= Solution
With downward position as the vertical coordinate, the full <potential energy> is
$$
\begin{aligned}
V={}&k(a+x-\ell)^2+\frac k2(b+y-x-\ell)^2+\frac k2(c+z-y-\ell)^2\\
&-3mg(a+x)-2mg(a+b+y)-mg(a+b+c+z).
\end{aligned}
$$
The equilibrium tensions are $2k(a-\ell)=6mg$, $k(b-\ell)=3mg$ and $k(c-\ell)=mg$. On expanding, these relations cancel every term linear in the displacements, leaving $V=kx^2+k(x-y)^2/2+k(y-z)^2/2+V_0$. <Newton's second law> therefore gives
$$
\boxed{3m\ddot x=-k(3x-y),\quad2m\ddot y=-k(2y-x-z),\quad m\ddot z=-k(z-y).}
$$
For a <normal mode> $\boldsymbol q=\boldsymbol v e^{i\omega t}$, set $s=m\omega^2/k$. The <generalized eigenvalue problem> is
$$
\left[\begin{pmatrix}3&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}-s\operatorname{diag}(3,2,1)\right]\boldsymbol v=0.
$$
Its determinant is $2(1-s)[3(1-s)^2-2]$, so the <angular frequencies> and mode directions are
$$
\boxed{\begin{array}{c|c}
\omega^2&\boldsymbol v\\\hline
(k/m)(1-\sqrt{2/3})&(1,\sqrt6,3)^T\\
k/m&(1,0,-1)^T\\
(k/m)(1+\sqrt{2/3})&(1,-\sqrt6,3)^T
\end{array}.}
$$
The square roots of these positive values give the requested frequencies. Only the lowest mode has all components of the same sign, so \b[all three particles oscillate in phase at the lowest frequency]. The mode shapes are shown below.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-2-unequal-mass-normal-modes.png]
{title=The three relative displacement patterns of the unequal-mass chain, from lowest to highest frequency}
For release from rest with $Y$ remaining at equilibrium, choose
$$
\boxed{y(0)=0,\qquad z(0)=-x(0).}
$$
This excites only the middle-frequency mode, so $y(t)=0$ for all time. It is a <stationary middle mass in a three-mass spring chain>: the equal and opposite adjacent spring forces cancel on $Y$.
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