Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 14B (Classical Dynamics)

With the pivot as origin and upward vertical coordinate,ThereforeThe potential energy isThus $L=T-V$ is

Put $\Delta ={\theta }_1-{\theta }_2$ and ${\omega }_0^2=g/l$. The Euler--Lagrange equations, after division by the appropriate powers of $l$, areTime-translation invariance conserves the total energy

Setting both angles and velocities to zero satisfies the equations, so the downward configuration is an equilibrium. Writing ${\theta }_i=z_i$ and retaining linear terms givesEquivalently, the mass and stiffness matrices are

Seek $z=v{e}^{i\Omega t}$ and put $\mu =m/M$. The generalized characteristic equation isorAs a quadratic in $X={\Omega }^2$, this issoA corresponding displacement vector isThus the four normal modes are $v_{+}{e}^{\pm i{\Omega }_{+}t}$ and $v_{-}{e}^{\pm i{\Omega }_{-}t}$. Both squared frequencies are positive for every $\mu >0$, proving linear stability.

For $\mu \ll 1$,Therefore, labeling the larger and smaller positive frequencies by $\omega$ and ${\omega }^{\prime }$,so $\alpha ={\omega }_0$.