 Solution

For small oscillations about equilibrium, put $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$. The potential has no linear term at equilibrium; to quadratic order the Lagrangian is $L_2={\dot{x}}^{\mathsf{T}}M\dot{x}/2-x^{\mathsf{T}}Kx/2$. Its equations are $M\ddot{x}+Kx=0$. A normal mode $x=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem $(K-{\omega }^{2}M)v=0$. For a positive-definite mass matrix, diagonalization $M^{-1/2}K{M}^{-1/2}$ gives the squared frequencies and independent normal coordinates. A strictly positive Hessian gives positive frequencies; zero eigenvalues require separate higher-order stability analysis.

Here the spring extensions are $x_1$, $x_2-x_1$, and $-x_2$. ThusThe stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and $M=mI$. Its eigenvectors $(1,1)$ and $(1,-1)$ have stiffness eigenvalues $k$ and $3k$, givingIn the first mode the masses move together with equal displacements and the middle spring keeps its length. In the second they move oppositely with equal amplitudes, stretching and compressing the middle spring twice as much as either end spring.