Lecture 25
Harmonic oscillator II: damped oscillator

In the previous lecture, I covered some basic aspects of simple harmonic oscillations. We considered the equation

and saw how its motion is described. A general solution of this equation is

I now make the system little more realistic and introduce damping into the system. Let us first look at what happens if we introduce friction into the system. I consider again our prototype spring-mass system and let there be a constant frictional force f on the mass. This force will always oppose the motion so the system will eventually come to a stop. Let us see when does it do that?

The simplest way of seeing when he system will stop is the through the consideration of energy. But I would like to solve the problem by employing the equation of motion. I will later solve it from energy considerations also. Here is one case where I will have to analyze motion step by step because as the velocity direction changes, so does the force direction. So let us pull the spring out to a distance A and let it move towards the equilibrium point (see figure 1).

When the block is moving towards the left, equation governing its motion will be

In the above, the frictional force f sign is positive because the mass is moving in the negative x direction and therefore the frictional force is in positive x direction. This equation can be recast into the form

We have encountered such kind of equation earlier. It has a homogeneous part and an inhomogeneous term on the right-hand side. So the general solution is

where

Thus

With the initial conditions , the solution is

This is the solution when the block is moving to the left. Since

so the block will come to a stop when . At that time

So by the time the block comes to a stop it has lost distance from its amplitude. And this loss is irrespective of the distance from where the block starts its motion from. This should then also happen when the block starts coming back. Let us find that out. On its way back (see figure 2), the block follows the equation

Notice that the sign of the friction force is now negative. This is because now the block is traveling to the right and therefore the friction force acts towards the left (see figure 2).

Now we have to solve this equation with the initial condition that

I leave it as an exercise for you to get the solution. It is

The corresponding velocity is proportional to ω₀t, and therefore goes to zero again after a time interval of . At that time . Thus every half time the block goes from one extreme to the other, it loses a distance of , and in each cycle it loses a distance of . Question is how many cycles does the block complete before it comes to a stop. The block stops when its final displacement is . If it completes n cycles before that, we have

The same result can also be obtained, as I said earlier, by energy methods. If stretched by A the total energy of the system is . Let us say that before stopping, the block it compresses the spring by A₁. Then its energy will be . The loss in the energy is caused by friction. Thus

The total distance moved by the block is (A+A₁) and so the energy lost against friction is f(A+A₁) . Thus the equation transforms to

and gives

which is the same loss in amplitude over half a cycle as obtained earlier. The rest of the analysis is the same as done earlier.

Having dealt with the constant friction case, we now consider the most common example of damped oscillations. This is the oscillator where damping force is proportional to the velocity i.e.,

In this case, the equation of motion is

Writing we get

This is the equation for a damped oscillator. The equation is homogeneous in x so we assume a solution and substitute it in the equation to get.

which gives

So the general solutions are

Except in the case when (we will deal with it later) the behavior of the solution depends on the relation magnitude of γ and ω₀. Let us first consider the case when. In that case

The general solution then is

This is known as a heavily damped oscillator. The coefficients C and D depend on the initial conditions. For example if I stretch the spring to a distance A and release the block, let us see what happen in this case. By initial conditions

This leads to

A t → ∞ this solution behaves like . The general solution is displayed in figure 3.

It is clear from the figure that there are no oscillations in this case the block slowly comes to rest at x = 0 , i.e. the equilibrium point. I now explore another situation. Suppose we give an implies (speed v ) at t = 0 then the boundary conditions are

Thus the solution would be

In this case the distance versus time graph looks as shown in figure 4.

The figure clearly shows that the block goes out to a maximum distance and then comes back and stops at the equilibrium point. So in both the cases studied above the mass does not cross the equilibrium point. Next I ask: what if we stretch the mass out to a distance A and give it an initial impulse from that point (in negative direction). Then the initial conditions will b

Solution in this case comes out to be

The solution is plotted schematically in figure 5.

It is clear that in this case the particle moves towards the equilibrium point, crosses it, goes a distance and comes back. However on its way back it slowly comes to rest at the equilibrium point and does not cross it. So in heavy damping cases, the block passes the equilibrium point at most once and its distance decays exponentially as .