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