Past exam of the mathematical methods II course of the University of Oxford / 2013 / 8

With angles measured from the downward vertical, the upper mass rises by $2l(1-\cos {\theta }_1)$ and the lower by $2l(1-\cos {\theta }_1)+l(1-\cos {\theta }_2)$. Taking zero gravitational potential energy at the hanging equilibrium therefore gives the exact resultBoth masses contribute to the elevation change caused by the upper rod.

Use positions ${\mathbf{r}}_1=(2l\sin {\theta }_1,-2l\cos {\theta }_1)$ and ${\mathbf{r}}_2={\mathbf{r}}_1+(l\sin {\theta }_2,-l\cos {\theta }_2)$. Their squared speeds are $4l^2{\dot{\theta }}_1^2$ and $4l^2{\dot{\theta }}_1^2+l^2{\dot{\theta }}_2^2+4l^2{\dot{\theta }}_1{\dot{\theta }}_2\cos ({\theta }_1-{\theta }_2)$. Summing their kinetic energies givesNo small-angle approximation has been used here.

Retain terms quadratic in the small displacements and velocities. The approximations $1-\cos \theta \simeq {\theta }^2/2$ and $\cos ({\theta }_1-{\theta }_2)\simeq 1$ yieldThus the correct quadratic forms areSource correction. The printed kinetic-energy matrix formula omits both time-derivative dots. It cannot hold as a function of angles alone: at $\mathbf{\theta }=0$, ${\dot{\theta }}_1\ne 0$, ${\dot{\theta }}_2=0$, the actual kinetic energy is $4m{l}^2{\dot{\theta }}_1^2>0$, whereas the printed undifferentiated quadratic form vanishes.

For the constant small-oscillation matrices, the total mechanical energy satisfiesThe specified conservative pendulum has Lagrangian $L=K-U$, whose Euler-Lagrange equations giveor $8{\ddot{\theta }}_1+2{\ddot{\theta }}_2+(4g/l){\theta }_1=0$ and $2{\ddot{\theta }}_1+{\ddot{\theta }}_2+(g/l){\theta }_2=0$. These equations make $\dot{E}=0$.

The scalar energy law alone is insufficient to infer two unique equations: energy conservation does not uniquely determine equations of motion. In particular, adding $\mathbf{G}\dot{\mathbf{\theta }}$ on the right with antisymmetric $\mathbf{G}$ changes the dynamics but contributes zero power. The conservative-force structure of the given pendulum excludes that extra gyroscopic force and supplies the information needed for the displayed matrix equations.

For $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\phi )$, a nonzero normal mode satisfies $(\mathbf{V}-{\omega }^2\mathbf{T})\mathbf{a}=0$. With $s={\omega }^{2}l/g$, its determinant is $4(s^2-3s+1)$. Hence the normal modes of a double pendulum with a two-to-one length ratio haveThe slow mode is in phase and the fast mode out of phase. The general small motion is their superposition with independent amplitudes and phases; ordinary frequencies are ${\omega }_{\pm }/(2\pi )$.