 Solution

The lower mass has coordinates $(x+l\sin \theta ,-l\cos \theta )$. Its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. Adding the bead's kinetic energy gives the kinetic part of the supplied Lagrangian. Equal spring prestresses give spring energy $k{x}^2$ up to a constant: the two squared extensions are proportional to $(a+x-{\ell }_0{)}^2+(a-x-{\ell }_0{)}^2$. The gravitational potential is $-m_{2}gl\cos \theta$. Thus $L=T-V$ has exactly the stated form, with constant energy offsets discarded.

Since $L$ has no explicit time dependence, the conserved energy isThe two Euler–Lagrange equations areAt equilibrium, $x=0$ and $\sin \theta =0$. There are two geometrically distinct positions, $\boxed{(x,\theta )=(0,0),(0,\pi )}$ modulo $2\pi$. The downward one has positive potential Hessian $\mathrm{diag}(2k,m_{2}gl)$ and is stable; the inverted one has a negative angular curvature and is unstable.

Keeping only quadratic terms about the downward equilibrium givesFor unit parameters, the mass and stiffness matrices are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. The equation $\det (K-{\omega }^{2}M)=0$ is ${\omega }^4-4{\omega }^2+2=0$. Therefore the normal angular frequencies are