 Hi, I'm Zor. Welcome to a new Zor education. Continue talking about oscillations. In this particular case, we will talk about forced oscillations. Well, forced oscillation, everybody knows about, for instance, the swing when we are trying to just push it at certain frequency and then it basically goes back and forth and we can maintain this particular back-and-forth movement on the swings and that's basically an example of forced oscillation. Now in our case, we will more concentrate on the spring, basically object on a spring. That's how it was for the previous lectures when we were considering just free oscillation and then oscillation related to friction or viscosity of the environment. So now it's the external force which acts on the object, which forces the oscillation. Okay. Now this lecture is part of the course called Physics for Teens, presented on Unisor.com. There is a prerequisite course called Mass for Teens. Mass is very important now. All of these lectures, for instance, about oscillations are like trick waters mass and only one quarter is some kind of scientific development, whatever it is. So the site is completely free. There are no advertisements. You don't even have to sign on. Unless you want to. Okay, so let's just go. Oh, yes, one more thing. If you found this lecture somewhere on, let's say, YouTube or somewhere else, well, you have to understand this is part of the course and the course is presented on Unisor.com. All the lectures, yes, they are stored at YouTube, for instance. But if you use the Unisor.com, you will have a complete menu of the course. So you can just sequentially go from one lecture to another because all the lectures are interrelated. Like, for instance, all these oscillation-related lectures, they're all about the same differential equations. Speaking about differential equations, just one more reminder that math is necessary absolutely for any kind of a physics course. In this particular case, I will definitely use differential equations, which are explained in the Maths 14's course, which is prerequisites for this one. Okay, let's go ahead. So we are talking about spring with some kind of an object of Math M, and the spring has elasticity K, and what we will be talking about is about certain external force, F, which is applied to this particular object. Now, first of all, let's just remind you what exactly happens if there is no external force. Well, if there is no external force, there is only one force, which is which is acting hooks low when the force is proportional to displacement. So this is the neutral position, and then object goes this way, this is my X of t. So when object goes back in force, X of t goes positive when we are stretching the spring, or negative when we are squeezing the spring. So that's what happens if the oscillations are completely free from all the external forces. So this is the hooks low, that the force is proportional to displacement, and it acts in the opposite direction. If positive displacement force goes to the negative side. If displacement is negative, the force of the spring is going to the positive direction. So that's why we have this. At the same time, we have the Newton's law when all the forces are supposed to be equal to mass times acceleration. Acceleration is the second derivative of distance. The first derivative being the speed. The second derivative is acceleration. So this is the Newton's law. So these two geniuses give us a differential equation mx second derivative of t is equal to minus k x of t. Now that's what we were considering in one of the previous lectures. This is free oscillation. Now the free oscillation has the equation, differential equation, has a solution. By the way, this is the end of the physics and start of the mathematics. So the differential equation has a general solution where omega zero is equal to square root of k over m. So this is something which we discussed before. A and B can be any coefficients, and if you will substitute this x of t to this differential equation, you will have identity, which means it's a general solution. So A and B describe all the possibilities, all the real numbers, and that encompasses all the general solutions. We were also talking about that this is a linear differential equation because it's all related to function and its derivatives with some coefficients. So it's a linear differential equation. There is one more interesting concept. This is a homogenous, I think I spelled it correctly. It's a homogenous equation, and it means that if function x of t is a solution, then function c times x of t, where c, any constant, also is a solution. Now without even looking at this general solution of this differential equation, you see that if x is a solution, which means if this is an identity for some function x of t, if instead of x of t you put c times x of t, where c is a constant, well, it will be multiplied by this, by c, and the derivative when you are multiplying function by a constant, it means the derivative also multiplied by constant, so it will be the same here. The first derivative and the second derivative, it will be the same c, and c will cancel out, so we will still have the identity. So it's a homogenous, linear differential equation, and this is the general solution. By the way, because of a second degree, which means it's a second derivative, linear differential equation, it depends on two variables. Now, these variables can be defined if you have initial conditions. Initial conditions can be, for instance, x of 0 and first derivative of x of 0, so these two derivatives will define these two constants. That's it about whatever was before. Now, what happens now? Well, what happens now is that our equation should be modified taking into account our external force. So external force is here, so we are adding external force. So one force is the force of the spring, that's the Hooke's law, and another is external force, which is also applied. Now, in our case, in case of oscillation, we will talk about periodic external force, and not just periodic, we're talking about external force, which can be expressed as this. So it's basically the same as you're pushing the swing with certain periodicity and certain efforts, which are probably a little bit bigger when you're really pushing, and then it's smaller when you really let the swing go, then you don't really push. So your force is, well, generally speaking, can be described by something like this. Now, obviously, it's a model. In real life, everything is much more complex, but for our purposes, this is a model which we are discussing right now. So this is the force, and that's why the equation will be mx of t. I will take the x of t here and equals to m0 times cosine omega t. Now, what's very important is, remember that general solution to homogeneous equation when it was equal to zero, where there is no force, was x of t is equal to a cosine omega 0 t plus b sine omega 0 t, where omega 0 is square root of k over m. Now, what's important in this particular case is that omega, this frequency which we are using when we are pushing the swings, this is the frequency of the function itself. Now, I assume that omega is not equal to omega 0. Now, why? Well, this is actually a subject of the second lecture when they are equal. So right now, we are not considering this primarily because you understand that if you are pushing with the same frequency, and if we are talking about an ideal situation, about a model, then you're swinging to a greater and greater degree, and basically the whole system might actually destroy itself because you will increase the intensity of the oscillation. Well, obviously we're assuming that there is no friction, there is no viscosity of the surrounding environment, and we are talking about the spring being basically of infinite lengths, that it can stretch infinitely. But if this is the case, which is absolutely not the case in real life, our spring will stretch more and more and more. If we are pushing with the same frequency as the inherent natural frequency of the spring actually is. So this is the case when we are talking about right now in this particular lecture. Next lecture will be when they are equal, and that's a completely different story. Okay, so we've got a differential equation. It's non-homogeneous because if I will multiply function by constant c, obviously this left part will be multiplied by c, but the right part will not. So it's no longer a solution. So it's a non-homogeneous equation. But what's interesting is let's consider you have two equations, two solutions to this particular equation. x1 is a solution and x2 is a solution. Well, let's consider function x3, which is difference between them. Well, if you will substitute x1 will be identity, and if you will substitute x2 will be identity. And if you will subtract one identity from another, you will get 1x1 of t minus x2 of t-second derivative plus kx1 of t minus x2 of t equals to zero, right? Because this is, we are subtracting one equation where x1 here and it's a solution. So it's identity and another equation x2 here. And we subtract one equation from another, the right part will cancel out and we will have what here? So, which is mx3 of t is equal to, sorry, plus kx3 of t equals to zero. Which means what? That x3 is a solution to a homogeneous, to a corresponding homogeneous equation, differential equation. So, two partial solutions, partial means just one particular solution. There are many, obviously, but if you will take two partial solutions and subtract them, then you will get a solution to a homogeneous equation. What does it mean? It means that if we know the general solution, we know all the solutions to a homogeneous equation, and by the way, we know. Over there, right? If we know, then it's sufficient to find only one solution to a non-homogeneous equation. And then if you will add this partial solution plus all the solutions to a general, you will have all solutions to a non-homogeneous, you will get basically a general solution to a non-homogeneous equation. Again, why? Because any other solution to a non-homogeneous equation can be obtained by adding our one particular which we have found and some general solution to a homogeneous equation. Now, if that is understood, and if not, send me an email and I will try to explain it later. Now, the every lecture has notes on the website and I explained it in the notes as well. It's really very easy. So, if you found one particular solution, then the result is always some solution to a homogeneous equation which you can add to get any other particular solution or partial solution, whatever the name is. So, all we have to do since we know general solution to a homogeneous equation when there is no external force, we just have to find one and only one solution which is a solution to a non-homogeneous equation. If we will find this one, then adding a general solution to a homogeneous equation, we will get a general solution to non-homogeneous. Okay, so how can we find one particular solution to this equation? Well, that's easy. I mean, you just have to guess. Look, this is a cosine. Now, if this is a cosine, the second derivative of cosine is also a cosine with some coefficients. So, we probably have to look for the function which partial solution which looks like some kind of a constant times cosine of a magnitude. If you will look in this particular format, all we have to do is to find constant C to basically satisfy this particular equation. Well, fine, let's do it. The first derivative partial solution is equal to C from cosine is minus sine, so it's minus C and then it will be inner function omega and sine of omega C. Okay, that's my first derivative. Constant, derivative of cosine is sine and then there is an inner function which is omega times T and omega should be multiplied. Second derivative is a derivative of the first, so we have minus C omega cosine of omega T and another omega, so it's square. Substitute into this equation, so what we will have? We will have this times m, this times k and equal to this and the cosine obviously cancels out, so we have an equation for C which is what? m times this, so it's minus m C omega square times cosine which will cancel out plus k C is equal to f0 and cosine again, cosine cancels out from which C is equal to f0 divided by k minus m omega square equals k, sorry, m0 mk over m minus omega square. Now you remember what k over m is? It's omega 0 square, omega 0 square minus omega square where omega 0 is inherent natural angular frequency of the oscillation of the spring when there is no external force. Now you see why I have this assumption because otherwise I will get zero in the denominator and the whole solution would be basically none. So we have found one particular solution, so it's f divided by m omega 0 square minus omega square times cosine omega t. This is a partial solution to my non-homogeneous equation, so if I will add this to a general solution where a and b can be anything I will get a general solution to this non-homogeneous equation, so we have solved the equation. The only thing is I would like to slightly modify how it looks and I did it already before in one of the lectures, it's just a simple trick. You see I don't like cosine and sine, I would like to have only one function trigonometric function, so how do we do it? a cosine omega 0 t plus b sine omega 0 t equals to a square plus b square times square root of a square plus b square cosine omega 0 t plus b divided by square root of a square plus b square sine. Okay, so basically I divide it and multiply it by square root of a square plus b square. Now let's consider you have a co-ordinated plane, so let's take a and b, this is angle five, well obviously a is a cosine in this particular a divided by square root, a divided by square root of hypotenuse basically this triangle would be a cosine of this angle and b which is this that would be a sine and doesn't really matter whether a or b are positive or negative it would be exactly the same thing. If for instance a is negative then this angle is my five. It's basically a definition of cosine and sine as functions. The only thing here is they're usually defined on a unit circle, but right now we're just stretching it to a radius of square root of a square plus b square. Okay, so that's true, I'm taking this angle five and so this would be my d, d is equal to square root of a square plus b square, a over this square root would be a cosine phi cosine omega zero t plus and this would be a sine, sine of phi times sine of omega zero t which is equal to, hope you remember the trigonometry, this is a cosine of difference between these angle doesn't really matter phi minus omega zero t or omega zero t minus phi because cosine is an even function so I will put it in this. Now this is easier so instead of constants a and b I'm using different constants d and phi which basically define the same thing it defines for any particular a and b there is a particular d and phi and for any particular d and phi there is a concrete a and b so it doesn't really matter so if a and b are spanning the whole real numbers d and phi are spanning the whole positive numbers for d and angles from zero to two pi for angle five so we still have this as a general solution to a homogeneous equation so the total solution would be for a non-homogeneous equation my total solution would be this which is a partial solution for non-homogeneous and this which is a general solution with different d and phi for homogeneous equation so let me just write it again so it's f divided by m omega zero minus omega square times cosine omega t plus d times cosine omega zero t minus phi so this is a general solution to a non-homogeneous equation this one all right so what's interesting is that this is periodic function right cosine is periodic regardless of what's in there so there is some kind of a period all right now how does it look well first of all to find out how it looks we have to find d and phi so let's just think about d and phi are defined by initial conditions now reasonable initial condition is zero when the spring is not stretched in the very beginning at moment t equal to zero spring is in a neutral position then we apply the force now we apply the force just the force we don't push anything so the initial speed would be zero so the force starts and the initial speed acceleration will be obviously not zero but the initial speed would be would be zero so it's not moving the in the very beginning the object at the end of the spring is at the neutral position and not moving and then they apply the force so what happens how can we define d and well first of all this is x of t we have to find the derivative as well right so derivative is equal to what well this coefficient f divided by m omega square times derivative of cosine it's a minus sign of omega t times this is f zero yes f zero times inner function which is omega times omega plus actually minus again because it's signed d sine of omega zero t minus phi and times of inner function times omega zero all right okay now what we are saying is that this is equal to zero at t equal to zero and this is equal to zero at t is equal to zero well this is easier because if t is equal to zero this thing is zero right because sine of is zero now and this is zero right now d is not so we have sine of t is equal to zero this disappears so it's minus phi must be equal to zero right d is not equal to zero d is square root of a square plus b square remember if we are not talking about trivial solution x of t is equal to zero always this is not an interesting there is no isolation obviously it's a solution by the way it's a solution to homogeneous equation but it's not interesting so if this if this is the case phi is equal to zero right well it can be pi as well but it doesn't really matter okay so we found uh phi so phi is zero so there is no phi here that's easier and if we will substitute zero to here so phi is zero this is zero cosine of zero is one so we'll have just d on the right so d is equal to um say the whole thing is equal to zero then d is equal to minus this minus f zero divided by m omega square zero minus omega square cosine of zero is one so that's basically d and this is phi and this is d so these are two concrete values in case we have these initial conditions i mean we have different conditions initial conditions it will be different values for d and phi uh for instance we will stretch it a little bit and even give it a push in the beginning well that would be different obviously uh solutions for d and phi this is just an easier case and more natural it's not it's not stretched or squeezed the spring and we don't really push it in any direction in the very beginning all right so this is the solution now how does it look well to to know how it looks we have to just design certain values to f zero to mass to uh to elasticity of the spring so we'll know omega zero and the uh angular um angular speed of our external force how frequently we are pushing the whole string well i have basically done that in the notes for this lecture so i assign certain concrete values to these variables and just graph the result and graph is as as you understand it's a composition of two different uh sinusoidal type uh graphs uh with with different frequencies you see this is different frequencies different very important so what happens if they are different well the graph you you can look at the exact graph uh on unison.com uh in notes for this lecture but basically it looks something like this it looks chaotic but it's still a periodical function so it's something like this so it's two different uh oscillations with different frequencies and that's what will be the result if these are rational numbers and there is some kind of a common denominator it's easy to find a common period right so if this is for instance p one divided by q one and this is an omega zero is p two divided by q two then multiplication two pi times q one times q two would be a period of this function right because if you will multiply t by by by this you will have exactly the same period as far as the periodicity is concerned the resulting angle will be two pi times something greater well if these are irrational numbers there might not be a period in this particular case but it's just i mean it's just pure mathematic and i don't think it's very interesting right now for for the people who trying to study physics but in any case if they are rational it's definitely something like this it all depends on how big the period is if it's irrational numbers then they will have a little bit more chaotic type of things it will not be any kind of periodicity in this particular case but that's besides the point anyway this is everything about um external force with with frequency acting different with different frequency than the inherent natural frequency of oscillation um of the spring itself without external forces all right so i do recommend you to read the notes for this lecture it's a new unit that comes you go to physics for t and scores it's the waves chapter and inside the waves it's mechanical oscillations there you will have the force force oscillations lecture that's it thank you very much and good luck