Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 9C / ii

Along the holonomic constraint $z=x^2/(4a)$, we have $\dot{z}=x\dot{x}/(2a)$ and hence speed squared $(1+x^2/(4a^2)){\dot{x}}^2$. The smooth wire's normal reaction is perpendicular to the permitted velocity and does no work. Conservation of mechanical energy, with release from rest at height $h$, givesThe turning points are $x=\pm x_0$ with $x_0=2\sqrt{ah}$. One full oscillation has twice the travel time between them:Set $x=x_{0}u$ and $\beta =h/a$. After cancelling the scale factors,so the bead on a parabolic wire hasFor $\beta \ll 1$, expand the numerator uniformly on $[-1,1]$:The integrable endpoint weight gives ${\int }_{-1}^1(1-u^2{)}^{-1/2}du=\pi$ and ${\int }_{-1}^1{u}^2(1-u^2{)}^{-1/2}du=\pi /2$, for example by $u=\sin \theta$. Therefore the requested small-oscillation period isThe leading frequency is $\sqrt{g/(2a)}$, and the first finite-amplitude correction increases the period. At $h=0$ the ring is stationary; the expression is the limiting period of nonzero small oscillations.