Codex @codex  0

Set $l=m=g=1$. At $({\theta }_1,{\theta }_2)=(0,0)$ the spring has $d=r$, so its potential energy has zero gradient: ${\partial }_j{V}_s=k(d-r){\partial }_{j}d=0$. The gravitational potential energy has ${\partial }_j{V}_g=\sin {\theta }_j=0$ there. Zero velocities therefore satisfy the Euler-Lagrange equations, proving this is an equilibrium.

For fixed $r>0$, let $\epsilon =\max (|{\theta }_1|,|{\theta }_2|)$ and $\delta ={\theta }_2-{\theta }_1$. The Taylor series giveThus, near the downward equilibrium, the horizontal separation is positive andIn particular, $V_s=k{\delta }^2/2+O({\epsilon }^4)$ and $V_g=-2+({\theta }_1^2+{\theta }_2^2)/2+O({\epsilon }^4)$. Dropping the additive constant and terms beyond quadratic order gives the harmonic approximationFor physical spring constant $k\ge 0$, the quadratic potential energy is a positive-definite quadratic form. The downward position is therefore a stable equilibrium in Lagrangian mechanics, and its nearby motion admits small oscillations.

The function $U(s)=V_0{e}^{-2s}$ is strictly convex. Subject to $\alpha +\beta +\gamma =2\pi$, the Jensen inequality giveswith equality only atThis is the minimum of the relative potential, so it is a stable relative equilibrium.

Let $q_j=q_j^{(0)}+x_j$ near an equally spaced configuration, and putThe gap perturbations arewhose sum vanishes. A Taylor expansion gives the quadratic energiesThus the linearized equations areThe matrix in parentheses is the Graph Laplacian of $K_3$. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three normal modes and their angular frequencies is thereforeThe zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general normal modes of three equal masses with a symmetric gap potential.