# The Physics Philes, lesson 78: Set Phasors to Physics Phun

Oh hello. I see you’ve swung back to read another Physics Philes. (Get it? It’s a periodic motion joke. Like a pendulum. Ha!)

We’re continuing our discussion of simple harmonic motion by comparing it something we’ve already discussed in detail: uniform circular motion. But first we need to talk about how we think about displacement in SHM. We need to think about the displacement in SHM as a function of time.

Listen, there was a time in my life when that phrase – a function of time – would have left me paralyzed with fear. Maybe that’s you, too. Don’t worry. I’m going to try to make this as painless as possible.

OK, so we need to think about displacement as a function of time. But why? Because the acceleration of the oscillating body is constantly changing. The way we keep track of that constantly changing acceleration is through calculus, specifically by finding the second derivative of the function that expresses how fast and in what direction the oscillating body is moving, aka the acceleration.

Don’t worry about knowing how to find the second derivative. That’s not terribly important right now. Just know that, in this case, it has to be equal to the negative of the force constant divided by the mass of the object, all times the value of the function itself.

This is all very abstract. Let’s think about this in a little more concrete way. Let’s think about uniform circular motion.

You know what a record player is right? With a turntable and stuff? It’s what old people and hipsters use to listen to something old people and hipsters call “vinyl.” Anyway, let’s pretend we put a little Army man or something on the turntable and turn it on so the Army man spins around. Now, pretend that we shin a light on the turntable from the side so that the Army man casts a shadow on a wall. Kind of like this:

Hm. I’m not sure how helpful that is. But maybe?

Anyway, as the Army man moves around the turntable on a disc in uniform circular motion, it will cast a shadow on the wall. But the shadow won’t spin in circles; it will bounce back and forth along the wall. Kind of like the oscillation of an ideal spring! More than that, actually. The motion of the shadow is exactly like the oscillation of an ideal spring if the amplitude of the oscillation is equal to the radius of the disc and if the angular frequency of the oscillating body is equal to the angular speed of the disc.

Kind of mind-blowing right? It’s one of those things that, now that I see it, I can’t unsee it. But in a good way.

Maybe the acceleration of the Army man shadow across the wall can tell us something about the acceleration of a body undergoing simple harmonic motion. First we need to define some terms. The disc on which the Army man stands is called the reference circle, and the Army man represents what we’ll call the reference point. (I’ll also refer to the reference point as point Q because math.) If we consider the center of the reference circle to be the origin O, then the line OQ makes and angle θ with the x-axis. As the Army man rotates with a constant angular speed, the line OQ rotates with the same angular speed. This is called the phasor.

If A is the radius of the reference circle, the x-component of the phasor is x=Acosθ. This is the x-coordinate of the shadow, which is a projection of point Q on the x-axis. In addition, the x-velocity of the shadow is equal to the velocity vector of point Q and the x-acceleration of the shadow is the x-component of the acceleration vector of point Q. The magnitude of the acceleration vector of Q (which, because it is in uniform circular motion, is always pointed toward the origin O) is constant and equal to the square of the angular speed times the radius of the disc:

The x-component of the acceleration vector of point Q is

If we combine these last two equations, we find that the acceleration of the shadow is

The acceleration of the shadow is directly proportional to the displacement x and has the opposite sign. This is what we’d expect from SHM. What’s more, this equation is the same as

if the angular speed of point Q is related to the force constant and mass of the oscillating body by

or

When a body starts oscillating, the angular speed is determined by the force constant and the mass.

The angular speed of the Army man and the frequency of the oscillating shadow are the same. Let’s say that the Army man makes one complete revolution in time T, then the shadow goes through one complete oscillation cycle in that same time. That time T is the period of oscillation. The Army man will move an entire circle – 2Π radians – the angular speed is 2Π/T. Which, incidentally, is the same as the angular frequency for the oscillating shadow. Now we can express the equations for frequency and period as:

and

The larger the mass of the oscillating body, the greater the inertia, the less acceleration, the more slowly it moves, and the longer it takes to complete a cycle. Or the stiffer the spring, the greater the force constant, the greater the force, the greater the acceleration, the greater the speed, and the shorter time it takes to complete a cycle.

This was a complicated topic this week. Hopefully I didn’t mess it up too badly. But if I did, please let me know in the comments.

Next week the journey continues with a look at the period and amplitude of a body in simple harmonic motion!

*Featured image credit: Devon Hollahan via Flickr*

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