= Solution
A <normal mode> is a small-oscillation motion with every coordinate proportional to one common sinusoid, at a fixed amplitude ratio and relative phase. Expanding an arbitrary small motion in <normal mode coordinates> decouples its linear equations into independent oscillators. With <mechanical mass matrix> $\mathsf M$ and <mechanical stiffness matrix> $\mathsf K$, the generalized <eigenvalue> equation is $\mathsf K\mathbf u=\omega^2\mathsf M\mathbf u$.
Horizontal gravitational restoring force on each bob is $-m_i(g/l)x_i$. The two springs add $k(x_2-x_1)$ on the first bob, $k(x_1-2x_2+x_3)$ on the middle bob, and $k(x_2-x_3)$ on the last. Dividing by the bob masses gives the stated equations with
$$
\boxed{\alpha=g/l,\qquad\beta=k.}
$$
<Reflection symmetry> separates an antisymmetric end motion from two symmetric motions. The <three coupled pendulums with equal end masses> have independent modes
$$
\begin{array}{c|c|l}
\text{amplitude vector}&\omega^2&\text{motion}\\\hline
(1,1,1)&g/l&\text{all bobs together; springs unstretched}\\
(1,0,-1)&g/l+k/m&\text{end bobs opposite; middle at rest}\\
(1,-2m/M,1)&g/l+k/m+2k/M&\text{ends together; middle opposite}.
\end{array}
$$
Call their positive frequencies $\omega_0,\omega_a,\omega_s$ respectively. The middle amplitude in the last mode ensures $m x_1+M x_2+m x_3=0$. The modes are orthogonal in the mass-weighted <inner product>, rather than generally in the unweighted coordinate dot product.
The initially stationary displacement decomposes as
$$
(0,0,d)=\frac{md}{M+2m}(1,1,1)-\frac d2(1,0,-1)+\frac{Md}{2(M+2m)}(1,-2m/M,1).
$$
There are no <sine> terms because every initial velocity is zero. Therefore
$$
\boxed{\begin{aligned}
x_1(t)&=\frac{md}{M+2m}\cos\omega_0t-\frac d2\cos\omega_at+\frac{Md}{2(M+2m)}\cos\omega_st,\\
x_2(t)&=\frac{md}{M+2m}(\cos\omega_0t-\cos\omega_st),\\
x_3(t)&=\frac{md}{M+2m}\cos\omega_0t+\frac d2\cos\omega_at+\frac{Md}{2(M+2m)}\cos\omega_st.
\end{aligned}}
$$
These give $(0,0,d)$ at $t=0$ and solve all three coupled equations. The <spring constant> is assumed positive; for $k=0$ the frequencies become degenerate and the same <superposition> reduces to uncoupled pendulum motion.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2011/paper-2-pendulum-modes.png]
{title=Three pendulum modes}
The original example uses $M=2m$ and $k/m=g/l$ to display the mode <vectors> and resulting <superposition>. Those illustrative parameter ratios are not additional supplied data.
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