= Three coupled pendulums with equal end masses
{title2=$\omega^2=g/l,\quad g/l+k/m,\quad g/l+k/m+2k/M$}
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 $\operatorname{diag}(m,M,m)$. Equal end masses, rather than equality of all three masses, suffice for this <symmetry> reduction.
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