These plots show the amplitude of two coupled pendula, connected by a spring. The oscillators may be "released" at different intial positions, but we assume zero initial velocity. The general motion of the coupled oscillators will be a linear combination of normal mode motions—thus, we show the normal mode motions and the combined motion.
Consider the following questions and possible actions:
The solutions for the two normal modes of coupled pendula are:
$$ \textbf{x}^{(1)}(t) = C_{1} \cos(\omega_{1}t + \phi_{1}) \textbf{A}^{(1)} $$ $$ \textbf{x}^{(2)}(t) = C_{2} \cos(\omega_{2}t + \phi_{2}) \textbf{A}^{(2)} $$where \(\omega_{1}\) and \(\omega_{2}\) are the normal mode frequencies (eigenvalues), and \(\textbf{A}^{(1)}\) and \(\textbf{A}^{(2)}\) are the normal mode amplitudes (eigenvectors).
The general solution is a linear combination of the normal mode solutions, and can be expressed as:
$$ \textbf{x}(t) = D_{1} \textbf{x}^{(1)}(t) + D_{2} \textbf{x}^{(2)}(t) $$Furthermore, by redefining constants such that \(\alpha=D_{1}C_{1}\) and \(\beta=D_{2}C_{2}\), we can express the general solution as:
$$ \textbf{x}(t) = \alpha \cos(\omega_{1}t + \phi_{1}) \textbf{A}^{(1)} + \beta \cos(\omega_{2}t + \phi_{2}) \textbf{A}^{(2)} $$Thus far, we have expressed everything very generally. For the specific example of coupled pendula (in this visualization), the eigenvectors are:
$$ \textbf{A}^{(1)} = \begin{pmatrix} 1\\ 1 \end{pmatrix} $$ $$ \textbf{A}^{(2)} = \begin{pmatrix} 1\\ -1 \end{pmatrix} $$The motion of each pendulum mass is the following:
$$ x_{1}(t) = \alpha \cos(\omega_{1}t + \phi_{1}) + \beta \cos(\omega_{2}t + \phi_{2}) $$ $$ x_{2}(t) = \alpha \cos(\omega_{1}t + \phi_{1}) - \beta \cos(\omega_{2}t + \phi_{2}) $$Copyright 2023 MIT All Rights Reserved.