 Solution

For $q=(x_1,{\eta }_1,{\eta }_2{)}^T$, the Euler-Lagrange equations giveIn components these are $m{\ddot{x}}_1=-\mu (19x_1/3-{\eta }_1)$, $m{\ddot{\eta }}_1=-\mu (2{\eta }_1-x_1-{\eta }_2)$ and $m{\ddot{\eta }}_2=-\mu ({\eta }_2-{\eta }_1)$. A normal mode has $q=v\cos (\omega t+\delta )$ and satisfies $Kv=(m{\omega }^2/\mu )v$. For eigenvalue $1/3$, the first and third rows give ${\eta }_1=6x_1$ and ${\eta }_2=3{\eta }_1/2$, and the second row then agrees. HenceThe particle positions are recovered by adding the equilibrium offsets to the last two entries. Their displacements are in phase with amplitude ratio $1:6:9$.