## Illustration 16.1: Representations of Simple Harmonic Motion

In 1610 Galileo discovered four moons of Jupiter. Each moon seemed to move back and forth in what we would call simple harmonic motion. What was Galileo really seeing? He was seeing the essentially uniform circular motion of each moon, but he was looking at the motion edge-on. We can use what Galileo was experiencing to hand-wave some properties of simple harmonic motion by using an analogy with uniform circular motion. Consider the above animation (position is given in meters and time is given in seconds). Restart.

First, let us look at position as a function of time. The point on the circle marked by the red ball is always at the same radius, R. If we look at the y position as a function of time we see that y = R cos(ωt), and the x position is x = R sin(ωt). How do we know this? We can decompose the radius vector into components.

What about the velocity? Well, we know it is tangent to the path of the ball and, since the motion is uniform, the magnitude of the velocity is constant and equal to ωR. We can break up the velocity vector into components. We get that vy = ωR sin(ωt)  and vx = -ωR cos(ωt) and that both are functions of time. Watch the animation to convince yourself that this decomposition is correct as a function of time. If we know a bit of calculus, we can take the derivative of the position with respect to time. We again get that vy = -ωR sin(ωt) and vx = ωR cos(ωt).

We also know that the acceleration is a constant, v2/R, and points toward the center of the circle. We can again decompose this acceleration as ay = -ω2R cos(ωt) and ax = -ω2R sin(ωt). Again, if we know a bit of calculus, we can take the derivative of the velocity with respect to time. We again find that ay = -ω2R cos(ωt) and ax = -ω2R sin(ωt). Note that since this is simple harmonic motion there must be a relationship between the position and the force. Since force must be a linear restoring force and, since force is also mass times acceleration, we must have that ma = - k x or that a(t) = - (k/m) and a(t) = - ω2 x(t), which is the case if we compare our functions for y(t) and x(t) to ay(t) and ax(t).

For simple harmonic motion we change two things, R → A where A is called the amplitude, and we only consider one direction, in this example the y direction. This yields: y = A cos(ωt), v = -ωA sin(ωt), and a = -ω2A cos(ωt). Simple harmonic motion requires a linear restoring force, an equilibrium position, and a displacement from equilibrium.

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