vault backup: 2024-09-23 13:26:12
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@ -50,3 +50,23 @@ These things relate directly to the eigenvectors and eigenvectors of the system.
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## Eigenvalues and Eigenvectors
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$z = \beta^2 - 4 \omega^2$
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Solving for lambda ($\det[A - I\lambda]$) will lead us towards the same expression for z. This is what's under the square root.
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# Nonlinear Things (Finally!)
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## Nonlinear Pendulum
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### Undamped
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$$ \ddot{\theta} = -\frac{g}{l} \sin(\theta) $$
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$\dot \theta = \zeta = P(\theta, \zeta)$
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$$ {\bf J} =
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\left[ \matrix{ \frac{\partial P}{\partial \theta} & \frac{\partial P}{\partial \zeta} \\ \frac{\partial Q}{\partial \theta} & \frac{\partial Q}{\partial \zeta}} \right] =
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\left[ \matrix{ 0 & 1\\ -\omega^2 \sin(\theta) & 0 } \right]
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$$
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What ar the equilibrium points?
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$$\left[ \matrix{ \dot \theta \\ \dot \zeta} \right ] = \left[ \matrix{ \zeta \\ -\omega^2 \sin(\theta) } \right] $$
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This system is in equilibrium when $\dot \theta = \dot \zeta = 0$, which is when $\zeta = \sin(\theta) = 0$ .
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For $\bf J$:
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- $\tau = 0$
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- $\Delta = \omega^2 \cos(\theta)$
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Then:
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- $\theta$ is 0, $\Delta = \omega^2 >0$, center, marginally stable
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- $\theta = n \pi$, $\Delta = - \omega^2 <0$, saddle. Unstable
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What does the phase plane look like?
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