 Solution

The lower particle has position $(x+l\sin \theta ,-l\cos \theta )$, so its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. The bead contributes $m_1{\dot{x}}^2/2$ to the kinetic energy. The two equal springs have combined potential energy $k{x}^2$ up to a constant: their lengths are $a+x$ and $a-x$, and equal natural lengths cancel the linear terms in the sum of their squared extensions. The gravitational potential is $-m_{2}gl\cos \theta$. ThereforeThere is no explicit time dependence, so the total energyis conserved. The Euler-Lagrange equations areAt rest these require $x=0$ and $\sin \theta =0$, giving the downward and upward positions $\theta =0,\pi$ modulo $2\pi$. The downward position has positive-definite quadratic potential and is stable, whereas the upward position has a negative angular potential curvature.

Keeping terms of second order about the stable position and dropping a constant givesFor unit parameters the mass matrix and stiffness matrix are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. A normal mode with squared frequency $s$ requiresThus the positive angular frequencies are