The following content isprovided under a Creative Commons license. Your support will helpMIT OpenCourseWare continue to offer high qualityeducational resources for free. To make a donation or toview additional materials from hundreds of MIT courses,visit MIT OpenCourseWare at ocw.mit.edu. YEN-JIE LEE: Andwelcome back, everybody, to this fun class, 8.03. Let’s get started. So the first thingwhich we will do is to review a bit whatwe have learned last time. And then we’ll goto the next level to study coupled oscillators. OK. Last time, we had learned a loton damped driven oscillators. So as far as the coursewe’ve been going, actually, we onlystudy a single object, and then we introducemore and more force acting on this object. We introduce dampingforce, we introduce a driving force last time. And we see thatthe system becomes more and more difficult tounderstand because of the added component.But after the classlast time, I hope I convinced you that we canunderstand driven oscillators. And there are two very importantthings we learned last time. The first one is thetransient behavior, which is actuallya superposition of the homogeneous solutionand the steady state solution. OK. One very good news is thatif you are patient enough, you shake thesystem continuously, and if you wait long enough,then the homogeneous solution contribution goes away. And what is actually left overis the steady state solution, which is actually much simplerthan what we saw beforehand. It’s actually justharmonic oscillation at driving frequency. Also, I hope that we also havelearned a very interesting phenomenon, which is resonance. When the driving frequency isclose to the natural frequency of the system, then thesystem apparently likes it. Then it would respondwith larger amplitude and oscillating up anddown at driving frequency.So that, we call it resonance. This is the equationof motion, which we have learned last time. You can see is theta doubledot plus Gamma theta dot, is a contributionfrom the drag force, and omega 0 squaredtheta is the contribution from the so-called spring force. And finally, that is equal tof0 cosine omega d t, the driving force. And as we mentionedin the beginning, if we prepare this systemand under-damp the situation, then the full solutionis a superposition of the steady statesolution, which is the left-hand side, thered thing I’m pointing to, this steady state solution. There’s no free parameter inthe steady state solution. So A, the amplitude, isdetermined by omega d. Delta, which is the phase isalso determined by omega d. There’s no free parameter. OK. And in order to make thesolution a full solution, we actually have to add inthis homogeneous solution back into this again.And basically, youhave B and alpha, those are the freeparameters, which can be determined by thegiven initial conditions. OK. So if we go ahead andplot some of the examples as a function oftime, so the y-axis is actually the amplitude. And the x-axis is time. And what is actuallyplotted here is a combination orthe superposition of the steady state solutionand the homogeneous solution. And you can see that theindividual components are also shown in this slide. You can see the red thingoscillating up and down harmonically, is steadystate solution contribution. And also, you havethe blue curve, which is decaying awayas a function of time. And you can see that if youadd these two curves together, you get somethingrather complicated. You will get some kind ofmotion like, do, do, do, do.Then the homogeneoussolution actually dies out. Then what is actually leftover is just the steady state solution, harmonic oscillation. And in this case, omega dis actually 10 times larger than the natural frequency. And there’s another examplewhich is also very interesting. It’s that if I make theomega d closer to omega 0– OK, in this case it’s actuallyomega d equal to 2 times omega 0, then you can producesome kind of a motion, which is like this. So you have the oscillation. And they stayedthere for a while, then goes back, andoscillates down, and stay there then goes back. OK, as you can see on there. The homogeneous solution partand steady state solution part work together and producethis kind of strange behavior. OK. And that’s just another example. And if you waitlong enough, again what is actually left overis the steady state solution. OK. So what are wegoing to do today? So today, we aregoing to investigate what will happen if we try toput together multiple objects and also allow themto talk to each other.OK. So if we have two objects, andthey don’t talk to each other, then they are stilllike a single object. They are still like simpleharmonic motion on their own. But if you allow themto talk to each other, this is the so-calledcoupled oscillator, then interesting thing happen. So in general, coupled systemsare super, super complicated. OK. So let me give youone example here. This is actuallytwo pendulums that are a coupled to eachother, they are actually connected to eachother, one pendulum, the second one is here, OK.And for example, I can actuallygive it an initial velocity and see what is going to happen. You can see that theresulting motion– OK. Remember, we are just talkingabout two pendulums that are connected to each other. The resulting motionis super complicated. This is one of myfavorite demonstrations. You can actually stare atthis machine the whole time. And you can seethat, huh, sometimes it does this rotation. Sometimes it doesn’t do that. And it’s almost likea living creature. So we are going tosolve this system. No, probably not, he knows. [LAUGHTER] But as I mentioned before,you can always write down the equation of motion. And you can solveit by computer. Maybe some of thecourse 6 people can actually try and write theprogram to solve this thing and to simulate whatis going to happen. So let’s take a look at thiscomplicated motion again. So you can see that the goodnews is that there are only two objects.And you can see– look at the green,sorry, the orange dot. The orange dot is alwaysmoving along a semi circle. But if you focuson the yellow dot, the yellow is doing allkinds of different things. It’s very hard to predictwhat is going to happen. So what I want to say is,those are interesting examples of coupled systems. But they are actually farmore complicated than what we thought, because theyare not smooth oscillation around equilibrium position. So you can see that nowif I stop this machine and just perturb it slightly,giving it a small angle displacement, then youcan see that the motion is much more easier to understand.You see. You even get one of thesequestions in your p-set. OK, that’s good news. So our job todayis to understand what is going to happen tothose coupled oscillators. Let me give you a fewexamples before we start to work on a specific question. The second example I wouldlike to show you is a saw and you actually connectit to two – actually a ruler, a metal ruler,which is connected to two massive objects. Now I can actually giveit the initial velocity and see what happens. And you can seethat they do talk to each other through thisruler, this metal ruler. Can you see? I hope you can see. It’s a bit small. But it’s really interestingthat you can see– originally, I just introducedsome displacement in the left-hand side mass. And the left-had sidething start to move or so after a while. There are two more exampleswhich I would like to give you, introduce to you.There are twokinds of pendulums, which I prepared here. The first one is there are twopendulums that are connected to each other by a spring. And if I try tointroduce displacement, I move both masses slightly andsee what is going to happen. And we see that the motionis still complicated. Although, if you stare atone objects, it looks more like harmonic oscillation,but not quite. For example, thisguy is slowing down, and this is actuallymoving faster. And now the right hand sideguy is actually moving faster. Motion seems to be complicated. Also, you can look at this one. Those are the two pendulums. They are connected to eachother by this rod here. And of course, youcan displace the mass from the equilibrium position. I’m not going to hit– not hitting each other.So you can displace themasses from each other. And you can see that theydo complicated things as a function the time. How are we goingto understand this? And I hope that by theend of this lecture you are convinced that you canas you solve this really easy, following a fixed procedure. In those examples, wehave two objects that are connected to each other. And therefore, theytalk to each other and produce coupled motion. Those are a coupleoscillator examples. There’s another veryinteresting example, which is called Wilberforce pendulum. So this is actually a pendulum. You can rotate like this. And it can alsomove up and down. It’s connected to a spring. The interesting thingis that if I just start with somerotation, you can see that it starts to alsooscillate up and down. You see? So initially I justintroduced a rotation. Now it’s actuallyfully rotating.And now it startsto move up and down. And you can see that theenergy stored in the pendulum is going back and forth betweenthe gravitational potential, between the potentialof the spring, and also between the kineticenergy of up and down motion and the rotation. They’re actually doing allthose transitions all the time. So you can see– so initially it’s just rotating. And then it startsto move up and down. And this one isalso very similar. But now the mass ismuch more displaced. And if I try to rotate thissystem without introducing a horizontal directiondisplacement, it still does is up and downmotion, like a simple spring mass system.So what causes thiskind of motion? That is because when we movethis pendulum up and down, we also slightlyunwind the spring. That can generate somekind of torque to this mass and produce rotational behavior. And you can seethat this is just involving one single object. But there’s a couplingbetween the rotation and the horizontaldirection motion. So that’s also specialkind of coupled oscillator. So after all this,before we get started, I would like to say thatwhat we are going to do is to assume all thosethings are ideal, without them being forced,without a driving force.We may introduce thatlater in the class. But for simplicity, we’lljust stay with this idea case, before the mass becomessuper complicated to solve it in front of you. And also, we can see thatall those complicated motion are just illusion. Actually, the realityis that all of those are just superpositionof harmonic motions. You will see that bythe end of this class. So that is really amazing.OK. Let’s immediately get started. So let’s take alook at this system together and seeif we can actually figure out the motionof this system together. So I have a system withthree little masses. So there are three littlemasses in this system. They are connected toeach other by spring. Those springs are highlyidealized, the springs. And they have spring constant k. And the natural length’s l0. And they are placed on Earth. And I carefullydesign the lab so that there’s no frictionbetween the desk and all those little masses. So once you get startedand look at this system, you can imaginethat there can be all kinds of differentcomplicated motions.You can actually, forexample, just move this mass and put the other two on hold. And they canoscillate like crazy. They can do verysimilar kind of motion. There are many,many possibilities. But if you stare atthis system long enough, you will be able to identifyspecial kinds of motion which are easier to understand. So what I would liketo introduce to you is a special kindof motion which you can identify from thesymmetry of this system. That is yourso-called normal mode. So what is a normal mode,a special kind of motion we are trying to identify? That is actuallythe kind of motion which every partof the system is oscillating at the samefrequency and the same phase. So that is yourso-called normal mode, and is a special kindof motion, which I would like you to identify with me. And we would later realize thatthose special kinds of motions, which are easier tounderstand, actually helps us to understand thegeneral motion of the system. You will realize that the mostgeneral motion of the system is just a superposition of allthe identified normal modes. And then we aredone, because we have a general solution already. So that’s very good news. That tells us thatwe can understand the system systematically,and step by step. And then we can write thegeneral motion of the system as a superposition ofall the normal modes. So let’s get started. So can you guess what arethe possible normal modes of this system? So that means eachpart of the system is oscillating at the samefrequency and the same phase.Can anybody and any one ofyou guess what can happen, each part of thisis an oscillating at the same frequency? Yeah? AUDIENCE: If the two masseson that side are displaced the same amountand then they’re — YEN-JIE LEE: Very good. So he was saying that nowI displace the right hand side two masses all togetherby a fixed amount, and also the left hand side, right, bya fixed amount and then let go. So that’s whatyou’re saying, right? OK. So the first mode we haveidentified is like this. So you have left hand sidemass displace by delta x. And the right handside two masses are also displaced by delta x. So basically you holdthis three little masses and stretch it by the same– introduce the same amplitude toall those three little masses, and let go. So that is actuallyone possible mode. And if we do this,then basically what you are going to seeis that this is actually roughly equal to this system. They’re connected to eachother by two springs. And the right hand side partof the system, both masses are oscillating at the sameamplitude and the same phase. They look like as if they arejust single mass with mass equal to 2m. And if you introduce adisplacement of delta x, then what is going tohappen is that if I take a look at the mass, lefthand side mass, and the force acting on thismass, the force will be equal to minus2k times 2 delta x, because that’s theamount of stretch you introduce to the spring.And that will giveyou minus 4k delta x. And we have already solvedthis kind of problem in the first lecture. So therefore you canimmediately identify omega, in this case,omega a squared will be equal to 4k divided by 2m. This is actually theeffective spring constant, and this is actually the mass. So that is actually thefrequency of mode A. Can you identify a second kindof motion which does that? So in this case, whatis going to happen is that the three masses will– OK, one, two, and three. The three masses willoscillate as a function of time like this with angular frequencyof square root of 4k over 2m. What is actually asecond possible motion? Yes? AUDIENCE: All masses beingstretched [INAUDIBLE] YEN-JIE LEE: Compressed. AUDIENCE: Compressed the same– YEN-JIE LEE: Very good. I’m very lucky that I’m in frontof such a smart crowd today. And we havesuccessfully identified the second mode, mode B.So what is going to happen is that the left handside mass is not moving. And you compress theupper one slightly and you stretch the lowerone, the lower little mass to the opposite direction.The displacement is deltax, and the displacement of the second mass is delta x. So what is going to happen? What is going to happenis that the left hand side mass will not move at allbecause the force, the spring force, acting on thismass is going to cancel. And apparently, thesetwo little masses are going to be doingharmonic motion. Since this left handside mass is not moving, it’s as if this is a wall andthis were a single spring, k, that’s connectedto a little mass. And it got displaced by delta x. So what will happen is that thismass will experience a spring force, which is F equalto minus k delta x. Therefore, we canimmediately identify omega b squared willbe equal to k over m. So you can see that we haveidentified two kinds of modes, which every part of thesystem is oscillating at the same frequencyand the same phase. Everybody agree? No not everybody agree. Look at this guy thisguy is not moving.How could this be? This is not the normal mode. Isn’t it? OK. I hope that willwake you up a bit. I can be very tricky here. I can say that this massis also oscillating, but with what amplitude? AUDIENCE: Zero. YEN-JIE LEE: Zero amplitude. Right So the conclusion isthat, aha, everybody is actually oscillating at thesame frequency, but these guy withzero amplitude. AUDIENCE: Are they oscillatingat the same phase as well? YEN-JIE LEE: Yeah. Oh very good question. Another objection I receive. So life is hard for me today. Hey. This guy is oscillatingout of phase. These two guys are out of phase. But I can argue that theamplitude of the first mass is actually has a minus signcompared to the second mass. Then they are again in phase. So very good. I like those questions. And I hope I have convinced youthat everybody is oscillating, although you cannot see it,because the amplitude is small, is zero.And they are all oscillatingat the same phase. Yes. AUDIENCE: How comethere’s only one mass? YEN-JIE LEE: Oh,the right hand side? AUDIENCE: Yeah. YEN-JIE LEE: Oh, yeah. That is because the lefthand side mass, the 2m one, is actually not moving. Because they aretwo spring forces, one is actuallypushing the mass, the other one’spulling the mass. And they cancel perfectly. Therefore, it’s as ifthose two guys are not– they don’t find each other. And then it’s like,they are just tools mass connected to a wall along. And then you can now identifywhat is the frequency. OK. Very good. So we make the made a lot of theprogress from the discussion. And now I would liketo ask you for help. What is the third oscillation? Yes? AUDIENCE: There’s nothird normal mode. YEN-JIE LEE: There’sno third normal mode. AUDIENCE: There’sno third normal mode because there arerestricted to one dimension.I can not imagine anothermode that would not displace the central mass. YEN-JIE LEE: Very good. That’s very good. On the other hand,you can also say, I also take thecenter mass motion as one of the normal mode. I think that’s alsofair to do that. Very good observation. You can see that the wholecan move simultaneously. I can also argue thatthey are oscillating at the same frequencyand the same phase, because they areall moving together. So these are the 2mconnected to mass one. All of them are movingin the same direction. So now I cancalculate the force. What is the force? F is 0. Therefore, omega c is 0. So you can the smalllimit of omega. So of course, I can pretendthat those mass are connected to a really, really small springto the wall with is a spring constant k’. And I have k’ goes to zero. And they are actuallygoing to oscillate with omega c goes to zero. So in this case,the amplitude is going to increase forever,because you have A sin omega c t. And this roughly A omega c t. And this is just vt. So what I want to argueis that this is actually also oscillation, but withangular frequency zero. And the general motion canbe in written as vt times c, for example, some constant. Any questions? So what I’m going todo next may amaze you. Very good. So we have identified threedifferent kinds of modes. We have mode A, which is withomega a squared equal to omega a squared. Where is omega a squared. There. It’s 4k over 2m. And also, themotion is like this. x1 equal t A cosineomega a t plus phi a. x2 is equal to minus A, becausethey have different sine. So if the motion is in theleft hand side direction, then the two massesare oscillating in the opposite direction. So therefore, I get a minussign in front of A. Cosine omega a t plus phi a. x3 will be also equal to minusA cosine omega a t plus phi a. Of course, I need to definewhat this x1, x2, x3. That’s why most of yougot super confused. So the x1, whatI mean is that is that the displacement ofthe mass 2m, I call it x1. The displacement of the uppermass, the upper little mass, I call it x2.And finally, the displacementof the third mass, I call it x3. Therefore, you cansee that mode A, you have this kind of motion. The amplitude of the firstmass is A. Therefore, if I define that to be A,then the second and third one, or the amplitude willbe defined as minus A. And you can see that allof them are oscillating at fixed angularfrequency, omega a, omega a, omega a; and alsofixed phase, phi a, phi a, phi a. Of course, we canalso write down what we get formode B. For mode B, the left hand side massis not moving, stay put. And the other twomasses are oscillating at the frequency of omega b. And amplitude, theydiffer by a minus sign. OK. Omega b squared isequal to k over m from that logical argument.And then we get x1 equalto 0 times cosine omega b t plus phi b. x2, I get B cosineomega b t plus phi b. x3, I get minus B cosineomega b t plus phi b. Any questions here? Finally, mode C.All the mass, x1 is equal to x2 is equal tox3, is equal to C plus vt. So you can see that we haveidentified three modes, mode A, mode B, and the mode C.And there are three angular frequencies which we identifiedfor all of those normal modes, omega a, omega b, and omega c. And you can see thatwe also identified how many free parameters. One free parameter, two,three, four, five, and six. If you careful, youwrite down the equation of motion of this system,you will have three coupled differential equations.And those are second orderdifferential equations. If you have three variables,three second order differential equations. If you manage it magically,with the help from a computer or from math departmentpeople, how many free parameter would you expect inyou a general solution? Can anybody tellme well how many? I have three second orderdifferential equations. Yes? AUDIENCE: 6? YEN-JIE LEE: 6. So look at what we havedone we identified already 1, 2, 3, three normal modes.By there are 1, 2, 3, 4,5 6, six free parameters. That tells me I am done. I’m done. Because what is thegeneral solution? The general solution is just asuperposition of mode A mode B and mode C. You have sixfree meters to be determined by six initial conditions,which I would like– I have to tell you what arethose initial conditions. So isn’t this amazing to you? I didn’t even solve thedifferential equation, and I already get the solution. And you can see anotherlesson we learned from here is that,oh no, you can imagine that themotion of the system can be super complicated. This whole thing can dothis, all the crazy things are all displaced,and the center of mass can move, as you said. But the result is actuallyvery easy to understand. It’s just three kinds of motion,the displacement, and two kinds of simple harmonic motion. We add them together. And then you get the generaldescription of that system. So everything is so nice. We understand themotion of that system. But in general,life is very hard. For example, now I dosomething crazy here.I change this to 3. What are the normal modes? Can anybody tell me? It becomes very, verydifficult, because there’s no general symmetry ofthat kind of system. So we are in trouble. One of the modesmaybe still there, which is actually modeB. But the other modes are harder to actually guess. So you can see that that alreadybrings you a lot of trouble. And you can see that I can nowcouple not just two objects, I can couple three objects,four objects, five objects, 10 objects. Maybe I will put that in yourp set and see what happens. And you can seethat this becomes very difficult to manage. So what I’m going to do inthe rest of this lecture is to introduceyou a method which you can follow in generalto solve the question and get the normal modefrequencies and normal modes. So we will take afour minute break. And we come back at 12:20. So if you have anyquestions, let me know. What we are going to doin the following exercise is to try to understand ageneral strategy to solve the normal mode frequenciesand the normal mode amplitudes, so that you can apply thistechnique to all kinds of different systems.So what I am going to do todayis to take these three mass system, and of course asusual, I try to define what is this coordinate system? The coordinate systemI’m going to use is x1 and x2 an x3 describingthe displacement of the mass from the equilibrium position. And the equilibriummeans that there’s no stretch on the spring. The string is unstretched. It’s at their ownnatural length, l0. So once I define that, I cando a force diagram analysis. So that starts fromthe left hand side mass with mass equal to 2m. I can write down the equationof motion, 2m x1 double dot. And this is going to be equalto k x2 minus x1 plus k x3 minus x1. So there are two springforces acting on this mass, the left hand side mass. The first one isthe upper spring. The second one is comingfrom the lower one. And you can seethat both of them are proportional tospring constant k, and also proportional tothe relative displacement.And you can see that the tworelative displacement, which is the amount ofstretch to the spring, is actually x2 minusx1, and the x3 minus x1. Am I going too fast? OK. Everybody’s following. And you can actuallycheck the sign. So you may not be sure. Maybe this is your x1 minus x2. But you can check that,because if you increase x1, what is going to happen? This term willbecome more negative.More negative in thiscoordinate system is pointing to theleft hand side. So that makes sense. Because if I move thismass to the right side, then I am compressingthe springs. Therefore, they arepushing it back. Therefore, this is actuallythe correct sign, x2 minus x1. The same thing also appliesto the second spring force. So that’s a way I doublecheck if I make a mistake. Now, this is actually thefirst equations of motion. And I can now alsowork on a second mass. Now I focus on amass number two. The displacement is x2. Therefore the left handside of Newton’s Law is m x2 double dot.And that is equalto the spring force. The spring force, there’sonly one spring force acting on the mass. Therefore, what I am goingto get is k x1 minus x2. Everybody’s following? You can actually checkthe sign carefully, also. And finally, I have thethird mass, very similar to mass number two. I can write down theequation of motion, which should k x1 minus x3. So that is my coupled secondorder differential equations.There are three equations. And all of them aresecond order equations. So this looks a bit messy. So what I’m goingto do Is no magic. I’m just collecting allthe terms belonging to x1, and put them together, allthe terms belonging to x2, putting all together, andjust rearranging things. So no magic. So I copied thisthing, left hand side. 2m x1 double dot. And the dot will be equalto minus 2k x1 plus, I collect all the timesrelated to x2, plus k x2, there’s only one termhere, then plus k x3. I’m just trying toorganize my question. So you can see that Icollect all the terms related to x1 to here. Minus k, minus k,I get minus 2k. And the plus k forx2, plus k for the x3. And I can also do thatfor m x2 double dot.That will be equal to k x1minus k x2 plus zero x3, just for completeness. I can also do the same thingfor the third mass, m x3 double dot. This is equal to k x1 plus 0 x2. There’s no dependence onx2, because x1 and x2– x3 and x2 are not talkingto each either directly. Finally, I have the third,which is minus k x3. Now our job is to solvethose coupled equations.Of course, you havethe freedom, if you know how to solveit yourself, you can already goahead and solve it. But what I amgoing to do here is to introduce technique,which can be useful for you and make it easier to follow. It’s a fixed procedure. So what I can dois the following. I can write everythingin them form of a matrix. How many of you heard thematrix for the first time? 1, 2, 3, 4.OK. Only four. But if you are not beingfamiliar with matrix, let me know, and I can help you. Let the TA know. And also, there’s a sectionin the textbooks, which I posted on announcement,which is actually very helpful to understand matrices. But sorry to thesefour students, we are going to use that. And maybe, you already learnhow it works from here. So one trick which wewill use in this class is to convert everythinginto matrix format. What I am going to dois to write everything in terms of M,capital X, capital M, capital X double dot equalto minus capital K capital X. Capital M, capitalX, and capital K, those are all matrices. Because I write thisthing I really carefully, therefore we can alreadyimmediately identify what would be M, capitalM and capital X and a K. So I can write down immediatelywill be equal to 2m, 0, 0, 0, m, 0, 0, 0, m. Because there’s only one in eachline, you only have one term. X1 double dot, x2 doubledot, x3 double dot.And also, you canwrite down what will be the X. Thisis actually a vector. X will be equal to x1, x2, x3. Finally, you have the K? How do I read off K? Careful, there’sa minus sign here, because I would like to makethis matrix equation as if it’s describing a simpleharmonic motion of a one dimensional system. So it looks the same, but theyare different because those are matrices. But therefore, Ihave in my convention I have this minus sign there. Therefore, when you read offthe K, you have to be careful. So what is K? K is equal to 2k. You have the minus 2khere in front of x1. But because I have a minussign there, therefore this one is actually taken out. So we have 2k there. Then you have minus k,minus k, minus k, k, 0. Minus k, plus k, 0. And finally, you canalso finish the last row. You get minus k, 0, k. K becomes minus k. Minus k becomes k. So we have read off allthose matrices successfully.So you may ask,what do they mean? Do they get the meaning? M, K, X, what those? M, capital M matrix, tellsyou the mass distribution inside the system. So that’s the meaningof this matrix. X is actuallyvector, which tells the position of individualcomponents in the system. Finally, what is K? K is telling you how eachcomponent in the system talks to the other components. So K is telling youthe communication inside the system. So now we understand abit what is going on.And as usual, I will goto the complex notation. So I have xj, the smallxj are the position of the mass, x1, x2, and x3. xj will be realpart of small zj. Small xj equal toreal part of zj. Therefore, I can now writeeverything in terms of matrices again. So now I can writethe solution to be Z, the capitol z is a matrix,exponential i omega t plus phi. This is the guessthe solution I have. A1, A2, and A3. Those are theamplitudes, amplitude A of the first mass, amplitudeof the second mass, amplitude of the third mass,in their normal mode. And all of those are oscillatingat the same frequency, omega, and the same phase, phi. Does that tell you somethingwhich we learned before? Oh, that’s the definitionof the normal mode.I’m using the definitionof the normal mode. Every part of thesystem is oscillating at the same frequencyin the same phase. And we use that toconstruct my solution. The complex version isexponential i omega t plus phi, oscillating at thesame frequency, oscillating at the same phase. And those are the amplitude,which I will solve later. OK, any questions? I hope I’m not going too fast. If everybody canfollow, now I can go ahead and solve theequation in the matrix format.So now I go to thecomplex notation. So the equation M X doubledot equal to minus KX becomes M Z doubledot equal to minus KZ. And also, I can immediatelyget the Z double dot will be equal to minusomega squared Z, because each time Ido a differentiation, I get i omega out of theexponential function. And I cannot kill thatexponential function, so it’s still there.Therefore, I getminus omega squared in front of Z. I hopethat doesn’t surprise you. So that’s very niceand very good news. That means I can replace thisZ double prime with minus omega squared Z. Then I getminus M omega squared Z. And this is equal to minus KZ. And I can cancel the minus sign. That becomessomething like this. So, I can now cancel theexponential i omega t plus phi, because I have Z inthe left hand side.And exponential i omega tplus phi is just a number. So therefore, I can cancel it. So what is going tohappen if I do that? Basically, whatI’m going get is I get M omega squaredA equal to K A. I’m trying to go extremelyslowly, because this is the first time wego through matrices. So now you have this expression. Left hand is a matrix, M, timessome constant, omega squared.I can actually get omegasquared in front of it, because this isactually just a number. A is just a vector, which isA1, A2, A3, also a matrix. K is actually how the individualcomponents talks to the others. So that’s there, times A. Now I would liketo move everything to the right hand side,all the matrices in front A to the right hand side. Then I multiply bothsides by M minus 1. So I multiply M minus 1to the whole equation. M minus 1, what is M minus 1? The definition isthat the inverse of M is called M minus 1. M minus 1 times M is equal toI, which is actually 1, 1, 1. Therefore, if I do this thing,then I would get omega squared. M minus 1 times Mbecomes I, unit matrix. And this is equal to Mminus 1 K A. And be careful, I multiply M minus1, the inverse of M, from the left hand side.That matters. So now I can moveeverything to the same side. I moved the left hand sideterm to the right hand side. Therefore, I get M minus1 K minus omega squared I. Those are all matrices. Times A, this is equal to 0. Any questions? So a lot of manipulation. But if you think about it,and you are following me, you’ll see that allthose steps are exactly identical to what we have beendoing for a single harmonic oscillator. Looks pretty familiar to you. But the difference is that nowwe are dealing with matrices. AUDIENCE: What is A? YEN-JIE LEE: Oh, A. Ais actually this guy. I define this to beA. And that means Z will be exponential i omegat plus phi times A. I didn’t actually write it explicitly.But that’s what I mean. Any more questions? Yes? AUDIENCE: [INAUDIBLE]is for [INAUDIBLE]?? YEN-JIE LEE: Canyou repeat that? AUDIENCE: So this wholeprocess, this is mode A, right? YEN-JIE LEE: Yeah. So this whole process isfor, not really the for mode A. So that A may be confusing. But in general, ifI have a solution, and I assume that the amplitudecan be described by a matrix. So it’s in general. And you’ll see thatwe can actually derive the angular frequencyof mode A, mode B, and mode C afterward. I hope that answersyour question. So you see that for ingeneral, what I have been doing is that now, allthose things are equivalent to the originalequation of motion.What I am doing ispurely cosmetic. You see, make it beautiful. So all those things, thisthing is exactly the equivalent to that thing, up there. Up to M X double dotequal to minus K x. Cosmetics. Beautiful. Looks– I like it. All right. Then what I have beendoing is that now I introduce using adefinition of normal mode. I guess the solution willhave this functional form. Z equals to exponential iomega t plus phi, everybody oscillating at the samefrequency, the same phase. Frequency omega, phase phi. And everybody can havedifferent amplitude. You can see from thisexample, normal modes, they can havedifferent amplitude.The amplitude is what? I don’t know yet. But we will figure it out. Then that’s my assumption. The definition of normal mode. And I plug in to theequation of motion. Then this is what weare doing to simplify the equation of motion. There’s no magic here. If I plug in the definition onnormal mode to that equation, this is actually going to bringyou to this equation, matrix equation. So if you have learned matricesbefore, you have something, some matrix, times Z.This is equal to zero. A is not zero. I hope. If it’s zero, then thewhole system is not moving. Then it’s not fun. So if A is not zero, thenthis thing should be– this thing times A shouldmake this equation equal to 0. So what is actuallythe required condition? I get stuck, and of course again, myfriend from math department comes to save me. That means if thisthing has a solution, this equation hasa solution, that means that determinant ofM minus 1 K minus omega squared I has to be equal to 0.So that is the conditionfor this equation to satisfy thisto be equal to 0. And just to make sure that Idon’t know what is the angular frequency omega yet. I don’t know whatis the phi yet. We can actually solve theangular frequency, omega. So now, turn everything around. And basically now,using this normal mode definition, and somemathematical manipulation, the condition we need for thisequation to satisfy equal to 0, is determinant M minus 1K minus omega squared I. I can write down Mminus 1 K minus omega squared I explicitly, justto help you with mathematics. M minus 1 K is equal to 1 over2m, 0, 0, 0, 1/m, 0, 0, 0, 1/m. It’s just the inversematrix of the M matrix. Therefore, now I can writedown the explicit expression of M minus 1 Kminus omega squared I. This will be equal to kover m minus omega squared, minus k over 2m, minus kover 2, minus k over m, k over m minus omega squared, 0. I will write down allthe elements first.Then I will explain to you howI arrived at the expression. OK. So this is M minus 1 k. The definition ofM minus 1 is that. And the definition ofK is in the upper right corner of the black board. Therefore, if youmultiply M minus 1 K, basically, the firstcolumn will get– wait a second. Did I make a mistake? No. OK. So basically, what youarrive at is k/m, k/m, k/m. And also, the minus kover 2m for the rest part of the matrix. And the minus omegasquared I will give you the diagonal component.Yes? AUDIENCE: Why do you haveto take the determinant and set it equal to0 instead of just setting that equal to zero? AUDIENCE: This is a matrix. So these are the matrix. So a matrix times Awill be equal to zero. The general conditionfor that to be satisfied is more general. It’s actually the determinantof this matrix equal to zero. Because this is actuallymultiplied by some back to A. So I think there aremathematical manipulation. Basically, you wouldjust collect the terms. And then calculateM minus 1 K first. And the minus omega squared Iwill give you all the diagonal and terms have a minusomega square there. And that is actually the matrix. And of course, I cancalculate the determinant. So if I calculatethe determinant, then basically I get thistimes that times that. So what you get is k over mminus omega squared times k over m minus omega squared timesk over m minus omega squared. So these are all diagonal terms. And the minus 1 over 2 ksquared over m squared, k squared over m squared. sorry. Minus omega squared.So that’s this off diagonalterm, this times this times that. OK. It will give youthe second term. And the third one,which survived because of those zeros, many,many terms are equal to 0. And then the third term, whichis nonzero, is again minus 1 over 2k squared over m squared,k over m minus omega squared. And this is actually equalto 0, because the determinant of this matrix is equal to zero. Everybody following? A little bit of a mess. Because I have been doingsomething very challenging. I’m solving a 3 by 3 matrixproblem in front of you right. So the math can geta bit complicated. But next time, I think we aregoing to go to a second order one, 2 by 2 matrix. And I think that willbe slightly easier. But the generalapproach is the same. So basically, you calculate Mminus 1 K minus omega squared.Then you get what is inside, allthe content inside this matrix. Then you would calculatethe determinant. And basically, you cansolve this equation. Now I can define omega0squared to be k/m. And I can actually make thisexpression much simpler. Then basically,what you are getting is omega0 squaredminus omega squared to the third minus 1/2 omega0to the fourth, omega0 squared minus omega squared. Minus 1/2 omega0 to thefourth, omega0 to the square, minus omega squared. And this is equal to 0. And you can factor outthe common components.Then basically,what you are going to get is, you canwrite this thing to be omega0 squared minusomega squared, omega squared. Because all of themhave omega squared. And omega squaredminus 2 omega0 squared. And that’s equal to 0. So I am skipping a lot of stepsfrom this one to that one. But in general, you can solvethis third order equation. And I can first combineall those terms together. And then I factor outthe common components. Then basically, what youare going to arrive at is something like this. A lot of math here. But we are close to the end. So you can see now what are thepossible solutions for omega. That is the omega,unknown angular frequency we are trying to figure out.You can see that there arethree possible omegas that can make this equation equal to 0. The first one isomega equal to omega0. The second one issquare root of omega 0, coming from thisexpression, that omega squared minus 2omega zero squared. If omega equal tosquare root 2 omega0, this will be equal to zero. And that will give you thewhole expression equal to 0. Then finally, I take this term. And then you will get zero. Omega squared, ifomega is equal to 0, then the whole expression is 0. I have defined omega0 squaredto be equal to k over m. Therefore, I can concludethat omega squared is equal to k overm, 2k over m, and 0. Look at what we have done,a lot of mathematics. But in the end, after yousolve the eigenvalue problem, or the determinantequal to zero problem, you arrive at thatthere are only three possiblevalues of omega which can make thedeterminant of M minus 1 K minus omegasquared I equal to 0. What are the three? k/m, 2k/m, and 0. If you look at thisvalue, then we’ll say, this is essentially what weactually argued before, right? Omega A squared is equalto 4k over 2m is 2k over m. Wow. We got it. The second one is,we think about really keep a straightquestion in my head and understand this system. The second identifiednormal more is having omega squared be equal to k/m. I got this also here magically,after all those magics. And finally, the third one,the math also knows physics. It also predicted that this isone mode which have oscillation frequency of 0. Isn’t that amazing to you? But that also givesus a sense of safety. Because I can now add10 pendulums, or 10 coupled system toyour homework, and you will be able to solve it. So very good. This example seemsto be complicated. But the what I want to say,I have one minute left, is that what wehave been doing is to write the equation ofmotion based on force diagram. Then I convert thatto matrix format, and X double dotequal to minus KX. Then I follow thewhole procedure, solve the eigenvalue problem. Then I will be able to figureout what are the possible omega values which can satisfythis eigenvalue problem or this determinant. M minus 1 K minus omegasquared I equal to 0 problem. And after solvingall those, you will be able to solve thecorresponding so-called normal mode frequencies.You can solve it. And of course, you can plugthose normal mode frequencies back in, then youwill be able to derive the relative amplitude,A1, A2, and A3. So what we havewe learned today? We have learned howto predict the motion of coupled oscillators. That’s really cool. And then next time,we are going to learn a special kind of motion incoupled oscillators, which is the big phenomena. And also, what willhappen if I start to drive the coupled oscillators? So I will be here ifyou have any questions about the lecture. Thank you very much.
