 Hi, I'm Zor. Welcome to a new Zor education. Let's solve a couple of problems related to oscillations. In the main lecture about mechanical oscillations we were talking about springs, but springs are not the only source of oscillations. And today I will present you two simple problems which also are related to oscillations. I mean, they result in the oscillating motions and very practical, by the way, examples. Now, the first one is about pendulum. We did talk about pendulum when we were talking about just plain mechanical rotation, kinematics, etc. I'll just address the same problem but from the position of oscillations. And then the second problem will be very much analogous to the first. Now, this lecture is part of the course called physics for team. The team is presented on Unizor.com. I do suggest you to watch this lecture from the website because every lecture on the website is accompanied by detailed notes. And then all the lectures are combined into a course. And there is a menu, the hierarchical menu, which you can actually go through. And that would allow you to go in some kind of logical sequence. So if you just by accident found this lecture on YouTube or somewhere else, just make sure that you know that this is just part of the course. Okay, so what we are considering right now is a pendulum as an example of oscillations. And obviously you understand that pendulum is oscillating back and forth. I'll just address it from the oscillating motion standpoint. So what's happening with pendulum? Okay, this is the vertical. This is our pendulum. Angle phi is function of time. And we will basically use phi as the main variable here. Now, what kind of forces are acting on the object which is moving back and forth on the thread? Well, the thread will be of lengths l. Let's say the mass of the object is m. Now, the first force, now we are assuming this is all happening in the gravitational field with constant gravity, constant g. So we have weight which is equal to mg. Okay, regardless of the position because the gravitational field is uniform. Now, what other forces are acting on this particular? Well, obviously this point object doesn't really fall through. It's kept on the thread. So there is some kind of tension. There is a reaction of thread. Let's call it t. And as a result of these two, the thread is tension. The thread has certain tension and the weight goes this way. The resulting force would be directed this way. Resulting force of these two, right? And that would actually force the object to move towards the vertical. Then it goes by inertia, it goes this way. But the same two forces on this side would go in the opposite direction, which is this way. Okay? Okay, now, how can we calculate these forces? All right, very simple. Let's forget about this part. In this particular case, we can always represent this force which is vertical as a sum of two forces. One goes in the same direction as the thread and another is perpendicular to the thread. Now, this and this are supposed to cancel each other because we are assuming actually the thread is not stretchable in any direction. So these forces are equal to each other. Now, what exactly is this force? Well, this angle is phi, so this angle is phi. So this force, this is rectangle, right? So this is mg times what? Cosine, phi of t. And this force, this angle of phi is the same as this angle of phi. So this force, which is f, and obviously they're all dependent on time. It's mg sine, phi of t. Okay. Now, we assume that movement towards this is a positive direction of changing the angle. So whenever our object is on this side of the vertical, our phi is positive. But the force always goes against the position towards the vertical. So if this is the positive direction, the force is here. So a real force which should participate in the equations is supposed to be negative in this case. And the same thing is true when object is in this position because this is a negative direction of the phi. And the force goes this way. So again, it's opposite to phi. So in this particular case, we will have exactly the same equation. But the force is equal to minus mg sine phi of t. When phi is positive, force is negative. When the phi is negative, like here, the force will be in the positive direction, which is opposite. Okay, so we have the force which drives the object along its trajectory. But trajectory is obviously circular in this particular case of the radius L. Okay, if trajectory is circular with the radius L, then the distance which object covers during the time t would be equal to L times angle. Right? This is a plane geometry. So the radius times angle, obviously in radians, gives you the distance. Speed along this trajectory would be L is constant. So it's L times derivative of phi. And acceleration would be L times second derivative of phi. So this is acceleration, this is the speed, and this is the distance. Now obviously we have to apply the second Newton's law. Remember this? So we have the force, we have the acceleration. Which is this? So minus mg sine of phi of t is equal to M times acceleration, L times second derivative of t. Now, M cancels, what it means is that all the oscillating parameters like amplitude, period are not dependent on the mass. They depend only on the length of the thread and the acceleration of the free fall in the gravitational field. Well, that's basically the property which all the pendulums are based upon. Obviously the period is independent. Sometimes we put the bigger mass at the end of the pendulum, sometimes smaller, but the period will be the same. Okay, now, let's just convert it into more traditional form of differential equation. And this is differential equation obviously, this is the second derivative. So we will have second derivative of phi. We will divide L, that would be g over L, and the minus will go here. So it's plus g divided by L sine of, sine of phi of t is equal to zero. Well, this is the differential equation which defines our angle phi as a function of time t. It's a second order because it's a second derivative. And obviously we need certain conditions, initial conditions. Well, let's consider that we will just move our mass at certain angle, initial angle. Phi zero and let it go, so no initial speed. So this is our initial condition. Well, based on these initial conditions and this differential equation, which obviously has some solution, we will have function phi as a function of time t. And as it happens in unfortunately most cases in reality, I cannot really solve this differential equation as it is in those functions which we are used to have. Well, maybe somebody else can, I cannot. It's very difficult differential equation and I checked actually on the internet whether there are any kind of normal solutions, I didn't find anything. And the physicists are doing always the same thing. They are kind of assuming certain other assumptions about the problem and they're saying that, okay, that would simplify the problem. Okay, how to simplify this? Well, consider angle phi is a small angle. Well, if it's a small angle, then the sine of the phi and the phi itself are very, very close to each other. So for a small angle, we will replace this equation with a different one. So instead of sine of phi, we'll just put phi. Well, again, it's approximation, but that's the only thing which allows us to derive the expression for function phi of t in normal kind of functions which we are used to. And now obviously this is the same equation which we were discussing in the previous lecture when the springs were and the Hooke's law was involved. So it's a regular harmonic oscillation and the solution to this, again, as I explained in the previous lecture, excuse me. This is C1 cosine omega t plus C2 sine omega t where omega square is equal to this coefficient G over L. So omega is square root of G over L. So this is typical harmonic equation. This is a general solution of this harmonic equation where the constants C1 and C2 depend on our initial conditions. So let's just find out what are they. And how I come up with this was actually in the previous lecture where I introduced the oscillations and that's kind of a simple thing. I just refer you to a previous lecture if you didn't really understand why I have this general solution to this differential equation. Now, if this is a general solution, we will use the first one, phi of zero is equal to phi zero of zero. This is zero. This is one. So C1 is supposed to be phi zero, right? If I substitute zero here and here, I will have only C1 times one and C2 times zero, so it will be just C1. So that's supposed to be phi zero. Now, the first derivative of this is equal to the derivative of cosine is minus sine, so it's minus C1 times omega times sine of omega t plus C2 omega cosine omega t. And if t is equal to zero, this is equal to zero. This is equal to one. And C2 times omega should be equal to zero and omega is not equal to zero, so C2 is equal to zero. And our equation is phi zero cosine omega t where omega is equal to square root of g over l. Now, from this, we derive that the period which is equal to 2 pi over omega. And again, I refer you to the previous lecture for explanation of what is a period. Well, that's actually 2 pi square root l over g. And obviously, you see that it depends on lengths of the thread. The longer the thread, the greater the period, the time which it actually makes the full cycle of oscillations will be greater. And the g, the heavier, so to speak, our object is in this gravitational field. So the stronger the gravity of the earth is, the smaller, obviously, because it will actually pull down this object at the end of the thread and it will not let it go too far, too slow, if you wish. So the longer thread contributes to a longer period, the greater the gravity constant contributes to a smaller period. Now, the amplitude, obviously, is phi 0. The initial deviation from the vertical actually is the maximum. After this, it will go no further than this phi 0, initial angle of deviation. Okay, that's basically it for this simple problem. So my point was that not only the springs contribute to oscillations, but also, for instance, the pendulum. And now I will consider another problem which also results in basically the same equations and the same results, but in a different physical arrangement. So what is this physical arrangement I'm talking about? Let's consider you have a semi-sphere, radius r, let's say, and you have some kind of a ball here. Now, if you will put it somewhere here and let it go, it will actually go back and forth, back and forth inside the surface of this semi-sphere. Now, the physics of this movement are exactly the same as if it was a pendulum of the same length. So let's just consider this again. So this is my P equals to mg. Now, here we have the reaction of the sphere. The sphere is supposed to be hard and it reacts and obviously this is the force. And the sum of these two forces is actually the one which forces the object to go inside the sphere, always. And the way how we will approach it is exactly the same as before. So if this is angle from the vertical, now this is phi, this is also phi, and this is phi. So this force which actually tangential to the surface will be equal to mg times sin phi of t. Exactly the same as with the pendulum. And they actually have to use again the minus sign here because the force is always opposite in its direction to a positive direction of the phi. If phi is positive, force is negative. If phi is negative, force is positive. That's why I have minus in front of this sign. And obviously I have to use exactly the same second Newton's law. And these formulas are exactly the same. Instead of lengths of the thread, I have to put radius of a circle. This is a circle, this is a radius, this is an angle. So the radius times angle is the distance along the circle which is covered by my rolling ball. And obviously the same with speed and acceleration. So that's equal to m times r times phi of t. We have exactly the same equation as in the case of a pendulum. And again m cancels out. So the big ball, heavy ball or small ball, they will all have exactly the same period of oscillation and the same speed, etc. All the characteristics will be the same. I put a massive ball here or I put a small ball here. They will do exactly the same kind of oscillations if the radius is the same and gravitational constant is the same. So what we do here is exactly the same phi of t plus g over r sine of phi of t is equal to zero. That's our differential equation which I cannot solve. So I will approximate it with sine approximately equal to angle itself. And without the sine I have a harmonic equation the same as I was using before for a pendulum. And the result will be the same which is phi of t is equal to phi zero times cosine omega t where omega is equal to square root of g over r. Period is equal to 2 pi square root of r over g. So the period is longer if the radius is greater. So if I have a bigger sphere with the same angle it will obviously go longer to cover the whole cycle. Well, that's basically it. I do suggest you to read the notes for this lecture on Unisor.com as I said. They are more accurate than whatever I'm just drawing on this board. And again, the problem is that the reality we are dealing with is much more complex than our models. Whatever I was just presenting to you right now is just a model obviously because I changed for instance sine of the angle with an angle itself. Now same thing was with Hooke's law in the previous lecture. The force is proportional to displacement if you have a spring. But again, it's an experimental fact and it's actually the fact only if these deviations from the equilibrium are very small. Same thing as this one. For a small angle I can do this. For a bigger angle I have to go through real equation which cannot be really simply solved. So just one more confirmation of my general concept that whatever we are modeling we are actually simplifying and we are deviating from the reality into our model. If it's a good model it corresponds to a certain degree to a reality. Now how can we do it a little bit better maybe? Well we can represent sine of phi of t if we can approximate it, not with just phi of t. It's a very kind of rough approximation but we can approximate it with a polynomial and if this is a polynomial we can probably solve it a little bit better. Okay, that's it for today. Thank you very much and good luck.