 End-displacement release of three coupled pendulums

Consider three coupled pendulums with equal end masses, with end masses $m$, middle mass $M$, equal lengths $l$ and connecting spring constants $k$. Give only the first bob a small horizontal displacement $a$, then release all three from rest. Write ${\omega }_0^2=g/l$, ${\omega }_1^2=g/l+k/m$ and ${\omega }_2^2=g/l+k/m+2k/M$. Decomposing $(a,0,0)$ into the three normal mode patterns givesEach cosine enforces zero initial velocity. At $t=0$, the three terms sum to the prescribed end displacement and cancel at the other bobs, which checks the coefficients independently of the time evolution.

The middle bob has no contribution from the antisymmetric normal mode $(1,0,-1)$, so a trace of that bob alone misses one frequency. This is a simple example of how a sensor's position determines which normal modes it can detect. The expression assumes the small-angle regime and horizontal coupling at the bobs.