= Solution
For a <central displacement in three freely ended coupled pendulums>, the initial vector decomposes as $(0,\epsilon,0)=(\epsilon/3)(1,1,1)-(\epsilon/3)(1,-2,1)$. The antisymmetric second mode is absent. Hence
$$
\boxed{x_1=x_3=\frac\epsilon3(\cos\omega_1t-\cos\omega_3t),\qquad
x_2=\frac\epsilon3(\cos\omega_1t+2\cos\omega_3t).}
$$
Let $\bar\omega=(\omega_3+\omega_1)/2$ and $\delta=(\omega_3-\omega_1)/2$. The outer displacement is $(2\epsilon/3)\sin\bar\omega t\sin\delta t$, explicitly exhibiting an amplitude modulation. The middle displacement is $\epsilon\cos\bar\omega t\cos\delta t-(\epsilon/3)\sin\bar\omega t\sin\delta t$, whose modulation envelope is
$$
\frac{|\epsilon|}{3}\sqrt{5+4\cos[(\omega_3-\omega_1)t]}.
$$
It ranges from $|\epsilon|/3$ to $|\epsilon|$, 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 $\omega_3-\omega_1\simeq3k/(2m\omega_1)$. The exact two-mode formulas remain valid without that additional weak-coupling assumption.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-three-pendulum-modes.png]
{title=Normal-mode displacement ratios and beating after releasing the central pendulum}
{description=Original eigenvectors and dimensionless time-domain solutions. The central initial displacement excites the uniform and highest-frequency modes, leaving the antisymmetric mode absent.}
Back to article page
