= Solution
The product-to-sum identity gives $\sin\Omega t\sin\omega t=[\cos(\Omega-\omega)t-\cos(\Omega+\omega)t]/2$. Thus the first <mass> must have equal and opposite cosine-mode amplitudes and no sine-mode amplitudes. Equivalently, release the second <mass> from a <displacement> $X$ while the first is at equilibrium, both initially at rest:
$$
\boxed{x_1(0)=0,\quad x_2(0)=X,\quad\dot x_1(0)=\dot x_2(0)=0.}
$$
The two <normal mode> coefficients in $x_1$ are $qX/(2D)$ and $-qX/(2D)$. Therefore, with
$$
\boxed{\Omega=\frac{\omega_++\omega_-}{2},\qquad\omega=\frac{\omega_+-\omega_-}{2},}
$$
the two <displacements> are
$$
\boxed{x_1(t)=\frac{qX}{D}\sin\Omega t\sin\omega t,\qquad
x_2(t)=X\left[\cos\Omega t\cos\omega t+\frac{\kappa}{2D}\sin\Omega t\sin\omega t\right].}
$$
These initial conditions are necessary for a nontrivial product with exactly the stated time origin, as well as sufficient: $x_1(0)=\dot x_1(0)=0$ follows from the product, and its zero third derivative at zero, together with the first coupled equation, forces $\dot x_2(0)=0$. Since both <normal frequencies> are positive, $\Omega>\omega>0$. This is the product form used to describe <beating>, although a well-separated slow envelope requires the two <normal frequencies> to be close. No such extra near-degeneracy is assumed in the exact formula.
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