= 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 gives
$$
\mathbf x(t)=\frac{am}{M+2m}(1,1,1)\cos\omega_0t+\frac a2(1,0,-1)\cos\omega_1t+\frac{aM}{2(M+2m)}(1,-2m/M,1)\cos\omega_2t.
$$
Each 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.
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