The Physics Philes, lesson 79: That’s So Derivative
Last week I said that this week I would write about the period and amplitude in simple harmonic motion. Well, I changed my mind. We’re doing displacement, velocity, and acceleration in simple harmonic motion, instead. (What? I don’t see you writing a weekly post about physics.)
Anyway, let’s begin.
We’ve learned how to find the frequency and period of SHM. Now we move on to the displacement x. The acceleration of body in SHM along the x-axis is the same as the acceleration of the reference point in uniform circular motion. The angular speed in uniform circular motion is the square root of the force constant divided by the mass. So the displacement x should be equal to the radius/amplitude A times the cosine of the angle created by the phasor and the x-axis. So, if the phasor makes an angle φ with the x-axis at time = 0, at any time after that the angle will be θ = ωt + φ. So we can get the displacement in simple harmonic motion by the equation:
The displacement of the body in SHM is a periodic function of time. The position of the body in SHM is a periodic, sinusoidal function of time. If you think back to your trig, you know that the cosine function oscillates between -1 and 1. So the displacement in our equation will be between -A and A.
The graph of cosine is basically never ending. The graph repeats every time the angle increases by 2Π. So the period of the function occurs every 2Π radians. So if we start at t=0, the time T it takes to complete one cycle is:
Notice that the period has nothing to do with the amplitude. It’s dependent on the force constant and the mass.
The angle φ is a constant, and it’s called the phase angle. It tells us the point in the cycle the motion was at at time 0, or the initial position.
If φ = 0, then cos φ = 1. So the equation above would equal A. The body starts at the maximum positive displacement. The converse is true if φ = Π. Then the equation above would equal to -A, the maximum negative displacement.
To find the velocity and acceleration, we need to take derivatives. We talked about derivatives a few weeks ago, but let me take a moment to try to explain what a derivative is with a graph.
Let’s say we have this function. It’s increasing.
We can figure out how quickly it’s increasing on average by picking a couple of points and finding the slope between them. But how do we find the rate of change at a particular point? In that situation, we need a derivative.
It sounds scary, but it’s actually kind of OK. The derivative is basically just the slope at a particular point. How can this be? A point doesn’t have a slope. It’s just…there. It’s actually the slope of the tangent line, which is a line that just grazes the line, like this:
The derivative tells us the slope of that line, which in turn tells us the rate of change at that particular spot. If this graph is a graph of distance with respect to time, that tangent line is the velocity at that particular moment. You can also take the derivative of the derivative (or take the second derivative). In the distance with respect to time situation, the second derivative would represent the acceleration at a particular point.
There are a million rules and shortcuts you can use to find derivatives of functions, which I don’t have time to go into right now. For the time being, just know that the derivative of x = Acos(ωt + φ) (aka the velocity) is:
The second derivative (aka the acceleration) is:
(All that dx/dt stuff is just Leibnitz notation for derivative stuff. Don’t worry about it too much.)
Wow! So now you basically know everything there is to know about the displacement, velocity, and acceleration of an object in SHM. Ha! No, you probably don’t. But I’m done for the day. Don’t worry, though. I’ll be back next week with more oscillating awesomeness.
Featured image credit: Alan Levine via Flickr