= Exact Lagrangian of two spring-coupled pendulums
{title2=$L=\tfrac12ml^2(\dot\theta_1^2+\dot\theta_2^2)+mgl(\cos\theta_1+\cos\theta_2)-\tfrac12k(d-r)^2$}
Two identical planar <simple pendulums> of length $l$ and bob <mass> $m$, with pivots separated horizontally by $r>0$, are coupled by a straight <spring> of natural length $r$ and <spring constant> $k$. In the ideal model without contact restrictions, their <configuration space> is the <torus> $S^1\times S^1$. Measure both <angles> from the downward vertical towards the same horizontal direction. Their bob separation is
$$
d=\sqrt{\bigl(r+l(\sin\theta_2-\sin\theta_1)\bigr)^2+l^2(\cos\theta_2-\cos\theta_1)^2}.
$$
The exact <Lagrangian> is the expression in the title. To derive it, use bob <positions> $(l\sin\theta_1,-l\cos\theta_1)$ and $(r+l\sin\theta_2,-l\cos\theta_2)$. <Kinetic energy from Cartesian position vectors> gives $T=ml^2(\dot\theta_1^2+\dot\theta_2^2)/2$. Gravitational <potential energy> is $-mgl(\cos\theta_1+\cos\theta_2)$ and <spring> <potential energy> is $k(d-r)^2/2$; subtracting these from $T$ proves the formula. Near the downward <equilibrium>, $d-r=l(\theta_2-\theta_1)+O(|\boldsymbol\theta|^3)$ for fixed $r$, giving the familiar <normal modes of two spring-coupled equal pendulums>.
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