= Bead on a parabolic wire
{title2=$T=2\sqrt{2a/g}\int_{-1}^1\sqrt{\frac{1+(h/a)u^2}{1-u^2}}\,du$}
For a smooth stationary wire $z=x^2/(4a)$ and release from rest at height $h$, the <normal reaction> does no work. The energy equation is $[1+x^2/(4a^2)]\dot x^2/2+gx^2/(4a)=gh$. Integrating the travel time between turning points gives the displayed period. Its small-amplitude expansion is $T=2\pi\sqrt{2a/g}[1+h/(4a)+O((h/a)^2)]$. The mass cancels because both gravitational and inertial forces are proportional to it.
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