= Solution
The quadratic <Lagrangian> gives $\partial L/\partial\dot q=T\dot q$ and $\partial L/\partial q=-Vq$. Thus the <Euler-Lagrange equations> are
$$
T\ddot q+Vq=0.
$$
A nonzero mode <vector> $v$ has a common <harmonic> time factor and satisfies the <generalized eigenvalue problem for small oscillations> $Vv=\omega^2Tv$; $\omega$ is its <normal frequency>. A zero-frequency mode instead admits a constant or linear time factor.
Measure the three coordinates as displacements from a configuration with both springs unstretched. The <kinetic energy> is $\tfrac12\dot x_1^2+\tfrac12M\dot x_2^2+\tfrac12\dot x_3^2$, and the <potential energy> is $\tfrac12(x_2-x_1)^2+\tfrac12(x_3-x_2)^2$. Expansion gives
$$
T=\begin{pmatrix}1&0&0\\0&M&0\\0&0&1\end{pmatrix},\qquad
V=\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}.
$$
For $M>0$, direct substitution yields the three <normal modes>:
$$
\boxed{\begin{array}{c|c|c}
&v&\omega\\\hline
\text{translation}&(1,1,1)^T&0\\
\text{antisymmetric outer motion}&(1,0,-1)^T&1\\
\text{symmetric outer motion}&(1,-2/M,1)^T&\sqrt{1+2/M}
\end{array}.}
$$
The three <vectors> are independent and orthogonal in the mass inner product $v^TTw$, so they exhaust the motion. Their squared mass norms are $M+2,2,2+4/M$; division by the corresponding square roots gives normalized modes. This is the unit-spring, unit-outer-mass case of the <normal modes of three masses connected by two equal springs>.
As $M\to\infty$, the central displacement in the symmetric oscillatory mode tends to zero and both nonzero frequencies tend to one. The two limiting shapes $(1,0,1)$ and $(1,0,-1)$ span independent motions of the outer masses against an effectively fixed centre; each unit mass has its own unit spring frequency. The translational mode remains at zero frequency. Under a bounded kinetic-energy condition its common velocity tends to zero as $M^{-1/2}$, while a static common displacement is still an equilibrium shift. One must specify that energy condition before concluding that an arbitrary freely translating heavy mass remains stationary.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2-three-mass-normal-modes.png]
{title=Mode shapes of a free three-mass chain, including the zero-frequency translation}
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