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With equal rod lengths $\ell$ and absolute downward-vertical angles $\theta ,\varphi$, the endpoint separation is $s^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. A zero-rest-length spring between the fixed support and the final mass contributes $-\kappa {\ell }^2[1+\cos (\theta -\varphi )]$ to the Lagrangian. It favours bending away from maximum extension and can reduce the gravitational bending stiffness.

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,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 $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

A zero-rest-length spring has quadratic potential energy in its actual length $s$, with no subtracted natural length. It is a useful idealization, not an ordinary unstretched spring of arbitrary length. For an end-to-end spring on a two-link pendulum, changing the natural length changes the quadratic bending stiffness about the straight configuration.