Normal Modes from the Generalized Eigenvalue Problem

Derivation

Normal Modes from the Generalized Eigenvalue Problem

D-085 Home PU-104 Threads waves · symmetry · energy Depends on Spectral Theorem for Self-Adjoint Operators, Gram–Schmidt Orthonormalization

Statement

For a conservative mechanical system with \(n\) degrees of freedom executing small oscillations about a stable equilibrium, with kinetic energy \(T=\tfrac12\,\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}}\) and potential energy \(V=\tfrac12\,\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q}\) (both quadratic forms, \(\mathbf{M}\) symmetric positive-definite, \(\mathbf{K}\) symmetric), the linearized equations of motion \(\mathbf{M}\ddot{\mathbf{q}}+\mathbf{K}\mathbf{q}=\mathbf{0}\) admit harmonic solutions \(\mathbf{q}(t)=\mathbf{a}\,e^{i\omega t}\) precisely when the amplitude vector \(\mathbf{a}\) and frequency \(\omega\) solve the generalized eigenvalue problem \(\mathbf{K}\mathbf{a}=\omega^2\mathbf{M}\mathbf{a}\). The \(n\) eigenvalues \(\omega_k^2\) are real, and the eigenvectors can be chosen \(\mathbf{M}\)-orthonormal, \(\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k=\delta_{jk}\) and \(\mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k=\omega_k^2\,\delta_{jk}\), so the normal modes simultaneously diagonalize both the inertia and the stiffness.

Why it matters

Almost every stable system — a crystal lattice, a bridge deck, a molecule's vibrational spectrum, a chain of coupled pendula — behaves near equilibrium as a set of independent harmonic oscillators once expressed in the right coordinates. The generalized eigenvalue problem is the machine that finds those coordinates. It converts a tangle of coupled second-order ODEs into \(n\) uncoupled ones, each a simple \(\ddot Q_k+\omega_k^2 Q_k=0\).

The \(\mathbf{M}\)-orthogonality is not a technicality: it is the statement that the normal coordinates diagonalize both the kinetic and the potential energy at once. That simultaneous diagonalization is why energy and the whole Hamiltonian split additively across modes, and it is the bridge from classical small oscillations to phonons and to the ladder operators of quantum field theory.

Assumptions

Motion is small about a stable equilibrium \(\mathbf{q}=\mathbf{0}\).If the excursions are not small, the neglected cubic and higher terms in the potential produce anharmonic coupling between modes: frequencies become amplitude-dependent, energy leaks between modes, and no fixed normal-mode decomposition exists.

The mass matrix \(\mathbf{M}\) is symmetric and positive-definite.Positive-definiteness is the statement that \(T=\tfrac12\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}}\) is strictly positive for any nonzero velocity. If dropped, the Cholesky factorization used below fails, the mass-weighted product is no longer an inner product, and \(\mathbf{M}\)-orthogonality is undefined.

The stiffness matrix \(\mathbf{K}\) is symmetric.Symmetry follows from \(K_{ij}=\partial^2 V/\partial q_i\partial q_j\) and equality of mixed partials for a smooth potential. If forces are non-conservative (velocity-dependent, e.g. magnetic or gyroscopic), \(\mathbf{K}\) loses symmetry, eigenvalues may become complex, and the real-orthogonal mode picture collapses.

The equilibrium is a strict minimum of \(V\), so \(\mathbf{K}\) is positive-definite.If \(\mathbf{K}\) is only positive-semidefinite, zero-frequency modes appear — rigid translations, rotations, or a marginal direction. If \(\mathbf{K}\) has a negative eigenvalue the equilibrium is unstable: that \(\omega_k^2<0\) gives imaginary \(\omega\) and exponentially growing motion, not oscillation.

No damping or external driving.Linear damping adds a term \(\mathbf{C}\dot{\mathbf{q}}\); unless \(\mathbf{C}\) is simultaneously diagonalizable with \(\mathbf{M}\) and \(\mathbf{K}\) (Rayleigh/proportional damping), the real normal modes no longer decouple the system and one must pass to complex modes.

Derivation

\[ T=\tfrac12\,\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}},\qquad M_{ij}=M_{ij}(\mathbf{0}) \]

Expanding the kinetic energy of generalized coordinates, the inertia metric \(M_{ij}(\mathbf{q})\) is evaluated at equilibrium; its \(\mathbf{q}\)-dependence contributes only at cubic order in small quantities, so \(\mathbf{M}\) is a constant symmetric matrix here. A

\[ V(\mathbf{q})=V(\mathbf{0})+\nabla V(\mathbf{0})\cdot\mathbf{q}+\tfrac12\,\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q}+\cdots,\qquad K_{ij}\equiv\left.\frac{\partial^2 V}{\partial q_i\,\partial q_j}\right|_{\mathbf{0}} \]

Taylor-expand the potential about equilibrium; the Hessian \(K_{ij}\) is symmetric by equality of mixed partials. A

\[ \nabla V(\mathbf{0})=\mathbf{0}\;\Longrightarrow\; V\simeq V(\mathbf{0})+\tfrac12\,\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q} \]

The defining property of an equilibrium is that the gradient vanishes; the constant \(V(\mathbf{0})\) produces no force and is dropped. A

\[ L=T-V=\tfrac12\,\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}}-\tfrac12\,\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q} \]

Form the Lagrangian from the quadratic kinetic and potential forms; this is the exact Lagrangian of the linearized system. A

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot{\mathbf{q}}}-\frac{\partial L}{\partial\mathbf{q}}=\mathbf{0}\;\Longrightarrow\;\mathbf{M}\ddot{\mathbf{q}}+\mathbf{K}\mathbf{q}=\mathbf{0} \]

Apply the Euler–Lagrange equations. Using symmetry of \(\mathbf{M}\) and \(\mathbf{K}\), \(\partial L/\partial\dot{\mathbf{q}}=\mathbf{M}\dot{\mathbf{q}}\) and \(\partial L/\partial\mathbf{q}=-\mathbf{K}\mathbf{q}\). B

\[ \mathbf{q}(t)=\mathbf{a}\,e^{i\omega t},\qquad \mathbf{a}\in\mathbb{R}^n\text{ constant} \]

Separation ansatz: because the coefficient matrices are constant, the system has solutions in which every coordinate oscillates in lockstep at one frequency with a fixed shape \(\mathbf{a}\). The physical motion is \(\operatorname{Re}[\mathbf{q}]\). B

\[ \ddot{\mathbf{q}}=-\omega^2\mathbf{a}\,e^{i\omega t}\;\Longrightarrow\;\big(\mathbf{K}-\omega^2\mathbf{M}\big)\mathbf{a}\,e^{i\omega t}=\mathbf{0} \]

Differentiate twice and substitute; the common scalar factor \(e^{i\omega t}\ne 0\) divides out. A

\[ \mathbf{K}\mathbf{a}=\omega^2\mathbf{M}\mathbf{a} \]

Rearranging gives the generalized eigenvalue problem: the eigenvalue is \(\omega^2\) and the "metric" of the problem is \(\mathbf{M}\) rather than the identity. A

\[ \det\!\big(\mathbf{K}-\omega^2\mathbf{M}\big)=0 \]

A nontrivial amplitude \(\mathbf{a}\ne\mathbf{0}\) exists only if the matrix is singular. This characteristic equation is a degree-\(n\) polynomial in \(\omega^2\), yielding \(n\) eigenfrequencies. B

\[ \mathbf{M}=\mathbf{L}\mathbf{L}^{\mathsf T}\;(\text{Cholesky}),\qquad \mathbf{a}=\mathbf{L}^{-\mathsf T}\mathbf{y} \]

Since \(\mathbf{M}\) is symmetric positive-definite it has a unique lower-triangular Cholesky factor \(\mathbf{L}\) with positive diagonal. Define \(\mathbf{y}=\mathbf{L}^{\mathsf T}\mathbf{a}\); the map is invertible. C

\[ \big(\mathbf{L}^{-1}\mathbf{K}\mathbf{L}^{-\mathsf T}\big)\mathbf{y}=\omega^2\mathbf{y},\qquad \tilde{\mathbf{A}}\equiv\mathbf{L}^{-1}\mathbf{K}\mathbf{L}^{-\mathsf T} \]

Substitute \(\mathbf{a}=\mathbf{L}^{-\mathsf T}\mathbf{y}\) into \(\mathbf{K}\mathbf{a}=\omega^2\mathbf{M}\mathbf{a}\) and left-multiply by \(\mathbf{L}^{-1}\), using \(\mathbf{L}^{-1}\mathbf{M}\mathbf{L}^{-\mathsf T}=\mathbf{I}\). This is now a standard eigenvalue problem. C

\[ \tilde{\mathbf{A}}^{\mathsf T}=\big(\mathbf{L}^{-1}\mathbf{K}\mathbf{L}^{-\mathsf T}\big)^{\mathsf T}=\mathbf{L}^{-1}\mathbf{K}^{\mathsf T}\mathbf{L}^{-\mathsf T}=\tilde{\mathbf{A}} \]

\(\tilde{\mathbf{A}}\) is real symmetric because \(\mathbf{K}=\mathbf{K}^{\mathsf T}\). By the spectral theorem for self-adjoint operators its eigenvalues \(\omega_k^2\) are real and it has an orthonormal eigenbasis \(\{\mathbf{y}_k\}\), \(\mathbf{y}_j^{\mathsf T}\mathbf{y}_k=\delta_{jk}\) (Gram–Schmidt supplies an orthonormal set within any degenerate eigenspace). C

\[ \mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k=\mathbf{y}_j^{\mathsf T}\mathbf{L}^{-1}\big(\mathbf{L}\mathbf{L}^{\mathsf T}\big)\mathbf{L}^{-\mathsf T}\mathbf{y}_k=\mathbf{y}_j^{\mathsf T}\mathbf{y}_k=\delta_{jk} \]

Map the orthonormal \(\mathbf{y}_k\) back through \(\mathbf{a}_k=\mathbf{L}^{-\mathsf T}\mathbf{y}_k\). The Euclidean orthonormality of the \(\mathbf{y}\)'s becomes \(\mathbf{M}\)-orthonormality of the physical mode shapes. C

\[ \mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k=\mathbf{a}_j^{\mathsf T}\big(\omega_k^2\mathbf{M}\mathbf{a}_k\big)=\omega_k^2\,\delta_{jk} \]

Use the eigen-relation on \(\mathbf{a}_k\) then \(\mathbf{M}\)-orthonormality from step 13. Thus the modes simultaneously diagonalize \(\mathbf{M}\) (to \(\mathbf{I}\)) and \(\mathbf{K}\) (to \(\operatorname{diag}\omega_k^2\)). C

Result

\[ \boxed{\;\mathbf{K}\mathbf{a}_k=\omega_k^2\,\mathbf{M}\mathbf{a}_k,\qquad \mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k=\delta_{jk},\qquad \mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k=\omega_k^2\,\delta_{jk}\;} \]

Reading. The system has \(n\) special patterns of motion — normal modes. In mode \(k\) every coordinate oscillates at the single frequency \(\omega_k\) with fixed relative amplitudes given by \(\mathbf{a}_k\). These patterns are "orthogonal" not in the ordinary sense but in the metric set by the masses: \(\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k=0\) for \(j\ne k\). Writing \(\mathbf{q}(t)=\sum_k Q_k(t)\,\mathbf{a}_k\) turns the coupled system into \(n\) independent oscillators \(\ddot Q_k+\omega_k^2 Q_k=0\).

Units check. \([\mathbf{M}]=\mathrm{kg}\) and \([\mathbf{K}]=\mathrm{N/m}=\mathrm{kg\,s^{-2}}\), so \(\omega^2=(\mathbf{a}^{\mathsf T}\mathbf{K}\mathbf{a})/(\mathbf{a}^{\mathsf T}\mathbf{M}\mathbf{a})\) has units \(\mathrm{(kg\,s^{-2})/kg}=\mathrm{s^{-2}}\), so \(\omega\) is a genuine angular frequency in \(\mathrm{rad\,s^{-1}}\). The normalization \(\mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{a}_k=1\) forces \(\mathbf{a}_k\) to carry units \(\mathrm{kg^{-1/2}}\), so the normal coordinate \(Q_k\) has units \(\mathrm{kg^{1/2}\,m}\) and \(\tfrac12\dot Q_k^2\) is an energy.

Limiting cases

Diagonal \(\mathbf{M}\) and \(\mathbf{K}\) (uncoupled): the modes are the coordinate axes, \(\mathbf{a}_k\propto\mathbf{e}_k\), and \(\omega_k^2=K_{kk}/M_{kk}\) — \(n\) separate springs.
\(\mathbf{M}=m\mathbf{I}\) (equal masses): the generalized problem collapses to the ordinary symmetric eigenproblem \(\mathbf{K}\mathbf{a}=m\omega^2\mathbf{a}\), and \(\mathbf{M}\)-orthogonality reduces to ordinary orthogonality.
Degenerate frequencies (\(\omega_j=\omega_k\)): any combination within the eigenspace is still a mode; Gram–Schmidt in the \(\mathbf{M}\)-inner product picks an \(\mathbf{M}\)-orthonormal basis. Degeneracy signals a symmetry of the system.
One zero eigenvalue (\(\omega_0=0\)): a rigid-body direction in \(\ker\mathbf{K}\); the "oscillation" \(\ddot Q_0=0\) is uniform drift, \(Q_0(t)=A+Bt\).
Weak coupling (off-diagonal \(\mathbf{K}\) small): frequencies split by an amount linear in the coupling; two nearly-equal oscillators show slow beats between the in-phase and out-of-phase modes.

Breaks when

Amplitudes leave the harmonic regime. Once cubic terms in \(V\) matter, frequencies become amplitude-dependent and modes exchange energy (mode–mode coupling, thermalization). The fixed \(\mathbf{a}_k\) decomposition is only the leading-order picture.
The equilibrium is unstable. If \(\mathbf{K}\) is not positive-definite, some \(\omega_k^2<0\); then \(\omega\) is imaginary and \(\mathbf{q}\propto e^{\pm|\omega|t}\) grows exponentially — a buckling or instability, not a mode.
Non-conservative or gyroscopic forces appear. Velocity-dependent forces add a term \(\mathbf{G}\dot{\mathbf{q}}\) with antisymmetric \(\mathbf{G}\) (Coriolis, magnetic). The effective operator loses symmetry, eigenvalues turn complex, and real \(\mathbf{M}\)-orthogonal modes no longer exist.
General (non-proportional) damping is present. With \(\mathbf{M}\ddot{\mathbf{q}}+\mathbf{C}\dot{\mathbf{q}}+\mathbf{K}\mathbf{q}=\mathbf{0}\), unless \(\mathbf{C}\) is simultaneously diagonalizable the three matrices cannot be diagonalized together; one doubles the state space and uses complex modes.

Failure modes

Using ordinary orthogonality. Computing \(\mathbf{a}_j^{\mathsf T}\mathbf{a}_k\) and expecting it to vanish. For unequal masses the modes are \(\mathbf{M}\)-orthogonal, not Euclidean-orthogonal; you must insert the mass matrix.
Diagonalizing \(\mathbf{M}^{-1}\mathbf{K}\) and treating it as symmetric. \(\mathbf{M}^{-1}\mathbf{K}\) has the right eigenvalues but is generally not symmetric, so its eigenvectors are not orthogonal in any obvious sense — students wrongly apply the spectral theorem to it.
Normalizing to \(\mathbf{a}_k^{\mathsf T}\mathbf{a}_k=1\). This does not give unit modal mass; the clean uncoupled equations require \(\mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{a}_k=1\).
Dropping the linear term without checking equilibrium. If \(\nabla V(\mathbf{0})\ne\mathbf{0}\) you are not at an equilibrium and there is a constant force; the mode analysis about that point is meaningless.
Taking \(\omega=\pm\sqrt{\text{eigenvalue}}\) and keeping both signs as distinct modes. \(+\omega\) and \(-\omega\) give the same standing mode; the two signs are the two real solutions \(\cos\) and \(\sin\), not two modes.
Forgetting to take the real part. The complex ansatz is a computational device; the physical displacement is \(\operatorname{Re}[\mathbf{a}\,e^{i\omega t}]\).

Discussion

The deep content of this result is a theorem of linear algebra dressed as physics: two quadratic forms, one of them positive-definite, can be simultaneously diagonalized. The kinetic energy supplies the positive-definite form and thereby a natural inner product on configuration space — the "mass metric." Once you accept that the geometry of small oscillations is set by \(\mathbf{M}\) rather than by the identity, everything falls out: eigenvectors are orthogonal in that metric, energy is a sum of independent modal energies, and the characteristic frequencies are the metric-adjusted eigenvalues of the stiffness.

Physically, a normal mode is a collective coordinate in which the entire system moves as one oscillator. The mode shape \(\mathbf{a}_k\) tells you the choreography — which masses move where, in what ratio, in or out of phase — and \(\omega_k\) tells you the tempo. Because the modes are complete, an arbitrary initial condition is just a superposition; each mode then rings on independently at its own frequency, and the complicated-looking motion of a coupled system is nothing but interference between simple oscillations.

The reach of this construction is enormous. Apply it to a lattice of atoms and the modes become phonons; the same \(\mathbf{M}\)-orthogonality that decouples the classical energies is what lets each phonon be quantized as an independent harmonic oscillator, giving the Debye and Einstein models of specific heat. Apply it to a continuous field and the discrete modes become Fourier modes; quantizing each gives the particle content of a free quantum field. The generalized eigenvalue problem is thus the classical seed of the entire ladder-operator formalism.

There is a variational face to the same result. The Rayleigh quotient \(R(\mathbf{a})=(\mathbf{a}^{\mathsf T}\mathbf{K}\mathbf{a})/(\mathbf{a}^{\mathsf T}\mathbf{M}\mathbf{a})\) is stationary exactly at the eigenvectors, with stationary values \(\omega_k^2\). Its minimum is the lowest squared frequency, and the Courant–Fischer min–max theorem characterizes every \(\omega_k^2\) as a constrained extremum. This is the engine behind the Rayleigh–Ritz method: minimizing \(R\) over a trial subspace gives a rigorous upper bound on the fundamental frequency, with monotone convergence as the subspace grows — the finite-element method for structural vibration is precisely this idea. Symmetry enters through the commutation of the symmetry group's representation with \(\mathbf{M}\) and \(\mathbf{K}\): modes organize into irreducible representations, degeneracies are forced by non-abelian symmetry, and group theory predicts which modes couple to a given perturbation before any matrix is diagonalized.

Common misconceptions. A normal mode is not a single particle's motion but a pattern of the whole system. "Orthogonal modes" does not mean the mode vectors are perpendicular on paper — they are perpendicular in the mass metric. And a higher mode number does not always mean a higher frequency unless the modes are explicitly ordered by frequency; the labeling is a convention, not a law.

Worked examples

Two equal masses, three identical springs (wall–mass–spring–mass–wall). Masses \(m=0.50\ \mathrm{kg}\) each; three springs of constant \(k=20\ \mathrm{N/m}\) (a wall spring at each end, one coupling spring in the middle). Coordinates \(x_1,x_2\). Find the normal frequencies and \(\mathbf{M}\)-orthonormal modes. A

\[ \mathbf{M}=m\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad \mathbf{K}=k\begin{pmatrix}2&-1\\-1&2\end{pmatrix} \]

Each mass feels its wall spring plus the shared middle spring (diagonal \(2k\)); the middle spring couples them (off-diagonal \(-k\)). A

\[ \det\!\big(\mathbf{K}-\omega^2\mathbf{M}\big)=(2k-m\omega^2)^2-k^2=0 \]

Characteristic equation from the singularity condition; expand the \(2\times2\) determinant. A

\[ 2k-m\omega^2=\pm k\;\Longrightarrow\;\omega_1^2=\frac{k}{m},\quad \omega_2^2=\frac{3k}{m} \]

Take square roots of the perfect square and solve each branch for \(\omega^2\). A

\[ \omega_1=\sqrt{\tfrac{20}{0.50}}=\sqrt{40}=6.32\ \mathrm{rad/s},\qquad \omega_2=\sqrt{\tfrac{60}{0.50}}=\sqrt{120}=10.95\ \mathrm{rad/s} \]

Insert numbers: \(k/m=40\ \mathrm{s^{-2}}\), \(3k/m=120\ \mathrm{s^{-2}}\). A

\[ \mathbf{a}_1\propto(1,\,1)\ \text{(in phase)},\qquad \mathbf{a}_2\propto(1,\,-1)\ \text{(out of phase)} \]

Substitute each \(\omega_k^2\) into \((\mathbf{K}-\omega_k^2\mathbf{M})\mathbf{a}_k=\mathbf{0}\) to read off the amplitude ratio. B

\[ \mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{a}_k=m(1^2+1^2)=2m\;\Longrightarrow\;\text{normalize by }1/\sqrt{2m} \]

Impose unit modal mass; divide each raw vector by \(\sqrt{\mathbf{a}^{\mathsf T}\mathbf{M}\mathbf{a}}\). B

\[ \omega_1=6.32\ \mathrm{rad/s},\ \mathbf{a}_1=\tfrac{1}{\sqrt{1.0\,\mathrm{kg}}}(1,1);\qquad \omega_2=10.95\ \mathrm{rad/s},\ \mathbf{a}_2=\tfrac{1}{\sqrt{1.0\,\mathrm{kg}}}(1,-1) \]

Reading. The low mode is the two masses swinging together (the coupling spring never stretches, so only the wall springs act, giving \(k/m\)); the high mode has them moving oppositely, stretching the middle spring hardest (giving \(3k/m\)). Here \(2m=1.0\ \mathrm{kg}\), so each normalized amplitude is \(\pm 1/\sqrt{1.0}\ \mathrm{kg^{-1/2}}\), and one checks \(\mathbf{a}_1^{\mathsf T}\mathbf{M}\mathbf{a}_2=m(1\cdot1+1\cdot(-1))=0\).

Units check. \(\sqrt{(\mathrm{N/m})/\mathrm{kg}}=\mathrm{rad/s}\). Consistent.

Unequal masses on a line of three springs (non-trivial \(\mathbf{M}\)). Mass \(m_1=1.0\ \mathrm{kg}\) and mass \(m_2=2.0\ \mathrm{kg}\) in a chain: wall–spring\((k)\)–\(m_1\)–spring\((k)\)–\(m_2\)–spring\((k)\)–wall, with \(k=12\ \mathrm{N/m}\). Show the modes are \(\mathbf{M}\)-orthogonal but not Euclidean-orthogonal. B

\[ \mathbf{M}=\begin{pmatrix}1.0&0\\0&2.0\end{pmatrix}\ \mathrm{kg},\qquad \mathbf{K}=k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\ \mathrm{N/m} \]

Same three-spring topology, but now the mass matrix is not a multiple of the identity. A

\[ \det\!\big(\mathbf{K}-\omega^2\mathbf{M}\big)=(2k-m_1\omega^2)(2k-m_2\omega^2)-k^2=0 \]

Expand the determinant, keeping the distinct diagonal mass entries. A

\[ m_1 m_2\,\omega^4-2k(m_1+m_2)\,\omega^2+3k^2=0 \]

Collect into a quadratic in \(\omega^2\); here \(m_1 m_2=2\), \(m_1+m_2=3\), \(k=12\). A

\[ 2\omega^4-72\omega^2+432=0\;\Longrightarrow\;\omega^2=\frac{72\pm\sqrt{72^2-8\cdot432}}{4}=\frac{72\pm\sqrt{1728}}{4} \]

Quadratic formula; discriminant \(5184-3456=1728\), \(\sqrt{1728}=41.57\). A

\[ \omega_1^2=7.61\ \mathrm{s^{-2}}\to\omega_1=2.76\ \mathrm{rad/s},\qquad \omega_2^2=28.39\ \mathrm{s^{-2}}\to\omega_2=5.33\ \mathrm{rad/s} \]

Evaluate both roots: \((72-41.57)/4=7.61\) and \((72+41.57)/4=28.39\). A

\[ (2k-m_1\omega^2)\,a^{(1)}=k\,a^{(2)}\;\Longrightarrow\;\frac{a^{(2)}}{a^{(1)}}=\frac{24-\omega^2}{12} \]

The first row of the eigen-equation gives the amplitude ratio for each mode. B

\[ \mathbf{a}_1\propto(1,\,1.366),\qquad \mathbf{a}_2\propto(1,\,-0.366) \]

Plug in \(\omega_1^2=7.61\to(24-7.61)/12=1.366\); \(\omega_2^2=28.39\to(24-28.39)/12=-0.366\). B

\[ \mathbf{a}_1\!\cdot\!\mathbf{a}_2=1+(1.366)(-0.366)=0.50\ne 0,\quad \mathbf{a}_1^{\mathsf T}\mathbf{M}\mathbf{a}_2=1\cdot1\cdot1+2\cdot1.366\cdot(-0.366)=0.00 \]

Compare the plain dot product with the mass-weighted one; only the latter vanishes. B

\[ \omega_1=2.76\ \mathrm{rad/s},\ \mathbf{a}_1\propto(1,\,1.37);\qquad \omega_2=5.33\ \mathrm{rad/s},\ \mathbf{a}_2\propto(1,\,-0.37);\qquad \mathbf{a}_1^{\mathsf T}\mathbf{M}\mathbf{a}_2=0 \]

Reading. The heavier mass moves less in the high-frequency out-of-phase mode. The ordinary dot product \(\mathbf{a}_1\!\cdot\!\mathbf{a}_2=0.50\) is emphatically nonzero, yet the mass-weighted product is exactly zero — a concrete demonstration that the correct orthogonality is in the metric of \(\mathbf{M}\).

Units check. To finish normalizing, divide \(\mathbf{a}_1\) by \(\sqrt{\mathbf{a}_1^{\mathsf T}\mathbf{M}\mathbf{a}_1}=\sqrt{1+2\cdot1.366^2}=\sqrt{4.73}=2.17\ \mathrm{kg^{1/2}}\); the result carries units \(\mathrm{kg^{-1/2}}\) as required.

Problems

Two identical pendula of length \(\ell=1.0\ \mathrm{m}\) and bob mass \(m=0.20\ \mathrm{kg}\) hang side by side, coupled by a spring of constant \(k=0.50\ \mathrm{N/m}\) attached at the bobs (\(g=9.8\ \mathrm{m/s^2}\)). Using small-angle displacements \(x_1,x_2\), find both normal frequencies.
Solution
Kinetic energy gives \(\mathbf{M}=m\mathbf{I}\). Each pendulum has gravitational stiffness \(mg/\ell\) and the spring couples them: \(\mathbf{K}=\begin{pmatrix}mg/\ell+k&-k\\-k&mg/\ell+k\end{pmatrix}\). In-phase mode \((1,1)\): spring never stretches, \(\omega_1^2=g/\ell=9.8\ \mathrm{s^{-2}}\), \(\omega_1=3.13\ \mathrm{rad/s}\). Out-of-phase mode \((1,-1)\): \(\omega_2^2=g/\ell+2k/m=9.8+2(0.50)/0.20=9.8+5.0=14.8\ \mathrm{s^{-2}}\), \(\omega_2=3.85\ \mathrm{rad/s}\). The two frequencies are close, so releasing one pendulum from rest produces slow beats at \(\omega_2-\omega_1=0.72\ \mathrm{rad/s}\) as energy sloshes between them.
For the system \(\mathbf{M}=\begin{pmatrix}2&0\\0&1\end{pmatrix}\ \mathrm{kg}\), \(\mathbf{K}=\begin{pmatrix}6&-2\\-2&4\end{pmatrix}\ \mathrm{N/m}\), compute the two eigenfrequencies and verify that the mode shapes are \(\mathbf{M}\)-orthogonal.
Solution
Characteristic equation \(\det(\mathbf{K}-\omega^2\mathbf{M})=(6-2\omega^2)(4-\omega^2)-4=0\). Expand: \(2\omega^4-14\omega^2+20=0\), i.e. \(\omega^4-7\omega^2+10=0\), so \(\omega^2=2\) or \(5\ \mathrm{s^{-2}}\). Thus \(\omega_1=1.41\ \mathrm{rad/s}\), \(\omega_2=2.24\ \mathrm{rad/s}\). Mode 1 (\(\omega^2=2\)): \((6-4)a_1=2a_2\to\mathbf{a}_1=(1,1)\). Mode 2 (\(\omega^2=5\)): \((6-10)a_1=2a_2\to\mathbf{a}_2=(1,-2)\). Check: \(\mathbf{a}_1^{\mathsf T}\mathbf{M}\mathbf{a}_2=2(1)(1)+1(1)(-2)=2-2=0\). The plain dot product is \(1+(1)(-2)=-1\ne0\).
A linear triatomic molecule modeled as masses \(m\)–\(M\)–\(m\) joined by two springs of constant \(k\) (like \(\mathrm{CO_2}\)). Along the axis, show that one normal mode has zero frequency and interpret it. Take \(m=16\ \mathrm{u}\), \(M=12\ \mathrm{u}\), \(k=1900\ \mathrm{N/m}\); give the two nonzero frequencies (\(1\ \mathrm{u}=1.66\times10^{-27}\ \mathrm{kg}\)).
Solution
With coordinates \(x_1,x_2,x_3\), \(\mathbf{K}=k\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}\), \(\mathbf{M}=\operatorname{diag}(m,M,m)\). The vector \((1,1,1)\) satisfies \(\mathbf{K}(1,1,1)^{\mathsf T}=\mathbf{0}\), so \(\omega_0=0\): uniform translation of the whole molecule, a rigid-body zero mode. Symmetric stretch \((1,0,-1)\) leaves the center still, \(\omega^2=k/m\); asymmetric stretch has the center recoil, \(\omega^2=k(1/m+2/M)=k(M+2m)/(mM)\). Numbers: \(m=16\cdot1.66\times10^{-27}=2.66\times10^{-26}\ \mathrm{kg}\). Symmetric: \(\omega=\sqrt{1900/2.66\times10^{-26}}=2.67\times10^{14}\ \mathrm{rad/s}\). Asymmetric factor \((12+32)/(16\cdot12)=44/192=0.229\ \mathrm{u^{-1}}\): \(\omega^2=1900\cdot0.229/1.66\times10^{-27}=2.62\times10^{29}\ \mathrm{s^{-2}}\), \(\omega=5.12\times10^{14}\ \mathrm{rad/s}\). The asymmetric stretch is higher because the light atoms move against a recoiling center, a smaller effective mass.
Prove in general that eigenvectors belonging to distinct eigenvalues of \(\mathbf{K}\mathbf{a}=\omega^2\mathbf{M}\mathbf{a}\) (with \(\mathbf{M},\mathbf{K}\) symmetric) are \(\mathbf{M}\)-orthogonal, without invoking the Cholesky factorization.
Solution
Let \(\mathbf{K}\mathbf{a}_j=\omega_j^2\mathbf{M}\mathbf{a}_j\) and \(\mathbf{K}\mathbf{a}_k=\omega_k^2\mathbf{M}\mathbf{a}_k\). Left-multiply the first by \(\mathbf{a}_k^{\mathsf T}\) and the second by \(\mathbf{a}_j^{\mathsf T}\): \(\mathbf{a}_k^{\mathsf T}\mathbf{K}\mathbf{a}_j=\omega_j^2\,\mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{a}_j\) and \(\mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k=\omega_k^2\,\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k\). Because \(\mathbf{K}\) and \(\mathbf{M}\) are symmetric, these scalars satisfy \(\mathbf{a}_k^{\mathsf T}\mathbf{K}\mathbf{a}_j=\mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k\) and \(\mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{a}_j=\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k\). Subtracting: \(0=(\omega_j^2-\omega_k^2)\,\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k\). Since \(\omega_j^2\ne\omega_k^2\), the factor vanishes: \(\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k=0\). For equal eigenvalues, orthogonalize within the eigenspace by Gram–Schmidt in the \(\mathbf{M}\)-inner product.
Using the modes and frequencies of Problem 2, a system starts at rest with \(\mathbf{q}(0)=(3,0)\ \mathrm{m}\), \(\dot{\mathbf{q}}(0)=\mathbf{0}\). Find \(\mathbf{q}(t)\) as a superposition of normal modes.
Solution
Write \(\mathbf{q}(t)=c_1\mathbf{a}_1\cos\omega_1 t+c_2\mathbf{a}_2\cos\omega_2 t\) (zero initial velocity kills the sines), with \(\mathbf{a}_1=(1,1)\), \(\mathbf{a}_2=(1,-2)\). Project using \(\mathbf{M}\)-orthogonality: \(c_k=(\mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{q}(0))/(\mathbf{a}_k^{\mathsf T}\mathbf{M}\mathbf{a}_k)\), with \(\mathbf{M}=\operatorname{diag}(2,1)\). Mode 1: numerator \((1,1)\!\cdot\!(6,0)=6\), denominator \(2+1=3\to c_1=2\). Mode 2: numerator \((1,-2)\!\cdot\!(6,0)=6\), denominator \(2\cdot1+1\cdot4=6\to c_2=1\). Therefore \(\mathbf{q}(t)=2(1,1)\cos(1.41\,t)+(1,-2)\cos(2.24\,t)\), i.e. \(x_1(t)=2\cos(1.41\,t)+\cos(2.24\,t)\) and \(x_2(t)=2\cos(1.41\,t)-2\cos(2.24\,t)\). Check at \(t=0\): \(x_1=3\), \(x_2=0\). Both satisfied.