Past exam of the classical mechanics course of the University of Oxford / 2023 / 1 / b

Set $l=m=g=1$. At $({\theta }_1,{\theta }_2)=(0,0)$ the spring has $d=r$, so its potential energy has zero gradient: ${\partial }_j{V}_s=k(d-r){\partial }_{j}d=0$. The gravitational potential energy has ${\partial }_j{V}_g=\sin {\theta }_j=0$ there. Zero velocities therefore satisfy the Euler-Lagrange equations, proving this is an equilibrium.

For fixed $r>0$, let $\epsilon =\max (|{\theta }_1|,|{\theta }_2|)$ and $\delta ={\theta }_2-{\theta }_1$. The Taylor series giveThus, near the downward equilibrium, the horizontal separation is positive andIn particular, $V_s=k{\delta }^2/2+O({\epsilon }^4)$ and $V_g=-2+({\theta }_1^2+{\theta }_2^2)/2+O({\epsilon }^4)$. Dropping the additive constant and terms beyond quadratic order gives the harmonic approximationFor physical spring constant $k\ge 0$, the quadratic potential energy is a positive-definite quadratic form. The downward position is therefore a stable equilibrium in Lagrangian mechanics, and its nearby motion admits small oscillations.

The Euler-Lagrange equations of the quadratic Lagrangian areThe mechanical mass matrix is the identity. A normal mode $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\varphi )$ therefore satisfies the eigenvalue problem $\mathsf{K}\mathbf{a}={\omega }^2\mathbf{a}$. Direct multiplication givesThese are normal modes of two spring-coupled equal pendulums. The corresponding normal mode coordinates $u={\theta }_1+{\theta }_2$ and $v={\theta }_1-{\theta }_2$ obey $\ddot{u}+u=0$ and $\ddot{v}+(1+2k)v=0$.

For the in-phase normal mode, the bobs undergo identical displacements, leaving their separation exactly $r$. The spring stays at its natural length and exerts no force, so this angular frequency is independent of $k$. The value $1$ is the individual simple pendulum angular frequency $\sqrt{g/l}$ in the chosen units.