= 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 $kx^2$ 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 $-m_2gl\cos\theta$. Thus $L=T-V$ has exactly the stated form, with constant energy offsets discarded.
Since $L$ has no explicit time dependence, the conserved energy is
$$
\boxed{E=\tfrac12(m_1+m_2)\dot x^2+m_2l\cos\theta\,\dot x\dot\theta+\tfrac12m_2l^2\dot\theta^2+kx^2-m_2gl\cos\theta.}
$$
The two Euler–Lagrange equations are
$$
\begin{aligned}
(m_1+m_2)\ddot x+m_2l\cos\theta\,\ddot\theta-m_2l\sin\theta\,\dot\theta^2+2kx&=0,\\
m_2l\cos\theta\,\ddot x+m_2l^2\ddot\theta+m_2gl\sin\theta&=0.
\end{aligned}
$$
At 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 $\operatorname{diag}(2k,m_2gl)$ and is stable; the inverted one has a negative angular <second derivative>[curvature] and is unstable.
Keeping only quadratic terms about the downward equilibrium gives
$$
L_2=\tfrac12(m_1+m_2)\dot x^2+m_2l\dot x\dot\theta+\tfrac12m_2l^2\dot\theta^2-kx^2-\tfrac12m_2gl\theta^2.
$$
For unit parameters, the <mass> and stiffness <matrices> are $M=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ and $K=\begin{pmatrix}2&0\\0&1\end{pmatrix}$. The equation $\det(K-\omega^2M)=0$ is $\omega^4-4\omega^2+2=0$. Therefore the normal angular frequencies are
$$
\boxed{\omega_-=\sqrt{2-\sqrt2},\qquad\omega_+=\sqrt{2+\sqrt2}.}
$$
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