 Solution

At $\alpha =1$, the supplied and physical systems agree. Their squared angular frequencies are $(g/l)(2\pm \sqrt{2})$. The upper row gives $a_2/a_1=3-\lambda$, soIn 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 $(\sqrt{2}-1)|a_1|$. These are horizontal displacement ratios. The rod angles instead satisfy ${\theta }_2/{\theta }_1=\pm \sqrt{2}$, because $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$; confusing these ratios changes the sketch.