= Solution
At $\alpha=1$, the supplied and physical systems agree. Their squared <angular frequencies> are $(g/l)(2\pm\sqrt2)$. The upper row gives $a_2/a_1=3-\lambda$, so
$$
\boxed{\frac{a_2}{a_1}=1+\sqrt2\ \text{for the slower mode},\qquad
\frac{a_2}{a_1}=1-\sqrt2\ \text{for the faster mode}.}
$$
In the slower <normal mode>, both masses move horizontally in phase and the lower has the larger <amplitude>. In the faster <normal mode>, the lower mass moves in the opposite direction, with magnitude $(\sqrt2-1)|a_1|$. These are horizontal <displacement> ratios. The rod angles instead satisfy $\theta_2/\theta_1=\pm\sqrt2$, because $x_1=l\theta_1$, $x_2=l(\theta_1+\theta_2)$; confusing these ratios changes the sketch.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2022-double-pendulum-modes.png]
{title=Equal-mass normal-mode snapshots with horizontal displacements enlarged for readability; arrow lengths show the relative displacement amplitudes}
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