Transverse degeneracy of unstretched spring-coupled pendulums (${\omega }^2=g/L\text{(three modes)},{\omega }^2=g/L+2k/m$)

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.