= Solution
Let $x_1,x_2,x_3$ be longitudinal <displacements> from equilibrium, with positive direction to the right. The fixed walls have zero displacement. The first mass feels forces $-kx_1$ and $k(x_2-x_1)$; the central mass feels $k(x_1-x_2)$ and $k(x_3-x_2)$; the last mass feels $k(x_2-x_3)$ and $-kx_3$. <Hooke's law> and <Newton's second law> therefore give
$$
\boxed{\begin{aligned}
m\ddot x_1&=-2kx_1+kx_2,\\
m\ddot x_2&=kx_1-2kx_2+kx_3,\\
m\ddot x_3&=kx_2-2kx_3.
\end{aligned}}
$$
Equivalently,
$$
\boxed{m\ddot{\boldsymbol x}+kA\boldsymbol x=0,\qquad
A=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}.}
$$
All four springs contribute: the wall springs supply one of the two diagonal terms in the first and third rows. This is a fixed-end chain, not a freely ended chain with a zero-frequency translation mode.
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