= Solution
Keep the constraint from part (b). For equal masses and spring constants, the <generalized eigenvalue problem for small oscillations> reduces to
$$
\frac km\begin{pmatrix}2&0&0\\0&2&-1\\0&-1&1\end{pmatrix}v=\omega^2v.
$$
The horizontal coordinate decouples, giving
$$
\boxed{\omega_x=\sqrt{\frac{2k}{m}},\qquad v_x=(1,0,0)^T.}
$$
In 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 $\lambda_\pm=(3\pm\sqrt5)/2$, and the first vertical equation gives $y_2/y_1=2-\lambda_\pm$. If $\varphi=(1+\sqrt5)/2$ is the <golden ratio>, the two vertical modes are
$$
\boxed{\begin{aligned}
\omega_-&=\sqrt{\frac{k}{m}\frac{3-\sqrt5}{2}}=\frac1\varphi\sqrt{\frac km},&v_-&=(0,1,\varphi)^T,\\
\omega_+&=\sqrt{\frac{k}{m}\frac{3+\sqrt5}{2}}=\varphi\sqrt{\frac km},&v_+&=(0,1,-1/\varphi)^T.
\end{aligned}}
$$
In 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 $1/\varphi$. Their <Euclidean inner product> is zero because $1+\varphi(-1/\varphi)=0$; 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.
Back to article page
