The lower mass has coordinates $(x+l\sin \theta ,-l\cos \theta )$. Its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. Adding the bead's kinetic energy gives the kinetic part of the supplied Lagrangian. Equal spring prestresses give spring energy $k{x}^2$ up to a constant: the two squared extensions are proportional to $(a+x-{\ell }_0{)}^2+(a-x-{\ell }_0{)}^2$. The gravitational potential is $-m_{2}gl\cos \theta$. Thus $L=T-V$ has exactly the stated form, with constant energy offsets discarded.

Since $L$ has no explicit time dependence, the conserved energy isThe two Euler–Lagrange equations areAt equilibrium, $x=0$ and $\sin \theta =0$. There are two geometrically distinct positions, $\boxed{(x,\theta )=(0,0),(0,\pi )}$ modulo $2\pi$. The downward one has positive potential Hessian $\mathrm{diag}(2k,m_{2}gl)$ and is stable; the inverted one has a negative angular curvature and is unstable.

Keeping only quadratic terms about the downward equilibrium givesFor unit parameters, the mass and stiffness matrices are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. The equation $\det (K-{\omega }^{2}M)=0$ is ${\omega }^4-4{\omega }^2+2=0$. Therefore the normal angular frequencies are

At a regular constrained stationary curve, choose a constant multiplier $\lambda$ and make the augmented functional $I-\lambda J$ stationary under endpoint-fixed variations. Its integrand is $H=F-\lambda G$ and the equation is $d{H}_{y^{\prime }}/dx-H_y=0$. This normal multiplier formulation requires the first variation of the constraint not to vanish identically; the usual abnormal case would require two multipliers. When $H_x=0$,so $\boxed{y^{\prime }{H}_{y^{\prime }}-H=c_0}$ is constant, proving the requested first integral.

For the chain, $ds=\sqrt{1+(y^{\prime }{)}^2}dx$, so the fixed physical length is exactly $J[y]$. Its gravitational potential, apart from the constant factor $\rho g$, is $I[y]={\int }_{-a}^{a}y\sqrt{1+(y^{\prime }{)}^2}dx$. Thus $H=(y-\lambda )\sqrt{1+(y^{\prime }{)}^2}$, and the Euler–Lagrange equation isThe first integral becomes $(y-\lambda )/\sqrt{1+(y^{\prime }{)}^2}=c$, after renaming the constant with the opposite sign. The differential equation also gives $(y-\lambda )y^{\prime \prime }=1+(y^{\prime }{)}^2$, so the nondegenerate solution integrates to $y=\lambda +c\cosh ((x-x_0)/c)$. Equal endpoint heights force $x_0=0$: the two cosh values agree only when their arguments are equal or opposite, and $a>0$ excludes the former. HenceFor the hanging energy minimum take $c>0$. Since $\sqrt{1+(y^{\prime }{)}^2}=\cosh (x/c)$,For $L>2a$ the length equation has exactly one positive solution: its right side decreases from infinity to $2a$ as $c$ increases from zero to infinity. For $L=2a$ the only feasible shape is the straight segment, obtained as a limiting case; for $L<2a$ no chain can span the endpoints. A negative $c$ gives the inverted stationary catenary, rather than the physically hanging minimum.