= Solution
For an ordinary conservative mechanical system, choose <generalized coordinates> $q_i$ measured from a static <equilibrium point>. Assume the <kinetic energy> is quadratic in velocities and there are no velocity-dependent forces. Expanding about equilibrium, the linear term in the <potential energy> vanishes. Retaining only terms of total degree two gives the <small oscillation> <Lagrangian>
$$
\boxed{L=\frac12\dot{\mathbf q}^{\,T}M\dot{\mathbf q}-\frac12\mathbf q^TK\mathbf q},\qquad
M_{ij}=\left.\frac{\partial^2T}{\partial\dot q_i\partial\dot q_j}\right|_0,\quad
K_{ij}=\left.\frac{\partial^2V}{\partial q_i\partial q_j}\right|_0.
$$
An irrelevant constant has been dropped. Coordinate dependence of $M$ contributes only terms of third or higher total degree. The matrices $M,K$ are symmetric, $M$ is positive definite for independent coordinates, and $K$ is positive semidefinite at a <stable equilibrium>. The <Euler-Lagrange equations> are
$$
\boxed{M\ddot{\mathbf q}+K\mathbf q=0}.
$$
A <normal mode> has a fixed <mode shape> $\mathbf v$ and one time factor: $\mathbf q=\mathbf v Q(t)$. Its <angular frequency> $\omega$ solves the <generalized eigenvalue problem> $K\mathbf v=\omega^2M\mathbf v$, with $\ddot Q+\omega^2Q=0$. Thus $\omega$ is the normal angular frequency, and the frequency in cycles per unit time is $\omega/(2\pi)$. A <zero mode> satisfies $K\mathbf v=0$; its time dependence is $Q=A+Bt$, describing an unopposed displacement or uniform drift rather than a restoring oscillation.
Back to article page
