**Problem1 :**

An object in circular motion has an easily defined period, frequency and angular velocity. Can circular motion be considered an oscillation?

Though circular motion has many similarities to oscillations, it can not truly be considered an oscillation. Though we can see circular motion as moving back and forth, in a sense, when we examine the forces involved in circular motion, we see that they do not meet the requirements of oscillations. Recall that in an oscillating system a force must always act to restore an object to an equilibrium point. In circular motion, however, the force always acts perpendicular to the motion of the particle, and does not act against the displacement from a particular point. Thus circular motion cannot be considered an oscillating system.

**Problem2 :**

What is the equilibrium point of a ball bouncing up and down elastically on a floor?

Though this type of oscillation is not a traditional one, we can still find its equilibrium point. Again, we use our principle that in an oscillating system the force always acts to restore the object to its equilibrium point. Clearly when the ball is in the air the force always points towards the ground. When it does hit the ground, the ball compresses, and the elasticity of the ball produces a force on the ball that causes it to rebound into the air. However, the instant the ball hits the ground, there is no deformation of the ball, and the normal force and the gravitational force cancel exactly, producing no net force on the ball. This point, the instant the ball hits the ground must be the equilibrium point of the system. Shown below is a diagram of the ball at equilibrium, and displaced in both directions from the equilibrium point:

a) The ball at equilibrium b) the ball in the air, with net downward force c) the ball deformed, with net upward force

**Problem3 :**

A mass on a spring completes one oscillation, of total length 2 meters, in 5 seconds. What is the frequency of oscillation?

The only piece of information we need here is the total time of one oscillation. 5 seconds is simply our period. Thus:

*ν*= = .2 Hz

**Problem4 :**

The maximum compression of an oscillating mass on a spring is 1 m, and during one full oscillation the spring travels at an average velocity of 4 m/s. What is the period of the oscillation?

Since we are given average velocity, and we want to find the time of travel of one revolution, we must find the total distance traveled during the revolution. Let's start our oscillation when the spring is fully compressed. It travels 1 meter to its equilibrium point, then an additional meter to its maximum extension point. Then it returns to its initial state of maximum compression. Thus the total distance traveled by the mass is 4 meters. Since

*t*=*x*/*v*we can calculate that*T*=*x*/*v*= 4 m/4 m/s = 1 second. The period of oscillation is one second.