= Transverse degeneracy of unstretched spring-coupled pendulums
{title2=$\omega^2=g/L\ \text{(three modes)},\quad\omega^2=g/L+2k/m$}
Two equal <simple pendulums> have small horizontal displacements $(x_j,y_j)$, with their support separation and connecting spring along $x$. If the spring is unstretched at equilibrium, its first-order extension is $x_2-x_1$; transverse differences change its length only at second order. The quadratic potential is $mg\sum_j(x_j^2+y_j^2)/(2L)+k(x_2-x_1)^2/2$. There is one higher antisymmetric longitudinal <normal mode> and a three-dimensional eigenspace at $\sqrt{g/L}$: the symmetric longitudinal mode and two independent transverse motions. Pre-tension would lift this transverse degeneracy.
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