= Solution
Set $L_i=\widehat L_i+y_i$. The equilibrium equations cancel the constant gravitational and elastic terms, leaving the <small-oscillation mass and stiffness matrices>
$$
\boxed{M\ddot y+Ky=0,\qquad M=\begin{pmatrix}m_1+m_2+m_3&m_2&m_3\\m_2&m_2&0\\m_3&0&m_3\end{pmatrix},\qquad K=kI_3.}
$$
The off-diagonal mass terms arise because rope-length coordinates are not absolute climber displacements. For any nonzero velocity vector, $\dot y^TM\dot y=m_1\dot y_1^2+m_2(\dot y_1+\dot y_2)^2+m_3(\dot y_1+\dot y_3)^2>0$, so $M$ is a <positive-definite matrix>.
Within the vertical Hookean model, the displacement equations are exactly linear while all ropes stay taut. Small amplitudes ensure that none reaches the slack regime $L_i\leq L_0$.
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