Normal Modes of Coupled Oscillation

Consider the coupled oscillations problem of two equal masses $m$

connected by springs of constant $k$

(see Figure).

**Figure 6.5:** Mass-Spring system.

The differential equation provided by each mass is written using Newton's law of motion as follows

$\displaystyle k(x_2-x_1)-kx_1$	$\displaystyle =$	$\displaystyle m \frac{d^2x_1}{dt^2}$
$\displaystyle -kx_2-k(x_2-x_1)$	$\displaystyle =$	$\displaystyle m \frac{d^2x_2}{dt^2}$

In this system, the frequencies ( $\omega$ ) that both masses oscillate as in common are called normal oscillation modes.

To find the normal mode frequencies, try a solution for both unknowns as:

$\displaystyle x_1=x_{10}cos \omega t ~~~~\&~~~~x_2=x_{20}cos \omega t$

Substitute these solutions into the system of linear equations above and simplify by removing $cos \omega t$ ,

$\displaystyle \frac{2k}{m}x_{10}-\frac{k}{m}x_{20}$	$\displaystyle =$	$\displaystyle \omega^2 x_{10}$
$\displaystyle -\frac{k}{m}x_{10}+\frac{2k}{m}x_{20}$	$\displaystyle =$	$\displaystyle \omega^2 x_{20}$

This system of equations can be written as the product of a matrix and a column vector as follows (replace $\omega^2$ by $\lambda$ ):

$\begin{indisplay} \left[ \begin{array}{rr} 2k/m & -k/m \\ -k/m & 2k/m \\ \en... ...ft[ \begin{array}{r} x_{10} \\ x_{20} \\ \end{array} \right] \end{indisplay}$

This structure can also be written as:

$\begin{indisplay} \left[ \begin{array}{rr} 2k/m-\lambda & -k/m \\ -k/m & 2k/m... ... \begin{array}{r} x_{10} \\ x_{20} \\ \end{array} \right] =0 \end{indisplay}$

The determinant must be zero for this linear system of equations to have a unique solution:

$\begin{indisplay} det\left\vert \begin{array}{rr} 2k/m-\lambda & -k/m \\ -k/m & 2k/m -\lambda \\ \end{array} \right\vert=0 \end{indisplay}$

$\displaystyle \longrightarrow (2k/m-\lambda)^2-k^2/m^2=0$

There are two oscillation frequencies ( $\omega$ ) and their corresponding amplitudes $\vec{x_0}=(x_{10}, x_{20})$ :

$\displaystyle \lambda_1=k/m \longrightarrow \vec{x_{0,1}}=\left( \begin{array}{r} 0.71 \\ 0.71 \\ \end{array}\right)$

$\displaystyle \lambda_2=3k/m \longrightarrow \vec{x_{0,2}}=\left( \begin{array}{r} -0.71 \\ 0.71 \\ \end{array}\right)$

The first of these solutions represents the mode in which the two masses oscillate in the same phase ( $--> --> $

)

and the second represents the mode in which they oscillate in the opposite phase ( $--> <-- $

Phyton Code:

    1 print("****************SymPy Solution for Characteristic Equation: ")
    2 from sympy import Matrix, symbols, pprint, factor
    3 M = Matrix([[2, -1], [-1, 2]])
    4 lamda = symbols('lamda')
    5 poly = M.charpoly(lamda) # Get the characteristic polynomial
    6 print(poly) # Printing polynomial
    7 pprint(factor(poly.as_expr())) # Prints expr in pretty form.
    8 print("****************NumPy Solution for Characteristic Equation: ")
    9 import numpy as np
   10 A = np.array([[2, -1], [-1, 2]])
   11 print(np.poly(A))
   12 print("****************NumPy Solution for Eigenvalues and Eigenvectors: ")
   13 w,v=np.linalg.eig(A)
   14 print('Eigenvalue:', w)
   15 print('Eigenvector1:', v[0])
   16 print('Eigenvector2:', v[1])
   17 print("****************SciPy Solution for Eigenvalues and Eigenvectors: ")
   18 import scipy.linalg as la
   19 w,v = la.eig(A)
   20 print('Eigenvalue:', w)
   21 print('Eigenvector1:', v[0])
   22 print('Eigenvector2:', v[1])