 Hi, I'm Zor. Welcome to Inezor Education. We will continue talking about mechanical oscillations. The previous lectures were about mechanical oscillations in the presence of friction, which is a constant force which prevents the oscillations of happening, basically always acting against the movement. Today is the first lecture of a set of three, actually, which deals with oscillations in the viscose environment. Now, an example of the viscose environment is, let's say, you have oscillations inside the water. So, water is resisting the movement of the object at the end of the spring. That's basically what we will be talking about. So, this lecture is part of the course. The course is called Physics for Teens, presented in Unizor.com. I do suggest you to watch this lecture from the website from Unizor.com rather than, let's say, from YouTube, where you might have found it using the search engine or something, primarily because this lecture is part of the course and the course, the whole course, is presented on the website. It's not just an isolated lecture, which means that there are some prerequisite lectures, including there is a prerequisite course, which is called Math for Teens. And that's very important because mathematics is very heavily utilized in physics. And in this particular lecture, you will see that I will heavily use the differential equations, for example. So, you have to know the calculus. Okay, so, viscosity. So, viscosity is a property of the environment to resist the movement. And, well, friction is also resisting the movement, right? But there is a fundamental difference between how the friction resists the movement, which we were analyzing in the previous two lectures, and how the viscosity of the environment surrounding the oscillation is resisting the movement. You probably remember that friction is basically dependent on, let's say, this is our spring, and this is our object, and this is some kind of a supporting surface, and there is, let's say, weight of this object. And obviously, since there is a weight, there is a friction against the surface. But this force depends on the weight, which means it's constant. No matter where it goes, it's always constant. It's always against the movement, but it's always constant. Now, in case of viscosity, let's say, the whole thing is underwater. There is no friction. Well, the viscosity is characterized by the fact that it increases as the speed of the object increases. So, in case of, let's say, spring, the force is proportional to displacement, right? The greater the displacement, the greater the force. That's the Hooke's law. In case of a friction, the force is constant. It depends basically on the weight and some kind of friction coefficient. In case of viscosity, so this is the whole thing is underwater, the force which resists the movement depends on the speed. Well, obviously, it's not exactly. We are talking about certain reasonable interval of speeds, where we can actually say that the force of viscosity as the function of t is proportional to the speed of the object. And not only it's proportional, it's always directed against the speed. So, that's why I see, and what is the speed? Speed is the first derivative of the distance from the neutral position. So, if this is some kind of a zero neutral position of the spring, it's not stretched, it's not squeezed, then whenever you are moving to the next location, that would be a function of time. Location, the distance from zero, it would be function of time. It will be positive if you are stretching the spring. It will be negative if the spring is squeezed. So, this is our function x of t which characterizes the position relative to the neutral. This is the first derivative which is speed and the force of resistance of the surrounding environment. If this environment is viscossos, then it's proportional with some kind of a coefficient c. Well, what c depends on? Well, it obviously depends on the shape of this object. Let's say, the bigger the object, the more resistance, obviously, the surrounding material actually, let's say, its water actually acts against it. It obviously depends on properties of the surrounding environment. Let's say, oscillations in the water would probably be with less resistance than oscillation inside honey, let's say. Honey is a very viscossos environment. So, well, at the same time, there is air. Air also has certain viscosity, but much less than the water, obviously. So, in any case, whatever surrounds the oscillation might have certain viscosity, and this coefficient is basically characterizing how strong the forces of resistance are acting. And again, it depends on the object itself, not on the mass, more on the shape of the object. And also it depends on the properties of surrounding. So, that's a very important force. Now, what other forces are acting on the object? Well, obviously, the spring itself, and we know the Hooke's law, and we know the expression for the force. It depends on the distance. So, the force of the spring is minus, again, some kind of coefficient multiplied by displacement. And again, it's a minus sign because if displacement is positive, the force is directing this way, which means negatively, right? And if x of t is negative, which means spring is squeezed, our object is here, the force is directed that way, which is the positive direction. So, that's why it's minus here. So, these are two forces which are acting. Now, what do we know from the second Newton's law that the sum of all these forces, the total force as a function of t, obviously, a function of time, is equal to mass times acceleration. Acceleration is the second derivative of the distance, right? So, this is the total force, and that's the result of acting of this total force. But this total force is equal to sum of these, which means this is equal to minus c x minus k x of t. So, what do we have now? We have differential equation, which basically describes function x of t, describes the position, which basically what we want. So, we want to find out what is the law of movement, what is x, the distance from the zero as a function of time? Well, now we can say that whatever the function x of c is, it's supposed to satisfy this differential equation. Now, this is, let me just rewrite this differential equation a little differently. Okay? This is my differential equation. Now, this is a linear differential equation, linear, because we have only the first degree of function, it's derivative, and it's second derivative with some coefficients. Okay? Now, it's a differential equation of the second order, because it's a second derivative here. Now, at this point, I would actually like you to maybe refresh your knowledge about differential equations. We did talk about this a few times, and again, in the prerequisite course, which is called mass routines in the same website, you can find the whole chapter dedicated to differential equations. There are different ways to basically solve them, etc. Now, the linear differential equation of the second order, it has certain general solution, because it's not only one function which can satisfy this, but let's say you have another function, which is, let's say you found one particular function, x1 of t, which satisfies this equation. Obviously, if you multiply by any coefficient, it will also satisfy, because this is just multiplication by a constant, and this will be a times x1, this will be a times x prime, and x the second derivative. So, a will basically cancel out, and obviously, the same equation will be satisfied. Now, what if you have two different functions, x1 and x2, which satisfy this particular differential equation? Well, then you can say that linear combination of these functions will also satisfy it, because again, it's a linear equation. So, if x1 satisfies and x2 satisfies, then a times x1 plus b times x2 will also satisfy, because you will have here a times x plus b times x, a times x prime plus b times x prime. So, you will open all the parenthesis, and each one, x1 will be equal to 0, and x2 will be equal to 0. So, obviously, they're linear combination also. Now, there is also a certain theorem in the theory of differential equations of the, let's say in this particular case, a second order. So, if you have a second order linear differential equation, then it's sufficient to find two functions, which satisfy, not proportional to each other, which satisfy this equation, and then linear combination of these functions will basically cover all the possible solutions, which is, this means it's a general solution. So, you have to find two partial solutions, and then any combination of these two partial solutions will also be a solution, but it will describe a general solution, which means no other solution but these, but described by this equation was different A and B, obviously. No other solution exists. So, it's sufficient to find this, the solution to this particular differential equation, obviously, 0 is important. That's why we can just do this type of thing. So, it's sufficient to find two partial solutions, two concrete functions, which satisfy this particular solution. Well, if it was a third degree, it would be three functions. I mean, the third derivative. If it's an nth derivative, it would be n function, which would be sufficient, not proportional to each other, nonlinearly dependent on each other. So, if you will find n particular partial solutions, then the linear combination of them will describe general solution. It's really a very nice theory, a very nice theory about differential equations. And it's kind of beautiful, actually. The same thing if you remember, there was a famous theorem that among complex numbers, the algebraic equation of the nth degree, whenever you have x to the nth degree, always has n complex solutions. That's very important and very beautiful theorem, actually. Well, this is kind of similar. So, you have a differential equation in the second order, linear differential equation, which is equal to 0 in this case. There are no constants here. Constant would be different story. But if it's something like this, then you have to just find two of them. Okay, so our purpose to solve this particular equation. So, we have to find two different solutions and then we will describe the general solution. How to choose which one of these is real solution? Well, you have initial conditions. For any differential equation, if you would like to have exact solution, you need initial conditions. And how many additional conditions? Again, two. It's a second order, so we need two initial conditions. So, we need two partial functions, which are solutions, and you need two additional conditions. That's all because it's a second degree equation. Again, kind of beautiful, I would say. So, do we have initial condition? Well, I didn't specify them, but initial condition means that you have to specify the function and the first derivative. So, function at moment 0 is equal to something, which means we have stretched or squeezed our spring to a certain degree. And first derivative. Do we push, after we stretch or squeeze, do we push in some direction our object? Well, for simplicity, I have decided that this we can just set to the A. Well, let's say it's greater than 0. It doesn't really matter. So, we're stretching by A, and then we'll let it go without any push, which means speed is equal to 0, initial speed. So, these are two conditions, which we will apply after we will find two particular, two partial solutions. We have to find A and B, and to find A and B, we will use this thing by substituting 0 to here and to here, and having the first derivative and also substituting 0. And that will be two equations for A and B with two unknown A and B, and that will be sufficient to find A and B. And that will be a real concrete solution to this particular differential equation with these initial conditions. Basically, our physics has finished here. Now, it's all mathematics. So, we have to know how to solve this differential equation and how to find A and B in this particular case. Okay. Solution to this differential equation can be found relatively easily by, it's a trick, if you wish. Yeah, it's a trick. We will look for solutions among certain category of functions, which are very kind of easy to use in this particular case. The functions which we are, which we will use are e to the, some kind of a gamma t, where gamma is some kind of a constant. I don't know this constant. But why have I chosen this? But let's think about it. The first derivative of this is gamma e to the gamma t. And the second derivative is gamma square e to the gamma t. Now, if you're not comfortable with first and second derivative, I don't think you really should go into physics quite frankly. You should go back to mathematics and learn calculus. So, I'm just using this. This is kind of a simple thing. And what happens if I will substitute x, x prime and the second derivative of x into this equation? Well, let's just think about what happens. It's really very easy. So, we will have 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. And what do I do next? e to the gamma t is never equal to zero. So, I can safely cancel it, divide the whole thing by e to the power of gamma t. And what do I have? I have a quadratic quadratic equation. From this quadratic equation, I can find two solutions, two different gammas. Generally speaking, complex, but it all depends on the discriminant of this quadratic equation. Remember, c square minus four mk. This is called discriminant. So, if it's positive, then I will have two different real solutions. Because solution to this thing is equal to what? One over two m minus c plus minus square root of c square minus four mk. You see, this is discriminant. If it's positive, it means I can extract the square root among the real numbers. And I will have two different real numbers. And that means that I will have two different solutions for gamma, not proportional to each other. So, one solution would be with plus, another solution will be with minus. So, these are two solutions with gamma, which means I will have two different functions. So, I will have gamma one and gamma two, both real. Not only real, but let's think about the following. This is negative, right? So, we are assuming all these coefficients c and k, and obviously m, they're all positive, right? So, minus c is negative. Now, this, well, from c square we are subtracting something. Result is still positive, but obviously result will be smaller than c square, right? Because m and k are positive, which means that square root will be smaller than c, which means whether we are plus or minus, this minus will be by absolute value greater than this one. So, the result will always be negative. So, both gamma one and gamma two will be negative. That's very important. Now, I have x of t is equal to e to the gamma one t, and that will be x one, and x two of t will be e to the gamma two t. So, these are gamma one and two, and both are negative. This. Now, at this particular moment, I would like to refer you to the notes for this lecture on unizor.com. Now, every lecture at unizor.com has very detailed notes. Basically, whatever I'm talking about right now is all written down, like a textbook. And I really strupulously put all the calculations in the text itself, which I don't want to hear. But in any case, you understand that I have already found my gamma. I found my two gammas, which means I have two different solutions. The formulas are rather long. That's why I'm not really doing it here. But it's all quite doable. I mean, from now on, you can just do it yourself, basically. Since I know these two functions, I know two partial solutions to my differential equation. So, now I know the general solution. All I have to do is to find a and b. And I will find it using my initial conditions. So, if I will substitute zero into this, so x1 is e to the gamma 1t and x2 is e to the gamma 2t. If I will substitute zero, I will have e to the power of zero, which means one. So, this thing results in a times one plus b times one, which is a plus b equal to a. So, what I do have right now is I have a plus b equals to a from this particular equation. Now, how about this one? Well, if I will do the same thing here, I will have a gamma 1 e to the power gamma 1t plus b to the gamma 1 gamma 2, sorry, e to the power gamma 2t. That's my first derivative, right? If x1 is this and x2 is this, then this is the derivative. And at point zero, which means t is equal to zero, which means I have to put this one and this one e to the power of zero is one, that's equal to zero. So, this is another equation. So, I have two equations with two unknowns a and b, and that gives me basically a and b. And well, knowing a and b, I know the exact solution to my particular problem. Now, here is a very interesting thing. You noticed when I said that both gamma 1 and gamma 2 are negative. So, what does it mean? Well, if this is negative, it means that the graph is like this. And if this is negative, graph is like this. They're both going through number one at zero and then they go to zero when t goes to infinity. So, obviously, we have to consider only positive t, that's the time. So, from time equal to zero forward. And as you see, both components, since gamma 1 and gamma 2 are negative, are going down. So, what does it mean? It means that the distance, this is the combination of these two. And obviously, since each one of them is going to zero, their linear combination goes to zero as well. So, it means that the distance goes to zero, distance from the neutral position. So, whenever we have stretched it and let it go, it will move towards initial position, neutral position. But it will never actually reach it, but it will be infinitely close as the time goes by. So, infinitesimally close to the neutral position. And that is the graph of the whole thing, actually. I mean, you can, the linear sum of these two will also have the same shape. And what's interesting is it always goes monotonically, goes towards the neutral position. It will never reach it. I mean, in this ideal situation, obviously, we're talking about ideal model. It's not the reality. Obviously, in reality, sooner or later, it will be somewhere in a very close neighborhood. But in this physical model, it will go into the neutral position and it will never actually cross it. So, the oscillation will not exist in this particular case. It will be movement from the stretched most position towards the neutral position. Speed will be a little bit greater in the beginning, but then slower and slower. But it will never reach the neutral position. It will never cross into the squeezing of the spring. So, there is no oscillation in this case. So, this particular condition, when this is positive, results in in non-existing oscillation in this particular case. So, spring will just push, well, pull rather. It will pull the object towards neutral position after I have stretched it initially. And it will be slower and slower and slower and will never basically reach it, never cross the neutral position. This is called over-damping. So, the whole lecture, this whole lecture is about this particular case of over-damping, which means our environment is so thick, so viscose, viscose, viscose or whatever. Viscosity is so high that it will be just slowly moving towards neutral position and never cross it and never squeeze the spring. This is the lecture number one. The second lecture will be when this is equal to zero, which means we don't have two different solutions. We have only one. We have to look for a different solution somehow differently. And the third lecture would be when this is negative. And if this is negative, the whole thing becomes complex. It's not a real number anymore. It's a complex number. And only in that case, we will have oscillations, well, obviously, with smaller and smaller amplitude, but still, we will have certain oscillations. In this particular case, and the whole lecture actually is devoted only to this case, when this is positive, the oscillations do not exist, and the object will move closer and closer. As the time goes, this is time, this is x of t. As the time goes, the object will be closer and closer to neutral position, never crossing it. Now, again, I do refer you to notes for this lecture because all the calculations, what exactly are gamma one and gamma two, how it's calculated into A and B, and final solution is written as a formula. It's rather complex formula, and that's why I'm not really saying it here. And you don't really need it to understand the whole concept. I mean, we have got to the point of just two equations with two unknowns. It's kind of a simple thing. So all you need is just some algebra, which I don't want to spend any time on the video. So you do read this particular notes for this lecture. And again, as I said, this is just the first out of three lectures. The second one will be when this is equal to zero, and then the third one will be when it's negative. Well, okay, that's it. Thank you very much. I appreciate it. If you will just go to the Unisor.com and just take a look at what exactly is available there. The site is completely free. There are no ads, so you can use any part of it. And I do suggest you, if you are not well familiar with mathematics, especially calculus, then just use the mass for teens course and refresh your memory. That's it. Thank you very much and good luck.