= 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 $\boldsymbol\theta(0)=(\alpha/2)\mathbf a_-+(\alpha/2)\mathbf a_+$. The solution from <linearization> is therefore
$$
\begin{aligned}
\theta_1(t)&=\frac\alpha2\left[\cos(\omega_-t)+\cos(\omega_+t)\right],\\
\theta_2(t)&=\frac\alpha{\sqrt2}\left[\cos(\omega_-t)-\cos(\omega_+t)\right].
\end{aligned}
$$
At 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 $(4+2\sqrt2)(2-\sqrt2)=(4-2\sqrt2)(2+\sqrt2)=4$. Equal coordinate <amplitudes> need not mean equal <energies> in general; here the modal inertias and squared <angular frequencies> happen to compensate. Releasing instead along $\boldsymbol\theta(0)=\alpha\mathbf a_-$ or $\alpha\mathbf a_+$, with zero <angular velocities>, excites just one <normal mode>.
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