 Welcome back. We are at lecture 35. We are near the end of the supplement to the textbook. I'd like to today look at some spring problems, see how they fit into the second-order differential equation situation. We have circuit problems. We'll probably get to those tomorrow. And then our task, our goal according to the syllabus is to get the first couple sections of chapter 8 this week back in the textbook. I'd actually like to get a little bit ahead because the following week what we're expected to cover is probably a little more ambitious than we could actually do in five days. So if we could get ahead this week, that would be a good thing. So let's pick up where we left off. We are looking at applications of second-order differential equations. We haven't actually gotten the equation in terms of getting the solution to the equation and what types of situations. You'll see some repetition here today, which I think is a good thing. We want to look at what that means into as far as the spring being set into motion and how do we dampen the motion so that it doesn't oscillate on the same pattern forever and ever and ever and what does that look like and what do we call those different categories. So we left off last class with this kind of a force equation. One side of the equation was the mass times acceleration, that particular version of force. And on this equation we had kind of the restoring force from Hooke's law that the force required to stretch or compress the spring is directly proportional to how far you stretch it. And I believe we added in, real briefly, the damping force, which the statement that I read I will read again. We assume, and this is an assumption, that the damping force is proportional to the velocity. So we're going to see some kind of coefficient, some c or some k in front of the velocity. Proportional to the velocity of the mass and x in the opposite direction to the motion. So that means we want to put it on this side of the equation. I think we did this. But if it's in opposite direction to the motion, so it's kind of slowing down the motion, you'll see that in our first couple of examples. And if it's proportional to the velocity and if x is the distance, then x prime or dx over dt. So there's what we're kind of adding in, so that we'll get every term, we'll get the second derivative term, the first derivative term, the original x of t function, and then we don't have anything on the right side now, we'll have zero, but it is possible to have something on the right side. So that stays at home, we'll bring both of these to this side, and we have a second order linear homogeneous differential equation. I think that's where we stopped. Is that correct? We got that. We were out of time on Friday. If you don't like the dx over dt, this is the same thing as x double prime, x prime, and x. So we're looking for an x of t solution to this equation. So it's a position, and a lot of times position is denoted by either s of t, or in this case, x of t. How would we solve this? Well, it depends on kind of the number of solutions. There should be three cases to this problem, just like there were three cases to any other second order differential equation. So we would categorize those three cases based on the number of roots. So if we have two real roots, then we would expect the solution. In your supplement, they categorize it according to the discriminant, the value underneath the square root, having used the quadratic formula. But we've kind of already done that. So we'll just categorize it based on the roots. It is an x function in terms of t time. So position in terms of time, c1 e to the r1 t, c2 e to the r2 t. So we've seen that. That's a category we've already looked at. If there are two distinct real roots, if there is a double root, so r1 equals r2 x of t, the position of the mass at the end of the spring would be this type of equation. c1 e to the, well, they're both the same, so I'll just put r. What's the other piece of that if it's a double root? c2 x, we're not having x's, x is r. So we want t. We want x in terms of t. c2 t e to the rt. So again, we've seen that double root situation. And if we have complex roots to the characteristic equation, then we should expect our solution to look like what in terms of t? e to the e? Alpha. What's e to the alpha x? Now, e to the alpha t. Either c1 or k1 doesn't matter. c1 beta t. So that's helpful because the same three categories we saw in section 7.7, the kind of the general non-homogeneous, excuse me, the homogeneous differential equation solution, we see those again. Now they're just equations of motion. So let's further categorize these according to what it means now in terms of a spring. So this first category, and you'll see that on page 1174, this is called over-damping. That means that we're throwing this term in there where the damping force is proportional to the velocity, and we're putting it in there a little too much. So it over-damps the motion that's happening. So we put a spring in some solution, some viscous material that slows the spring down way too much. So it doesn't oscillate at all. In fact, for lack of a better term, it kind of oozes back toward equilibrium. So that would be an over-damping, and that is this situation. A picture of what over-damping looks like, there are two possible situations. So you can see that it's not oscillating just because of the viscosity of the stuff that's doing the damping. It just kind of oozes back toward equilibrium, or it could go on the other side of equilibrium and kind of ooze back toward equilibrium from that side. So these are the kind of the two pictures that go along with the motion of the mass at the end of the spring that have to do with this first situation. Two distinct real roots, we've got an over-damping situation. This one, where it's a double root, is called critical damping. You're not going to see a whole lot of difference. Critical damping is where it is just enough viscosity in the liquid to keep it from oscillating. If it were any less viscous, it would, in fact, oscillate. So it's kind of that last value that causes it to not oscillate. So critical damping looks a little bit like this in the situation. So if you just look at the picture, sometimes it's hard to say, is that over-damping or is that critical damping? But it was just enough that would keep it from oscillating. In fact, let me read what the authors say here about critical damping. It is similar to case one, over-damping, but the damping is just sufficient to suppress vibrations. Any decrease in the viscosity of the fluid leads to the vibrations of the following case. So it's kind of that last value for C times dx over dt that you throw into the equation that keeps it from oscillating. And if it's not damped enough, it will suppress the vibrations to the point where they maybe have a decreasing amplitude, but it does have vibrations. So it does oscillate on either side of equilibrium. So that's under-damping. As you would expect, under-damping is going to look something like this. It does oscillate, but if you'll notice the amplitude here decreases as we work our way to the right. So it is oscillating, but it's got that decreasing amplitude. So it is bounded above. By this value, you'll see what those values are when we get the equation, and bounded underneath by that. This would be kind of an exponential decay type function, and this would be its mirror image on the other side of the x-axis. So same three categories. They kind of have different names, is all. All right, let's look at two examples, one of which has the damping thrown into the problem, and the first one then it does not, but we're going to use the same kind of conditions on the problem. A spring with a mass of 2 kilograms has a natural length 0.5 meters. A force of 25.6 Newtons, which seems like kind of an odd value, but it actually makes the problem work out pretty nicely, is required to maintain the spring with the mass at the end, maintain it stretched to a length of 0.7 meters. So we want to know how much it stretched it beyond its natural length, which that's kind of the elongation involved in that Hooke's Law equation. So it went from 0.5 meters to 0.7 meters, so when we look at this particular equation, the Hooke's Law equation, the force is 25.6 Newtons. This constant that we don't know is k, and then the elongation or the distance that it stretched the mass at the end of the spring is 0.2 meters. So we've got to have meters with Newton, so we're okay as far as the units are concerned. So k would be 25.6 divided by 0.2, which is 128, am I remembering the right number? Yes. I'll read the rest of the problem, we'll see what it is we have and we don't have in this problem, and then we'll go about solving it. If the spring is stretched to a length of 0.7 meters and then released, well, if it's released, that's a pretty key word that tells us its initial velocity was zero, okay? It wasn't thrown downward or pushed downward to start it in motion, so it was just released. So the initial velocity is zero, it actually says that, it's then released with initial velocity zero, find the position of the mass at any time t. So they say nothing in this particular problem about a damping coefficient of such and such or we're throwing it into the problem, it's in some viscous fluid that's going to slow down the oscillation. So we're to assume that it in fact doesn't have that in this problem. So we're going to need mass times acceleration. That'll be the left side of our equation, the right side of the equation will be that we don't have that c times dx over dt piece in this particular problem, good first problem. So the mass is 2 dx over dt, second derivative of x with respect to t, both times I'll call that x double prime, got negative 128x. So no first derivative term in the first example, we will take a second problem, throw something in there with these same kind of conditions, slightly different initial conditions and see what it looks like. So we have nothing that's going to dampen the motion of this spring once we set it in motion. Nothing that oscillates forever at the same amplitude in the same period. It seems like it would without anything acting on it. So it's not a realistic situation, but it's a good first example so we can see how it alters when we throw in that damping force. All right, I'm just writing it down again at the top of the page. So we can divide everything by two, characteristic equation. So r is what? What squared is negative 64, 8i plus or minus 8i. Well we need the real part, which is the alpha part, and we already have the beta part, so that would be 0 plus or minus 8i. So alpha is 0 and beta is 8. So this has two complex roots. So this would be our third category where it's under-damped. We know that because there's no damping at all in this problem. So what should x of t be in this situation? E to the 0. So that's just going to be e to the 0, which eventually is just going to be 1. C1 cosine of 8t plus C2 sine of 8t. Okay, thank you. Everybody okay with that? 0 for alpha, 8 for beta? Probably is helpful to actually write it rather than just say it, right? Because then it's not actually there if you don't write it. All right, we can get rid of the first term because that's e to the 0, which is 1. All right, let me reread the kind of initial conditions and we'll see if we can solve for C1 or C2 or potentially both. If the spring is stretched to a length of 0.7 meters, so we're kind of starting this thing in motion, what would we call that? If the spring is stretched to a length of 0.7 meters, then released. So the first one is its initial position, right? So this is a position function, the x of t is a position function. So the position at time 0 is what? 0.2, 2 tenths of a meter below, down is positive in this particular model. So it's 0.2 beyond equilibrium. Spring is stretched to a length of 0.7 meters, that's its initial position, and then released with an initial velocity of 0. So x prime of 0 is 0. There's our initial conditions for this problem. Initial position, two meters further down from equilibrium, down is positive. X prime of 0 is dx over dt, which is velocity, initial velocity is 0. So let's plug the first one in to the equation. So x of 0 is 0.2, we'll plug that in on the left side. Cosine of 8 times 0 is cosine of 0, which is 1. Sine of 0, sine of 8 times 0 is sine of 0, which is 0. So it looks like C1 is 0.2. So there's our equation so far. So our second initial condition is its initial velocity is 0 because it was just released rather than kind of thrown into motion one direction or the other. So we need x prime of t. Derivative of 0.2 cosine 8t would be 0.2 negative sine 8t times 8. C2 derivative of sine of 8t, cosine 8t times 8. We're going to put in 0 for t and we're going to get 0 for the velocity. So there's our velocity, 0.2, sine of 8, 0, that's sine of 0, which is 0. So it doesn't really matter what the rest of that is, that term drops out. Cosine of 8 times 0 is cosine of 0, which is 1. So 1 times 8 is 8, 8 times C2, which doesn't leave a whole lot for C2. C2's got to be 0 for that to be true. So we allowed for the other term possibility of its appearance in the oscillatory motion equation. It in fact is not there. So C2 is 0. So we'll come back to this equation and we'll eliminate that term. So our x of t is 0.2 cosine of 8t. Does that seem to be a reasonable model for the situation that we encounter? It gets set into motion, 0.2, I mean if we made a picture of it, 0.2 would be its starting position at t equals 0. We go up here to 0.2 meters and 8t, how does the 8 affect the normal period of the cosine function? It affects it inversely. So our normal period is 2 pi, so now it's 2 pi over 8. So there's the period. So every pi over 4, which I'll kind of stretch it out just a little bit, this thing goes through a full cycle. So it's halfway through its cycle at pi over 8 and it's a cosine graph. So it's already through one full cycle at pi over 4. Now pi over 4 is a t value, right? What is an approximation for pi over 4? The number pi divided by 4. How many seconds would that be? 0.78, what's the next digit, 5? So for every basically three quarters of a second, it's gone through its full cycle. We pulled the spring down to 0.2 meters beyond equilibrium, it was released, so it goes all the way up to 0.2 and all the way back down to 0.2 in that first nearly three quarters of a second. Pretty springy little spring, right, with a period of pi over 4. It's going to oscillate pretty rapidly, it doesn't have a descending amplitude because we don't have anything that's acting on this other than just this kind of the spring constant and the mass itself. Now we don't need to convert this, by the way, to the other form, the phase amplitude form, because it already is a single trig function. It's not a sine plus a cosine, so there'd be no reason to convert it to another form. Question on that one. All right, let's take a similar problem, this time with a damping force. This actually is example two, but it has the same kind of stuff, same spring, same mass. Suppose that the spring from the previous example is immersed in a fluid with damping constant c equals 40. Find the position of the mass at any time t if it starts from the equilibrium position, that's slightly different, and is given a push to start it with an initial velocity of 0.6 meters per second, that's different. So our initial conditions are different, but the spring itself is the same. So it's got the same mass of two, it's the same spring with the same mass, so it's got the same spring constant which was 128. So this is the new part, we're going to add in this damping force with the constant of 40 and it's proportional to the velocity. Is that too viscous? Is it going to critically damp it, is it going to over-damp it? We don't know how that 40 is going to affect this equation, but that's what's handed to us. Question? Is the problem saved at this again? Is immersed in a fluid with damping constant c equals 40? So that's in opposition to the motion, so that's why we put it on this side and give it a negative coefficient. So if we move things over to the left side, we've got this equation drastically different from the first one because we have an x prime term. We don't know what that's going to do until we actually solve it. If we took everything and divided it by two, we get that. What is the characteristic plus 20R? Yeah. I think that factors, doesn't it? Figures of 64 that could somehow be arranged to get us plus 20 in the middle, 16 and 4. So we have two distinct real roots to the characteristic equation, so we know what kind of equation results, so this is going to be over-damped, right? C1, e to the negative 4t, C2, e to the negative 16t. Does that work so far? So let's go back to the initial conditions. If it starts from the equilibrium position, what do we call that? How do we translate that English phrase into a mathematical equation? Good, x of 0 equals 0. 0 is the equilibrium position, so that's its position at time 0. And is given a push to start it with the initial velocity of 0.6 meters per second. So that's different than just releasing it, like on the previous example, so its initial velocity was 0. Now we have an initial velocity of 0.6 meters per second. So there's our initial conditions on this problem. So x of 0 equals 0, so we're going to get 0 on the left side. So e to the 0 is 1, and e to the 0 is again 1. So we don't get a solution, but we at least get an interrelationship between C1 and C2 that we can use again. Let's go on to the second piece here, x prime of 0. So let's get the derivative of our solution thus far. What's the derivative of C1, e to the negative 4t? Thank you. You know, these two examples I had on Friday's plan that we could do these on Friday, I'm just a foolish person. I mean, it's taken us, we could not only get to them Friday, it's probably going to take us all day today to get to these. Not a very good planner, but at least we're getting to it. Derivative of the other one, derivative of that one. All right, so the x prime of 0 is 0.6, so 0.6 is what, negative 4 C1? Because that's e to the 0, which is 1, and we've got a minus 16 C2 and another e to the 0, which is 1. So this equation, and this equation, two equations, two unknowns, we should be able to use substitution or some version of linear combination to come up with a solution. So how do you want to approach this from here? Four, okay? Or we could do substitution, right, from the first one, because it's such a simple equation, C1 would equal negative C2 or the reverse of that, right, and then substitute into this one. So if we multiply through the top equation by four, which was recommended, and if we add, we get 0.6, if we add those two drop out, and what do we get here? Negative 12. So C2 is, what, does the arithmetic give us there for C2? Negative 4.05. And if we go back to our first equation, C2, if we solve that for C2, it's the, or negative C1 equal to, it's the negative of C2, so that should be C1, so our equation, where is our equation? C1. So we know what category this is, this is overdamped, so we should expect the picture to be one where it either kind of oozes back toward equilibrium or it crosses through toward equilibrium and comes through toward equilibrium on the other side. If you picture this, let me see the conditions here, starts from equilibrium. If you were to graph this and if you have a graphing calculator, take a few seconds and see what you get. When you picture that, it should look something like this. So this is over, overdamped, because whatever the viscosity of the fluid was, way too much, way too much for the oscillatory motion and it kind of started here, it was given this push downward, which down is positive, remember? And it reached its lowest point, which for us on the picture is the highest point and then it oozes back toward equilibrium. Questions on that? Now there is one other type because this is a second order homogenous differential equation. There is a way you could have something that is not zero on the other side and that's what they would call a forced vibration. So you have some external force. Other than the typical forces here of the viscosity of the fluid that's doing the damping, the spring constant itself that kind of gives us a restoring force and then the mass itself, which the mass times its acceleration, is also kind of the one way to examine or look at the force. So if you have some external force, then it becomes a non-homogeneous equation. And we already know how to solve those too, so it just fits in the category. We've got a few minutes. Is there a web assigned question? Okay, let me see if I can diffuse a web assigned question. And then we have time, we'll go ahead and set up the circuit problems and then we can just work on those tomorrow. Minus 2y equals x plus sine of 2x. And then y of zero is equal to 1. y prime of zero is equal to zero. How about the characteristic equation and then the homogenous solution. Is that okay? You okay with that? Plus r minus 2. So the equation is e to the negative 2x. C1. Yeah, e to the negative 2x times c1 cosine square root of 2x. Square root of 2x. Yeah, plus c2 sine. Oh, we've got a c1 here? Yeah. Sorry, let me start again. So out in front we've got e to the negative 2x. C1 cosine square root of 2x. C2 sine square root of 2x. Does that look familiar? You had success with this problem? So what's the characteristic equation? R squared plus r minus 2. Yeah, something's not right there. Because this doesn't have, this has two real solutions, right? So this should be r plus 2, r minus 1, I think. So we need to scrap that. Yeah, I just got really confused. Yeah, and you've been gone for a couple of days. It's like, ah, that's not fair. All right, let's try again. So was that the correct equation or the correct solution? That was neither. That was the wrong level of science. Okay. All right, so let's start again. Does anyone have a web assigned question? Oh, wait, wait, wait, wait. That wasn't the right equation or the right thing. I just, that wasn't the right homogeneous equation. Okay, so you want me to rewrite the equation? Yeah. Okay. Well, I was trying to get through the first part kind of quickly because I know with these, you've got two pieces to the particular solution and we've got some initial conditions so it's kind of a lengthy problem, but that kind of backfired. All right, homogeneous equation or the characteristic equation r squared plus r minus 2, that factors into r plus 2 minus 1. So our homogeneous solution is c1 e to the negative 2x c2 e to the x. Does that look better? All right, did you do the particular solution in pieces? Yes. Because we have a linear part and a sine or cosine or sine plus cosine part. So what's going to generate something like this? A x plus b. Generic linear. So its derivative is a and the second derivative is 0. So our equation should be 0 plus a minus 2 a x plus b. This seems really familiar. Have we done some of this? Part of it. We did this? Yeah. Good. Another efficient use of time. Let's keep going because I don't know that we did it all the way through. I don't think we found. Okay. Okay, so coefficient of x on the left side is what? Negative 2a. And everything else that doesn't have an x is a constant. So what? Minus 2b. So it looks like negative 2a must be equal to 1. So therefore a is negative a half. A minus 2b is the constant on the left side which must be equal to the constant on the right side which is 0. A is already negative a half. So negative a half equals 2b. So b is what? Negative a fourth? Because that looks familiar. I think we did that part. We stopped there? Yeah. All right. So particular solution 1 that's going to generate the linear term is a x plus b which is negative 1 half x minus 1 fourth. Okay, what's the particular solution look like that's going to generate sine of 2x? C1 or C cosine 2x plus d sine 2x. How about let's use d and e. Since we already have c's in the problem. Is that going to make a difference? I don't know. Let's just try to keep u's, c's and d's. Yep. Okay, c. Cosine 2x plus d sine 2x. Everybody feel comfortable that if this is the thing we're trying to generate sine of 2x's and we've got y's and y primes and y double primes that we can do that with sine of 2x and also cosine of 2x. Now when it's all said and done we want the cosine 2x terms to disappear and we want 1 sine of 2x. All right. C comes along. Cosine 2x is negative sine 2x times 2. D comes along. Derivative of sine 2x is cosine 2x times 2. Second derivative we're going to have a negative 2c out in front. Derivative of sine 2x is cosine 2x times 2. Here we're going to have a 2d out in front. Derivative of cosine 2x is negative sine 2x times 2. So what do we have? Negative 4c minus 4d. So if we plug that into the left side y double prime let's get it set up and then we're going to have to stop. There's y double prime. We have 1y prime added in. Minus 2 of the originals. And we're supposed to generate 1 sine 2x. So what would be the next step? Let's just kind of talk it through for a few seconds. We want to group up the cosine of 2x terms. Right? See how many of those we have. Then gather up the sine of 2x terms. See how many of those we have. What should the coefficient of the cosine of 2x on the left side be equated to? Zero. And the coefficient of sine of 2x on the left side should be equated to one. We're out of time. We can pick up from this point if we need to tomorrow. But tomorrow we'll spend most of our time on circuits.