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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])

yolo
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1 Answers1

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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

$\underline{x}(t) = A_1 \underline{\hat{x}}_1 e^{j( \omega_1 t +\phi_1)} + A_2 \underline{\hat{x}}_2 e^{j( \omega_2 t+\phi_2)}$

$x_1(t) = A_1 \hat{x}_{11} e^{j( \omega_1 t +\phi_1)} + A_2 \hat{x}_{21} e^{j( \omega_2 t+\phi_2)}$
$x_2(t) = A_1 \hat{x}_{12} e^{j( \omega_1 t +\phi_1)} + A_2 \hat{x}_{22} e^{j( \omega_2 t+\phi_2)}$

basics
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  • I am aware of all this. The OP asks how the normal modes being (anti)-symmetric being in (anti)-phase – yolo Nov 03 '22 at 22:14
  • Is it not clear enough? – basics Nov 03 '22 at 22:15
  • It answers the question upfront. However it doesn't explain why, by definition, phase and symmetry imply the same thing – yolo Nov 03 '22 at 22:27
  • I don't understand "phase and symmetry imply the same thing". What do you mean? – basics Nov 03 '22 at 22:29
  • You can interpret "in-phase oscillations" taking the same positive direction for the displacement of both the points, as you can see in the low-frequency mode with amplitudes $[1, 1]$. On the other hand, you can interpret "anti-phase oscillations" taking the same positive direction for the displacement of both the points, as you can see in the low-frequency mode with amplitudes $[1, -1]$. – basics Nov 03 '22 at 22:36
  • I mean to say that, both normal modes satisfy both the definition (of which I'm not 100% certain of both) of having (anti)symmetry and being in (anti)phase – yolo Nov 03 '22 at 22:41
  • But in general, modes of mass-spring systems without damping have real eigenvectors and complex eigenvalues, and represent solutions that oscillate in phase, i.e. the dofs reaches the maximum amolitude of the oscillations at the same time. Then, we can interpret them to be in phase or in anti-phase depending on the convention that defines the positive directions of the dofs – basics Nov 03 '22 at 22:43
  • in general, mass-spring systems may represent lots of mechanical systems, and the dofs may represent different variables, like displacement of nodes, rotation of nodes,...once you choose the positive direction of these dofs, than you can say that two dofs are in phase if they have components with the same sign in the eigenvector, in antiphase if they have components of with opposite signs in the eigenvector. But for undamped systems, there is no other phase beside $0$ or $\pi$ – basics Nov 03 '22 at 22:49
  • What do you mean by dofs? – yolo Nov 03 '22 at 22:56
  • degrees of freedom. Are you fine with the answer? If you have any other doubts about mass-spring systems as you go into their study, I'll try to give an answer to those doubts – basics Nov 03 '22 at 22:56
  • I feel you should add that to your answer to make it clearer – yolo Nov 03 '22 at 22:58