Central displacement in three freely ended coupled pendulums

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.