Normal modes of three freely ended coupled pendulums (${\omega }^2=g/L+(k/m)\lambda ,\lambda =0,1,3$)

For three equal simple pendulums in a straight open chain with springs only between neighbours, the spring stiffness matrix in displacement coordinates is $k\left(\begin{array}{ccc}1& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right)$. Its eigenvectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2,1)$ have respective dimensionless eigenvalues $0,1,3$. Adding the gravitational restoring term yields the displayed squared eigenfrequencies. These are free spring ends: fixed wall springs would change the matrix and mode shapes.

Releasing the central bob of three identical spring-coupled simple pendulums from displacement $d$ excites only the symmetric normal modes. In displacement ratios, $(0,d,0)=(d/3)(1,1,1)-(d/3)(1,-2,1)$. Thus the outer coordinates are $(d/3)(\cos {\omega }_{1}t-\cos {\omega }_{3}t)$ and the central coordinate is $(d/3)(\cos {\omega }_{1}t+2\cos {\omega }_{3}t)$, where ${\omega }_1^2=g/l$ and ${\omega }_3^2=g/l+3k/m$. The antisymmetric mode is absent by reflection symmetry. The central modulation envelope is $|d|\sqrt{5+4\cos [({\omega }_3-{\omega }_1)t]}/3$, so it never vanishes completely. Slow beats require a small frequency difference; the decomposition itself is exact in the linear model.