Three coupled pendulums with equal end masses (${\omega }^2=g/l,g/l+k/m,g/l+k/m+2k/M$)

For equal lengths, end masses $m$, middle mass $M$, and equal springs between adjacent bobs, reflection symmetry gives mode vectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2m/M,1)$. Their squared frequencies are displayed above. These vectors are orthogonal in the inner product weighted by $\mathrm{diag}(m,M,m)$. Equal end masses, rather than equality of all three masses, suffice for this symmetry reduction.