As this wheel turns at a constant speed, the shadow of the peg sweeps back and forth on the surface below. Notice that the spot moves fastest near the center of its sweep, while at the ends of its sweep, it slows to a stop and changes direction. The center is the spot's equilibrium position. The motion of the shadow is an example of simple harmonic motion. The spot gains speed fastest as it begins its trip from a side. In other words, its acceleration is greatest at the extremes. Its speed is fastest at the center of the trip from one end to another, but at this point its acceleration has dropped to zero. Simple harmonic motion occurs when the motion's acceleration is directed toward the equilibrium position and when the acceleration is stronger with increased distance from the equilibrium position. In simple harmonic motion, acceleration is proportional to distance from the equilibrium point. We will find out more about this motion later, but first let's derive an equation that relates the position of the spot on the x axis to the peg's angular velocity, omega, as it moves on the rotating wheel, and the time, t. As the diagram illustrates, cosine theta equals x divided by A. Thus, x equals A times cosine theta. As we discovered in the lesson on circular motion, the angular displacement, theta, equals the angular velocity, omega, times the time, t. This is a form of the expression displacement equals velocity times time. Now we can write the equation as X equals A times cosine omega t. A is the amplitude. It represents the greatest distance on either side of equilibrium that the spot reaches. Suppose the spot in a previously described apparatus moves with an amplitude of 5.0 meters, and the peg moves around the center at a constant rate of 1.5 radians per second. Where is the spot 2.5 seconds after it leaves the extreme positive point on the x axis? ... Displacement to the left of the equilibrium position is a negative term. Let's take a closer look at the motion we described earlier. The time that it takes for the peg to make a complete circle and for the shadow to complete one cycle is called the period, and is symbolized by the variable T. The period has units of seconds. Frequency, f, is the number of cycles per unit time, which is the inverse of period. The SI unit of frequency is the hertz, where one hertz equals one cycle per second. Remember from our study of kinematics that distance equals velocity times time. Here, the analogous expression is circumference, C, equals velocity times time, where the velocity of the light as it moves around the circle is v equals omega r and C equals 2 pi r. The time, t, is one period. T can be written as t equals two pi divided by omega. In one complete cycle, the peg moves a total of two pi radians in time T. Thus, if the peg moves one-fourth of a cycle, it moves one-fourth of two pi or pi over two radians which is equivalent to 90 degrees. Remember, theta is the angle in radians, measured counterclockwise around the circle from the positive x axis. One quarter of a circle is covered in one quarter of the time. It makes sense. The time, (t) in seconds, corresponds to the position, pi over two in radians. Where would the peg be if it traveled around the circle for half the total period, one-half t? Answer: One-half t would be the time required to travel around one-half a circle, 180o or p radians. Take a moment to figure out how many radians around the circle you would travel in three-fourths T and in T. In three fourths of the time, three-quarters T, the spot will move three pi over two radians. The full circle is completed in one full period, T, and in that time the peg moves two pi radians.

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