Symmetry and Finite Coupled Oscillators

[Ibraheem]

For an infinite system of coupled oscillators of identical mass and spring constant k. The matrix equation of motion is $\ddot{X}=M^{-1}KX$

The eigenvectors of the solutions are those of the translation operator (since the translation operator and $M^{-1}K$ commute). My question is, for the case of a large BUT finite number of coupled oscillators, does $M^{-1}K$ still commute with the translation operator? and if not, is there a way to find the eigenvectors of the solutions, besides directly finding them by diagonalizing?

 

[jambaugh]

To your first question, "Yes, of course" since you specify "identical" with respect to masses and spring constants, thus there's no positional dependency. Does that also resolve your second question?