Expand about the equilibrium and keep only terms quadratic in displacement and velocity. Writing $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$, the linearized Lagrangian is $L_2={\dot{x}}^{T}M\dot{x}/2-x^{T}Kx/2$, up to a constant. Its equations are $M\ddot{x}+Kx=0$. With positive-definite mass matrix $M$, the normal frequencies and mode vectors solveFor a nondegenerate stable equilibrium, $K$ is positive definite and all these frequencies are positive. A zero eigenvalue indicates a zero-frequency linearized mode requiring higher-order analysis if nonlinear stability is in question.

The three spring extensions are $x_1$, $x_2-x_1$ and $-x_2$. HenceThe matrices are $M=mI$ and $K=k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. The two independent modes areIn the first mode both masses move together with equal displacement, leaving the middle spring unchanged. In the second they move in opposite directions with equal magnitude; the midpoint stays fixed and the middle spring changes length by twice either displacement. Arbitrary small motion is a superposition of sinusoidal motions in these two modes.

These are the normal modes of two equal masses between three springs.

For fixed endpoint values, independent variations of $u$ and $v$ give the Euler-Lagrange equationsThe sphere parametrization has metric $d{s}^2=a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$, obtained by differentiating its position vector and taking its Euclidean squared length. The displayed functional is therefore one half the integral of squared speed, the geodesic energy. Its fixed-endpoint extremals are affinely parametrized geodesics; energy stationarity chooses a constant-speed parametrization.

Its explicit equations, after dividing by $a^2$, areThe second gives ${\sin }^2\theta \dot{\varphi }=h$. Since the Lagrangian has no explicit time dependence, its energy $\dot{q}\cdot L_{\dot{q}}-L=L$ is conserved. ThusThe same second identity follows by differentiating it and substituting the two equations. These polar-coordinate formulas hold on a nonsingular coordinate patch; the geodesic curve itself can pass through a pole in another chart.

Burnside's lemma gives the number of orbits asFor $m=2n-1\ge 3$, the polygon symmetry group contains the $m$ rotations through angles $2\pi j/m$, $0\le j<m$, and $m$ reflections. Each reflection axis passes through a vertex and the midpoint of the opposite edge. These are all the symmetries, so $|G|=2m=\boxed{4n-2}$.

For the pentagon, apply the same group element simultaneously to vertex colours and edge colours. The identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations is a 5-cycle on vertices and also on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two swapped vertex pairs, and one fixed edge and two swapped edge pairs. It therefore fixes $v^3{e}^3$ colourings. Averaging yieldsThe two colour palettes are chosen independently, but rotations and reflections act on both parts of the same pentagon.