= Solution
Expand about the equilibrium and keep only terms quadratic in displacement and velocity. Writing $M_{ij}=T_{ij}(0)$ and $K_{ij}=\partial_i\partial_jV(0)$, the linearized Lagrangian is $L_2=\dot x^TM\dot x/2-x^TKx/2$, up to a constant. Its equations are $M\ddot x+Kx=0$. With positive-definite <mass matrix> $M$, the normal frequencies and mode vectors solve
$$
\boxed{Kv=\omega^2Mv,\qquad\det(K-\omega^2M)=0.}
$$
For 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 $-x_2$. Hence
$$
\boxed{T=\frac m2(\dot x_1^2+\dot x_2^2),\qquad
V=\frac k2\big[x_1^2+(x_2-x_1)^2+x_2^2\big].}
$$
The matrices are $M=mI$ and $K=k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$. The two independent modes are
$$
\boxed{\begin{aligned}
v_1&=(1,1)^T,&\omega_1&=\sqrt{k/m},\\
v_2&=(1,-1)^T,&\omega_2&=\sqrt{3k/m}.
\end{aligned}}
$$
In 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.
\Image[/past-exam-of-the-options-i-course-of-the-university-of-oxford/2010-two-mass-spring-modes.png]
{title=The two normal modes of equal masses joined to fixed walls by three equal springs}
{description=The in-phase mode leaves the middle spring unchanged; the opposite-phase mode stretches or compresses it by twice either mass displacement.}
These are the <normal modes of two equal masses between three springs>.
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