So far, we have looked at an oscillator with a force that depends on the separation, basically to the equilibrium point, Hooke's law.

Then we added some damping, which was a force that was actually trying to reduce, was against the velocity, so a force that was proportional to the velocity but with a minus sign.

And finally, we want to see how we can actually include energy into the system, so how we can inject energy into the system.

And we are going to do this by including a sinusoidal driving force.

So basically what we are doing, we are including energy into the system in an oscillatory manner.

So now that we've said this, we can write the equation, Newton's law.

And what we see, the three terms on the left, the terms that we saw before for the damped oscillator, so just as a reminder, the acceleration, Hooke's law, and the damping factor, and the right-hand side will be a driving force.

So we are starting to accumulate the frequencies and symbols and so on.

So I'd like to insist on the fact that omega, in this case, is a parameter, if you will.

It's an external, it's the frequency at which you're actually applying the force.

So omega is a parameter of the problem.

And you see that the force is actually time-dependent.

So it's an oscillatory force, F0 being a constant.

Now we are going to rewrite this equation again so that we put an x dot dot without a pre-factor.

So we end up with this equation.

We've done this a number of times already.

And we need to solve this problem.

And we have to find a solution to this differential equation.

Well, the solution to this equation is the usual approach that we learn in ordinary differential equation.

We have two parts.

The first is a solution.

Sorry about that.

The first is a solution to the equation without the right-hand side, which we actually know already because this is the damped oscillator that we saw in the previous screencast.

And two, we want a solution that reproduces the right-hand side.

So we call that a complementary function and then a particular function.

So the complementary function, we've discussed it at length in the damped oscillator screencast, where we have the under-damped, over-damped, and critically damped.

The solution is here.

So this is a solution of the equation in the blue box without the right-hand side.

And the particular solution, we are going to try as a solution a cosine.

And we are going to see that the cosine actually should work.

But we are going to study this.

So the particular solution, let's try to see how this particular solution that we suggest here with d and delta being elements that we have to determine, let's see if it's a possible solution to the equation in blue.

So we are going to introduce the solution xp in that equation and see if it's possible.

So as always, I know that students always remember trigonometry very well.

But it's probably a good idea to remember that trigonometry is always your friend.

So some equations that you might find useful.

And then what we do, we insert xp as a possible solution on the equation on the top.

And then we end up with those two solutions, that equation.

And so what we see here is something that's interesting, is that we have a term in front of cosine omega t and a term in front of sine omega t.

However, cosine omega t and sine omega t are linearly independent functions.

Therefore, in order for this equation to be always true for any time, we need the two coefficients in front of cosine omega t and in front of sine omega t to be zero.

And in fact, by doing so, we will be able to determine the two unknowns, which are d and delta.

So let's start with the simplest one, which is this term.

So this term should be equal to zero.

In order to ensure that the function is zero, the other term will also need to be zero.

We'll discuss that in a few minutes.

So we see directly that for this to be zero, if we divide by cosine delta left and right, we see that we find that the tangent of delta should be two omega beta divided by omega zero squared minus omega squared.

Just to remind you, omega zero is the natural frequency of the undamped oscillator that we've discussed since the beginning of this chapter.

Omega is the frequency of the driving force that we have introduced.

And delta is really, when you think about it, when you go back to the equation that we used as xp, delta is a de-phasing between the application of the driving force and the response.

So delta is really kind of a delay, if you will, between the application of the force and the maximum of the response.

So we have this solution.

Again, it's a good idea to remember some trigonometry.

And that allows us to calculate sine delta and cosine delta.

This is pretty elementary from trigonometry.

So we had this equation that has to be equal to zero.

We already focused on the second part.

Now let's look at this first part.

So we want this to be equal to zero.

And that allows me to find the value of a, actually the value of d as a function of a.

And then when we see this, we see that in the denominator, we have a cosine delta and a sine delta.

But we already calculated sine delta and cosine delta in the previous slide.

So when we substitute, we find that the value of d, which is the amplitude of the solution, the particular solution, is going to be given by this equation here.

So the amplitude is going to depend on a number of things.

It's going to depend on the damping, it's going to depend on the driving frequency, and also on the natural frequency.

So when we put everything together, we see that the particular solution of the problem is completely known now.

The particular solutions are completely known because we just described delta and we described d.

So this is the solution, this exact solution.

And then as I discussed already, delta will be the phase difference between the maximum of the driving force and the maximum of the resultant motion.

Interestingly enough, this phase difference, delta, actually depends on the driving frequency.

Now, for example, I'm sorry, for example, just an example on the last slide, on the last line of this slide, we see that if there is no driving frequency, of course, there is no delay, obviously, there is a delay of pi over 2 at the frequency omega equal omega 0 and a dephasing of pi when the frequency is extremely large.

So that's going to give, you know, like the response of the system is actually delayed compared to the application of the force.

So what have we done in the past seven minutes and a half?

Well, we found a solution to the damped oscillator with a driving force.

And we saw that the solution is the sum of a particular solution and of the complementary solution.

So Xc is what we did in the previous screencast and Xp is the particular solution.

So here is something that's very important to notice.

What I'm going to argue about now is that in steady state, so in the long time scales, what matters is Xp.

In fact, Xc, the complementary, represents a transient effect.

So transient effect, I affect the die out quickly.

And let me explain to you why.

So I reproduce the plot there that we saw in the previous screencast into the damped oscillator.

And we see that in each case, the amplitude after a while goes to zero.

So Xc, after a sufficiently long time, Xc will go to zero.

So after you start the motion, if you wait long enough, the solution in blue will actually no longer matter.

So these effects will die out.

And so the only thing that's going to survive is Xp.

So this is the situation in the steady state.

And you can ask, what is long enough time?

Well, the time that's long enough will be the one where the exponential, which remember the exponential was e to the power minus beta t, when that exponential is low enough.

So in other words, when the time, when beta t is large enough, so in other words, when the time is much larger than 1 over beta.

So if you have a very damp situation, so beta is very large, the time it takes to get to steady state is very small.

So at that time, we are going to suppose that the blue solution is actually not going to matter anymore because we waited long enough so that the exponential got to zero.

So we only have to worry about Xp.

And when we do that, we see that depending on the frequency at which we drive the system, we are going to see that Xc is going to go to zero.