= Solution
The <Euler-Lagrange equations> of the quadratic <Lagrangian> are
$$
\ddot{\boldsymbol\theta}+\mathsf K\boldsymbol\theta=0,\qquad
\mathsf K=\begin{pmatrix}1+k&-k\\-k&1+k\end{pmatrix}.
$$
The <mechanical mass matrix> is the identity. A <normal mode> $\boldsymbol\theta=\mathbf a\cos(\omega t+\phi)$ therefore satisfies the <eigenvalue problem> $\mathsf K\mathbf a=\omega^2\mathbf a$. Direct multiplication gives
$$
\boxed{\begin{array}{c|c|c}
&\text{angular frequency}&\text{mode vector}\\\hline
\text{in phase}&\omega_+=1&(1,1)\\
\text{opposite phase}&\omega_-=\sqrt{1+2k}&(1,-1)
\end{array}}
$$
These are <normal modes of two spring-coupled equal pendulums>. The corresponding <normal mode coordinates> $u=\theta_1+\theta_2$ and $v=\theta_1-\theta_2$ obey $\ddot u+u=0$ and $\ddot v+(1+2k)v=0$.
For the in-phase <normal mode>, the bobs undergo identical <displacements>, leaving their separation exactly $r$. The <spring> stays at its natural length and exerts no <force>, so this <angular frequency> is independent of $k$. The value $1$ is the individual <simple pendulum> <angular frequency> $\sqrt{g/l}$ in the chosen units.
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