Editing Normal mode (section)

=== Coupled oscillators ===
Consider two equal bodies (not affected by gravity), each of [[mass]] {{mvar|m}}, attached to three springs, each with [[spring constant]] {{mvar|k}}. They are attached in the following manner, forming a system that is physically symmetric:

[[File:Coupled Harmonic Oscillator.svg|300px|center]]

where the edge points are fixed and cannot move. Let {{math|''x''{{sub|1}}(''t'')}} denote the horizontal [[displacement (distance)|displacement]] of the left mass, and {{math|''x''{{sub|2}}(''t'')}} denote the displacement of the right mass.

Denoting acceleration (the second [[derivative]] of {{math|''x''(''t'')}} with respect to time) as {{nowrap|<math display=inline>\ddot x</math>,}} the [[equations of motion]] are:

<math display="block">\begin{align}
  m \ddot x_1 &= - k x_1 + k (x_2 - x_1) = - 2 k x_1 + k x_2 \\
  m \ddot x_2 &= - k x_2 + k (x_1 - x_2) = - 2 k x_2 + k x_1
\end{align}</math>

Since we expect oscillatory motion of a normal mode (where {{mvar|ω}} is the same for both masses), we try:

<math display="block">\begin{align}
  x_1(t) &= A_1 e^{i \omega t} \\
  x_2(t) &= A_2 e^{i \omega t}
\end{align}</math>

Substituting these into the equations of motion gives us:

<math display="block">\begin{align}
  -\omega^2 m A_1 e^{i \omega t} &= - 2 k A_1 e^{i \omega t} + k A_2 e^{i \omega t} \\
  -\omega^2 m A_2 e^{i \omega t} &= k A_1 e^{i \omega t} - 2 k A_2 e^{i \omega t}
\end{align}</math>

Omitting the exponential factor (because it is common to all terms) and simplifying yields:

<math display="block">\begin{align}
  (\omega^2 m - 2 k) A_1 + k A_2 &= 0 \\
  k A_1 + (\omega^2 m - 2 k) A_2 &= 0
\end{align}</math>

And in [[matrix (mathematics)|matrix]] representation:

<math display="block">\begin{bmatrix}
  \omega^2 m - 2 k & k \\
                 k & \omega^2 m - 2 k
  \end{bmatrix} \begin{pmatrix}
    A_1 \\
    A_2
  \end{pmatrix} = 0
</math>

If the matrix on the left is invertible, the unique solution is the trivial solution {{math|1=(''A''{{sub|1}}, ''A''{{sub|2}}) = (''x''{{sub|1}}, ''x''{{sub|2}}) = (0, 0)}}. The non trivial solutions are to be found for those values of {{mvar|ω}} whereby the matrix on the left is [[singular matrix|singular]]; i.e. is not invertible. It follows that the [[determinant]] of the matrix must be equal to 0, so:

<math display="block"> (\omega^2 m - 2 k)^2 - k^2 = 0 </math>

Solving for {{mvar|ω}}, the two positive solutions are:

<math display="block">\begin{align}
  \omega_1 &= \sqrt{\frac{k}{m}} \\
  \omega_2 &= \sqrt{\frac{3 k}{m}}
\end{align}</math>

Substituting {{math|''ω''{{sub|1}}}} into the matrix and solving for {{math|(''A''{{sub|1}}, ''A''{{sub|2}})}}, yields {{math|(1, 1)}}. Substituting {{math|''ω''{{sub|2}}}} results in {{math|(1, −1)}}. (These vectors are [[eigenvector]]s, and the frequencies are [[eigenvalue]]s.)

The first normal mode is:
<math display="block">\vec \eta_1 = \begin{pmatrix}
    x^1_1(t) \\ x^1_2(t)
  \end{pmatrix} = c_1 \begin{pmatrix}1 \\ 1\end{pmatrix} \cos{(\omega_1 t + \varphi_1)}
</math>

Which corresponds to both masses moving in the same direction at the same time. This mode is called antisymmetric.

The second normal mode is:

<math display="block">\vec \eta_2 = \begin{pmatrix}
    x^2_1(t) \\ x^2_2(t)
  \end{pmatrix} = c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} \cos{(\omega_2 t + \varphi_2)}
</math>

This corresponds to the masses moving in the opposite directions, while the center of mass remains stationary. This mode is called symmetric.

The general solution is a [[Superposition principle|superposition]] of the '''normal modes''' where {{math|''c''{{sub|1}}}}, {{math|''c''{{sub|2}}}}, {{math|''φ''{{sub|1}}}}, and {{math|''φ''{{sub|2}}}} are determined by the [[initial condition]]s of the problem.

The process demonstrated here can be generalized and formulated using the formalism of [[Lagrangian mechanics]] or [[Hamiltonian mechanics]].