= Coupled torsional pendulums
{title2=$\omega_\pm^2/\omega_0^2=1/2+1/R\pm\sqrt{1/4+1/R^2}$}
Two identical torsional loads of <moment of inertia> $I$ hang in series from massless fibres of lengths $\ell_1,\ell_2$ and common <filament torsional stiffness> $K_t$. Put $R=\ell_2/\ell_1$ and $\omega_0^2=K_t/(I\ell_1)$. The elastic <energy> is $K_t\phi_1^2/(2\ell_1)+K_t(\phi_2-\phi_1)^2/(2\ell_2)$ and the kinetic <energy> is $I(\dot\phi_1^2+\dot\phi_2^2)/2$. Their <normal modes> have the displayed squared <angular frequencies>. Writing $\lambda=\omega^2/\omega_0^2$, their amplitude ratio is $A_2/A_1=1/(1-R\lambda)$. The lower mode twists in phase and approaches a combined inertia $2I$ as $R\to0$, giving $\omega_-^2\to\omega_0^2/2$. The upper mode approaches the original single-load oscillation as $R\to\infty$, giving $\omega_+^2\to\omega_0^2$.
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