So, we know what resonances is. But then most physical systems have some damping associated with it. We consider the three physical problems. We consider the RLC circuit. The LC circuit is a nice resonance circuits, but the resistor adds damping. We considered a mass on a spring. A mass on a spring with forcing can have a very strong resonance, but here we assume of some friction in the problem, so that gives us damping. We consider the pendulum, let's say a girl on a swing with her father pushing her. Low amplitude is the same problem. We showed that the dimensionless equation is the same equation at low amplitude, but that one also has frictional damping. So, the differential equation we derived for those three problems is the same dimensionless differential equation, which I write here as x double dot plus Alpha x dot, plus x equals cosine beta t. A key feature of this equation is that alpha is positive and alpha represents the damping, the damping being proportional to the velocity. So, let's solve this equation and then we'll look at the solution at resonance. Resonance here means that beta would be equal to one. How do we solve this equation? We look at the homogeneous solution first. We try x equals e to the rt. Remember we get the characteristic equation, r squared plus alpha r, plus one equals zero, for the homogeneous equation. Quadratic equation, so r plus or minus equals negative b, negative alpha plus or minus the square root of b squared, alpha squared minus four ac, minus four divided by two a, divided by two. Of course, there are three cases here, but they all share a common characteristic. The first case is alpha squared minus four is positive. Then we have minus alpha plus the square root of alpha squared minus four which is smaller than alpha, so minus alpha plus something smaller in magnitude than alpha will be negative, minus alpha minus the square root will also be negative. So, in the case where you have two distinct real roots, both of the roots are negative. In the case where you have complex conjugate roots, so alpha squared minus four is negative, so this becomes the square root of a negative number. So, this is the imaginary part, right? So, the real part is minus alpha over two, is also negative. Then in the case of degenerate roots, you have r equals minus alpha over two which is negative. So, in all cases here, the real part of r plus and minus is going to be negative. What that means is you're going to have exponential functions which decay in time, decay to zero. So, the homogeneous solution decays in time to zero to limit as t goes to infinity of the homogeneous solution will be zero, decays to zero. We say that the homogeneous solution in this equation is just a transient. Eventually, it will become small and you can neglect it. What does that mean in terms of the physics? Remember that the solution of this equation is the homogeneous solution plus the particular solution. The two free constants are multiplying the two homogeneous solutions. The two free constants are determined from the two initial conditions. The homogeneous solution will decay to zero, taking with it the two free constants so that the final solution will be independent of the initial conditions, right? The initial conditions represented by the two free constant, say C1 and C2, decays to zero with the homogeneous solution. That means that the solution to this type of damped equation is a slave to the forcing function. The solution goes to the particular solution. The daddy pushing the child on the swing, the child can start anyway she wants. She can run and start moving, however she starts, eventually her motion is going to be completely determined by the way her father pushes her. That's the particular solution. So, let's find the particular solution. We need an ansatz. So, let's try the particular solution. How should we do this? Well, we can convert first to a complex equation, right? So, we should do it that way other than doing cosines and sines. So, let's go first to a complex equation. Let me put that up here. We can have the equation to solve then is going to be z double dot plus alpha z dot, plus z and then cosine beta t, I'll take the e to the i, beta t. So, we convert to a complex equation and remember that x here is going to be the real part of z, right? Because the real part of e to the i beta t is cosine beta t, it matches the cosine beta t. So, to solve this equation, we're going to try xp equals constant times matching the right-hand side, e to the i beta t. So, we're going to try xp of t equals a times e to the i beta t. Okay, substitute in the second derivative z double-dot. Sorry. So, because we converted to a complex differential equation, we're actually looking for z here not x. So, we're looking for z. So, we're substituting into our complex differential equation. The second derivative will bring down i Beta squared, which is minus Beta squared. So, we have minus Beta squared A, e to the i Beta t will cancel, plus Alpha z dot. Z dot will bring down an i Beta. So, we have plus Alpha Beta A plus z, that'll be plus A, that will be equal to the right-hand side which is one. I cancel the e to the i Beta t. We put this, we combine terms. So, this is one minus Beta squared. One minus Beta squared. Then plus i Alpha Beta times A equals one. We have here then a solution for A. So, A is equal to one over one minus Beta squared plus i Alpha Beta. Okay. We're looking for our particular solution and we found A. We can put that together, but I think now I want to consider the resonance case. So, for resonance, we're going to have Beta equal to one. If Alpha was zero, that would be the resonance we previously considered. Now, we're considering resonance when there is a damping term, Alpha is not zero. So, we put Beta equal to one here. Then, what do we get for A? Then, one minus Beta squared becomes zero, Beta equals one. So, we have one over i Alpha. So, A equals one over i Alpha. Then, what did we find for the particular solution? So remember, the particular solution will be the long time solution. So, Zp is one over Alpha e to the i Beta t, and x is the real part of that. So, we get the particular solution for x is going to be the real part of A times e to the i Beta t where Beta is equal to one. So, it's the real part of one over i Alpha times e to the i t. Okay. That will be our resonance solution, e to the i t is Cosine t plus i Sine t. We're dividing by i, so that will be i Sine t. So, we get Sine t divided by Alpha. Okay. That's the particular solution for resonance, damped resonance. That means that this will be the long time solution after the homogeneous term decays to zero. How do we interpret this? There's two points to make note of. The forcing term is Cosine t when Beta equals one, Cosine t. The response, the solution for x is proportional to Sine t. So, that means the forcing is Pi over two out of phase with the oscillator. You can imagine a father pushing the daughter on the swing, the maximum push occurs not when the daughter hits the maximum height. The maximum of push occurs at the bottom of the swing when the daughter has zero amplitude. So, you can imagine pushing, you don't just push as hard as you can when the daughter gets to the top of the swing. You push slowly and then maximum at the bottom. That's the Pi over two out of phase here. The second point is that when the damping becomes small, the amplitude becomes large. The amplitude is proportional to one over Alpha. So, the smaller the amount of friction, the lesser the friction, the larger the amplitude. Certainly, you would expect that under damped resonance. So, let me review. We're considering this dimensionless governing equation for the three application problems that we previously considered. The defining feature of this equation is that there are two dimensionless parameters. Furthermore, this Alpha is greater than zero. The fact that Alpha is greater than zero means that there is damping. The homogeneous solution decays to zero, you are left with the particular solution. We can find the particular solution most easily by converting to a complex differential equation where x will be the real part of z. We make our onslaughts here. We solve for A and then we consider what is the solution at resonance. So, Beta equal to one and we get the particular solution is Sine t over Alpha. Pi over two out of phase with the force and with an amplitude that is inversely proportional to the damping coefficient. I'm Jeff Jasanoff. Thanks for watching, and I'll see you in the next video.