The analytical solution of an Inverted Pendulum system that behaves as a Harmonic Oscillator is obtained using the following equations:
Ordinary Equation in form of: $$\ddot\theta(t) + a\cdot \theta(t) = 0 \quad(1)$$
where $a \gt 0$, has the solution $$ \theta(t) = A \cos (\omega_n \cdot t - \phi)\quad (2)$$ and $$\omega_n = \sqrt{a}\quad$$
I have the ordinary equation form as: $$0 = \ddot\theta +220.725\theta$$
Giving me $\omega_n = \sqrt{220.725} = 14.86$
Given Information includes:
- Initial Position $\theta$ at t = 0 is $10^{\circ}$
- Initial Angular Velocity $\dot\theta$ at t = 0 is 0.01 $rad\cdot s^1$
- Length of Pendulum Arm is $0.4 m$
- Gravity is assumed as $9.81 m\cdot s^2$
I used this information and equation (2) to find an Amplitude (A) = 10.
I am trying to find the additional information: Natural Frequency, Period and Phase Angle.
I understand simple harmonic motion equations for a simple pendulum but am not sure if these carry over and I can't find the other characteristics. How can these other characteristics be found.
$$\theta(0) = A \cos (-\phi) = 10^\circ =0.174 \text{ rad}$$
$$\dot \theta(0) = -A \omega_n \sin(-\phi)=0.01\text{ rad s}^{-1}$$
Soving these two simultaneous equations gives $$\tan(-\phi) = - \frac{0.01}{0.174\omega_n} \implies \phi = (-3.85 \times 10^{-3}) \text{ rad}$$
Assuming that this pendulum is lightly damped, $\omega_n \approx \omega_0$ so you already have the natural frequency.
The time period $T$
$$T = \frac{2 \pi}{\omega_0} = 0.422 \text{s}$$