 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 $r_{\ast }$, their potential is $k[(r-r_{\ast }{)}^2+(l-r-r_{\ast }{)}^2]/2$. 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 asThis uses the illustrated symmetric two-spring arrangement without assuming an unstated natural length.