## 16.2 Linear Restoring Force

An ordinary spring has behavior described by a linear restoring force. The spring possesses a normal length, xe, and if stretched or compressed, it experiences a force of strength

k |x-xe|

pushing or pulling it back toward its normal position. This can be described by the following force law

We can describe such a force by a potential energy function V given by , and so again have conservation of energy.

If the motion is characterized by mass m, the differential equation obeyed by the system is

which can be rewritten as

This is exactly the differential equation obeyed by (and cos wt as well).

We may therefore conclude that

where c and d are constants that depend on the initial values of position and velocity of the spring .The general solution can also be written as the sum of a sine and a cosine term.

This system has the interesting property that the parameter w that appears here, which by the way is a measure of the frequency of the sinusoidal motion, and its period, depends only on m and k, and not on the initial conditions.

The Potential Energy function for this system is .

You can again use the conservation of energy to deduce the relation between speed and extension of the spring once you know the energy in the system.

The solution here, that the spring oscillates on for ever is obviously unrealistic.

Springs stop moving. This is because the model we have just used obeys conservation of energy in the spring motion, while in reality there is air resistance and some friction in the spring motion, and some of the energy in it when you start gets converted into heat.

Similarly air resistance changes the motion of a thrown ball, for example in the constant gravity case.

We will consider force laws that model friction and air resistance soon. First we consider an important reformulation of Newton's Laws of motion that can be applied to conservative systems, that is those that are like the two we have considered here, in which energy is conserved.

To understand a physical system you want to develop a feeling for how it behaves when left alone, and how that depends on its parameters, and also how it responds to outside stimuli, that is outside forcing functions.

At first glance you might think that there are so many different possible stimuli that this is an impossible task.

The standard approach to investigating this response is to examine response to a periodic forcing function at a single frequency as a function of that frequency.

We then hope to describe response to more general stimuli by using this information. This is done by a process called Fourier Analysis, which is briefly discussed in Section 30.6.

In Section 33.3 we describe how to solve the second order differential equation next mentioned, on a spreadsheet. It is quite easy to do, and you can use the method discussed there to investigate behavior of an oscillator.

The differential equation for the oscillator can be written as

Mx" = -kx - f x' + csin wt

where x is the variable hitherto called x-x0, M is the spring mass, k its spring constant, f measures the frictional loss experienced by the system, w is the frequency of the forcing function and c its amplitude.

In the free system, c is 0.

Exercises:

16.1 First follow the scheme described in Section 33.3 to examine what happens when there is no friction so that f = 0.
You can choose your scale for x such that M = 1 and suppose your scale for t is such that k = 1 as well.
You can chart x and x' vs t and x vs x' and look at the following questions:

16.2 What happens when you multiply k by 4?

16.3 Now what happens when you introduce a small positive value for f?

16.4 What happens as you increase f with k and M fixed?

The oscillation with f > 0 and c = 0 is said to be transient.

When f and c are both positive, then there is a transient behavior not unlike that for c = 0, but also a steady state response to the forcing function.

An interesting parameter is the ratio of the amplitude of the steady state response to c, as a function of w (with other parameters fixed).

16.5 Find the value of w which makes this parameter maximum, for M - k = 1 and f = .1.

16.6 What happens to it when you increase f to .2? .3? .5? 1?