 Solution

For a central displacement in three freely ended coupled pendulums, the initial vector decomposes as $(0,ϵ,0)=(ϵ/3)(1,1,1)-(ϵ/3)(1,-2,1)$. The antisymmetric second mode is absent. HenceLet $\bar{\omega }=({\omega }_3+{\omega }_1)/2$ and $\delta =({\omega }_3-{\omega }_1)/2$. The outer displacement is $(2ϵ/3)\sin \bar{\omega }t\sin \delta t$, explicitly exhibiting an amplitude modulation. The middle displacement is $ϵ\cos \bar{\omega }t\cos \delta t-(ϵ/3)\sin \bar{\omega }t\sin \delta t$, whose modulation envelope isIt ranges from $|ϵ|/3$ to $|ϵ|$, so it does not vanish completely. These are beats between the two excited frequencies. A slow envelope well separated from the carrier requires $k/m\ll g/l$; in that limit ${\omega }_3-{\omega }_1\simeq 3k/(2m{\omega }_1)$. The exact two-mode formulas remain valid without that additional weak-coupling assumption.