-103 Threads waves · symmetry Depends on Standing Waves and Boundary Quantization, function-space-orthogonality

Statement

For a uniform string of length \(L\) fixed at both ends, the normal modes \(\phi_n(x)=\sin\!\left(\frac{n\pi x}{L}\right)\), \(n=1,2,3,\dots\), form a complete orthogonal set on \([0,L]\). Any admissible transverse configuration \(y(x)\) obeying the boundary conditions expands uniquely as \(y(x)=\sum_{n=1}^{\infty} b_n\,\sin\!\left(\frac{n\pi x}{L}\right)\), and each modal amplitude is recovered by projection, \(b_n=\frac{2}{L}\int_0^L y(x)\,\sin\!\left(\frac{n\pi x}{L}\right)dx\).

Why it matters

The wave equation is linear, so its general motion is a superposition of independently oscillating normal modes. Separation of variables supplies the modes but not the recipe for matching an arbitrary starting shape; decomposition converts that intractable boundary-value problem into a countable list of decoupled harmonic oscillators, one per mode, each evolving as \(\cos(\omega_n t)\) with \(\omega_n=\frac{n\pi c}{L}\).

The projection formula is the physical content of orthogonality: the modes do not "leak" into one another, so the amplitude and energy carried by each harmonic can be read off the initial data alone. This same machinery — expand over eigenfunctions, extract coefficients by inner product — recurs in quantum mechanics (states over energy eigenstates), optics, and signal analysis. The string is the cleanest place to watch the whole machine run.

Assumptions

Fixed ends (Dirichlet boundary conditions).If the ends are free or loaded, the mode set becomes cosines or a mixed set with shifted eigenvalues, and the sine basis no longer spans the admissible configurations.

Uniform linear density and tension.Non-uniform \(\mu(x)\) or \(T(x)\) gives a Sturm–Liouville problem whose eigenfunctions are not sines, though they remain orthogonal under a weighted inner product.

Linear, small-amplitude transverse displacement.Large amplitude introduces nonlinearity and mode coupling, so superposition fails and the coefficients cease to be constant in time.

The configuration lies in \(L^2([0,L])\) and the eigenset is complete.If \(y(x)\) is not square-integrable the projection integral may diverge; if the eigenset were incomplete, some configurations could not be represented and the equality would fail. Completeness is a theorem for the Dirichlet Laplacian, not a fact verifiable by inspection.

Derivation

\[ y(x)=\sum_{n=1}^{\infty} b_n\,\sin\!\left(\frac{n\pi x}{L}\right) \]

Posit the expansion over the normal modes; legitimate because the fixed-end modes are a complete orthogonal basis of \(L^2([0,L])\) (completeness theorem for the Dirichlet Laplacian). B

\[ \int_0^L \sin\!\left(\frac{n\pi x}{L}\right)\sin\!\left(\frac{m\pi x}{L}\right)dx=\frac{L}{2}\,\delta_{nm} \]

Orthogonality of the modes on \([0,L]\) (prior result); the self-adjoint operator \(-\frac{d^2}{dx^2}\) with Dirichlet conditions has orthogonal eigenfunctions. The delta \(\delta_{nm}\) is \(1\) if \(n=m\), else \(0\). A

\[ \int_0^L y(x)\,\sin\!\left(\frac{m\pi x}{L}\right)dx=\int_0^L\left[\sum_{n=1}^{\infty} b_n\,\sin\!\left(\frac{n\pi x}{L}\right)\right]\sin\!\left(\frac{m\pi x}{L}\right)dx \]

Project onto mode \(m\): multiply both sides by \(\sin(m\pi x/L)\) and integrate over the string — the exact analogue of dotting a vector with a basis vector. A

\[ \int_0^L y(x)\,\sin\!\left(\frac{m\pi x}{L}\right)dx=\sum_{n=1}^{\infty} b_n \int_0^L \sin\!\left(\frac{n\pi x}{L}\right)\sin\!\left(\frac{m\pi x}{L}\right)dx \]

Interchange sum and integral. Legal because the partial sums converge in \(L^2\) (uniformly for continuous, piecewise-smooth \(y\)), so the integral of the sum equals the sum of the integrals (dominated / uniform convergence). C

\[ \int_0^L y(x)\,\sin\!\left(\frac{m\pi x}{L}\right)dx=\sum_{n=1}^{\infty} b_n\,\frac{L}{2}\,\delta_{nm}=b_m\,\frac{L}{2} \]

Apply orthogonality from Step 2; the Kronecker delta collapses the infinite sum to the single \(n=m\) term. A

\[ b_m=\frac{2}{L}\int_0^L y(x)\,\sin\!\left(\frac{m\pi x}{L}\right)dx \]

Divide by \(L/2\) and relabel \(m\to n\). Symbols rearranged; no numbers introduced. A

\[ \left.\frac{\partial y}{\partial t}\right|_{t=0}=v_0(x)=\sum_{n=1}^{\infty} a_n\,\omega_n\,\sin\!\left(\frac{n\pi x}{L}\right)\;\Rightarrow\; a_n=\frac{2}{L\,\omega_n}\int_0^L v_0(x)\,\sin\!\left(\frac{n\pi x}{L}\right)dx \]

Differentiate the full solution in time, evaluate at \(t=0\) (now the sine time-factors survive), and project identically. This fixes the velocity family of coefficients from the initial velocity. C

Result

\[ y(x)=\sum_{n=1}^{\infty} b_n\,\sin\!\left(\frac{n\pi x}{L}\right),\qquad b_n=\frac{2}{L}\int_0^L y(x)\,\sin\!\left(\frac{n\pi x}{L}\right)dx \]

Reading. The string's shape is a weighted sum of standing modes, and the weight of each mode is its overlap with the shape, measured by the inner product and normalised by the mode's own norm \(L/2\). A mode "hears" only the part of \(y(x)\) that shares its spatial rhythm; everything orthogonal to it is invisible to the projection. Nothing about the shape is lost — the coefficients carry complete information.

Units check. With \([y]=\text{m}\) and the sine dimensionless, \(\int y\sin\,dx\) has units \(\text{m}\cdot\text{m}=\text{m}^2\). Multiplying by \(2/L\) (units \(\text{m}^{-1}\)) gives \(b_n\) in metres — correct, since \(b_n\) is a displacement amplitude.

Limiting cases

Single-mode initial shape. If \(y(x)=A\sin(k\pi x/L)\), orthogonality gives \(b_n=A\,\delta_{nk}\) — the projection returns only the mode present.
Smooth shape. For an infinitely differentiable \(y\) vanishing at both ends, coefficients decay faster than any power of \(1/n\); the series converges rapidly and truncation is safe.
Kinked shape (plucked string). A corner makes \(b_n\sim 1/n^2\); a jump in \(y\) itself gives \(1/n\) with Gibbs overshoot near the discontinuity.
Node-plucked mode absent. Exciting exactly at a node of mode \(n\) (e.g. \(x=L/2\) for \(n=2\)) gives \(b_n=0\); that harmonic is never sounded — the basis of guitar harmonics.

Breaks when

Non-Dirichlet or moving boundaries. Free ends, an attached mass, or a time-varying length break the fixed sine eigenset; the correct basis differs and the \(2/L\) normalisation no longer holds.
Nonlinear or large-amplitude motion. When tension varies with amplitude, modes exchange energy; the coefficients \(b_n\) become time-dependent and superposition is invalid.
Non-\(L^2\) data. A configuration with a non-integrable singularity has a divergent projection integral, so no finite \(b_n\) exists.
Dispersive or stiff strings. Bending stiffness adds a \(\partial^4 y/\partial x^4\) term; true modes shift and are no longer pure sines, so the sine expansion no longer diagonalises the dynamics with \(\omega_n=n\pi c/L\).

Failure modes

Dropping the \(2/L\). Writing \(b_n=\int_0^L y\sin\,dx\) forgets to divide by the mode norm \(L/2\), giving amplitudes too large by \(L/2\).
Full-period normalisation. Using \(1/L\) (as for a \(2L\)-period Fourier series on \([-L,L]\)) instead of \(2/L\); the sine series on \([0,L]\) is the odd half-range extension and carries the factor \(2\).
Wrong basis function. Projecting fixed-end data with free-end cosines or a shifted sine yields spurious nonzero overlaps.
Assuming all modes contribute. For a symmetric centre pluck, even modes have a node there and vanish identically; students often "find" small nonzero even coefficients from arithmetic slips.
Confusing shape and velocity coefficients. Feeding \(v_0(x)\) into the \(b_n\) integral, or dropping the \(1/\omega_n\) that must accompany the velocity projection.
Including \(n=0\). A constant term is impossible for a doubly-fixed string since \(\sin 0=0\), yet the cosine-series \(a_0/2\) is copied reflexively.

Discussion

The decomposition is an eigenfunction expansion of a self-adjoint operator. The spatial operator \(-\frac{d^2}{dx^2}\) acting on functions that vanish at \(x=0,L\) is Hermitian under the inner product \(\langle f,g\rangle=\int_0^L f g\,dx\). Hermiticity forces distinct eigenvalues \((n\pi/L)^2\) to have orthogonal eigenfunctions, and Sturm–Liouville theory guarantees the set is complete. Projection is then nothing more than resolving a vector along an orthogonal basis — the string configuration is a vector in an infinite-dimensional function space, and \(b_n\) are its components, \(b_n=\langle y,\phi_n\rangle/\langle\phi_n,\phi_n\rangle\).

Inserting the time dependence, each mode rides as \(\big[b_n\cos(\omega_n t)+a_n\sin(\omega_n t)\big]\sin(n\pi x/L)\) with \(\omega_n=\frac{n\pi c}{L}\) and \(c=\sqrt{T/\mu}\). The full solution \(y(x,t)\) is thus fixed entirely by the initial shape (setting the \(b_n\)) and initial velocity (setting the \(a_n\)). Because the modes are orthogonal, the total energy is a sum of independent modal energies with no cross terms — Parseval's theorem, \(\int_0^L y^2\,dx=\frac{L}{2}\sum_n b_n^2\), is the Pythagorean theorem of the function space and makes energy bookkeeping exact.

The rate at which \(b_n\) decays with \(n\) encodes the smoothness of the shape — a spectral-analysis principle. A kink (discontinuous slope) gives \(1/n^2\); a genuine jump gives \(1/n\) and audibly bright harmonics. This is why a string plucked with a hard plectrum near the bridge sounds brighter than one bowed gently near the centre: the excitation's spatial sharpness and location directly set the overtone spectrum, and hence the timbre, at fixed pitch.

Rigorously, convergence is subtle. The sine series converges to \(y\) in the \(L^2\) norm for any square-integrable configuration, \(\lim_{N\to\infty}\big\|y-\sum_{n\le N} b_n\phi_n\big\|_2=0\), but this does not imply convergence at every point. Pointwise convergence requires Dirichlet-type conditions (bounded variation), and at a jump the series converges to the midpoint value with a Gibbs overshoot of about \(9\%\) that does not vanish as terms are added. The interchange in Step 4 is licensed by \(L^2\) completeness for coefficient extraction even when pointwise reconstruction misbehaves — a distinction that matters when the "shape" is defined only almost everywhere.

Common misconceptions. The expansion does not require \(y(x)\) to be periodic or even continuous — only square-integrable and consistent with the boundary conditions; periodicity is a property of the odd extension, not a prerequisite. And a mode being absent (\(b_n=0\)) is a statement about overlap with the initial shape, not about whether the string "can" vibrate at \(\omega_n\).

Worked examples

Centre-plucked string: \( y(x)=\dfrac{2h}{L}x\) for \(0\le x\le L/2\), and \(y(x)=\dfrac{2h}{L}(L-x)\) for \(L/2\le x\le L\).

Triangular initial shape, peak height \(h\) at the midpoint. Symbols first, then numbers. A

\[ b_n=\frac{2}{L}\int_0^L y(x)\,\sin\!\left(\frac{n\pi x}{L}\right)dx=\frac{8h}{n^2\pi^2}\sin\!\left(\frac{n\pi}{2}\right) \]

Integrate by parts on each half and add; \(\sin(n\pi/2)\) is \(0\) for even \(n\) and \(\pm 1\) for odd \(n\), so only odd modes survive. B

\( h=5.0\ \text{mm}=5.0\times10^{-3}\ \text{m}\), \( L=0.65\ \text{m}\).

Guitar-string values. Note \(L\) cancels in \(b_n\); amplitudes depend only on \(h\). A

\[ b_1=\frac{8(5.0\times10^{-3})}{\pi^2}=4.05\times10^{-3}\ \text{m},\quad b_3=-\frac{8(5.0\times10^{-3})}{9\pi^2}=-4.50\times10^{-4}\ \text{m},\quad b_5=\frac{8(5.0\times10^{-3})}{25\pi^2}=1.62\times10^{-4}\ \text{m} \]

Insert numbers for \(n=1,3,5\). The \(1/n^2\) decay reflects the corner at the pluck point; signs alternate. B

\[ b_1\approx 4.05\ \text{mm},\quad b_2=0,\quad b_3\approx -0.450\ \text{mm},\quad b_5\approx 0.162\ \text{mm} \]

Reading. The fundamental carries about \(81\%\) of the peak height as its amplitude; overtones fall as \(1/n^2\) and alternate in sign. Even modes have a node at the centre, so the symmetric pluck cannot excite them — a mellow tone.

Units check. Every \(b_n\) is a pure multiple of \(h\), so all are in metres — correct for amplitudes.

Parabolic release: \( y(x)=\dfrac{4A}{L^2}\,x(L-x)\), peak \(A\) at \(x=L/2\).

A smooth, corner-free initial displacement — contrast with the plucked case. A

\[ \int_0^L x(L-x)\,\sin\!\left(\frac{n\pi x}{L}\right)dx=\frac{2L^3}{n^3\pi^3}\big[1-(-1)^n\big] \]

Integrate by parts twice; the bracket is \(0\) for even \(n\) and \(2\) for odd \(n\). B

\[ b_n=\frac{2}{L}\cdot\frac{4A}{L^2}\cdot\frac{2L^3}{n^3\pi^3}\big[1-(-1)^n\big]=\frac{32A}{n^3\pi^3}\quad(n\ \text{odd}) \]

Collect factors; \(L\) cancels again. Coefficients decay as \(1/n^3\) — faster than the plucked \(1/n^2\), because the shape has no kink. B

\( A=3.0\ \text{mm}\): \(\;b_1=\dfrac{32(3.0\times10^{-3})}{\pi^3}=3.10\times10^{-3}\ \text{m},\;\; b_3=\dfrac{32(3.0\times10^{-3})}{27\pi^3}=1.15\times10^{-4}\ \text{m}\).

Insert numbers. The fundamental slightly exceeds \(A\) because the parabola is "fuller" than a half-sine. B

\[ b_1\approx 3.10\ \text{mm},\quad b_3\approx 0.115\ \text{mm},\quad b_5\approx 0.0248\ \text{mm};\quad \text{even modes zero} \]

Reading. The \(1/n^3\) decay makes the third harmonic about \(27\times\) weaker relative to the fundamental than in the plucked case — smoother shape, purer tone.

Units check. Each \(b_n\) is a dimensionless number times \(A\) (metres), so all amplitudes are in metres.

Problems

Show that a string in the pure shape \(y(x)=0.4\sin(3\pi x/L)\) mm has \(b_3=0.4\) mm and all other coefficients zero.
Solution
Project: \(b_n=\frac{2}{L}\int_0^L 0.4\sin(3\pi x/L)\sin(n\pi x/L)\,dx\). By orthogonality the integral is \(\frac{L}{2}\delta_{n3}\), so \(b_n=0.4\,\delta_{n3}\) mm. Only \(b_3=0.4\) mm survives; every other mode is orthogonal to the shape and expands to itself.
A string of length \(L=1.20\) m is plucked at its centre to height \(h=8.0\) mm. Find \(b_1\) and \(b_2\).
Solution
Use \(b_n=\frac{8h}{n^2\pi^2}\sin(n\pi/2)\). For \(n=1\): \(\sin(\pi/2)=1\), so \(b_1=\frac{8(8.0\times10^{-3})}{\pi^2}=6.48\times10^{-3}\) m \(=6.48\) mm. For \(n=2\): \(\sin(\pi)=0\), so \(b_2=0\). The even mode has a node at the centre and cannot be excited by a centred pluck. (\(L\) does not enter.)
Using Parseval's theorem \(\int_0^L y^2\,dx=\frac{L}{2}\sum_n b_n^2\), find the fraction of the mean-square displacement carried by the fundamental of the centre-plucked string (odd \(n\) only, \(|b_n|=8h/n^2\pi^2\)).
Solution
Modal weights \(b_n^2\propto 1/n^4\) over odd \(n\). Fraction in the fundamental \(=1/\sum_{n\ \text{odd}}(1/n^4)\). The odd sum \(\sum_{n\ \text{odd}}1/n^4=\pi^4/96=1.01468\). Fraction \(=1/1.01468=0.9855\), i.e. about \(98.6\%\) of the mean-square displacement is in the fundamental.
A uniform initial displacement is approximated by \(y(x)=d=2.0\) mm (constant) on \(0<x<L\). Find the modal amplitudes and comment on convergence.
Solution
\(b_n=\frac{2}{L}\int_0^L d\sin(n\pi x/L)\,dx=\frac{2d}{L}\cdot\frac{L}{n\pi}[1-\cos n\pi]=\frac{2d}{n\pi}[1-(-1)^n]\). Odd \(n\): \(b_n=\frac{4d}{n\pi}=\frac{4(2.0\times10^{-3})}{n\pi}\), so \(b_1=2.55\) mm, \(b_3=0.849\) mm, \(b_5=0.509\) mm. Even \(n\): zero. The slow \(1/n\) decay arises because the constant shape violates the fixed-end conditions (a jump at \(x=0,L\)), producing Gibbs overshoot near the ends.
A string (\(L=0.50\) m, \(c=3.0\times10^2\) m/s) is released from rest in the shape \(y(x)=0.010\sin(\pi x/L)+0.004\sin(2\pi x/L)\) metres. Write \(y(x,t)\) fully and give the midpoint displacement at \(t=2.0\) ms.
Solution
By inspection \(b_1=0.010\) m, \(b_2=0.004\) m, all others zero; released from rest means the velocity coefficients vanish, so time factors are cosines. Frequencies: \(\omega_1=\pi c/L=\pi(300)/0.50=1885\) rad/s, \(\omega_2=2\omega_1=3770\) rad/s. Thus \(y(x,t)=0.010\cos(1885t)\sin(\pi x/L)+0.004\cos(3770t)\sin(2\pi x/L)\). At \(x=L/2\): \(\sin(\pi/2)=1\), \(\sin(\pi)=0\), so only the fundamental contributes: \(y=0.010\cos(1885\times2.0\times10^{-3})=0.010\cos(3.77)\). Since \(\cos(3.77\ \text{rad})=-0.795\), \(y(L/2,2.0\ \text{ms})=-7.9\times10^{-3}\) m \(=-7.9\) mm.