 Hi, I'm Zor. Welcome to Inizor Education. Today we continue talking about damping of oscillations in case these oscillations occur in discosso's environment. So, viscosity is an important factor which damps the oscillations. Now, the previous lecture was also about viscosity, but in that case we were talking about so-called over-damping. When viscosity is so high that the object, after we stretch the spring and let it go, it will start moving towards the neutral position, but it will never reach it and never cross it, obviously, so there is no oscillation. It will just slow down, slower and slower, and it will be slower, slower moving towards the neutral position, never reaching it. Well, in this ideal, obviously, situation, this is our model, obviously. Okay, now let me just remind you that the main differential equation which we were dealing with was, which describes basically the movement of the object in the discosso's environment was as follows. mx plus kx of t is equal to zero. Now, why is that? Well, because the spring is exhorting certain force which is described by the Hooke's law, which is minus kx of t, where x is displacement from the neutral position. That's the Hooke's law. Then we have the viscosity of the environment, and we were talking that in our model, in our ideal situation, the viscosity force is always proportional to the speed, where c is some kind of a coefficient of proportionality, and the speed is the first derivative of displacement with a minus sign, because it's always against the speed. Speed goes this way, the force which resists the viscosity goes the opposite way. And the sum of these two forces is the total force which is acting on the object, and it's supposed to obey the second Newton's law, so fs plus fv, the spring force plus viscosity force should be equal to the total force which is m times acceleration. That's the second law of Newton. And from this, you get this. This is minus kx, this is minus c, and if it's equal to mx second derivative, you will transfer everything to one side, and that's how you will get this equation. Now this equation needs to be solved, and since it's a differential equation of the second order, we must have two initial conditions, and because it's a linear equation, I did explain it in the previous lecture, it's sufficient to find two partial solutions, two concrete functions which satisfy this. Let's say x1 of t and x2 of t. So let's say we found these two solutions, then any linear combination of those guys, where a and b are any real constants, well actually even complex constants, would satisfy this equation, and this will describe a general solution. So because it's a second derivative, we need two functions and two initial conditions, and linear combination of these two partial solutions will give the general solution. So how did we do it in the previous lecture? We just tried to find these two solutions using a substitution e to the gamma t, because if we will do that, we will have the first derivative equals to gamma e to the gamma t. The second derivative would be equal to gamma square e to the gamma t. This equation would look like m gamma square e to the gamma t plus c gamma e to the gamma t plus k e to the gamma t equals to zero. e to the gamma t never equals to zero, it's an exponent, so we will have a quadratic equation. We just cancel e to the power of gamma t and we will have a quadratic equation. And quadratic equation has two roots, two solutions gamma 1 and 2, which is 2m minus c plus minus square root c square dot minus mk. So these are two solutions of the quadratic equation. So 1 would be gamma 1 and my x1 would be e to the gamma 1t and the second one would be e to the gamma 2t. So it's two different functions, non-linearly dependent on each other. So this would describe where gamma 1 and gamma t are these. Now in the previous lecture, we used only the case when c square minus 4mk is greater than zero. It's very, very important because in this case we have two different values and these two values are real. So we have two different real values and gamma 1 and gamma 2, they're different. If this is positive under the radical and that's why we have two linearly independent solutions and that's how we get a general solution. Then to describe to find a and b we used the initial conditions. Well, let's just have initial conditions in this case. The initial stretch at moment t is equal to zero would be a and no initial speed. These are our initial conditions. It can be different but let's just say that these are initial conditions. From these two initial conditions and having the general expression of the solution to differential equation, we can find a and b, two independent variables, two conditions. We have two linear equations with two different variables, very easily to find. That's what we did last time. Today we will have a different case. Why is it completely different case? Well, because if this is equal to zero, it means I have only one root. Gamma 1 is equal to minus c over 2m and that's it. Two solutions basically coincide but I need two functions, two independent, two linearly independent functions and I have only one. So where can I find the second one to basically come up with a general solution? So that's why this is a completely separate case and that's why I have devoted completely separate lecture about this. So we need something new, a new ingenious guess but whenever I was trying to find the solution in the form of e to the gamma t, that was kind of an intelligent guess. Let's put it this way. I mean it's not my guess, obviously. Somebody else guessed it for me that these types of equations are very easy to solve with this exponential function. So I will try to do something similar but again it's not mine. Somebody else was smarter than me and many many years ago came up with the way how to find another solution to another partial solution to this particular equation using some other function, not this type of function but a different function and that's exactly what I will just present to you and you will see how we can find another function which will together with this one deliver, so I don't have it this way. This is not. So another function which we will use will be the following and I will try to find another partial solution in this form t times e to the gamma t. It's very close to this one. I just multiply it by t. Now, well, let's see what happens. How to find such a gamma that this would satisfy this equation? Well, let's see. First derivative is equal to, now this is a multiplication of two functions. So the derivative of two functions, well if you remember u times v derivative is equal to u, v derivative plus u derivative v. I mean, you have to remember this formula from the calculus. So u is a t and v is e to the gamma t. So derivative of the first function times the second. Derivative of t is one so it's e to the gamma t plus the first without any change by derivative of e to the gamma t. So it's t times gamma times e to the gamma t. That's my first derivative. My second derivative is derivative of this function. This is a sum. So derivative of the sum is sum of derivative. So derivative of this is gamma e to the gamma t. Derivative of this is, well, gamma is just a multiplier, right? So it's t times gamma times derivative of this, which is gamma square e to the gamma t plus derivative of this function times this. So it's gamma e to the gamma t. So that's my second derivative. Now, let's substitute it into this equation. So what happens? m times this, well, by the way, this and this, I can just combine together into two gamma e to the power of whatever. Okay, so m times this, m times two gamma e to the power of gamma t plus t gamma square e to the gamma t plus c to the first derivative. First derivative of this. So it's c e to the power of gamma t plus t gamma e to the power of gamma t plus k x plus k and x is this, t e to the power of gamma t. Now, this is supposed to be equal to zero for any t. So you have to find such a gamma where this thing is supposed to be equal to zero identically for any t. That means it's a solution, right? Well, okay. So let's just summarize all things which are related to t and then all the three stuff. So with t, what do I have with t? I have m. Well, by the way, I immediately cancel this. It's not equal to zero, so I divide everything by zero, so it's not dependent on e to the power of gamma t. So what do I have with t? I have m gamma square. I have c gamma and I have k. Now, what do I have as a free member? I have 2m gamma and I have c times 1 plus c. Now it's supposed to be equal to zero identically, which means this is supposed to be equal to zero and this is supposed to be equal to zero. Okay? So we have to find gamma such that that is equal to zero and this is equal to zero. Well, let's see if it exists. For this to be equal to zero, I have to choose gamma equals minus c divided by 2m, right? If gamma is equal to minus c divided by 2m, this would be zero. Now, what happens with this if this is true? Well, I will have m times gamma square, which is c square divided by 4m square, plus c times gamma, which means minus c square divided by 2m, right? If gamma is this multiplied by c, I will have this. Plus k. Now, speaking about k, let me remind you that we are talking about this particular discriminant, which is c square minus 4m k equals to zero. This is only the case we are talking about. This is why we are actually going into all these troubles because our quadratic equation, the first method of finding the solution using quadratic equation, gave only one root instead of two. So if this discriminant is equal to zero, I have only one solution to m gamma square plus c gamma plus k is equal to zero. Now, from this, follows that k is equal to c square divided by 4m. So, in this plus k, I will just have plus c square divided by 4m. So, what happens here? This reduces this. I have c square divided by 4m and another c square divided by 4m. So, it is c square divided by 2m and minus c square divided by 2m. So, in this case, my this part is also equal to zero, the same thing as this one. So, this, under these circumstances, whenever I have this kind of a relationship between viscosity, mass and elasticity of the spring, if these are in this particular condition, then this gamma delivers me a solution which looks like this function, where gamma is this. So, I have a second solution, only if our, the whole, our system, which can, which consists of spring, which has elasticity, an object, mass, and viscosity of the environment, only in this particular condition when they are in this relationship, I have troubles looking for solutions during, using the first methodology. That's why I have invented the second methodology, which function, which is proposed solution looks like this. And I did indeed find it, it gamma is equal to minus c over 2m. So, now I have two solutions. So, the first solution was, I was trying to find in this way. And I did find that m gamma square plus c gamma plus k is equal to zero would be an equation and gamma would be equal to minus c divided by 2m. Right? Because the square root of something is equal to zero in this case. Now, let's put it as a gamma one. Now, another way of finding the function, I proposed this. And I found that gamma 2 is equal to minus c to m. Same thing as this one, by the way. So, let's just forget about gamma 1 and gamma 2. It's still the same gamma. But we have two different functions. This is x1 and this is x2. So, my general solution looks like a times x1, which is e to the power of minus c divided by 2m t plus b second function, which is t times e to the power same thing. The same exponent. Or, let's say, x1 is equal to minus c divided by 2m if you wish a plus bt times e to the power minus c over 2m t. So, this is a general solution which I can find a and b using my initial conditions on our system. Okay, so, if I will substitute zero, this will go down. This will be one. So, we'll have only a. So, in this case, a is equal to walk is a. Now, what's the derivative of this at zero? And we will find b. So, derivative of this is equal to x, derivative is equal to the first function times the derivative of the second, which is a plus bt times derivative of this, which is minus c over 2m e to the power minus c over 2m t plus derivative of this function times that derivative of this function is b. So, it's b times e to the power minus c over 2m t. So, that's my derivative. Now, it's t is equal to zero. It's supposed to be equal to zero, right? So, the exponent goes to 1. So, it doesn't really matter what it is. This disappears. a is equal to lowercase a. That's a given one, right? Just wipe out this thing. I don't get this equation here. x of zero is equal to a plus bt. So, a is lowercase a. So, it's a. This is zero times this minus c to m times 1 plus b times 1. And it's supposed to be equal to zero, from which b is equal to minus a, I mean, plus a times c divided by 2m. So, our general solution is a is equal to lowercase a. It's initial stretch. And b is equal to a c over 2m e to m e is c. Or equals to a times 1 plus c over 2m e to the power minus c to mt. So, this is a solution to our differential equation with our initial conditions. So, this is a solution to this. Now, how does it look? Oh, I'm sorry. I think I have to put t here. Yeah. How does it look? Well, this is linear function. This is exponential function which goes to zero. Well, exponentially goes to zero. This is linearly increasing 1 plus something. It's like this. And this is decreasing to zero very fast. So, what is their product? Well, obviously exponential function goes to zero much faster than this function is growing. So, the result would be this is, this curve has a horizontal tangential line because at zero we have first derivative is equal to zero. So, we will have it horizontal. So, it's like, it starts maybe like sinusoid more or less. So, it goes this way. And then it very, very fast goes to zero. And that is a graph of displacement from the initial position. This is a graph a. Right? This is a. So, initially we stretched by a. And then it goes down all the way this way. Actually, it's very close to the graph which you would receive in a similar, with similar initial conditions, the same a, let's say, in case of over-damping, whatever we were talking before. Because in this case, as we see, we never reach zero. We never reach the neutral position. It goes closer and closer to the neutral position. And speed would be less and less because the tangential line would be more and more horizontal here. That's the speed. So, what's the difference between over-damping and this case? And this case by the way is called critical damping. Well, for the same a and the same spring. So, initial stretch is the same and the spring is the same, which means k is the same and mass is the same. So, what's the difference actually? Remember, c square minus 4 mk. This is discriminant of the original differential equation. So, if this is the same mass and this is the same spring, it all depends on this. If this viscosity is significantly greater than mass and the strength of the spring, well, it means it will be slower. Whenever it's equal to zero, that's basically as fast as we can get without crossing the neutral position, without real oscillations. So, the critical damping also produces no oscillations. But this is as fast as we can go without crossing. That's why it's called critical. So, in case of critical oscillations, damping is the fastest among non-oscillating systems. So, the smaller the c, the closer the thing is to zero, and that's why the faster it will be reaching the neighborhood of the neutral position, but still it will not cross it. So, if you want the fastest way of damping the oscillation, which means without real oscillation, if you would like to bring the object as close to neutral position as possible without really oscillating, then you have to really choose the viscosity of the surrounding environment in such a way that it's equal to m times k. Now, why is it proportional to m? I mean obviously it's proportional to k because the stronger the spring, the more viscosity we need to prevent the bouncing, the real oscillation. Why m is important? Well, let's put it this way. Consider you have two different balls, a heavy one and a light one. And then you drop both of them into water. Which one would go to the bottom first? Well, the heavier one. Right? Because it's really kind of difficult for viscosity to stop the movement of more massive. So, if it already has certain speed, it will maintain this. To reduce the speed of the massive object is more difficult for Biscousel's environment than to reduce the speed of the light object. So, that's why it's proportional to m, basically. So, it has more inertia. So, if it has certain speed, to slow it down by viscosity is more difficult. So, you need more viscosity. So, that's why it's proportional to both. So, if our task, our engineering task, to damp my, for instance, the car jumped over some kind of a bump, you don't want the car to be bouncing back and forth, right? You want it to bring basically to the neutral position all these springs which are inside. So, you have to bring it into neutral position as fast as possible, right? So, that's why you have to, you know, think about whatever the dampers are used inside the car to be as close to this equal to zero as possible. So, now, in the notes for this lecture, I'm using a concrete example with concrete numbers and exact graph how it looks. I don't remember all these numbers and basically I'm referring you to notes for this lecture. It's always very useful to go to the notes of the lecture to find exactly what's there. Maybe I'm missing something. I don't think I missed something in this case, but maybe I have something interesting in the notes as well. But, in particular, I do have a concrete example with concrete numbers. What's m? What's k? What's c? And I draw a graph and present it in the notes for this lecture. Well, basically that's it. I refer you to Unizor.com's site because it has basically the whole course of physics for teens. And also it has prerequisite course which is called math for teens. So, it's very useful to refresh your math knowledge, especially calculus in this case, like I'm operating with derivatives and differential equations. So, you need all these apparatus to be at your hands without any problems. That's it. Thank you very much and good luck.