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If the mass and stiffness matrices commute with a permutation representation, each symmetry subspace is invariant under the generalized eigenvalue problem for small oscillations. On an irreducible subspace, Schur lemma forces a commuting operator to be scalar, producing a degenerate normal mode multiplet.

An eigenfrequency is the natural frequency associated with a free normal mode. Always specify whether it means the angular frequency $\omega$ or the frequency $f=\omega /(2\pi )$ in hertz. For a uniform fixed-end wave equation on a string, ${\omega }_n=n\pi \sqrt{\tau /\mu }/L$ and $f_n=n\sqrt{\tau /\mu }/(2L)$; the amplitude does not change either within the linear model. In a generalized eigenvalue problem for small oscillations, the eigenvalue is ${\omega }^2$, rather than $\omega$ itself.

For a first-order system $\dot{z}=iAz$, a real eigenvalue $\lambda$ of $A$ instead gives a signed eigenfrequency in $e^{i\lambda t}$. A positive imaginary part of $\lambda$ gives exponential decay with this sign convention. Thus a complex damped eigenfrequency encodes both oscillation and decay; it is not just the number of cycles per second.

For equal rod lengths $l$, upper mass $m$ and lower mass $\alpha m>0$, use horizontal displacements $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$ at first order. The quadratic kinetic energy is $m{\dot{x}}_1^2/2+\alpha m{\dot{x}}_2^2/2$, while the stiffness matrix isThe generalized eigenvalue problem for small oscillations is $Ka={\omega }^{2}Ma$, not $Ka=m{\omega }^{2}a$. In particular the lower Euler-Lagrange equation is ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, independent of the lower mass factor after division by that mass. This reproduces the normal modes of a double pendulum and distinguishes the mechanical mass matrix from its stiffness matrix.

Suppose a two-coordinate linear oscillator has independent normal modes $v_1,v_2$ as displayed, with positive angular frequencies ${\omega }_1,{\omega }_2$. For initial displacement $d=(d_1,d_2{)}^T$ and velocity $w=(w_1,w_2{)}^T$, its solution iswhereThese formulas follow by inverting the two-by-two matrix with columns $v_1,v_2$. Each coefficient evolves by its own scalar harmonic-oscillator equation, and substituting $t=0$ recovers both prescribed vectors. Euclidean orthogonality of the mode shapes is unnecessary; with unequal masses, the natural orthogonality is in the mass inner product of the generalized eigenvalue problem for small oscillations.

The symmetric coefficient matrix of kinetic energy to second order in generalized velocities. For independent horizontal point-mass coordinates it is diagonal, with the particle masses on the diagonal. Together with the mechanical stiffness matrix it gives the generalized eigenvalue problem for small oscillations.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

For three collinear masses $m,\mu m,m$, with $m>0$, $\mu >0$, connected consecutively by two springs of stiffness $k>0$ and with free ends, let $x$ denote displacements from an equilibrium. The kinetic energy and potential energy are $T={\dot{x}}^{T}M\dot{x}/2$, $V=x^{T}Kx/2$, whereThe generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$ hasDirect substitution verifies each pair; the three distinct eigenvalues exhaust the three-dimensional problem. The zero normal mode translates the complete system without extending either spring. The second holds the middle mass fixed while the end masses move oppositely. The third moves the end masses together and the middle mass oppositely, leaving the centre of mass fixed.

For mass-matrix orthogonality of normal modes, the squared norms $v^{T}Mv$ are respectively $m(\mu +2)$, $2m$ and $m(2+4/\mu )$. Dividing each shape by its corresponding square root gives $q_i^{T}M{q}_j={\delta }_{ij}$. The general motion is a linear combination of the two oscillatory modes and $(A+Bt)v_0$; the latter describes free translation with constant velocity.

The kinetic energy of the three unit masses is $\frac{1}{2}{\sum }_i{\dot{x}}_i^2$. Expanding the potential givesThus the small-oscillation mass and stiffness matrices arewhere $J$ is the all-ones matrix.

The vector $(1,1,1{)}^T$ is an eigenvector of the stiffness matrix with eigenvalue $1$. Every vector orthogonal to it has coordinate sum zero and is an eigenvector with eigenvalue $4$. Since $T=I$, the generalized eigenvalue problem for small oscillations gives ${\omega }^2$ equal to these eigenvalues. One orthonormal choice of normal modes isThe last two may be replaced by any orthonormal basis of the sum-zero plane because the frequency is degenerate. Hence the normal frequencies are $\boxed{1,2,2}$.

The Euler--Lagrange equations for the quadratic Lagrangian areFor a normal mode $x(t)=a{e}^{i\omega t}$, they become the generalized eigenvalue problem for small oscillationsPut $\lambda =m{\omega }^2/k$. With the stated spring constants,The final matrix has characteristic polynomialso its eigenvalue are ${\nu }_{\pm }=\pm \sqrt{1+{\delta }^2}$. HenceThe two angular frequencies therefore satisfyThis is the spectrum recorded by normal modes of two equal masses between three springs.

Taking downward distance as positive, the Lagrangian is

At equilibrium the lower spring supports only the lower mass, while the upper spring supports both masses:Hence

Let $q_i=z_i-z_i^{(0)}$. The linear terms cancel by the equilibrium equations, and the quadratic Lagrangian iswith the small-oscillation mass and stiffness matrices

The generalized eigenvalue problem for small oscillations isIts characteristic equation issoCorresponding displacement eigenvectors are

They are not orthogonal in the ordinary Euclidean inner product. They are orthogonal in the kinetic-energy, or mass-matrix, inner product:This is mass-matrix orthogonality of normal modes.

General definitions. The Euler-Lagrange equation gives $m_i{\ddot{q}}_i+{\sum }_j{V}_{ij}{q}_j=0$. Write $M=\mathrm{diag}(m_i)$ and $K=(V_{ij})$. A normal mode has fixed displacement pattern $v$ and common time factor, $q(t)=vQ(t)$; its normal frequency satisfiesHere $\omega$ is angular frequency. A normal coordinate is the coefficient of one pattern in a complete modal expansion, so these coordinates decouple the equations. Since $M$ is positive definite, the change $z=M^{1/2}q$ turns the mode equation into the ordinary symmetric eigenvalue problem for $M^{-1/2}K{M}^{-1/2}$.

Three-mass equations. Let $x_1,x_2,x_3$ be horizontal displacements from the unstretched equilibrium, ordered left to right. The kinetic energy and potential energy areDifferentiating the potential energy givesThus $M=m\mathrm{diag}(1,2,3)$ and

Frequencies and patterns. Put $\lambda =m{\omega }^2/k$. Expansion of the generalized eigenvalue problem for small oscillations givesSolving at each root givesThe zero normal mode is uniform translation. In the lower nonzero mode, the central mass stays fixed, the two end masses move oppositely, and their unequal displacements balance both the end-mass momenta and the forces on the central mass. In the upper mode the end masses move together and the centre oppositely. Both vibrating modes leave the centre of mass fixed.

Orthogonality. Because $K$ is symmetric, two mode equations implyDistinct eigenvalues therefore give the mass-matrix orthogonality of normal modes:The Euclidean dot product need not vanish for unequal masses; for example $v_0^T{v}_1=-2$, while $v_0^{T}M{v}_1=m(-3+3)=0$. The modal squared norms are $6m,12m,12m$.

Normal coordinates. Expand $x=Q_0{v}_0+Q_1{v}_1+Q_3{v}_3$. Project using the mass-weighted inner product and divide by each modal norm to obtainFor example, the inverse equations are $x_1=Q_0-3Q_1+Q_3$, $x_2=Q_0-2Q_3$, $x_3=Q_0+Q_1+Q_3$. They reduce the energies toand henceThe first coordinate is the centre of mass and has free linear motion. Multiplying $Q_0,Q_1,Q_3$ by $\sqrt{6m},\sqrt{12m},\sqrt{12m}$ respectively gives mass-normalized normal coordinates if desired.

For a conservative system near a stable equilibrium, the quadratic Lagrangian has the form $L=\frac{1}{2}{\dot{q}}^{T}M\dot{q}-\frac{1}{2}{q}^{T}Kq$, with symmetric mass matrix $M$ and stiffness matrix $K$. A normal mode $q=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem for small oscillations $(K-{\omega }^{2}M)v=0$, so the frequency equation is $\det (K-{\omega }^{2}M)=0$.

To quadratic order, vertical velocities contribute only fourth-order corrections to the kinetic energy, leaving $T=({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)/2$. A string of length $\ell$ with small horizontal difference $d$ shortens its vertical projection by $d^2/(2\ell )$. The upper segment carries all three masses, the middle segment two, and the lower segment one. Hence their increases in gravitational potential energy sum toSubtracting $V$ from $T$ gives the stated approximate Lagrangian. Here $M=I$ andFor a mode with middle displacement zero, write $v=(X,0,Z{)}^T$. The middle row requires $X/b+Z/c=0$, so $Z=-cX/b$. A nonzero mode then has $X,Z\ne 0$, and the outer rows require ${\omega }^2/g=a^{-1}+b^{-1}=c^{-1}$. Therefore the necessary and sufficient condition isWhen it holds, the mode shape is $(1,0,-c/b)$ with $\omega =\sqrt{g/c}$. This is the force-cancellation mechanism of the stationary middle mass in a three-mass spring chain, applied to the equivalent quadratic stiffnesses here.

The quadratic Lagrangian gives $\partial L/\partial \dot{q}=T\dot{q}$ and $\partial L/\partial q=-Vq$. Thus the Euler-Lagrange equations areA nonzero mode vector $v$ has a common harmonic time factor and satisfies the generalized eigenvalue problem for small oscillations $Vv={\omega }^{2}Tv$; $\omega$ is its normal frequency. A zero-frequency mode instead admits a constant or linear time factor.

Measure the three coordinates as displacements from a configuration with both springs unstretched. The kinetic energy is $\frac{1}{2}{\dot{x}}_1^2+\frac{1}{2}M{\dot{x}}_2^2+\frac{1}{2}{\dot{x}}_3^2$, and the potential energy is $\frac{1}{2}(x_2-x_1{)}^2+\frac{1}{2}(x_3-x_2{)}^2$. Expansion givesFor $M>0$, direct substitution yields the three normal modes:The three vectors are independent and orthogonal in the mass inner product $v^{T}Tw$, so they exhaust the motion. Their squared mass norms are $M+2,2,2+4/M$; division by the corresponding square roots gives normalized modes. This is the unit-spring, unit-outer-mass case of the normal modes of three masses connected by two equal springs.

As $M\to \infty$, the central displacement in the symmetric oscillatory mode tends to zero and both nonzero frequencies tend to one. The two limiting shapes $(1,0,1)$ and $(1,0,-1)$ span independent motions of the outer masses against an effectively fixed centre; each unit mass has its own unit spring frequency. The translational mode remains at zero frequency. Under a bounded kinetic-energy condition its common velocity tends to zero as $M^{-1/2}$, while a static common displacement is still an equilibrium shift. One must specify that energy condition before concluding that an arbitrary freely translating heavy mass remains stationary.

The Lagrangian is $\mathcal{L}=T-\mathcal{V}$, and the Euler-Lagrange equations areAt a rest equilibrium $q=0$, ${\partial }_i\mathcal{V}(0)=0$ for all three coordinates. The Taylor expansion therefore starts asThe symmetric Hessian matrix is the matrix $\mathsf{V}$ of the quadratic potential energy. If a quadratic coefficient vanishes along a direction, higher-order terms can be the first nonconstant terms there; equilibrium by itself does not imply strict stability.

Write ${\mathsf{T}}_0=\mathsf{T}(0)$. Terms involving the variation of $\mathsf{T}(q)$ are cubic in small displacements and velocities, so the quadratic small oscillation Lagrangian isA normal mode has fixed displacement ratios, $q(t)=v\cos (\omega t+\varphi )$. Its angular frequency $\omega$ is an eigenfrequency, or normal frequency, and satisfies the generalized eigenvalue problem for small oscillationsHere ${\mathsf{T}}_0$ is positive definite for independent physical coordinates and positive masses. A stable quadratic minimum makes ${\omega }^2\ge 0$; a negative value would describe exponential instability rather than oscillation.

For the suspended bar, let ${\kappa }_0$ and ${\kappa }_1$ denote the actual incremental spring constants of the upper and each lower spring. The physical displacement of the bar is $q_1$, while the lower spring extensions change by $q_2-q_1$ and $q_3-q_1$. ConsequentlyThe linear gravitational and elastic terms cancel at the loaded equilibrium. Expanding the remaining squares gives the small-oscillation mass and stiffness matricesThusSource convention: the printed matrix is obtained by setting ${\kappa }_0=\lambda /L$ and ${\kappa }_1=\mu /l$. Although the PDF prose calls $\lambda ,\mu$ spring constants, that matrix treats them as elastic parameters whose ratios to the stated lengths are the incremental stiffnesses. If they are read literally as Hooke-law spring constants, the divisions by $L,l$ should instead be omitted. The numerical case has $L=l=1$, so this discrepancy does not affect its results.

With the numerical parameters, ${\mathsf{T}}_0=\mathrm{diag}(2,1,1)$ andUse the exchange symmetry of the lower masses to find the vertical modes of a horizontally constrained suspended bar. The antisymmetric vector $v_a=(0,1,-1{)}^T$ obeys $\mathsf{V}{v}_a={\mathsf{T}}_0{v}_a$, giving ${\omega }_a=1$. In the symmetric sector put $v=(A,B,B{)}^T$ and $s={\omega }^2$. The two independent equations areTheir determinant gives $s^2-4s+2=0$, so $s=2\pm \sqrt{2}$. Taking $A=1$, the full set isIn the lowest normal mode, all three masses move in phase, with the lower masses moving about $2.414$ times as far as the bar. In the antisymmetric normal mode, the bar stays still and the lower masses move equally in opposite directions; their changing forces on the bar cancel. In the highest normal mode, the lower masses move together against the bar, with magnitude about $0.414$ of its displacement. Any small motion is a superposition of these three modes.

Use displacements $x=(x_1,x_2,x_3{)}^T$ of the left oxygen, carbon and right oxygen along a common positive direction. In this harmonic model the kinetic energy and elastic potential energy areThus the generalized eigenvalue problem for small oscillations is $Kv={\omega }^2\mathsf{M}v$, whereThe normal modes of three masses connected by two equal springs give the three independent solutionsThe second row follows by holding the centre fixed and moving the end masses oppositely. In the remaining internal mode the end displacements are equal, and zero displacement of the centre of mass gives $2m{v}_1+M{v}_2=0$, yielding the third row; substitution verifies its frequency. Both nonzero modes have fixed centre of mass. In the symmetric stretch the two bond-length changes are equal; in the antisymmetric stretch they are opposite.

The zero mode occurs because a uniform translation changes neither spring length, so $K{v}_0=0$. It represents free center-of-mass motion, not an additional restoring vibration. Its general time amplitude is $A+Bt$. Only one translational zero mode exists in the restricted one-dimensional model; transverse and rotational motion are outside its specified coordinates.

For the initial impulse, all displacement amplitudes at zero time vanish. Resolve the initial velocity into these three mode shapes:Writing ${\omega }_s=\sqrt{k/m}$ and ${\omega }_a=\sqrt{k/m+2k/M}$ gives the impulse on an end mass of a free symmetric three-mass chain:In particularwhere $V_{\mathrm{cm}}=mu/(M+2m)$. The initial derivatives are $(u,0,0)$, as required. The carbon dioxide model therefore drifts uniformly while undergoing both internal stretches. Its total momentum is the initial $mu$; omitting the zero mode would incorrectly hold the centre of mass fixed.

Take both displacement coordinates positive to the right. The kinetic energy and elastic potential energy areThus the equations of motion are $m_1{\ddot{x}}_1=-2k{x}_1+k{x}_2$ and $m_2{\ddot{x}}_2=k{x}_1-2k{x}_2$. With $x=v\cos (\omega t+\phi )$, the generalized eigenvalue problem for small oscillations isIts determinant gives $m_1{m}_2{\omega }^4-2k(m_1+m_2){\omega }^2+3k^2=0$. Put $D=\sqrt{m_1^2-m_1{m}_2+m_2^2}$. The two-mass three-spring normal modes haveBoth squared frequencies are positive because $(m_1+m_2{)}^2-D^2=3m_1{m}_2>0$. Also $D>|m_2-m_1|$, so $r_{-}>0$ and $r_{+}<0$: the lower mode moves the masses in phase, and the higher mode moves them oppositely. A general mode is an arbitrary sinusoidal amplitude multiplying its fixed shape.

For equal masses these reduce to $v_{-}=(1,1)$, ${\omega }_{-}^2=k/m$ and $v_{+}=(1,-1)$, ${\omega }_{+}^2=3k/m$. For unequal masses the vectors need not be Euclidean-orthogonal; they are orthogonal in the mass inner product, $m_1+m_2{r}_{-}{r}_{+}=0$.

For equal masses, write $M=mB$ with $B=\left(\begin{array}{ccc}3& 1& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right)$. The generalized eigenvalue problem for small oscillations is $kv={\omega }^{2}mBv$.

Exchange symmetry of climbers 2 and 3 gives $v_a=(0,1,-1{)}^T$, for which $B{v}_a=v_a$. HenceThe physical displacements are $(y_1,y_1+y_2,y_1+y_3)$, so climber 1 is stationary and the other two move vertically with equal amplitudes in opposite directions. Their changes in force on climber 1 cancel.

For the remaining symmetric modes, put $v=(a,b,b{)}^T$ and let $\mu =k/(m{\omega }^2)$. The eigenproblem becomes $3a+2b=\mu a$, $a+b=\mu b$. Eliminating $a,b$ gives $(3-\mu )(1-\mu )-2={\mu }^2-4\mu +1=0$. Therefore $\mu =2\pm \sqrt{3}$. Since $(2+\sqrt{3})(2-\sqrt{3})=1$,For clarity, their relative physical climber displacements, rather than rope-length amplitudes, areThe lower mode moves all climbers in phase; the higher mode moves the lower pair against the top climber. These are the normal modes of three vertically suspended masses with a branched spring connection.