 Release of a double pendulum into both normal modes

Release the upper rod at ${\theta }_1(0)=\alpha =0.06$ radians while keeping the lower rod vertical, ${\theta }_2(0)=0$, with both angular velocities zero. The initial condition is the sum $\mathbf{\theta }(0)=(\alpha /2){\mathbf{a}}_{-}+(\alpha /2){\mathbf{a}}_{+}$. The solution from linearization is thereforeAt release the lower angle vanishes by cancellation between the two normal modes. Their different angular frequencies then destroy that cancellation. Differentiating twice gives ${\ddot{\theta }}_1(0)=-2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)=2g\alpha /\ell$, agreeing with the small-angle limit of the exact double pendulum equations of motion.

The quadratic mechanical energy at release is $E_2=mg\ell {\alpha }^2$. Each normal mode contributes half, because $(4+2\sqrt{2})(2-\sqrt{2})=(4-2\sqrt{2})(2+\sqrt{2})=4$. Equal coordinate amplitudes need not mean equal energies in general; here the modal inertias and squared angular frequencies happen to compensate. Releasing instead along $\mathbf{\theta }(0)=\alpha {\mathbf{a}}_{-}$ or $\alpha {\mathbf{a}}_{+}$, with zero angular velocities, excites just one normal mode.