= 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> $kx^2$ 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 $-m_2gl\cos\theta$. Therefore
$$
L=\frac12(m_1+m_2)\dot x^2+m_2l\cos\theta\,\dot x\dot\theta+\frac12m_2l^2\dot\theta^2-kx^2+m_2gl\cos\theta.
$$
There is no explicit time dependence, so the total <energy>
$$
E=\frac12(m_1+m_2)\dot x^2+m_2l\cos\theta\,\dot x\dot\theta+\frac12m_2l^2\dot\theta^2+kx^2-m_2gl\cos\theta
$$
is conserved. The <Euler-Lagrange equations> are
$$
(m_1+m_2)\ddot x+m_2l\cos\theta\,\ddot\theta-m_2l\sin\theta\,\dot\theta^2+2kx=0,
$$
$$
m_2l\cos\theta\,\ddot x+m_2l^2\ddot\theta+m_2gl\sin\theta=0.
$$
At 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 gives
$$
L_2=\frac12(m_1+m_2)\dot x^2+m_2l\dot x\dot\theta+\frac12m_2l^2\dot\theta^2-kx^2-\frac12m_2gl\theta^2.
$$
For unit parameters the <mass matrix> and <stiffness matrix> are $M=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ and $K=\begin{pmatrix}2&0\\0&1\end{pmatrix}$. A <normal mode> with squared frequency $s$ requires
$$
0=\det(K-sM)=(2-2s)(1-s)-s^2=s^2-4s+2.
$$
Thus the positive <angular frequencies> are
$$
\boxed{\omega_-=\sqrt{2-\sqrt2},\qquad\omega_+=\sqrt{2+\sqrt2}}.
$$
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