# Which normal mode/combination of normal modes actually ensue?

For a classic problem consisting 2 coupled oscillators, I did the usual and found the normal mode frequencies, and constructed a general solution of the form:

$$\begin{bmatrix} x_1(t) \\ x_2(t)\end{bmatrix} = \begin{bmatrix}1\\v_1\end{bmatrix}Aexp(i\omega_1t) + \begin{bmatrix}1\\v_2\end{bmatrix}Bexp(i\omega_2t)$$

My issue is, given that one of the masses is displaced a distance d from equilibrium, what exactly would be the expression for the motion that ensues? It could be given exclusively by the first normal mode, second normal mode, or the combination of them. A single boundary condition (i.e, $$x_1(0)=d$$) does not seem like it is enough to determine expressions for the motion of both masses, does it? Since all three are equally valid (either normal modes, or a combination of them), which one is the one that actually ensues?

EDIT: What I mean is that, since all I am given is that $$x_1(0)=d$$, either could be the solution: $$x_1(t)=d*exp(i\omega_1t)$$

or,

$$x_1(t)=d*exp(i\omega_1t)$$

or,

$$x_1(t)=A*exp(i\omega_1t)+B*exp(i\omega_1t), A+B=d$$

• Would you mind putting an image (sketch) of the system you are trying to understand? Commented Jan 14, 2022 at 17:09
• If the question says, or supposes, that the mass is relaased from rest then $x_1'(0)=0$ gives your second condition Commented Jan 14, 2022 at 17:16