 Hi, I'm Zor. Welcome to Unisor Education. This lecture is the third in a series of oscillation within certain environment which has certain viscosity. So, viscosity is damping the oscillation. The previous two lectures were dedicated to the cases when viscosity is so high that actually there is no oscillation. We are talking about example of a spring and some kind of an object. So, it's all in some kind of an environment which has certain viscosity. So, whenever we are stretching the object to a certain initial position and let it go, in case viscosity is very high, it will not really oscillate at all. It will just move slower and slower back to original position, never actually reaching the original position. It will be exponentially closer and closer. So, these are two previous lectures and this lecture will be when viscosity is not as high and the object will actually go to the neutral position and further and then back and forth and back and forth and back and forth, gradually diminishing the amplitude. So, that's what will be in today's lecture. Now, this lecture is part of the course called Physics 14. It's presented on Unisor.com. I suggest you to watch this lecture from this website because it's a course which means there is a menu. There are certain parts of this course interrelated and within each part there have certain topics again interrelated to each other. Plus, every lecture has not only the video part which you might actually find, let's say, on YouTube, but also a textual part notes and the notes for the lecture is basically like a textbook. So, whatever I'm saying right now, more or less in the same kind of fashion, explained in a textual format as a textbook. Certain things I'm not really going to talk about, but I will put it in the textual part. Like, for instance, complex calculations. I'm not going to do it on the board, but they are presented in the textual part of this lecture. So, I suggest you always to go to the website and it has parallel lecture and the text related to this lecture. And the website is completely free. There are no advertising, no financial strings attached. You don't even have to sign on if you don't want to. But there are some exams, for instance, which are presented on this website. There is also a prerequisite course called Math for Teens and Math is mandatory. Today I'm talking about differential equations and complex functions. It's all presented in the Math for Teens and whatever else you can use as the source of Math knowledge. But you do need mathematics. Okay, let's get closer to this. Now, again, this is the third lecture in the series and previous two lectures must actually be familiar with. Because I will be much briefer in this lecture about the same topics which I addressed in the previous lecture. So, the movement of the object on this particular, on the spring. If it's within the viscose environment depends on certain factors. Well, obviously it depends on the spring. Spring has coefficient of elasticity K. Well, it depends on the mass of the object, which is M. And it also depends on viscosity of the environment, which is also some kind of a constant C, which actually affects the movement of the object. There are certain forces which are acting. First of all, there is a force of the spring itself, which is equal to minus Kx of t, where x of t is displacement from the neutral position minus, because if it's positive, force goes to the negative direction. If displacement is negative, the force goes to the positive direction. So, that's why there is a minus here. So, K is just coefficient which characterizes this particular spring. Now, there is also a viscosity. Viscosity is, viscosity depends on properties of the environment. It can be water or oil or anything, whatever. And also, it depends on the speed. And that's a very important characteristic of viscosity. It depends on the speed of the object and obviously on the shape of the object itself as well. So, it depends on the object, how it's basically geometrically structured and its speed. And it's more or less corresponds to proportionality between the speed and some kind of a coefficient C, which characterizes the viscosity of the environment and also a shape of the object itself. So, C is kind of a combined coefficient. Now, these two forces are acting on the object. So, this is proportional to displacement and this is proportional to speed of the object. And both are negative because this is negative because displacement can be positive or negative and this negative because it's always against the speed. Whatever the vector of speed is, viscosity resists this particular movement. So, and obviously, we know that the resulting movement of the object obeys the Newton's second law, mass times acceleration, this is the second derivative of the displacement, is equal to sum of the forces. Oops, sorry. And this is a differential equation which we were dealing with in the previous two lectures as well. So, this is the main differential equations. I will rewrite it in this format. And we are interested in solution to this differential equation. We were solving this differential equation in the previous two lectures as well. So, how did we solve it? Well, we just made an intelligent guess that if you have a function e to the power of gamma t and use this function as x of t, what happens? First derivative would be gamma e to the power of gamma t. Gamma is some kind of a constant. And second derivative would be derivative of the first derivative. So, it's gamma square e to the power of gamma t. And if you will substitute it to this equation, you will get m gamma square e to the power of gamma t plus c gamma e to the power of gamma t plus k e to the power of gamma t equals to zero. e to the power of gamma t never equals to zero, so we can just cancel it out. And we have a quadratic equation for gamma. We can find gamma and we can find quadratic equation has, generally speaking, two different solutions. And we will have two different functions, x1 of t is equal to e to the gamma 1 and gamma 2. So, two solutions. And two solutions for a linear differential equation are sufficient to come up with the general solution. General solution is a linear combination of two partial solutions. So, that's the theory of differential equations. I'm not going to talk about this, but again, if you have this type of differential equation, you need two partial solutions which you just guess or whatever. It doesn't really matter how. In this case, we will find it this way. And then, the linear combination of these two partial solutions gives a general solution. Now, what are the solutions to this? Well, quadratic equation has the following solutions, 2m minus c plus minus square root c square minus 4mk. So, that's the two solutions to our equation. Now, in the previous two lectures, we were considering two different cases. One case when this under this square root is called discriminant. Discriminant equals c square minus 4mk. Discriminant of this quadratic equation is positive. It means we will have two positive solutions. And then, since we have two positives, I mean positive and negative, sorry, positive and negative, two real solutions. And since we have two real solutions, we will have two different real functions. And these functions are solutions to our quadratic equation. So, their linear combination is a general solution. That was our first lecture. Second lecture was when this is equal to zero. Well, that's a problem because now we have only one solution minus c over 2m. And we need some other solution, partial solution. And we found this partial solution in this case. And that was the second lecture. Now, this is the third lecture when this is negative. Well, since it's negative, there is no real root of this expression. There is only imaginary complex root. So now, back to mathematics, you need the theory of complex numbers. And by the way, in the textual part of this lecture, I have a reference to corresponding chapter of Math 14's course. So now we are assuming that this is less than zero. And now, what do we do with this? What kind of oscillations are if basically what it says is c is really small? You remember that in the previous cases, we did not really have oscillations when this is positive or even zero. Well, it's positive when c squared is greater than 4mk. So it's greater, so it's big. The viscosity of the environment is big. Now, we are talking about when viscosity is small. And since it's small, then our oscillations have less resistance. And we might actually have the case when the object goes to the neutral and overcrossing the neutral position to the squeezing of the spring and then back and forth, back and forth. I mean, there is still resistance, c is not equal to zero, but resistance is not as big as in the previous two cases. So this is basically a characterization of whether we do or we don't have oscillations. And now we expect that oscillations will happen because c is small since this is negative. So now from this time on it's just pure mathematics. Physics is basically finished. C is small, the viscosity is small. That's why we will have some kind of oscillations. But obviously we expect that these oscillations will be with a smaller amplitude. But let's just get to analytics of this. All right, so let me just transform it a little bit. Minus c over 2m plus minus square root. So 2m goes here. So it's c over 2m square minus 2m. It would be 4m square, so minus k over m. So what I will do, I will put omega equals square root of k over m minus c over 2m square. I changed, so now it's positive. Since it's positive, I can actually use a square root and I have only a arithmetic value, so the positive, omega is a positive real number. It's convenient. Now using the omega, I can put this as equal to minus c over 2m plus minus omega i where i is imaginary unit in complex numbers. So these are two solutions. They are complex. And that's why our function would be a function which takes complex values, which is not really what we're looking for, because we're looking for real values, right? We are talking about x of t means the displacement. The displacement is a real value. So we need the real thing. So now our first problem is how to basically deal with real versus complex solutions. So we found the two solutions which are complex. So what do we do now? Now, for instance, I have a function, let's call it y of t, which is a solution to this particular differential equation. Now, this is a complex function. It means that it has real part and i complex part. I can always imagine it. I can always represent any complex number as a combination, this and this. Now, in our case, our complex numbers are e to the power of minus c over 2m plus omega i and e to the power of minus c to the power of minus omega i c. So the gamma is this, omega i c. So these functions are not easily represented in this way. But it's possible and it's very easy actually. Sorry. Well, first of all, this is e to the power of minus c over 2m. This is the real part multiplied by e to the power omega i t. Similarly, this is also, and this is real. Now, remember the Euler's formula. e to the power i, I put a is equal to cosine alpha is better, plus i sine alpha, where alpha is any kind of a real number. This is the Euler's formula. And again, I present it in the mass 14's course. In the textual part of this lecture, I have a reference to a specific chapter of the course, mass 14, where this is presented and explained. So I'm just using it as is. Now, using this formula, I can always represent this part. I'm sorry, this should be minus. And this part is e to the power of omega i t is equal to cosine omega t plus i sine omega t. And e to the power minus omega i t is equal to cosine of minus omega t. Now, cosine of negative angle is exactly the same as cosine of the positive. So I can always put minus t plus i sine of minus omega t. But again, sine of minus is minus sine. That's trigonometry, which I hope you still remember. So I can represent each of these two complex solutions as a combination of these two functions. And what happens is I will have the corresponding representation in this format. Right? And now we have a very interesting part. Look, if this y of t is a solution of our linear differential equation. Remember, mx of t plus cx of t plus kx of t equals to zero. If this is a solution and there is a real and imaginary part with this i, then each of these real part and separately imaginary part must be solutions as well. Why? Well, obviously, if you have a plus bi is equal to zero in the complex number, what it means? It means a is equal to zero and b is equal to zero. Right? So whenever we put i, we will have, I mean, whenever we will have a real part, we will substitute it and we will have only real functions here. If we will substitute this, since this is linear equation, I will always go outside. So the whole thing would be only with i's, which means only with, will be imaginary part. And that's why coefficient of i, which will be my second derivative plus c, my first derivative plus ky, it will be with multiplier i. So it will be this part. So this must be equal to zero and this must be equal to zero. Which means this is a solution and this is a solution of this differential equation. So again, if you have a complex function, which can be represented in this way, and you know that this complex function is a solution, then each one, real part and imaginary part, are solutions as well. So in this case, our real part is in this case. So instead of this, I have to put, multiply by cosine omega t plus i sine omega t, cosine omega t minus i sine omega t. So what's the real part? Real part of this solution is this. And real part of this solution is this, which is the same thing. So I have real part as one particular partial solution. So the partial solution, which I'm talking about right now, is this one. This is one function, x1 of t. Now the imaginary part is the one which has a coefficient i. So it's x2 of t is equal to e to the power minus c over 2m times sine omega t. In this case, it's with a plus sign, in this case with a minus sign. But it doesn't really matter because what we are talking about is any linear combination. So whether it's a plus or a minus, doesn't really matter because when the general solution comes up, you will have any kind of a, any kind of a linear combination. x of t would be equal to c1 times x1 of t plus c2 x2 of t. So this is the solution and this is the solution of this linear equation. Then any linear combination of these two solutions will give me general solution. So that's why it doesn't really matter whether it's plus or minus. This is any multiplier. So this is a general solution. x3 is equal to c1 times this plus c2 times this. Okay, that's encouraging because it's already in the format which looks like oscillations. Sine, cosine, as the time goes, it looks like it's oscillating, right? But I would like to slightly change this particular format and it would be even better understood that this is a real oscillations. Well, let's talk about this expression. c1 cosine omega t plus c2 sine, sine omega t. That's what we have here. The only thing is there is a multiplier e to the power of minus. Oh, by the way, it's supposed to be t here. I always forget this. Okay. Now, how can I transform this into something which seems to be better understood as oscillations? Very simply. Let's multiply and divide it by the square root of c1 square plus c2 square. Nothing's changed. I multiply and divide it by this, right? Now, with any two numbers, c1 over square root of c1 plus c2 square and c2 over, I can find an angle when this is equal to cosine of this angle. And this is equal to sine of this angle. How can I do it? Well, obviously it's very simple because I'll just take a point on the unit circle. Now, this and these two are basically, well, it's better to have it square root of c1 square plus c2 square. And then we have c1 and c2. So, the radius is square root of c1 square plus c2 square. And this point corresponds to c1 and c2 coordinates. Whatever these coordinates are, minus can be minus or plus, positive or negative. But obviously, if this is the quality, then this point belongs to the circle with this radius, right? So, I'll just take this angle as an angle, as a phi. Then c1 would be cosine and c2 would be sine of this angle. No matter what c1 and c2 are, because square root of c1 square plus c2 square is always positive, obviously, number. And both c1 and c2 by absolute value are smaller than this root. Because it's always, it's addition to c1 square would be greater, would be less than c1 square plus c2 square. Same thing here. So, they're always somewhere inside this circle. So, I can always find a point. For instance, c1 is negative, then we'll do this. And the point will be here. So, whatever it is, I can always find an angle which has these particular qualities. And what do I have now? Cosine times cosine plus sine times sine times sine is nothing but cosine of omega t minus phi. If you remember trigonometry, that's the function for cosine of difference between two angles. So, this is a multiplier. So, what do I have as a result? As a result, I have my general solution written in a different format. I put some multiplier D, which is actually square root of c2 square plus c2 square times, not one more, times e to the power of minus ct over 2m and times cosine of omega t minus phi. So, now we see basically the character of this movement. As the t goes, it's basically a sinusoidal kind of oscillations around the neutral point zero. Now, we are multiplying also on something which is a negative exponent. And as the t goes to infinity, this exponent goes to zero. So, we have to basically multiply two graphs. One graph is a cosine and another graph is exponentially going down. So, what happens as a multiplication? Well, if we will multiply, it would be something like this. So, the amplitude goes smaller and smaller because we are multiplying something which is between minus one and one. We are always multiplying by something which is infinitesimally small. As the t goes by, our oscillations exist. And that's very important because in the previous two lectures, when discriminant of that quadratic equation, if you remember, was positive for zero, we did not have oscillations. Now, we do. But they are diminishing as the time goes by. Well, basically, that's it. Now, d and phi, no matter what they are, will be solution to our differential equation. So, how can I find concrete values for d and phi? Well, we do have initial conditions. Now, I'm using condition like this. This is initial stretch at moment t is equal to zero. And initial push of zero is zero, which means there is no push. So, if you will use these, then you can find d and phi. It will be two equations with two variables, d and phi. t is equal to zero in both cases. So, first, you have to have this, which means point zero. This e to the power of zero is one. Cosine of zero minus phi would be cosine of phi. So, we will have d times cosine of phi is equal to a. That's the first equation. Then we have to have the derivative of this. Derivative is, so this is x of zero. Now, derivative is d e to the minus ct over 2m times derivative of this, which is minus sine of omega t minus phi times derivative of the inner function plus derivative of this, which is minus c over 2m e to the power minus ct over 2m cosine omega t minus phi. And then you substitute zero and you have to have zero here. So, if you have zero, that means x of zero is equal to d minus sine of minus phi. So, it's sine of phi with a plus times omega. And this would be minus, this is one. So, it's g c over 2m cosine of minus phi same as phi. And this is equal to zero. So, this is one equation and this is the second equation. Equations are really simple. I have solved it inside the textual part in the detail, so you can just take a look at this. So, that's technical problem. The most important problem was to create the differential equation, find two complex solutions, from these two complex solutions derive two real solutions and the combination of these two real solutions, which are this and this, gives you a general solution. And the coefficients, well, this is one format, this is another format. This format is kind of easier to understand because it's really visible that this is oscillations. So, the only thing is you have to find these two based on initial conditions and that's just simple algebra, trigonometry. Well, that's it basically and that's it for viscosity. So, damping using friction was considered in two lectures and damping of the oscillations using the viscosity was considered in three lectures. And again, in some cases when either friction is too big or viscosity is too strong, we do not have oscillations, but under certain conditions we do and all these conditions were explained in these lectures. That's it for today. Thank you very much. I do suggest you to read the textual part of this lecture. There are more, including one example actually and I built a graph for that particular example with concrete numbers A and whatever, C, M. So, I just assigned certain numbers and I come up with certain concrete results and I graph it. So, basically I suggest you to take a look at this. That's it for today. Thank you very much and good luck.