 Solution

Keep the constraint from part (b). For equal masses and spring constants, the generalized eigenvalue problem for small oscillations reduces toThe horizontal coordinate decouples, givingIn this normal mode, mass 1 oscillates horizontally, mass 2 has its horizontal coordinate fixed, and neither mass has a vertical displacement.

For the vertical block, write $\lambda =m{\omega }^2/k$. Its characteristic polynomial is $(2-\lambda )(1-\lambda )-1={\lambda }^2-3\lambda +1$. Thus ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, and the first vertical equation gives $y_2/y_1=2-{\lambda }_{\pm }$. If $\phi =(1+\sqrt{5})/2$ is the golden ratio, the two vertical modes areIn the lower-frequency mode, the masses move vertically in phase, with the lower mass having the larger amplitude. In the higher-frequency mode they move oppositely, with lower-to-upper amplitude ratio $1/\phi$. Their Euclidean inner product is zero because $1+\phi (-1/\phi )=0$; since $M=mI$, this is also mass-matrix orthogonality of normal modes.

The three frequencies satisfy ${\omega }_{-}<{\omega }_x<{\omega }_{+}$. Each mode has a scalar coordinate $A\cos \omega t+B\sin \omega t$, and their superposition gives the general constrained motion. No zero-frequency mode remains because the first spring anchors the system. Gravity fixes the equilibrium extensions but drops out of these frequencies for zero-natural-length springs.