= Solution
Measure both <angles> from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,
$$
\mathbf r_B=\ell(\sin\theta,\cos\theta),\qquad
\mathbf r_C=\ell(\sin\theta+\sin\phi,\cos\theta+\cos\phi).
$$
Thus $|AC|^2=2\ell^2[1+\cos(\theta-\phi)]$. The requested expression implicitly uses a \b[<zero-rest-length spring>]: its <energy> is $\kappa|AC|^2/2$. The exact <Lagrangian> for equal <masses> is
$$
L=m\ell^2\dot\theta^2+\frac{m\ell^2}{2}\dot\phi^2
+m\ell^2\dot\theta\dot\phi\cos(\theta-\phi)
+mg\ell(2\cos\theta+\cos\phi)
-\kappa\ell^2[1+\cos(\theta-\phi)].
$$
Expand the <cosines> and count <angles> and <velocities> as first order. The cosine correction multiplying $\dot\theta\dot\phi$ is fourth order and is dropped. After removing constants,
$$
\boxed{L_2=\frac{m\ell^2}{2}(2\dot\theta^2+2\dot\theta\dot\phi+\dot\phi^2)
-\frac{mg\ell}{2}(2\theta^2+\phi^2)
+\frac{\kappa\ell^2}{2}(\theta-\phi)^2.}
$$
The positive <spring> term in $L_2$ means the <spring> lowers its <energy> when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa\ell(2\ell-s_0)(\theta-\phi)^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.
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