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Let a finite permutation group act on mechanical coordinates through permutation matrices $P$, and let the mechanical mass matrix $M$ be a positive-definite matrix and the mechanical stiffness matrix $K$ a symmetric matrix. If $PM=MP$ and $PK=KP$, thenEvery transformed normal mode therefore stays in the same squared-frequency eigenspace. A nontrivial irreducible representation of dimension $d$ carried by that eigenspace forces at least $d$ independent modes at that squared frequency.

For a multiplicity-free symmetry decomposition, Schur lemma makes the mass-normalized operator $M^{-1/2}K{M}^{-1/2}$ scalar on each irreducible representation subspace. If equivalent irreducible representations occur several times, a commuting operator can mix their copies. One diagonalizes the operator on this copy space; symmetry does not force all copies to have one common frequency.

For three equal masses with full exchange symmetry, $M=mI$ and $K=aI+bJ$, where $J$ is the all-ones matrix. The collective normal mode $(1,1,1)$ has squared frequency $(a+3b)/m$. The two-dimensional sum-zero vector subspace has squared frequency $a/m$, providing a degenerate pair. Stability additionally requires the relevant squared angular frequencies to be nonnegative.

Mechanical dynamic condensation eliminates selected coordinates from a frequency-dependent mechanical response while retaining their effect on the remaining coordinates. It is exact algebraic elimination at the chosen transform variable when the eliminated block is invertible. It is not equivalent to clamping those coordinates. Starting from the partitioned mechanical dynamic-stiffness matrix,solve the internal equation as ${\mathbf{X}}_i={\mathbf{D}}_{ii}^{-1}({\mathbf{F}}_i-{\mathbf{D}}_{ir}{\mathbf{X}}_r)$. Substitution givesThe effective matrix is a Schur complement; the right side is the condensed mechanical load vector. Both matter when the internal coordinate is externally forced. The coupled dynamic equations and measurement boundary conditions are the starting model, rather than independent scalar stiffness reciprocals.^[ref]

For the two-mass example at $\omega =5rad/s$, take $D_{11}=D_{22}=175N/m$ and $D_{12}=D_{21}=-100N/m$. With $F_1=0$ and $F_2=1\mathrm{N}$, eliminating coordinate2 gives effective stiffness $117.8571N/m$ and effective load $100/175\mathrm{N}$. Hence $X_1=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Omitting the load correction would incorrectly predict no response at coordinate1.

For constant original mass matrix, damping and mechanical stiffness matrix, the condensed matrix is generally a rational function of frequency, not simply a smaller constant mass–damping–stiffness triple. A low-frequency approximation requires its own justification. The elimination can also become singular at an internal-block frequency even when the complete system matrix remains invertible. For example, $\left(\begin{array}{cc}0& -k_c\\ -k_c& 0\end{array}\right)$ has invertible full matrix for $k_c\ne 0$, but its lower-right scalar block cannot be inverted. One must then retain that coordinate or choose a different elimination.

A mechanical dynamic-stiffness matrix maps displacement coordinates to applied forces in a linear structural dynamics model at a specified transform variable or frequency. Starting with constant mass matrix $\mathbf{M}$, damping matrix $\mathbf{C}$ and mechanical stiffness matrix $\mathbf{K}$,zero initial conditions give $\mathbf{F}(s)=\mathbf{D}(s)\mathbf{X}(s)$ with $\mathbf{D}=s^2\mathbf{M}+s\mathbf{C}+\mathbf{K}$. For the $e^{i\omega t}$ convention, $\mathbf{D}(i\omega )=\mathbf{K}-{\omega }^2\mathbf{M}+i\omega \mathbf{C}$. Its matrix inverse, when it exists, is the complete mechanical receptance matrix. Models and measured responses must use matching coordinates and boundary conditions.^[ref]

Consider two $1\mathrm{kg}$ masses, each grounded by a $100N/m$ spring and joined by another $100N/m$ spring, with no damping. At angular frequency $5rad/s$,A force amplitude $F_1=1\mathrm{N}$ with $F_2=0$ gives $X_1=175/20625\mathrm{m}\approx 8.485\mathrm{mm}$ and $X_2=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Clamping coordinate2 instead gives $X_1=1/175\mathrm{m}\approx 5.714\mathrm{mm}$, because the clamp supplies a reaction force.

Partitioning retained and internal coordinates gives mechanical dynamic condensation. With ${\mathbf{F}}_i=0$ and invertible ${\mathbf{D}}_{ii}$,The effective matrix is a Schur complement, generally frequency dependent even when the original mass matrix, damping and mechanical stiffness matrix are constant. In the example its scalar value is $175-100^2/175\approx 117.857N/m$, rather than175. A reduced force-free internal coordinate and a clamped coordinate describe different experiments. At a singular block, that particular elimination is unavailable; it does not establish that the full mechanical system lacks a valid solution.

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.

A normal mode is a small-oscillation motion with every coordinate proportional to one common sinusoid, at a fixed amplitude ratio and relative phase. Expanding an arbitrary small motion in normal mode coordinates decouples its linear equations into independent oscillators. With mechanical mass matrix $\mathsf{M}$ and mechanical stiffness matrix $\mathsf{K}$, the generalized eigenvalue equation is $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$.

Horizontal gravitational restoring force on each bob is $-m_i(g/l)x_i$. The two springs add $k(x_2-x_1)$ on the first bob, $k(x_1-2x_2+x_3)$ on the middle bob, and $k(x_2-x_3)$ on the last. Dividing by the bob masses gives the stated equations withReflection symmetry separates an antisymmetric end motion from two symmetric motions. The three coupled pendulums with equal end masses have independent modesCall their positive frequencies ${\omega }_0,{\omega }_a,{\omega }_s$ respectively. The middle amplitude in the last mode ensures $m{x}_1+M{x}_2+m{x}_3=0$. The modes are orthogonal in the mass-weighted inner product, rather than generally in the unweighted coordinate dot product.

The initially stationary displacement decomposes asThere are no sine terms because every initial velocity is zero. ThereforeThese give $(0,0,d)$ at $t=0$ and solve all three coupled equations. The spring constant is assumed positive; for $k=0$ the frequencies become degenerate and the same superposition reduces to uncoupled pendulum motion.

The original example uses $M=2m$ and $k/m=g/l$ to display the mode vectors and resulting superposition. Those illustrative parameter ratios are not additional supplied data.