= Solution
Assume small planar oscillations about the vertical equilibrium, with the connecting <spring> relaxed in that equilibrium. Let $x_j=l\theta_j$ be the first-order horizontal <displacements> of the two <simple pendulums>. The gravitational restoring <force> on bob $j$ is $-(mg/l)x_j$; the spring extension to first order is $x_2-x_1$. Thus the linearized <equations of motion> are
$$
\begin{aligned}
m\ddot x_1&=-\frac{mg}{l}x_1+k(x_2-x_1),\\
m\ddot x_2&=-\frac{mg}{l}x_2+k(x_1-x_2).
\end{aligned}
$$
Adding and subtracting them diagonalizes the system. For $q_+=x_1+x_2$ and $q_-=x_1-x_2$,
$$
\ddot q_++\frac glq_+=0,\qquad
\ddot q_-+\left(\frac gl+\frac{2k}{m}\right)q_-=0.
$$
The <normal modes> and <angular frequencies> are therefore
$$
\boxed{\begin{array}{c|c}
\text{mode shape}&\text{angular frequency}\\\hline
(1,1)^T&\omega_1=\sqrt{g/l}\\
(1,-1)^T&\omega_2=\sqrt{g/l+2k/m}
\end{array}}
$$
The in-phase mode does not stretch the spring, while the opposite-phase mode adds a restoring force of magnitude $2k$ times each bob's displacement. These are the small-oscillation frequencies, not exact amplitude-independent frequencies for arbitrary pendulum angles.
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