Question 1a asks us to relate a physical pendulums motion to the magenta trajectories
For each of these, I will provide a GIF that illustrates a physical pendulum and a trajectory in phase space
But Ill answer the question using the phase-space portrait so that you can get familiar with this powerful representation
The first option is that the pendulum never becomes inverted, but this cannot be the case
Notice that each magenta trajectory passes through theta = -pi and pi, which are the unstable fixed points
Because it passes through the unstable fixed points, we know that at some point in time the pendulum has to become inverted, and this answer cannot be correct
Notice also that each magenta trajectory also passes through theta = 0
In fact, its the case that the magenta trajectories have theta values of both even and odd multiples of pi
This means that the pendulum arm regularly passes by both the stable and the unstable fixed points
So the second answer is correct
For part b, we need to characterize the blue trajectories
First notice that, for all the blue trajectories, the angular velocity, or the omega coordinate, is constantly changing
The fact that the angular velocity is not equal to zero for all time means that the pendulum is in fact moving and The pendulum never moves cannot be the answer
Now notice, on the T axis, that each of the blue trajectories is bound between, but not touching, -pi to pi
This means that the blue trajectories never pass through the unstable fixed point
Another way of saying this is the pendulum never becomes inverted
This makes the first answer wrong and the second answer correct
So the answer to this question is The pendulum never becomes inverted
Parts c and d ask us to classify the red and black dots as stable or unstable fixed points
The black dot corresponds to a pendulum in this position
Notice that theta is equal to 0, so the pendulum is in the bottom position, and omega is also equal to 0, so the pendulum is not moving
We know from lecture that this fixed point is stable
In the undamped pendulum, this is called an elliptic fixed point
Well learn much more about this in unit 4
For the red dots, notice that theta is an odd multiple of pi
That means the pendulum is in this position
Also notice that omega is zero, so the pendulum is not moving
We know from lecture and real life that this pendulum is unstable
Even blowing on this pendulum would cause the pendulum to fall over
This type of fixed point is referred to as a hyperbolic fixed point, or saddle point
Well learn much more about this type of fixed point in unit 4
For these reasons, the black dot is a stable fixed point, and the red dots are unstable fixed points
Part e asks how the trajectories in the state space figure would change if damping was introduced
This is equivalent to asking, How would these trajectories change if this was a real-world pendulum? since in the real world we have friction
In this example, all the parameters are the same as the trajectories that were generated in the phase-space portrait, except that I have made the drag coefficient, or the coefficient of friction, equal to 0.5
The left-hand pane shows a pendulum and the right-hand pane will show a phase-space portrait
As we would expect, the pendulum slows down; that is, our angular velocity decreases
So omega should be getting smaller and smaller
As the angular velocity decreases, the angle theta also decreases
Since both the angular velocity and the angular position are decreasing, the pendulum will spiral in to the stable fixed point at (0,0)
Notice that, if adding friction caused the trajectories to spiral towards the unstable fixed points, that would mean if you let a pendulum go at pi/2, it would attract towards the inverted position, rather than the hanging position
We all know that this is not the case
From the video we just presented, we know that the answer is The trajectories would spiral towards the stable fixed points, which lines up with our intuition about real pendulums
With this in mind, if we were to re-generate this figure, but instead setting ß equal to 0.5, we would get the following state-space portrait
This answers question 4: the parameter we have to change is the coefficient of friction, or ß