
Title:
0754 The Simple Pendulum

Description:

Okay, so Huygens wanted to look at a pendulum to measure time.

Let's analyze what's going on exactly with a pendulum

Well, a pendulum is just a mass on a rod.

When we pull it back, like we have here, away from the equilibrium position,

which I've drawn in dotted lines, it's going to swing back and forth.

If there's no friction or air resistance, it'll just do that forever.

We expect this pendulum to give us simple harmonic motion,

which means we expect that when we work out the forces in the acceleration

we want to get acceleration that looks like minus something times x.

Let's look at the forces and see if we can get that,

because we want to understand what's going on here.

Let's imagine pulling our pendulum by some angle θ.

Well, of course, our pendulum has some mass.

I said simple pendulum so that what meansI should explain

is that all of the mass is in the bob at the end,

which is just some really tiny little point mass at the end,

and there's no mass in the rod, and everything is going to swing without friction.

The pendulum has mass or weight pulling down, and there's only one other force.

That's the force of tension supplied by the string or rod.

I know that when I release this pendulum, if I've pulled it back here and I release it,

it's going to swing this waydown back towards the equilibrium.

Actually, what we want to do is break this mg into a component this way

that will balance the tension and a component in this direction and a component in this direction.

This picture is starting to get a little cluttered, so let me pull it off to the side and make it a little bigger.

I can see that this angle is the same as this angle over here.

If that's the case, well, I can do my trig.

This is a right triangle with a hypotenuse,

and there is this side, which is mgcosθ.

This side is pretty uninteresting, because I know it's going to exactly balance tension.

How do I know that?

Well, because the mass never accelerates towards the pivot point or away.

It only accelerates perpendicular directly back towards the equilibrium point.

That force that's doing this acceleration is mg times sin θ.

Okay, this is the force.

Actually, I should say that it's negative,

because if I pull it back to the left I would normally call that a negative angle.

The force is to the right, so this negative sign is actually important.

So, we have this force doing our acceleration minus mgsinθ.

Let's use our good old friend F = ma.

In this case mgsinθ = ma.

When we check it out, the m's cancel,

and we're left with this.

The acceleration is equal to g times sinθ.

We're in a big of a pickle here becauseuh, ohthis does not look like a = 1 something x.

Have no fear. We're going to do some sly massaging of this equation

to figure out what's going on.