Sure, let's examine the equation provided,
[tex]\[ r = 0.1 \cdot \cos \left( \pi t - \frac{3 \pi}{2} \right) \][/tex]
The equilibrium position of the pendulum is when the angle [tex]\( r \)[/tex] is 0. Therefore, we need to find the values of [tex]\( t \)[/tex] for which [tex]\( r = 0 \)[/tex].
Given:
[tex]\[ 0 = 0.1 \cdot \cos \left( \pi t - \frac{3 \pi}{2} \right) \][/tex]
This means,
[tex]\[ \cos \left( \pi t- \frac{3 \pi}{2} \right) = 0 \][/tex]
The cosine function [tex]\( \cos(\theta) = 0 \)[/tex] when [tex]\( \theta = \frac{\pi}{2} + k\pi \)[/tex], where [tex]\( k \)[/tex] is any integer. So we set the inside of the cosine function to these values:
[tex]\[ \pi t - \frac{3 \pi}{2} = \frac{\pi}{2} + k\pi \][/tex]
[tex]\[ \pi t - \frac{3 \pi}{2} = \frac{\pi}{2} + k\pi \][/tex]
Let's solve for [tex]\( t \)[/tex]:
[tex]\[ \pi t - \frac{3 \pi}{2} = \frac{\pi}{2} + k\pi \][/tex]
[tex]\[ \pi t = \frac{\pi}{2} + \frac{3 \pi}{2} + k\pi \][/tex]
[tex]\[ \pi t = \pi + k\pi \][/tex]
[tex]\[ t = 1 + k \][/tex]
The points where the pendulum is at its equilibrium position occur at:
[tex]\[ t = 1 + k \][/tex]
This can be written as:
[tex]\[ t = 1, 2, 3, 4, \ldots \][/tex]
Thus, the pendulum is at equilibrium position at [tex]\( t = 1, 2, 3, \ldots \)[/tex].
So when we plot these coordinates on a graph [tex]\( (t, r) \)[/tex]:
- At [tex]\( t = 1 \)[/tex], [tex]\( r = 0 \)[/tex]
- At [tex]\( t = 2 \)[/tex], [tex]\( r = 0 \)[/tex]
- At [tex]\( t = 3 \)[/tex], [tex]\( r = 0 \)[/tex]
- And so on...
These points can be illustrated on a graph as points along the [tex]\( t \)[/tex]-axis where [tex]\( r = 0 \)[/tex].
