 Hi, I'm Zor. Welcome to Inusor Education. We continue talking about propagation of waves in medium. Today's lecture is a direct continuation of the previous one where we have introduced basically physical concepts behind propagation of waves. Today we will primarily deal with mathematics of this. So I will remind you what kind of an equation we have derived with in the previous lecture, and then basically solve this equation. I call this wave equation one actually. Well, there are many different equations which can be called wave equations. This is the most, well, not the most, but very, very simple form for only like two molecules oscillating and basically transferring the oscillations from one to another. It's as simple as it can be whenever we are dealing with medium. Okay. Now, this lecture is part of the course called Physics for Teens. It's presented on Unisor.com. I do suggest you to watch this lecture and all other lectures of the course from the website because the website, well, it has manual, which means you can actually direct yourself in the logical sequence lecture by lecture. Every lecture has notes. Notes are basically like a textbook and they put, well, I put in writing basically whatever I'm talking about during the lecture time. And also the website has exams and there are some other functionalities. So I do suggest you to go to this lecture from the website. You go to Unisor.com, then choose the course Physics for Teens. The subject will be waves. And then when the new menu will open, you will see the particular topic called waves in medium. And this is the second lecture in this topic. Now, the first lecture, the previous one, was basically about deriving some kind of differential equation which describes propagation of waves. And I have decided to simplify the whole problem to only like two molecules you have. Let's say this is a thin metal rod which contains only two molecules. That's a very short one. And there are springs which connect to the sides and in between these two molecules. So I assume that the molecules have mass m. Springs are all the same and they have coefficient of elasticity k. And then how can we derive the equation for motion of these molecules? Well, that's very simple. This molecule has basically two springs attached on both sides. If x1 of t is a displacement of this molecule from initial, we are assuming that initially the position is neutral, which means strings are not stretched nor squeezed. So they're all in neutral position of the same length. And x1 of t is a displacement of this molecule from the neutral position as a function of t. And this molecule will be x2 of t, obviously. So depending on this displacement, the forces of the springs will start acting. So the force from the left would be minus kx1 of t. That's the Hooke's law. And the force from the right depends on both displacements, basically. So it's basically kx2 of t minus x1 of t. So this is basically the total force which acts on this molecule. And that's why, according to the second Newton's law, this is the mass times acceleration. x1 is acceleration with two slots. Okay. Now, this is equal to minus 2kx1 of t plus kx2 of t. Now, the second one, again, we will do second law of Newton, mass times acceleration, total force. It's equal to force from the right. It's minus kx2 of t, right? That's the Hooke's law. And the Hooke's law from the left, it depends on whether the x1 and x2 are related among themselves. So basically, it would be the other way around. x2 of t minus x, x1 of t minus x2 of t. And that's why it's equal to kx1 of t minus 2kx2 of t. So these are differential equations which were derived in previous lecture. And this lecture will be devoted to solution to this problem. Now, obviously, as with all differential equation, what's important is initial conditions. And I will not talk much about this, but initial conditions is basically position and velocities of these two molecules. Position, you can say that since I was talking about initially, they're all in a neutral position. So x1 of 0 and x2 of 0 are 0, no displacement in the initial position. And we can talk about, let's say, some kind of a push, which means that x1 of t can have the first derivative equal to something like b, whatever the speed is. And then we can basically, on this information, we can not only solve the general solution to the system of equation, but also concrete for this particular configuration and initial conditions. I will not talk much about initial conditions. It's only about just general solution. Now, this lecture actually would be more mathematical than physical. Physics was in the previous lecture, and this is the end of it. Now I will talk about basically how to solve this particular system. Now it's system of two equations. Now let me just remind you that in the very beginning, when we were talking about harmonic oscillations, our main equation was, and then k over m, we actually symbolized as omega square, where omega was angular speed. And solution to this was d cosine omega t plus phi, where d is any kind of amplitude and phi is any kind of a phase shift. So this is the general solution to this particular equation. And we were discussing this in the lecture about harmonic oscillations. Now, keep this in mind. This is something which we know, and it's basically very easy to derive. If you don't remember, you can just go back to that lecture. And I will be actually using this idea in solution to this system. Now this system doesn't look like this one, obviously, because there are x1 and x2. It's a system of two equations, differential equations. And x1 and x2 are mixed together. So what can we do about that? How can we approach solving this? Well, again, this is more mass than physics. Okay, so first of all, let me just change this equation slightly. I will divide it by m. So I will have k over m in both cases, which I will replace with omega square. And then my equation will look like this. So I will have x1 of t. I will put everything on the left side. So it's plus 2 omega square x1 of t minus omega square x2 of t equals to 0. And the second equation would be minus omega square x1 of t plus 2 omega square x2 of t equals to 0. Okay, so this is the system of two differential equations with two unknown functions of time. And we will talk about how to solve. First of all, I would like to remind you that there is a mass of teens course, which is prerequisites for this, where in particular I'm talking about something which I'm going to use extensively in this lecture, which is matrices. So matrices, how to multiply matrix by vector, or vector by matrix, their properties. And basically that's all we need is just basically rules of multiplication. Not just in case you forgot multiplication of constant by vector. In this case, I will use the vector of two components, a and b. If I multiply constant, that's basically a new vector which is multiplication of this. So that's one property which I'm going to use. Another property is if I have a matrix, let's say c11, c12, c21, c22. Multiplied by vector. Well, in case you forgot about how to multiplication work, let me just remind you, it's the first row times this column, which is c11e plus c12b. And then the second row, c21e plus c22b. That's the result of multiplication. This is how it's all defined and it has all the nice properties, et cetera, et cetera. And I will be using these type of things. Now, using this, let me just express this in a vector format. Vector and matrix format. I will use the vector, I will call it x. It's a function vector. It depends on t. It's x1 of t, x2 of t. It's a vector. Okay, so at any given moment time t, I have the two values which you can interpret as a point on a plane with Cartesian coordinates. One is x and another is y, or in any other way. But let's consider this particular geometrical interpretation of a vector as, so this would be my x1, this would be my x2. Absolutely, this is x1, not x2. Just keep it in mind. Okay. Now, this is my vector. Now, if I would like to differentiate a vector, well, I will differentiate component by component. So, if the vector is a vector function, then its derivative is basically a derivative of each component, which means x. The first and the second derivative is basically vector of derivatives. I mean vector components of which are derivatives of components. Okay, now, using this, let me just express this particular equation as vector equation. Now, what is x1 and x2 with the second derivative? That's second derivative of the vector. Okay. Now, consider this is a small case and this is upper case. It's kind of difficult to do it on the board. But this, if without index, that's the vector in this case. Sometimes I'm using maybe this just to make sure it's a vector. Okay. Okay. Now, how can express this in vector form? Well, very easily. Consider the matrix 2 omega square minus omega square minus omega square 2 omega square. Multiply by my vector x of t, which is x1 and x2. x1, x2. Multiplication of this times this would be 2 omega square x1 minus omega square x2. Right? Okay, let me write it down. x1 of t, x2 of t. So multiplication of matrix by vector is a new vector. So this is a vector and a new vector. The top component would be rho by column, which is 2 omega square x1 minus omega square x2, which is exactly this. And the bottom component would be this minus omega square x1 plus omega square x2. So instead of this, I can say some kind of a matrix omega, which does not depend on time. It's a constant, matrix of contents, omega square is a constant, times vector x of t equals to 0. Now, what is 0? 0 is actually also a vector, which means 0, 0, 2 components. So this is a vector. Matrix times vector is also a vector, which means this has 2 components, this has 2 components like this, and this has 2 components, 0, 0. And whenever we are talking about equality between 2 vectors, that means each component is equal. Whenever we are talking about plus vector, plus vector, we are adding by component again. So basically this is the same as this, just shorter. However, it's not just shorter. It's actually multiplication, but with different objects, not with numbers, but with matrices. This is addition, not with numbers, but with vectors. This is a derivation, derivative of the vector, not with vector function, not just a regular function. And this is a vector. So it's still the same meaning and operations plus and multiplication with vectors and matrices have very similar properties. So I can really treat it as multiplication and addition, because the properties are basically the same, almost the same. Now, why did I do this? Well, because this looks like this. Let me just rewrite this. Instead of this, I will put omega square. Now it looks exactly like this one, but instead of omega square, we have a matrix, which contains four components, 2 by 2, and instead of x, we have vector x, and instead of 0, we have vector 0. So it looks the same, but it's quite different. I can just easily write this particular formula as a solution. It's not that easy. However, I would like to somehow reduce this one to this. How can I do it? And that's a very, very important moment. It's important for mathematics and for physics. What matrix is doing with the vector? Let me just give you an example. What matrix, let's say, 5, 0, 0, 5 doing when it's multiplied by vector v1, v2? Let's just think about what happens. Rho by column. So it's 5 by v1 plus 0 by v2. So it's 5 v1. 0 times v1 plus 5 times v2, which is 5 v2, which is equal to 5 times v1, v2. So this matrix basically stretches this vector. Whatever it is, v1, v2, we convert it into 5 v1, v2. So there are certain matrices which are stretching the vectors. Now, this particular matrix will stretch any vector. Now, there are some matrices which are not really stretching any vector, but there are certain vectors which they are stretching. Now, there are matrices which are doing something else, like turning the vectors. For instance, let's give you another example. For instance, you have a matrix 0 minus 1, 0. What happens then? We have 0 times v1, which is 0, minus v2. So we have minus v2. And 1 times v1 plus 0 times v2 is v1. So what is the relationship between the initial vector and the result? Well, let's just think about it. If the initial vector is this, v1, v2, the new vector is minus v1. So this is v2. This is v2. It goes to here. This is minus v2. And v1, instead of horizontal, it goes to up. This is v1. So this is a new point. So this is a new vector. So this vector is converted to this, which means it's turned by 90 degrees. So this particular matrix is turning any vector by 90 degrees. And now there are some matrices which are turning some vectors or stretch another vector, the same matrix can do differently. This always turns. The previous example 5500 was always stretching, but there are matrices which are stretching one vector but turn another vector or turn and stretch maybe or something. So there are more complicated cases, basically mixture of these two cases, pure stretching and pure turning. These are just two examples, but there are some examples which are more complicated and they have both. But if I will find certain set of vectors which this matrix just stretches, so instead of multiplication by matrix, I can put multiplication by number, whatever the number is. Then our equation would look like c times. This is a number. This is a number. Basically, I can solve this equation component by component, which means x1, x2 plus c times x1, x2 equals x0, 0. Now c can be here, c times. And now I can component by component solve these two equations. And they are by the way the same, so it's basically one equation. So if I will be able to find certain set of vectors, set of vectors where omega matrix acts only as a stretching mechanism, then I will be able among these vectors at least attempt to solve the equation. Maybe it will have some solution in that set of vectors which are stretched by the matrix. Okay, now let me introduce terminology. If the vector is stretched by a matrix, only stretch, just the same direction but longer or shorter, then this vector is called Eigenvector. And the coefficient by which we are stretching this vector is called Eigenvalue. Okay, so my purpose is to find certain vectors where omega times x of t, well just let's forget about x of t, just d, some kind of vector, is equal to some kind of lambda times z. That's my purpose. For this given matrix, matrix is a constant that's given, I would like to find this matrix Eigenvectors. And then if I will have a whole family of Eigenvectors among these Eigenvectors, I can then try to find the solution to this differential equation. Okay? Now, what is this? Let me just describe it in more, well, matrix format. It's 2 omega square minus omega square minus omega square 2 omega square matrix times V1 V2 equals 2 lambda times V1 V2. So this is where I have to find all V1 V2 which basically give me this particular kind of stretch. I have to find all these vectors. Well, first of all let me give you the answer. If, what happens if I will put the same values V and V, whatever the V is, look at this, 2 omega square times V minus omega square V what will be as a result omega square V. Now, minus omega square V plus 2 omega square V would be what? Again, omega square V. So I can put here omega square as a lambda and this is a correct equation. Check it out again. 2 omega square times V minus omega square times V is equal to omega square V minus omega square V plus 2 omega square V is equal to omega square V. So we have a correct answer. So any vector which has both components the same would be the vector, would be Eigenvector with Eigenvalue equals 2 omega square. Okay? That's the answer. Now, let me give you another answer. Let me put minus here. Here is what I'm basically telling. Let's check it out. 2 omega square times V minus omega square times minus V would be 3 omega square V minus omega square V plus 2 omega square minus V would be minus 3 omega square V. So again, from V minus V, we get V minus V as a coefficient with a factor. So this is also an Eigenvector. So any vector which has two values which are exactly equal to absolute value but opposite and sine also is Eigenvector with a different Eigenvalue. The first one was omega square. The second Eigenvalue is 3 omega square. How did I receive it? How did I guess the answer to this? It's very simple actually. Let's consider again this is V1 V2 and this is lambda and this is V1 and V2. Let me rewrite this slightly differently. Instead of lambda times V1 V2, I will put a matrix lambda 0 0 lambda times V1. Why can I do it? Well, for a very simple reason. This matrix times this vector is what? Lambda times V1 plus 0 times V2 which is lambda V1. 0 times V1 plus lambda times V2 is lambda V2 which is exactly equal to lambda times V1 V2. Now why did I express it as a matrix? For a very simple reason. Because now I can write it instead of this. I can write this matrix here. Now I have a little bit more uniform. Matrix times vector, matrix times vector. And now all these manipulations with equations which we are doing are valid for vector and matrix type. So I will transfer everything on the left and I will take V1 V2 out of parenthesis and I will have 2 omega square minus omega square. This is not minus. This is too different. Minus omega square. 2 omega square. Matrix minus matrix lambda 0 0 lambda. This is a new matrix multiplied by vector V1 V2 equals to 0. What is the difference between two matrices? It's by component by component. So instead of this I can write 2 omega square minus lambda minus omega square minus omega square minus 2 omega square plus minus lambda. This is my matrix. So the new matrix, let me put it in parenthesis. New matrix times this vector is exactly the same as this. Now the matrix times vector is equal to 0. Well obvious solution is V1 and V2 are 0. But we don't need these trivial solutions. We need real eigenvector to find. So this is basically a system of two equations with two unknowns. Now obvious solution is 0 so we have one solution. Now I need other solutions. When do we have other solutions if we have a system of two equations with two unknowns? Well, when both equations are basically like proportional to each other. You know, you have an equation like x plus 5 equals 0 and 2x plus 10 equals to 0. Sorry, plus 10 y. These are the same equations, right? Because this one is just this multiplied by 2. So we have not 2, we have actually one equation. And that's the only way how we can get multiple nonzero solutions. Because if it's something like 2x plus y is equal to 0, then this system has only 0 as a solution. x0 and y0, the only solution. So whenever you have 0 on the right and you have this linear type of equations on the left, the only non-trivial solution is when the determinant of this matrix is equal to 0. Now what is the determinant of this matrix? If you forgot about the determinants, again I can refer you to previous lecture. But it's kind of obvious. Whenever you have equation that say ax plus dy equals 0, cx plus dy equals to 0. Two equations with two unknowns. How do you solve it usually? Well, you multiply for instance this by d and this by b and subtract from each other. And what happens? bd, bd, y, it will cancel out. And what will be ad minus bc is equal to 0, right? Times x. Now if you are not interested in solution x equal to 0, ad and bc must be 0. And that's what actually is ad minus bc. This is a determinant of this 2 by 2 matrix. So my idea about having non-trivial solution basically results in a very simple equation for lambda. 2 lambda square minus lambda times 2 lambda square minus, which is square, minus, minus omega square times minus omega square, which is omega to the fourth. And that's supposed to be equal to 0, right? So 2 omega square minus lambda is equal to plus minus omega square, right? Omega to the fourth goes to the right. Square root of both sides. And then we have two solutions. If it's a plus, then lambda is equal to lambda square. If it's a minus, then lambda is equal to lambda first, lambda second, 3 omega square. Remember when I was talking about I guessed the results of eigenvalue equation, my first case was coefficient was omega square, and the other eigenvalue was 3 omega square. That's where I from. 1 and 2. Okay, so we found the possible eigenvalues. Let's find possible eigenvectors in both cases. Okay, so the first one is, okay, let's consider this case. Okay, what happens in this particular case? Our omega matrix times some kind of vector v1 v2 is equal to omega square v1 v2. Well, let's find out what is v1 and v2. So instead of omega matrix, I will put exactly what it is, 2 omega square minus omega square, minus omega square, 2 omega square matrix. Must multiply. 2 omega, 2 omega square v1 minus omega square v2 is equal to the first one. Now this one minus omega square v1 plus 2 omega square v2 equals omega square v2. Now from this, this omega square and this omega square. So we go this to the left, this to the right, and what do we have? Omega square v1 is equal to omega square v2. Here, this goes to the right, this goes to the left, and we have omega square v2 is equal to omega square v1, which is basically the same thing. Obviously, we have only one condition. v1 is equal to v2. So if v1 is equal to v2, if both components of the vector are the same, then this is an eigenvector with eigenvalue omega square, which means that any vector which describes the displacement of both molecules are such that x1 is equal to x2, if then this is an eigenvector, and these vectors actually are such that our matrix is just stretching them. So we can actually find solution among these vectors, solution to our differential equation. Now, what happens if we start from this? Okay, let's do it this way. Let's do it this way, this way, and what do we have? This goes to the left, this goes to the right, and I will have minus omega square v1 is equal to omega square v2. Here, this goes to the right, this goes to the left, I will have minus omega square v2. So as a result, again, these are the same. v2 and v1 are opposite in sign, but exactly the same. So what it means? It means that all vectors which are this type also are eigenvalues, but with different eigenvectors, but with a different eigenvalue. This with eigenvalue omega square, this with eigenvalue 3 omega square. So being as it may, we will actually consider solving our differential equation not for all the different possible, this is our differential equation. If I will solve it not for all the possible vectors x, which has component x1 x2, but only for those vectors x where components are the same, or components are the same by absolute value but different in sign, then all these vectors are eigenvalues, and instead of multiplying by matrix, I can multiply by corresponding eigenvalue. So let me solve this equation in the first case. So x1 x2 plus omega square x of t x of t. Sorry, I said they are the same components. Only if they are the same, then my equation actually can be expressed as a multiplication is equal to zero. So if I'm looking for a solution x of t and x1 of t and x2 of t are the same, which means our molecules are synchronously moving left and right back to the picture. Two molecules springs, so they are synchronously moving. I'm looking only for this particular solution. Is there a solution? Is our differential equation would satisfy in this particular case? Well, the answer is obviously yes, because this is basically two exactly identical differential equations, the first component and the second component, right? They are the same, because these are the same, these are the same. Multiplication by omega square is multiplication by a component. So I have exactly this type of equation, and I know the result. I know the solution. So this is a particular solution when x1 of t is equal to x2 of t equals to d cosine omega t plus 5, where d amplitude and phi phase shifts are basically anything, and it will satisfy. So if both are the same, you will see that the equation is completely satisfied, because then multiplication by omega square is the same as multiplication by omega square, and everything falls from there differently. If I will do this and this, I know that this is an eigenvector with eigenvalue 3 omega square. No, I have to put minus here as well, right? Because if this is vector, its first derivative and second derivative are the same, plus and minus would be here. So, and the solution is, again, exactly the same. This equation on the first component and the second component are exactly the same, because minus and minus and zero, so it's still the same thing as plus and plus and zero. And I will have yet another solution. x1 of t is equal to minus x2 of t is equal to some kind of other amplitude and cosine. Now, instead of omega, now I have to have square root of 3 omega, right? t plus some other psi phase. Again, e and psi are any amplitude and phase shift. Now, square root of 3 omega basically comes from here, right? If this is omega square, this is omega. If it's 3 omega square, it's supposed to be square root of 3 omega. So, I have another solution. So, let me just summarize it. What kind of two solutions I have? One solution is a vector with both components the same, and it's d, d cosine omega t plus phi, d cosine omega t plus phi. This is my vector which represents a solution to differential equation for any d and any phi, any amplitude, any phase shift. Now, another solution is another vector, e cosine square root of 3 omega t plus psi e minus, sorry, now it's minus e. They are opposite. So, I have two solutions, two partial solutions. Now, if I have two partial solutions to a system of linear differential equations, obviously any linear combination of them would also be a solution, right? So, if I multiply this by alpha and this by beta and add them together, that would be also a solution. Now, basically it's another story if this represents all the possible solutions. Well, let me just express this as one vector actually. So, it would be alpha times d, but d is any and alpha is any. So, it doesn't really matter whether I multiply two any numbers. I can put just one coefficient and we'll have alpha. Cosine omega t plus phi plus beta. Again, beta times e, both are any numbers. So, I can just leave just beta. It would be the same thing, cosine square root. And the second component of this vector would be, so this vector, this is x1, this is x2. So, this represents displacement of the first molecule. This represents displacement of the second molecule. So, this is a general solution to our problem. Now, whether these two components of this particular vector where alpha and beta and phi and psi are any amplitude and phase shifts, it's an open question right now. I'm not addressing this if this is really all of the solutions. This is kind of beyond the complexity of this course. It's complex enough. But in any case, yes, this is a complete solution. And let's just now think about it. If beta, for instance, is equal to zero, then I have the same components here. They're always synchronously moving left and right. There are two molecules. If alpha is equal to zero and beta is not, then they are opposite in case the phase are equal to zero. But phase might be not equal to zero. And that complicates the issue. So, we have different phases like this and different amplitudes, alpha and beta. And the combination is complex. So, these two molecules on springs in a thin road are actually making, depending on initial conditions, by the way, they are making, might make very complicated movements. Because both initial position can be not equal at zero. Both initial speeds can be not equal to zero. But basically, for example, if you will take these two molecules and just move it to the same displacement, initial displacement, and then let it go, most likely they will be together oscillating. Or if you will put it into different directions by the same displacement and let it go, they will probably will move like this. So, it all depends. But if you will start doing some shifts, etc., that would be obviously a problem. But this represents a general solution. And if you want, for those of you who are really inquisitive minded, you can just try to put these two into our differential equation, into our system of differential equations. Not big deal to differentiate couple of times this thing. And you will see that this is a solution regardless of alpha, beta, phi and psi. So, that's very important. And again, whether it's complete solution or not is beyond of this course. It actually belongs to advanced course of differential equations, not physics. Now, all things which I was just talking about are, I think, very nicely presented in notes for this lecture on Unisor.com. And I encourage you to basically take a look at this. Well, it's a little longer lecture than usual. Thank you very much. And good luck.