Coupled Oscillators and Normal Modes

Definition Let $t$ be time, let primes denote derivatives with respect to $t$, and let the vector $q(t) = (q_1(t), \dots, q_n(t))^{T}$ list the displacements of $n$ bodies from an equilibrium configuration. A linear undamped system of coupled oscillators is a system

where the mass matrix $M$ is a constant symmetric positive definite $n \times n$ matrix and the stiffness matrix $K$ is a constant symmetric $n \times n$ matrix. When $K$ is also positive definite, the equilibrium is stable and every motion is a combination of oscillations, as the normal modes below show. The oscillators are coupled when $M$ or $K$ is not diagonal, so that the equation for one coordinate contains other coordinates and the equations cannot be solved one at a time. The model describes small motions with no damping and no outside force. For given $q(0)$ and $q'(0)$ it has exactly one solution, and that solution is defined for every real $t$. With the state $X$ that stacks $q$ on top of $q'$, a vector with $2n$ entries, it becomes the first-order system

but the second-order form keeps $M$ and $K$ in view and leads to a problem of size $n$ instead of $2n$.

Definition The mass matrix of a linear system in coordinates $q$ measured from equilibrium is the symmetric positive definite matrix $M$ for which the kinetic energy is

When body $i$ is a point mass $m_i$ moving along a line and $q_i$ is its own displacement, the mass matrix is $M = \operatorname{diag}(m_1, \dots, m_n)$ in kilograms, and positive definiteness says that every mass is positive. Other coordinates, such as the two angles of a double pendulum, can give a mass matrix that is not diagonal. Because $M$ is invertible, the system can be solved for the accelerations as $q'' = -M^{-1}Kq$.

Definition The stiffness matrix is the symmetric matrix $K$ for which the spring forces on the bodies form the vector $-Kq$, or equivalently for which the energy stored in the springs is $\tfrac{1}{2}q^{T}Kq$. Its entry $k_{ij}$, in newtons per meter, is minus the spring force on body $i$ per unit displacement of body $j$, with every other body at its equilibrium position. It is symmetric because all the spring forces come from one stored energy. For the chain made of a wall, a spring $k_1$, body 1, a spring $k_2$, body 2, a spring $k_3$ and a wall, the stored energy is

so

The matrix $K$ is positive definite when every nonzero displacement stores energy. For this chain with $k_2 > 0$ and $k_1, k_3 \ge 0$, that happens exactly when $k_1 + k_3 > 0$. With $k_1 = k_3 = 0$ the vector $(1, 1)^{T}$ satisfies $K(1, 1)^{T} = 0$, and the pair can slide together without stretching any spring.

Definition Try $q(t) = v\,s(t)$ with a constant vector $v \ne 0$ and a scalar function $s$, not identically zero, with $s'' = -\omega^2 s$. Substituting gives $(K - \omega^2 M)v\,s(t) = 0$, so the trial solves $Mq'' + Kq = 0$ exactly when $Kv = \omega^2 Mv$. The generalized eigenvalue problem, or generalized eigenproblem, for the pair $(K, M)$ asks for numbers $\lambda$ and vectors $v \ne 0$ with

The eigenvalues are the roots of $\det(K - \lambda M) = 0$, a polynomial equation of degree $n$ because $M$ is invertible. If $M$ is symmetric positive definite and $K$ is symmetric, there are $n$ real eigenvalues $\lambda_1, \dots, \lambda_n$, counted with multiplicity, and eigenvectors $v_1, \dots, v_n$ that form a basis of $\mathbb{R}^n$ with $v_i^{T}Mv_j = 0$ for $i \ne j$. Each eigenvalue equals

so every $\lambda_i$ is positive when $K$ is positive definite, and none is negative when $K$ is positive semidefinite. The problem has the same solutions as the ordinary eigenvalue problem for $M^{-1}K$, but that matrix is usually not symmetric. When $M$ is a multiple of $I$, as for equal masses, the eigenvectors of $K$ alone are the mode shapes, but in general they are not.

Definition Let $\lambda > 0$ and $v \ne 0$ satisfy $Kv = \lambda Mv$, and put $\omega = \sqrt{\lambda}$. The corresponding normal mode is the family of solutions

with real constants $A$ and $B$, not both zero. The number $\omega$ is a natural angular frequency of the system, and $v$ is the mode shape, fixed only up to a nonzero factor. Modes with different eigenvalues $\lambda_i \ne \lambda_j$ are $M$-orthogonal, meaning $v_i^{T}Mv_j = 0$, because symmetry of $K$ and $M$ gives

and $v_j^{T}Mv_i = v_i^{T}Mv_j$, so $(\lambda_i - \lambda_j)\,v_i^{T}Mv_j = 0$. For a repeated eigenvalue an $M$-orthogonal basis of its eigenspace can be chosen, but two arbitrary eigenvectors in that eigenspace need not be $M$-orthogonal. If $K$ is positive definite, every solution of $Mq'' + Kq = 0$ is a sum of the $n$ terms $v_i(A_i\cos\omega_i t + B_i\sin\omega_i t)$, one for each eigenvector of the basis above, with $\omega_i = \sqrt{\lambda_i}$ and some terms possibly zero. An eigenvalue $\lambda = 0$ gives the motion $v(a + bt)$ instead, which stretches no spring, and a negative eigenvalue gives solutions that grow exponentially, so the equilibrium is then unstable.

Definition Let $v_1, \dots, v_n$ be $M$-orthogonal eigenvectors of $Kv = \lambda Mv$ that form a basis, and let $V$ be the invertible matrix with these columns. The modal coordinates of a displacement $q$ are the entries of $\eta = V^{-1}q$, so that

By $M$-orthogonality both $V^{T}MV$ and $V^{T}KV$ are diagonal, with modal masses $\mu_i = v_i^{T}Mv_i > 0$ and entries $v_i^{T}Kv_i = \lambda_i\mu_i$. Substituting $q = V\eta$ into $Mq'' + Kq = 0$, multiplying on the left by the invertible matrix $V^{T}$ and dividing row $i$ by $\mu_i$ gives the separate equations

for $i = 1, \dots, n$. The initial data become $\eta(0) = V^{-1}q(0)$ and $\eta'(0) = V^{-1}q'(0)$, and $M$-orthogonality gives each entry directly, for example $\eta_i(0) = v_i^{T}Mq(0)/\mu_i$. The energy separates too,

and each term of the sum is constant along every solution.