Periodic Motion

    What do Maxwell's equations tell us about the change in electric and magnetic fields? To answer that question, we need to look at periodic motion and waves. An object is said to undergo periodic motion when after some set amount of time it returns to its starting point and begins to repeat its motion. Periodic motion is an example of motion about a stable equilibrium point. We have already encountered one example of periodic motion, the spring.
    Recall that a mass attached to a spring that is stretched past its equilibrium point will experience a force F = -kx back towards the equilibrium position. This force will cause an acceleration, and the mass will have a velocity of formula at the equilibrium position. This will cause the mass to overshoot the equilibrium position, and it will be decelerated until it comes to a stop and is pulled back towards the equilibrium position again. Again it will overshoot the equilibrium point and return to its original stretched location. At that point the motion will start over again. If there were no friction, then there would be no energy loss, and the mass would return exactly to its starting point and oscillate forever.
    We call motion under the influence of a linear restoring force (and with no friction) simple harmonic motion. The motion of the object from a starting position back to its starting position is called a cycle. The time it takes to complete one cycle is called the period, t. The frequency, f, is the number of cycles per unit time. We can see that it is related to the period by f = 1/t. The SI unit of frequency is one cycle per second, called a hertz, Hz. The maximum displacement of the mass from its equilibrium position is called the amplitude, A.

Equations of Motion

    Let us now write down the equations of motion for a mass on a spring. Using Hooke's law, we have that
F = -kx

which yields

a = -kx/m

Thus we see that the acceleration is proportional to the negative of the displacement. In order to obtain the velocity at any point, recall that (119) is an example of a conservative force, so that it has a potential energy function

U = ½kx2

From conservation of energy, we then find that the energy at any point is

½mv2 + ½kx2 = E

where E is a constant. We can relate the total energy E to the amplitude by noting that at the point where the amplitude is greatest, the velocity is zero and so (121) reduces to

½kA2 = E

Substituting this into the right hand side of (121) and solving for v yields


We can see this relationship easier by looking at a plot of energy verses position


The potential energy yields a parabola centered on the origin, and the total energy yields a line at a constant height above the axis. The region contained inside these two curves is the physically allowed region for the oscillator. The amplitude can be read off directly by noting the positions where the two curves intersect, and the difference in energy at any point between the total energy and the potential energy is just the kinetic energy at that point.

Time Dependent Description of Periodic Motion

    We still need to find the position and velocity as functions of time. In order to do this, we need calculus. Fortunately, a theorem of calculus states that one valid method of solving differential equations is to guess the solutions. If the guess satisfies the equation, then it must be the only correct solution. So we want to guess at functions that have an acceleration that is proportional to the displacement. Trigonometric functions satisfy this requirement, so we are led to a guess of the form

where A and d0 are arbitrary constants. Calculus shows us that this is the correct solution. We can see immediately that the constant A is just the amplitude of the cycle. d0 is called the phase angle. Its purpose is to describe the position of the oscillator at t = 0. What is the period of (123)? Recalling the definition of the period, we see that


Similarly, we find that the frequency is


It is usually more convenient to work with the angular frequency, w. This is defined to be w = 2pf so


Thus we can rewrite (123) as

x(t) = Acos(wt - d0)

Notice that the velocity is

v(t) = -wAsin(wt - d0)

    A block of mass 5 kg is connected to a spring. A force of 150 N is required to stretch it 1.5 m, at which time it is released to undergo simple harmonic motion. Assuming no friction, find the period, frequency and angular frequency.


    First we must calculate the spring constant k. Recall that


then we have that


Simple Pendulum

    As an example of periodic motion, consider a simple pendulum. A simple pendulum is an idealized point mass suspended by a weightless, unstretchable string in a uniform gravitational field. Consider the following drawing

The forces acting on the pendulum are the tension, T, and gravity. Decomposing gravity into a radial and tangential component, we see that the restoring force is

F = -mgsinq

    This is not simple harmonic motion, since the force is proportional to the sine of the angle and not to the angle itself. Recall, however, that we can expand sinq as


so, if the angle is small, we can approximate sinq as q. Then (129) becomes


This is simple harmonic motion, so we immediately see that








    Notice that all these are independent of mass. This observation was first made by Galileo. We must now ask, how small must the angle be for this approximation to be valid? (129) can be solved using special functions called elliptical integrals, and then we find that the period is given by


where now q is the maximum angular displacement. By comparing this with (133), we can determine the difference in the periods and find the accuracy of the approximation. For example, for 15 degrees, the true period differs from the approximate one by less that 0.5%.