Relation with circular motion

If you've been keeping your eyes open, you might have noticed a similarity between the lingo used here and that for rotational motion. This quantity $\omega$ the angular frequency, looks quite similar to the angular velocity.

Is this just a choice designed to maliciously mislead you, or a random choice, or a helpful hint? Well I'll let you decide. Read on!

A way of seeing the relation between two dimensional motion and simple harmonic motion can be made by rotating a wheel with a bit protruding from the rim, and looking at the shadow of this on a screen.

By defining the angle $\theta$ in the usual way:

we see that

$$\theta ~=~ \omega t +\theta_0$$ (1.16)

But let's call $\theta_0 \equiv \delta$ (can you guess why?). So if the radius of the wheel is $r$, then

$$x ~=~ r cos\theta ~=~ r\cos(\omega t + \delta)$$ (1.17)

Let's call $A \equiv r$,

$$x ~=~ A\cos(\omega t + \delta)$$ (1.18)

Great! This is just what we guessed earlier, so you see that some how it's related to two dimensional motion. Let's try to figure out why.

Let's ``two dimensionalize" the mass on the spring! Eqn. 1.1 can be written in terms of vectors.

$${\bf a}~=~ - {k\over m}{\bf r}$$ (1.19)

Here ${\bf r}$ is the displacement vector and ${\bf a}$ is the acceleration. How do we interpret this? Instead of a spring going back and forth along a line, imagine a bungey cord with zero equilibrium length. You can describe the force that it exerts on you when you pull it, but saying ${\bf F}~=~ -k{\bf r}$. So now imagine you have a bungey cord with one end tethered to the middle of a frictionless table. It's just like the one dimensional case, except now you can pull it in any direction. What happens if you start it off doing a circular ``orbit"? Well we can figure out what happens just like we did for gravitational orbits. The acceleration is $-\omega^2 {\bf r}$ so applying ${\bf F}~=~ m{\bf a}$

$$-k{\bf r}~=~ m\omega^2 {\bf r}$$ (1.20)

Solving for $\omega$

$$\omega^2 ~=~ {k\over m}$$ (1.21)

Hmm, this looks just like what we found in the one dimensional case. So two dimenionalizing the problem didn't change this. Why not? Let's think about the problem now in component form. The x component of eqn. 1.19 is

$$a_x ~=~ {k\over m} x$$ (1.22)

This is the same as eqn.1.1, the one dimensional problem! So it's not surprising that the results should be the same, if it's the same equation. Right?

Well if we go further, we can derive what the x-component actually is. We just did this with the shadow and wheel example. We saw that this gave the right general form for the solution, but from above we just got the correct value of $\omega$. This time we derived the whole thing, with a bit of dimensional trickery.