 Hi, I'm Zor. Welcome to Inizor Education. We continue talking about waves, about oscillations, to use a little bit more scientific term. In this case we will talk about waves or oscillations inside some kind of a medium. Medium is basically used as a term in this particular case as some kind of environment which basically can propagate these oscillations. For instance, air. We hear the sound just because the molecules of air are delivering that sound inside our ear. Or if you, for instance, hit the bell with a hammer, you will actually even see sometimes how the whole bell is vibrating. Well, it means that inside this bell the molecules are oscillating. So waves in the medium, that's the term basically this particular lecture. Well, this lecture is part of the course called Physics for Teens, presented on Inizor.com. That's where I suggest you to watch this lecture and read notes for every lecture which are basically parallel to the video. I'm asked a couple of times actually about what is the sequence of studying your course. Well, the sequence is here because everything is menu-driven. You have some kind of a menu-driven. First is a Physics for Teens as a course because there is a prerequisite course, Mass for Teens, for instance. Now the course, if you will open that menu, will open up into some kind of a subject like in Physics, for instance, its mechanics, maybe electromagnetism, etc. And then every subject, when you open that particular menu, will present you a certain number of topics. Again, it's top-down, left-to-right. These topics must be addressed and each topic contains a certain number of lectures devoted to this particular topic. So it's like a multi-layer menu. I think it's quite conveniently presented on the web. And that's basically the way how it's supposed to be studied. Now the website is completely free. There are no advertising, no strings attached. You don't even have to sign in if you don't want to. Okay, so let's go back to oscillating molecules. So when we're talking about medium, as the environment where waves are propagated, the question is how they are propagated, these waves inside the medium. Well, obviously we're talking about oscillations of molecules. Now, if it's something like air, well, the molecules of air are chaologically moving inside whatever room or wherever they are. And then if there is some kind of a sound, which means basically I'm forcefully trying to move certain molecules, which are very close to the source of the sound. So they are moving a little bit more intensely and they have some kind of a direction from the sound to the sound, from the sound to the sound, because that's how the sound actually is produced, right? Now that movement superposes onto the chaotic movement of the molecules themselves, which creates a certain more probability to move back and forth from the source of the sound in all directions, than just playing a chaotic movement. So if a chaotic movement has, let's say, 50% probability to move here or there, 50% here or there, and 50% here or there. If there is a sound, like 80% probability to move from the sound and then towards the sound, then the other way around, and that's supposed to be done in sync with the movement of the source of the sound, whatever that source is. Like a bell for instance is ringing and the sound is propagating from the bell to all the sides. So it gives a little more probability to move back and forth from the sound towards the sound in all directions, than in all other tangential, let's say, or perpendicular to the direction to the source direction. So this more probability creates pressure, so there is a little more pressure from and to the sound in the air than across, let's say, perpendicular to this direction. And that basically propagates the waves because eventually it reaches our ear and there is a probability of hitting the ear back and forth rather than across, which doesn't produce any sound. Now if we're talking about sound inside the metal route, for instance, you hit the hammer with the hammer one end of the metal, thin metal rod, then obviously the oscillation of the crystalline structure at the end of the hit will be deformed a little bit and then it will return back. But this deformation will deform the next layer of the crystalline structure and then the next one, the next one and that's how the sound is propagated towards the other end of the road. Okay, this is all nice. But as you understand, the real physical essence of this propagation is extremely complex. If we are talking about molecules, it's definitely very, very complex. Well, first of all, the sound is propagated towards all the different direction inside the medium. So it's not like a one-dimensional string when we were talking about just a string and it goes back and forth, back and forth along one dimension. That one was quite primitive, relative to whatever is happening inside the medium. Let's say it's inside this crystalline structure of the metal. This really kind of deforms in a very, very complicated way. Okay, so we know that. So we have to somehow research it. We have to somehow analyze it. Well, what businesses do in this particular case and in many other cases when the structure is really complex, they build a model which is mathematically not as complex and just research the model, see if the model produces the results which are kind of comparable with experimental results and if they do, the model is supposed to be good and good for predicting, for instance, some effects or qualities for the future. So that's what we are going to do. We will build a relatively simple model of how the sound is propagated inside the medium. Here is what we will do. First of all, we will reduce our three-dimensional problem to one-dimensional along a ray which goes from the source of the sound or source of any kind of oscillations to any direction because it goes to all the direction inside the medium. So we will choose one particular direction and we are thinking that the movement will be only along this ray from the sound towards the sound like a spring, basically. So we have basically many springs which are radially going out from the source of the oscillations inside the medium. Now, since we are talking about independence of these directions, let's just forget about all directions except one. Let's just consider we have one particular thin metal rod, for instance, and the width of this thin metal rod is so small that we can consider that one single molecule actually is inside on every certain distance from each other and basically it looks like this. So this is our rod, very thin, which has molecules, one molecule per any kind of a distance. And now, since we are talking about propagation of the oscillations, so if it's a crystalline structure or if it's air and then there is a chaotic movement and they are bumping into each other, we are considering that there are springs between them. So molecules have certain weight or mass, let's say, springs have certain coefficient of elasticity between them. And that's the model which we are talking about right now. So we will talk about this particular model of how the oscillations, how waves are propagated inside any kind of a medium. So they are propagated in one particular direction from the source to whatever they go along a straight line. And that straight line consists of molecules, identical molecules, connected to identical strings. And whenever we are basically moving one particular molecule, it squeezes this string, then it actually acts on this model, on this molecule, it squeezes this string and that's how the oscillations propagate it. Okay, so this is our model and now let's investigate this model using the apparatus which we already know. The Hooke's law and the second Newton's law. Exactly the same way as we were analyzing the movement of the object on the string in previous lectures related to purely mechanical oscillations. Well in this case it looks like we are kind of forcing the first molecule with the source of the sound and then it forces the other and the other and the other. And at certain moment there is no more source of the sound. There is only propagation. So let's say we hit the bell with a hammer and then just let it go. Which means we basically made an initial speed basically of one particular molecule and that molecule transfers these oscillations further and further. Okay, so this is our model and we just investigate it in some kind of a relatively simple case. Now the simple case is to analyze the propagation is when I have only two molecules. I will tell you a little bit later the situation with N molecules along the rod which means we are talking about propagation through N molecules is very very similar. Let's just consider that we have only two molecules to examine how they act on each other. So this is my left part, this is my right part, this is my x-axis and let's assume that left. Okay, let's assume that initially all the springs are in their neutral state. So if there is no sound I just assume that there is some kind of a neutral state. The springs are not squeezed, not stretched. Okay, and then so if L is the length of the spring then the coordinate of this will be L coordinate of this 2L and coordinate of this 3L. So all springs are of equal lengths and equal elasticity and molecules have the same mass. So that's my that's my basically model which we are talking about right now. And let's assume that in the very beginning everything is at rest basically and the strings are in neutral position. Right, fine. Now let X1 be a displacement of molecule number one and X2 will be the same for number two. This is number one, this is number two. Now initially both of them are at zero because we are talking about displacement which means it's deviation from the original position because we are basically talking about oscillation so they are going back and forth around initial position. So we are talking about well in this case coordinate of the first molecule will be L plus X1 of t coordinate of the second molecule will be 2L plus X2 of t coordinate on the X axis. But we are not talking about actual coordinates we are talking about only about displacement from the initial position. So initial position initial displacement is zero that's not displacement that's derivative and initial displacement of the second molecule is zero but now we have to imitate initial hit of the hammer on the bell. I will put a speed of the first molecule to V and the second to zero. So in the very beginning we give it a push so it moves and then we will see what happens with the other molecule. So these are my initial conditions now we have to talk about differential equations which basically describe the movement. Let's talk about the first molecule. What kind of forces are acting on this molecule? The force of the left spring and the force of the right spring. The right spring is the one which is in the middle. Let's talk about force towards the first molecule from the left boundary. Now this is fixed, we are considering, right? So this force as a function of time according to the Hooke's law depends on how far my displacement of this particular model actually moves from the initial position, right? So it's minus k times x1 of t. That's regular Hooke's law. So if x1 is positive so if molecule goes here it stretches this particular spring over the neutral state and then it means it pulls it to the left and that's why it's minus. And obviously the force is proportional to displacement and that's the Hooke's law. What other force? Spring. Now middle spring is not from the fixed but from the variable position of the second. So my second force which acts on the first model from the second. So you see these indices this is from the left side and this is from the second molecule of t. Well it depends again on the length of the spring. Now what is the length of the spring? Well there is initial length and there is a displacement. So what is a displacement? Displacement is the difference between displacements of the molecules. So if this molecule is moving by x1 at moment t and this one is moving at distance x2 from the initial position then the difference between these two numbers is increase or decrease of the spring between them. So let's just think about and it's obviously it's supposed to be proportional to the elasticity of the spring. Now let's think about the sign. Well if x1 is greater than x2 so this moves to the right greater than this one moves to the right. The spring will be squeezed, right? If this moves further than this the spring will be squeezed and the force will be towards the left. So if this is positive the force is negative. Now if this is negative it means that my increment of this molecule is less than increment of that molecule which means spring will be stretched and the force in this case would be towards the positive direction of the x. So if this is negative times minus would be positive. So this is the correct formula. Great. So what is the force which acts on the first molecule? Well it's sum of these two forces, right? So if I sum of these two forces I will get force acting on the first molecule and it's equal to minus 2kx1 of t plus kx2. Okay? Now let's do exactly the same with the second molecule. Second molecule has two different again springs on both sides. Now the spring which is on the right it's attached to to the right side right edge of the thin metal rod, right? So from the right second molecule from the right. Well that's usual. Minus k times x2 of t. So if x2 is positive it squeezes the spring so it will push it to the left. That's why it's minus. Now if this molecule moves to the left which means x2 is negative it will stretch the spring and it will push to a different direction. This is the Hooke's law. Now how about force from the first molecule? How from the first? Well obviously it also depends on this on difference between displacement and its proportional. Question is the sign. Well let's just think about it again. If x1 is greater than x2 then the spring is squeezed. If it's squeezed it attempts to expand and the left hand forces this way and the right hand moves that way. So if this is positive the sign should be positive here on the second molecule. And again if it's negative if x1 is less than x2 which means we are basically stretching the spring then the force will be towards the left. So if this is less than this it's negative. So this is the correct thing. And as a result we can have that f2 of t is some of these two forces, right? Which is what? k times x1 of t minus 2k minus x2 of t. k is 1 and minus 2k for x2. So these are our equations and now we can attach second derivative. Let me just put it x of t, right? mx first, second derivative of time and this is mx2 of t, second derivative. So this is acceleration. m times acceleration that's the second Newton's law should be equal to the force, right? Well, so we have these two differential equations. So instead of one differential equation which we had for spring when mechanical oscillation we have two because we have two molecules. Now, if we will have three molecules or four molecules well, let's just talk about three I will tell you without basically going into the details you see the third molecule doesn't really act on the first and the first molecule only the second and the second molecule only the first and the third are acting so only neighbors are acting on any particular molecule. So the whole thing will be basically very, very much like this one but when I will just express it somehow differently I will tell you how the expression will be for n molecules. Okay, so let's forget about how we derived it so we have this system of two differential equations linear equations by the way. Now let's just remember that this whole course is the mathematicians view onto physics. How would mathematicians view this? They don't like the systems they are using whatever the apparatus they have to simplify the whole thing. Now, what can we do to simplify? Well, let's consider a vector which is basically a vector of displacement. So this is a displacement vector which basically has two components displacement of first and displacement of second and let's consider a matrix minus 2k divided by m k divided by m k divided by m minus 2k divided by m. Well, if I divide m it will be k over m and nothing here. Now, in terms of this vector and this matrix if you remember what is the matrix what is the vector and what is the multiplication of vector and matrix you can say that the second derivative of this vector which is the vector of second derivatives derivative of the vector is vector of derivatives is equal to matrix times well, I'm using the vectors okay? So, as we see we have basically an equation which very much looks like equation which we have created when analyzing the let's do it slightly differently let me put this as a plus and this as a minus and then I will use plus here equals to 0 0 is vector 2 Now, this is exactly like equation which we have Now, this equation was the main equation for free oscillations of the object on the spring where x of t is basically displacement of the object mechanical oscillations lectures that's the previous topic had this as the main homogeneous differential equation of the second order linear by the way and we were using actually the deviation omega square is equal to k over m and then we had basically a solution to this was equal to d times cosine of omega t plus phi now this is all part of the mechanical oscillations now you see equation is very much like this one the only thing is we are using scalars here it's a scalar function and scalar product here we are using vector and matrix and this is vector derivative and this is the multiplication of matrix by vector if the behavior of this properly defined is similar to the behavior of regular operations on scalars then we can probably solve it in more or less the same way how we solved this differential equation basically we were looking for a solution and we found it we found a general solution to this equation which looks like this with d and phi depending on initial conditions and initial conditions we have so here in this particular case when we are talking about vectors and matrices we kind of expect that we might proceed the same way and obviously and well as you see in this particular case mathematics is really serving physicists quite well actually one very smart and very accomplished mathematician Russian mathematician Arnold said once that mathematics is supposed to be basically part of the physics because everything mathematicians did it was not only for their own satisfaction it was actually used well almost everything used in certain very very practical situations like this one ok now how we will approach this particular differential equation with vectors and matrices well that would be a subject of the next lecture it's like tales of Arabian knights finishing one particular tale she was selling she was saying that Shekhar is out she was saying that ok she did not finish this tale and said that the continuation will be tomorrow and her life was spared because of that so I hope my life will be spared as well and the next lecture will be devoted to this particular equation and how can we deal with it it's purely mathematics actually and we might have basically have this lecture in the math 14th course but I think it's more appropriate here even it's more like mathematics than physics but it's very very direct application to physics ok that's it thank you very much I suggest you to read all the notes for this lecture on theunisor.com you go to physics 14th you go to subject and this is waves in the medium topic first lecture of this topic actually there will be the next one will be for instance a continuation of this alright thanks very much and good luck