= Solution
Use $r\in(0,l)$, the bead's distance from the pivot, and $\theta$, the rod angle from downward vertical. The <kinetic energy> is $m(\dot r^2+r^2\dot\theta^2)/2$. If the identical springs have common natural length $r_*$, their potential is $k[(r-r_*)^2+(l-r-r_*)^2]/2$. Its $r$-dependent part is $k(r-l/2)^2$; the unspecified common natural length changes only an additive constant. Gravity contributes $-mgr\cos\theta$. Consequently the <Lagrangian> may be taken as
$$
\boxed{L=\frac m2(\dot r^2+r^2\dot\theta^2)+mgr\cos\theta-k(r-l/2)^2}.
$$
This uses the illustrated symmetric two-spring arrangement without assuming an unstated natural length.
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