The Harmonic Oscillator
Derivation of Equation (1) for the Simple Harmonic Oscillator
A spring with a mass m attached to its lower end is suspended vertically from a mounting and allowed to comed to rest in an equilibrium position. The system is then set in motion by releasing the mass with an initial velocity v0 at a distance x0 below its equilibrium position and by simultaneously applying to the mass an external force F(t) in the downward direction. See Figure 1 above.
For convenience, we choose the downward direction as the positive direction and take the origin to be the center of gravity of the mass in the equilibrium position. Furthermore, we assume that air resistance is present and is proportional to the velocity of the mass. Thus, at any time t, there are three forces acting on the system:
where a is the constant of proportionality.
Note that the restoring force Fs always acts in a direction that will tend to return the system to the equilibrium position: if the mass is below the equilibrium position, then x is positive and -k x is negative; whereas if the mass is above the equilibrium position, then x is negative and -k x is positive. Also note that because a > 0 the force Fa due to air resistance acts in the opposite direction of the velocity and thus tends to retard, or damp, the motion of the mass.
It now follows from Newton's second law that m x'' = -a x' - k x + F(t), or
m x'' + a x' + k x = F(t) (1)
Since the system starts at t = 0 with an initial velocity v0 and from an initial position xo, we have along with (1) the initial conditions
x(0) = x0 x'(0) = v0 (2)
mass m measured in kilograms, length x is measured in meters, time t is measured in seconds.
The force of gravity does not explicitly appear in (1), but it is present nonetheless. We compensated for this force by measuring distance from the equilibrium position of the spring.
If gravity were to appear in the model, it would take on the form,
m x'' + a x' + k x = F(t) + mg
Notice that if the spring is in equilibrium when it is stretched the length , then the upward force must equal the downward force. Hence,
k l = m g
Let x = 0 be the equilibrium position of the spring with the weight w attached to it. If the spring with this weight attached is now stretched an additional distance x, then the following forces will be acting on the spring.
(i) An upward force due to the tension of the spring which, by Hooke's law, is now k(l + x).
(ii) A downward force due to the weight attached to the spring which is equal to m g.
Equation (1), with a=0 and F(t)=0, is the differential equation for what is often called a simple(or undamped) harmonic oscillator.
Note that x'' means and x' means as x is a function of t, that is x(t).
In this lab, we will consider the case where F(t) = 0; no external forcing. During the course of the lab, we will need to convert (1) into a first-order system of differential equations. Here's the method for conversion:
Let u = x and v = x', then write m x'' + a x' + k x = 0 as .
We now have the following system
Notice that u' = x' and v' = x''.