Assume an infinite string of masses $m$ connected by springs with constant $\kappa$. The masses in equilibrium are evenly spaced by $a$. Then the equation of motion for the $j$-th mass is $$ \ddot{\psi}_j=\Omega^2(\psi_{j+1}-2\psi_j+\psi_{j-1}) $$ where $\Omega^2 = \kappa/m$. It is well known that the normal modes are $$ \psi_j^k=A_ke^{i(kaj-\omega t)},\ \ \omega^2=4\Omega ^2\sin^2\left(\frac{ka}{2}\right) $$

Now assume I have an initial condition $$ \psi_j(0)=B_j,\ \ \dot{\psi}_j(0)=0 $$ How do I find the evolution of $\psi_j(t)$? I have a continuum of available $k$ numbers in the range $-\pi/a<k<\pi/a$. In a continuous infinite rope, the solution would be given by the Fourier transform. What is the analog here? Is it something like

$$ \psi_j(t) = \sum\limits_k A_ke^{i(kaj-\omega(k) t)} $$ Or some sort of integral?