= Solution
For equal masses, write $M=mB$ with $B=\left(\begin{smallmatrix}3&1&1\\1&1&0\\1&0&1\end{smallmatrix}\right)$. The <generalized eigenvalue problem for small oscillations> is $kv=\omega^2mBv$.
Exchange symmetry of climbers 2 and 3 gives $v_a=(0,1,-1)^T$, for which $Bv_a=v_a$. Hence
$$
\boxed{\omega_a=\sqrt{k/m},\qquad y(t)=A(t)(0,1,-1)^T.}
$$
The physical displacements are $(y_1,y_1+y_2,y_1+y_3)$, so climber 1 is stationary and the other two move vertically with equal amplitudes in opposite directions. Their changes in force on climber 1 cancel.
For the remaining symmetric modes, put $v=(a,b,b)^T$ and let $\mu=k/(m\omega^2)$. The eigenproblem becomes $3a+2b=\mu a$, $a+b=\mu b$. Eliminating $a,b$ gives $(3-\mu)(1-\mu)-2=\mu^2-4\mu+1=0$. Therefore $\mu=2\pm\sqrt3$. Since $(2+\sqrt3)(2-\sqrt3)=1$,
$$
\boxed{\omega_-^2=\frac km(2-\sqrt3),\qquad\omega_+^2=\frac km(2+\sqrt3).}
$$
For clarity, their relative physical climber displacements, rather than rope-length amplitudes, are
$$
q_-\propto\left(1,\frac{1+\sqrt3}{2},\frac{1+\sqrt3}{2}\right),\qquad q_+\propto\left(1,\frac{1-\sqrt3}{2},\frac{1-\sqrt3}{2}\right).
$$
The lower mode moves all climbers in phase; the higher mode moves the lower pair against the top climber. These are the <normal modes of three vertically suspended masses with a branched spring connection>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2-2015-ib-climber-modes.png]
{title=Relative physical climber displacements in the three vertical <normal modes>; branch spacing is schematic}
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