 Solution

The diagram has a connection between every pair of indices $0\le i<j\le n$, including the fixed beam $0$. Two parallel strings, each with spring constant $k/2$, together have energy $k(X_i-X_j{)}^2/2$. Thus the printed $E_{ij}$ is the energy of a connecting pair; calling it the energy of one string would introduce a factor-of-two discrepancy.

Take vertical coordinates positive upwards, with $X_0$ fixed. Each bar has gravitational potential energy $m_{i}g{X}_i$ and kinetic energy $m_i{\dot{X}}_i^2/2$. The Lagrangian isDifferentiating gives the Euler-Lagrange equationsEquivalently, with $S={\sum }_{j=1}^n{X}_j$, the spring term is $k[(n+1)X_i-S-X_0]$. Gravity changes the equilibrium but not the stiffness matrix. Subtracting a static equilibrium and writing ${\eta }_i=X_i-X_i^{\mathrm{eq}}$ gives the normal mode systemThis is the spring system underlying normal modes of an anchored complete spring graph.