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Write the displacement as $\mathbf{x}=\mathbf{a}\cos (\omega t+\varphi )$. The generalized eigenvalue problem isThe symmetry of the two end atoms suggests an end-antisymmetric shape $(1,0,-1)$ and two end-symmetric shapes. Direct substitution gives the normal modes of three masses connected by two equal springs:These three independent eigenvectors exhaust the three-dimensional system. The ordinary frequencies are $\omega /(2\pi )$; the values $\omega$ above are angular frequencies.

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.

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.

Use displacements $x_i$ from an equilibrium, so the two spring extensions are $x_2-x_1$ and $x_3-x_2$. The kinetic energy and elastic potential energy areThus $T={\dot{x}}^{T}M\dot{x}/2$ and $V=x^{T}Kx/2$, with the small-oscillation mass and stiffness matricesIf $x_i$ instead denote absolute positions, the appropriate natural spring lengths must first be subtracted from the differences in $V$. They do not affect the equations for displacements.

Euler-Lagrange equations give $M\ddot{x}+Kx=0$, so a normal mode $x=v{e}^{i\omega t}$ obeys the generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$. Reflection symmetry separates antisymmetric shapes $(A,0,-A{)}^T$ from symmetric shapes $(A,B,A{)}^T$. In the antisymmetric sector the end equation is $kA=m{\omega }^{2}A$, giving ${\omega }^2=k/m$. In the symmetric sector the two independent equations areTwice the first plus the second gives $m{\omega }^2(2A+\mu B)=0$. If $\omega =0$, then $B=A$. Otherwise $B=-2A/\mu$, and substitution into the first equation gives ${\omega }^2=(k/m)(1+2/\mu )$. Thus the three independent shapes are:For $v_0$ both springs retain their lengths, so this is uniform translation. For $v_1$ the end masses move oppositely and their forces on the middle mass cancel. For $v_2$ the end masses move together, while the middle mass moves oppositely so the centre of mass remains fixed. Direct substitution verifies all three eigenpairs; for positive $m,k,\mu$ the three eigenvalues are distinct and exhaust the three degrees of freedom. These are the normal modes of three masses connected by two equal springs.

A useful normalization is the weighted inner product induced by $M$. Since the squared mass norms are $m(\mu +2)$, $2m$ and $m(2+4/\mu )$, respectively, takeIf ordinary Euclidean unit eigenvectors are intended instead, divide $v_0,v_1,v_2$ by $\sqrt{3},\sqrt{2},\sqrt{2+4/{\mu }^2}$, respectively. The frequencies and physical motions are independent of this normalization choice. The zero-frequency coordinate has general motion $A+Bt$ rather than an oscillation.

Substitution of $\mu =2$ into the normal modes of three masses connected by two equal springs givesThe corresponding shapes are $v_0=(1,1,1{)}^T$, $v_1=(1,0,-1{)}^T$, $v_2=(1,-1,1{)}^T$. For example, $K{v}_2=k(2,-4,2{)}^T=2(k/m)M{v}_2$, verifying the third eigenvalue directly.

Their relevant orthogonality uses the kinetic matrix $M=m\mathrm{diag}(1,2,1)$:This is mass-matrix orthogonality of normal modes. It also follows by subtracting $v_i^{T}K{v}_j={\omega }_j^2{v}_i^{T}M{v}_j$ from its transpose and using the symmetry of $K,M$. Ordinary Euclidean inner products need not vanish: in fact $v_0^T{v}_2=1$.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.