Difference between Symmetric and Phase difference? For a simple coupled oscillator system such as the one here, with equal spring constants and equal masses (with a displacement from equilibrium of $x_1$ and $x_2$), it follows that:
$(\ddot{x}_1+\ddot{x}_2)\propto x_1+x_2$
$(\ddot{x}_1-\ddot{x}_2)\propto x_1-x_2$
Wikipedia refers to this as the 'normal modes' of the system and the former as antisymmetric and the latter as symmetric. Does this imply that these proportionalities demonstrate the masses are in phase and in antiphase?
(Note for answering: this is my first attempt at understanding coupled oscillators; I am not familiar with modes in a generic sense either. If it helps, I visualised these equations as the dot-products of the displacements and acceleration vectors of $x_1$ and $x_2$ with the eigenvectors generated with their coupled differential equations, as a way of projecting the transformation of $(x_1,x_2)$ to $(x_1,x_2)''$ onto components along lines of invariance [if that makes sense])
 A: IMHO, the best approach to mass-spring systems is via matrix formulation,
$\underline{\underline{M}}\underline{\ddot{x}} + \underline{\underline{K}}\underline{x} = \underline{F} $,
where the vector $\underline{x}(t)$ collects the degrees of freedom of the problem, and finding the eigensolution of the homogenous system,
$\left[s^2  \underline{\underline{M}} + \underline{\underline{K}} \right]\underline{\hat{x}} = \underline{0} $.
In your example,
$\underline{\underline{M}} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}$
$\underline{\underline{K}} = \begin{bmatrix} 2 k & -k \\ -k & 2 k \end{bmatrix}$
and the determinant of the matrix $s^2\underline{\underline{M}} +\underline{\underline{K}} $ reads
$\det (s^2\underline{\underline{M}} +\underline{\underline{K}} ) = (s^2 m + 2k)^2 - k^2 = m^2 s^4 + 4 km s^2 + 3 k^2$
the eigenvalues reads
$s_{1,2}^2 = - \left[ 2 \pm 
1 \right]\dfrac{k}{m} $
and thus
$s_1^2 = - \dfrac{k}{m} $$\quad \rightarrow \quad$$s_{1\pm} = \pm j \sqrt{\dfrac{k}{m}}$
$s_2^2 = - 3 \dfrac{k}{m} $$\quad \rightarrow \quad$$s_{2\pm} = \pm j \sqrt{3\dfrac{k}{m}}$,
while the corresponding eigenvectors are:
$\underline{\hat{x}}_{1} =\begin{bmatrix} 1 \\ 1 \end{bmatrix}$,$\quad$$\underline{\hat{x}}_{2} =\begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
The first eigensolution is the mode with the two masses oscillating "in phase" with the same amplitude and pulsation $\omega_1 = \sqrt{\frac{k}{m}}$.
The second eigensolution is the mode with the two masses oscillating "in anti-phase" with the same amplitude and pulsation $\omega_2 = \sqrt{3\frac{k}{m}}$.
We can use these eigensolutions to build the general solution of the homogeneous system as
$\underline{x}(t) = A_1 \underline{\hat{x}}_1 e^{j( \omega_1 t +\phi_1)} + A_2 \underline{\hat{x}}_2 e^{j( \omega_2 t+\phi_2)}$
$x_1(t) = A_1 \hat{x}_{11} e^{j( \omega_1 t +\phi_1)} + A_2 \hat{x}_{21} e^{j( \omega_2 t+\phi_2)}$
$x_2(t) =  A_1 \hat{x}_{12} e^{j( \omega_1 t +\phi_1)} + A_2 \hat{x}_{22} e^{j( \omega_2 t+\phi_2)}$
