Coupled Oscillators (Normal Modes and Frequencies)

hi in today's video we're going to talk about coupled oscillators and normal modes so whenever you combine two or more than two oscillating systems together then you end up getting a coupled oscillator coupled oscillators are quite interesting because even though the nature of their motion is not necessarily harmonic not necessarily sinusoidal it can be represented as the superposition of distinct sinusoidal frequencies and these sinusoidal frequencies which are hidden within the general solution of coupled oscillators can be extracted by studying what are known as normal modes and normal frequencies so normal modes are those setups in which the entire coupled oscillator executes simple harmonic motion which is only possible for few restricted initial conditions so in this video first of all what i'm going to do is i'm going to discuss these ideas of coupled oscillators by taking a very simple example of a two mass three spring system so in this particular setup first of all i am going to obtain the equations of motion and from the equations of motion i am going to obtain the general solution and i'm going to show how the general solution of this kind of a coupled oscillator can be written as a linear superposition of distinct frequencies and i'm going to obtain those distance frequencies by looking at normal modes these configurations in which the entire setup will execute shm and the frequencies associated with them which are normal frequencies and i'm also going to discuss normal coordinates which are those combinations of the coordinates that executes shm all right and finally i'm going to look at the very specific situation of the general solution in which the coupled oscillator exhibits beats in which one point mass exchanges energy with the other point mass slowly over a period of time so let us begin [Music] all right so what we have here is a system of two point masses so the first point mass let's suppose its mass is m is connected to a spring which has a spring constant of k let's suppose which is attached to the fixed ceiling or a wall on the other end and this mass is attached to another spring let's suppose it has a spring constant of k dash which is attached to another point mass let's suppose it has the same mass m and this mass m is on the other hand attached to another spring having a spring constant let's suppose k whose other end is attached to some kind of a wall now let's suppose the first diagram is the diagram of equilibrium in which the entire system is at rest and the springs have their equilibrium lengths associated with it however if i displace these two point mass by some kind of a distance so that there is a net effective displacement created in the springs compared to its equilibrium lengths then the entire system starts executing motion so let me say that the first mass is displaced by an amount of x1 and the second mass is displaced by an amount of x2 here i am considering that this is the x-axis all right so the entire motion is going to be along the x-axis it is restricted to one dimensions so x1 and x2 here are the variables which are the instantaneous displacements of mass m and m for the first and the second mass respectively and if we can create some kind of equations of motion which governs how x1 and x2 varies with time and from there we can understand how the motion is going to take place so to do that i'm going to apply the newtonian approach so let me write the force equation for the first and the second mass so the force equation is simply the newton's second law which says that whatever net force is acting on a given mass at any given point in time results in the acceleration experienced by that particle at that instant all right so let's suppose first we look at the first mass so the first mass has a mass m and its acceleration is restricted to the x-axis and i'm defining its instantaneous coordinate to be x1 so its acceleration is x1 double dot so x1 double dot simply represents the second order derivative of x1 with respect to time all right now what is the net amount of force experienced by the first mass the net amount of force experienced by the first mass is associated with the first spring as well as the second spring right so what is the extension experienced by the first spring the extension experience by the first spring is x1 so the force associated with that is kx1 and since the force is acting opposite to x1 i'll have to write minus kx1 quite simple now what about the extension experienced by the second spring now the second spring is different from the first spring in the sense that in the second spring both the ends are experiencing some kind of an extension so the overall extension experience by the second string can be simply given as x1 minus x2 so x1 minus x2 will give us the extension of the second spring compared to its equilibrium and since the spring constant of the second spring is k dash i write the force as k dash x1 minus x2 now let's suppose if x1 is greater than x2 in that case the direction of force is opposite to the extension that is created in that sense i'll apply a negative sign here and this gives us the force equation for the first mass so the force equation for the first mass is mx1 double dot plus k x 1 plus k dash x 1 minus x 2 is equal to 0 this is the force equation for the first mass similarly we can also obtain the force equation for the second mass all right so the force equation for the second mass is if the second mass has an instantaneous displacement of x2 then its acceleration is x2 double dot now the second mass will be affected by the extensions of the second spring as well as a third spring if we are only looking at the force due to the third spring then the extension in the third spring is given by x two right and since its spring constant is k the force is k x 2 and since this is acting opposite to the direction of x 2 so x x 2 is in this direction the force will be acting in the opposite direction so i apply a negative sign here and the second force is due to the middle spring which causes an extension of let's suppose x 2 minus x 1 and it has a spring constant of k dash so now if x 2 is greater than x1 in that case the force is again opposite to x2 so i apply a negative sign here so the force equation for the second mass comes out to be mx2 double dot plus kx2 here i'll have to write k x dash x 2 minus x 1 so if i write minus this simply becomes k x dash x 1 minus x 2 is equal to 0. so these are the two force equations corresponding to the both these two masses let's suppose i name these point number one and point number two now if i am interested in finding out the solutions for x1 and x2 from here what i'm going to do is i'm going to apply a couple of tricks so first of all i'm going to add these two equations 1 and 2 and this is going to help us in obtaining the solutions for x1 and x2 this gives us mx1 double dot plus k x 1 plus m x 2 double dot plus k x 2 plus k dash x 1 minus x 2 minus x 1 plus x 2 is equal to 0 here x 1 x 1 x 2 x 2 gets cancelled so i am left with the equation m x 1 double dot plus x 2 double dot plus k x 1 plus x 2 is equal to 0 or finally i have x 1 double dot plus x 2 double dot plus k upon m x 1 plus x2 is equal to 0. now some of you may be familiar with this kind of a differential equation this is simply the differential equation corresponding to some kind of a sinusoidal solution or this kind of a system usually gives us simple harmonic oscillations but for the variable x1 plus x2 so let me write it in different manner so if i have a second order derivative of the variable x1 plus x2 plus k upon m x1 plus x2 so this is a second order differential equation with respect to the variable x1 plus x2 can you see that and the standard solution for this kind of a differential equation can simply be written as x1 at a time t plus x 2 at a time t can simply be written as let's suppose cos omega t plus d sine omega t where omega is simply equal to root over k upon m all right i am sure that you must have seen differential equations of this form that give this kind of a standard solution now here the entire solution of x1 plus x2 okay these are not the solutions individual of x1 and x2 this is a solution of x1 plus x2 is c cos omega t plus d sine omega t where omega here is the frequency root over k upon m and the sum of cos omega t and sine omega t basically can be written as some kind of a sinusoidal function so let me show you that this expression is very much equivalent to 2 a 1 cos omega t plus phi 1 okay i can write this expression and this expression in the same manner let me show you how so if i have some expression of 2 a 1 cos omega t omega 1 t plus phi 1 so here let's suppose instead of this being omega this is omega 1 okay because i will get another omega later on so let's suppose this is omega 1 in this case if you look at this particular expression why i have written 2 this will also become clearer in just a moment so 2 a 1 cos omega 1 t plus phi can be written as cos a plus b okay what is cos a plus b so this is nothing but 2 a 1 cos a plus b is cos omega 1 t cos phi 1 minus sine omega 1 t and sine phi 1 right so this is equal to 2 a 1 cos omega 1 t now what is cos phi 1 so phi 1 here is just some kind of a constant here phi 1 and a1 are constants okay these all depend upon the initial conditions of the given system so this constant can be written as maybe something like b1 okay and again minus 2 1 sine omega 1 t and again sine phi 1 is another constant so which can simply be written as let's suppose b 2 so these constants being multiplied can be written as let's suppose another constant c cos omega 1 t while minus 2 a 1 b 2 can be written as some constant let's suppose d sine omega 1 t which is nothing but this particular expression right this is the same so c cos omega 1 t plus d sine omega 1 t is the same as let's suppose 2 a 1 cos omega t plus phi 1 here capital c and capital d are some constants and capital a 1 and phi 1 are also some kind of a constant that depends upon the initial configurations and if we want we can find out the relationships between ca1 and d51 but that's not necessary here the reason i have written both these two expressions is because i may use one of these expressions depending upon what i'm discussing so we have obtained the solution for x1 plus x2 right now let me do one more thing here i've added equations 1 and 2 let me subtract equation number 2 from equation 1 and see what we get so if we subtract equation number 2 from equation 1 then i will simply get the expression that m x 1 double dot plus k of x 1 minus m x 2 double dot minus k of x 2 plus k dash x 1 minus x 2 plus x 1 minus x 2 which is equal to 0 so this equation can be simplified as m x 1 double dot minus x 2 double dot plus k x 1 minus x 2 plus k dash if i take 2 out of it it becomes x 1 minus x 2 because there is 2 x 1 minus 2 x 2 right which is equal to 0 this can be simplified as m x 1 double dot minus x 2 double dot plus k plus 2 k dash x 1 minus x 2 is equal to 0 which can be further simplified as x 1 minus x 2 its second order derivative with respect to time plus k plus 2 k dash upon m x1 minus x2 now this is the same standard differential equation of x1 minus x2 with respect to time as was this one and it has a similar kind of a standard differential equation solution so the solution of x 1 t minus x 2 t is simply equal to some kind of a constant e cos omega 2 t plus f sine omega 2 t where e and f these are just constants that depend upon the initial conditions where omega 2 is simply equal to k plus 2 k dash upon m root over okay so the omega 2 is just equal to root over k plus 2 k dash upon m this can in a similar fashion as i showed you here can also be written as just some kind of a sinusoidal variation let's suppose 2 a 2 cos omega 2 t plus phi 2 here the constants are the phase phi 2 and the amplitude a2 so as you can see the difference of these coordinates x1 minus x2 and the sum of these coordinates x1 plus x2 both have sinusoidal variations now this is something very interesting that i'm going to come back to later on these coordinates are given a very special name it is called normal coordinates you see the individual point masses may not necessarily execute simple harmonic oscillations but a linear combination of these coordinates may execute simple harmonic oscillations and these linear combinations of the coordinates that is x1 plus x2 and x1 minus x2 are known as normal coordinates but before i come to normal coordinates let me use these expressions let's suppose this is point number three and point number four to obtain a general solution for x1 and x2 separately again i can do that by let's suppose adding point number 3 and 4. so if i add point number 3 and 4 that is x1 plus x2 and x1 minus x2 what do i get so x1 plus x2 plus x 1 minus x 2 this simply gives us point number 3 is 2 a 1 cos omega 1 t plus phi 1 and point number 4 contains plus 2 a 2 cos omega 2 t plus phi 2 so this is nothing but if you cancel out the x 2 here it simply becomes 2 x 1. so 2 x 1 the 2 and the 2 these gets cancelled and we are left with x 1 t is equal to a 1 cos omega 1 t plus phi 1 plus a 2 cos omega 2 t plus phi 2. this is quite interesting we have obtained the general solution for the first mass now let us quickly do a subtraction of point number 3 from 4 or other point number four from three so what i'm going to do is x 1 plus x 2 minus x 1 minus x 2 which becomes x 1 plus x 2 is simply equal to 2 a 1 cos omega 1 t plus phi 1 minus 2 a 2 cos omega 2 t plus 5 to here again the x 1 cancels out right and you're left with 2 x 2 the 2 gets cancelled on both hand sides and finally you end up getting x 2 t is equal to a 1 cos omega 1 t plus phi 1 minus a 2 cos omega 2 t plus phi 2 so this is the general solution for the second mass now it's interesting let us just do a little bit of a discussion of these two expressions so this is let's suppose point number five and point number six so we have obtained the general solution for both these two masses so if you look at this system both the masses the first mass and the second mass is experiencing some sort of oscillation and whatever nature of oscillations these two masses are expressing can be mathematically written using these two expressions that we have just now obtained x 1 is a 1 cos omega 1 t plus 5 1 plus a 2 cos omega 2 d plus phi 2 and x 2 is a 1 cos omega 1 t plus 5 1 minus a 2 cos omega 2 t plus phi 2 here these expressions uh that you see that is a 1 a 2 these are just some kind of a constants that depend upon the initial conditions similarly the phases that you see phi 1 phi 2 these are also constants that depend upon the initial conditions however the omega 1 and omega 2 these are some distinct frequencies right we already showed that omega 1 is equal to root over k upon m while omega 2 is root over k plus 2 k dash upon m so these are distinct frequencies both omega 1 and omega 2 are distinct frequencies so what you see here is that the general solution of the coupled oscillator is actually a superposition of two distinct frequencies you see the general solution of x1 is a superposition of these two distinct frequencies while the general solution of x2 is a superposition of these two distinct frequencies now x1 and x2 are not necessarily executing some sinusoidal variation but they're executing some kind of a motion that can be represented by the linear superposition of two distinct sinusoidal frequencies now what are these frequencies these frequencies are known as normal coordinates to appreciate these frequencies better let us discuss two very specific scenarios in which the entire system will exhibit either one frequency or the other frequency so to do that we'll consider two special cases so let us take a look at that so we'll start with the normal modes and normal coordinates so let us discuss two very interesting configurations in which when we provide some kind of initial condition to the coupled oscillatory system then the entire oscillator is going to exhibit simple harmonic oscillations so the first case is when we provide some kind of an initial condition such that the constant a 2 comes out to be 0 all right so if we provide some kind of initial condition where the constant a 2 comes out to be 0 let's suppose if we are able to create a setup where the constant a2 comes out to be zero in that case both the systems x1 and x2 are going to oscillate with the same frequency omega 1. so in that kind of a situation both x1 and x2 are going to exhibit some kind of a sinusoidal variation with frequency omega 1 and constant phase difference phi 1 as you can see from the equation here now what does this kind of a configuration represent let me draw a diagram and explain it to you so here you see we have this simple schematic which represents that situation in which it is possible to find a set of initial conditions where the second constant a2 goes to 0 such that now both x1 and x2 oscillates with the same frequency and the same constant phase difference this represents what is known as the normal mode so here as you can see both these two masses are oscillating in this kind of a manner all right so if i give as equal displacements to both the masses in the same directions then what is going to happen is that both the masses are going to oscillate in this kind of a fashion so as you see here this inner spring is unstretched this inner spring is unstretched and both the masses and their extensions associated with x1 and x2 are in phase and exactly equal at any given point in time they are both oscillating with the same frequency and the same constant phase difference of zero this is known as the first normal mode the first normal mode is characterized by the entire system executing simple harmonic oscillations of same frequency and a constant phase difference of zero and same amplitude as is quite evident from the equation the frequency is the angular frequency of omega 1 is root over k upon m so as you can see it is only the spring constant of the outer springs that is contributing towards the angular frequency and not the spring constant of the inner spring because the inner spring remains unstretched due to this kind of a motion so the normal mode is in generally defined as that kind of a state of coupled oscillators in which under specific initial conditions the entire system starts executing simple harmonic oscillations of a distinct frequency now the other kind of configuration is also possible if we can give some kind of an initial condition where a1 comes out to be 0 in that case you can see from the general solution that if a1 comes out to be 0 only the second two terms are left the second term is just 2 cos omega 2 t plus phi 2 so x 1 t is basically equal to minus x 2 t which is nothing but a 2 cos omega 2 t plus phi 2 yes x 1 t is just equal to x 2 or minus x 2 which is equal to a 2 cos omega 2 t plus phi 2 here there is some kind of a constant phase difference phi 2 and a constant frequency of omega 2 so under those situations where capital a1 constant is 0 both the configurations will oscillate with the same frequency but because there is a minus sign here both of them will be out of phase with respect to each other by let's suppose 180 degrees let me draw a schematic to show you the motion that can be visualized so here we have the schematic for the second normal mode so here as you can see the configuration is such that if you displace the entire setup in such a manner that you displace the first mass by a certain amount and the second mass by the same amount but in the opposite directions the oscillations that you end up getting is something like this see both the two masses will be exhibiting the same frequency with a constant phase difference of 180 degrees that means if one mass is going outwards the other mass is going outwards in the opposite direction and then both the masses are coming towards each other going away from each other towards each other going away from each other at every point their amplitudes are same and they're oscillating with the same frequencies but they are out of phase by 180 degrees so as you can see here if one mass is let's suppose uh going to the right with some kind of a displacement of x2 the other is going towards the left with some kind of a displacement x1 where the magnitude of x1 and x2 are exactly equal but they are directed in opposite direction so this is known as the second normal mode so in the second normal mode what you end up getting is you have the same frequency with which the system is oscillating or all the points in the system is oscillating that is omega 2 is equal to k plus 2 k dash upon m but the amplitudes are equal but out of phase by 180 degrees so these are two distinct configurations which are possible in this kind of a coupled oscillator setup when the entire system will execute simple harmonic oscillations of a distinct frequency one is the setup where the entire system is oscillating with respect to a frequency omega one in phase other is the setup in which the entire system is oscillating with the same frequency omega 2 but out of phase now this must be kind of obvious to you that this configuration is possible when i provide some initial displacement to x1 that is let's suppose some displacement a2 and initial initial displacement to x2 that is equal to minus a2 and i provide 0 velocities to x1 x2 at time t is equal to zero so under these initial conditions both the masses will be exhibiting distinct frequencies different from each other now why are normal modes important normal modes are important because they reveal to us that there are distinct frequencies in that particular setup because the general solution as it turns out is a linear superposition of these distinct frequencies so if you remember the general solution here the general solution is a linear superposition of these distinct frequencies this is the one frequency this is the other this is the one frequency this is the other so whenever you deal with coupled oscillators even though it may be a very complicated looking coupled oscillator in our case we have taken a very simplistic coupled oscillator but you may have coupled oscillators with many number of masses but if you are able to find out the normal frequencies associated with the normal modes then it just so happens that the general solution is usually a linear superposition of those distinct frequencies therefore these frequencies are given the name of normal frequencies so this coupled oscillator has two normal frequencies it has two degrees of freedom and two normal frequencies and two normal modes which brings me to something that i derived just some moments back these uh quantities x1 plus x2 you remember this x1 plus x2 is a quantity that is also oscillating with uh a one frequency omega 1 while x1 minus x2 is something that is oscillating with frequency omega 2. so it is also something unique in these kind of coupled oscillators that the general solutions may not exhibit shm but we can find certain kinds of linear combinations of those general coordinates in this case i have x1 plus x2 and x1 minus x2 which will have some kind of a sinusoidal variation of a fixed frequency so this is kind of interesting isn't it the general coordinates will exhibit shm only under very particular initial conditions not always it is usually a linear superposition of the frequency but we can find linear superpositions of the coordinates that will exhibit some kind of a sinusoidal variation these coordinates are given a special name they are known as normal coordinates so normal coordinates are those coordinates or rather they are the linear combination of the coordinates of the masses that at all times will undergo some kind of a sinusoidal variation and the frequency associated with those coordinates are known as normal frequencies and these normal frequencies can be physically visualized when we give such initial conditions that the entire system starts exhibiting one or the other normal frequency and these states are known as normal modes so this is all very very interesting now i want to before i end the video uh show something very unique about the nature of the motion of the system the system is moving in such a manner that these masses are not necessarily executing simple harmony oscillations but they are exchanging energy with respect to each other and i can show that by looking at the beats exhibited by these coordinates let me give some kind of an initial condition to the entire configuration such that x1 at time t is equal to 0 is let's suppose 0 that means the first mass is not displaced and x2 at time t is equal to 0 is let's suppose i give it some kind of an initial displacement let's suppose capital a i do not provide them any velocities at time t is equal to zero so x one dot and x two dot at time t is equal to zero is zero under such initial configuration what is the nature of the general relationship that is what i want to look at so here are the general solutions that we have already obtained and i want to find out the values of the constants involved based on these initial conditions because i want to show a very interesting phenomena that happens in this motion but i am i would be more comfortable with uh the expression that involves the signs and the cosines both so i showed just some time back that an expression of the form a cos omega t plus phi can be written as let's suppose some constant a cos omega 1 t plus some constant b sine omega 1 t just some time back i showed to you that both these two expressions are kind of equivalent to each other so i'm going to use that expression just because i i would feel more comfortable solving the constants there so the first expression can be written let's suppose with respect to constants small a and small b and the second expression also i would like to write it in this fashion let's suppose this is simply equal to c cos omega 2 t plus let's suppose some constant d sine omega 2 t here also c and d are constants similarly the expression for x 2 can also be written as a cos omega 1 t plus b sine omega 1 t minus c cos omega 2 t minus d sine omega 2 t so i want to find out what these values a b c d are after applying these initial conditions so let us apply that and see what we get so at time t is equal to 0 what does x 1 become x 1 simply becomes 0 because x 1 at time t is equal to 0 is 0 is equal to a cos 0 plus b sine 0 plus c cos 0 plus d sine 0 here of course the sine terms vanishes and we are left with simply a plus c is equal to 0 this is one equation we have similarly let us apply it to x is equal to two equation and if i applied x is equal to two equation i simply get zero is equal to a cos zero plus b sine zero minus c cos 0 minus d sine 0 again here the sine terms vanishes and i am left with a minus c is equal to 0. sorry i made a mistake x 2 was equal to a x 2 was not equal to 0 x 2 was equal to a so this must be equal to a so this must be equal to a so here in this equation if i apply the fact that a is equal to minus c so this simply gives us minus c minus c is equal to a or c is equal to minus a upon 2 and a is equal to minus minus a upon 2 which is a upon a as in capital a so we have found out the constants the small a and the small c now let we have to find out the constants uh b and d to find out the constants b and d what i'm going to do is i'm going to do the time derivative that is i'm going to look at the velocities so if i'm interested in finding out the velocities i can find out what is x 1 t dot and x 2 t dot okay so x 1 t dot i can obtain from this expression as a omega 1 minus sine omega 1 t right and b omega 1 cos omega 1 t and minus c omega 2 minus c omega 2 sine omega 2 t and we finally have plus d omega 2 cos omega 2 t similarly for x 2 dot i will end up getting minus a omega 1 psi omega 1 t plus b 1 omega 1 cos omega 1 t minus minus plus c omega 2 sine omega 2 t plus or rather minus d omega 2 cos omega 2 t now what happens to these values at time t is equal to 0 x 1 dot at t is equal to 0 and x 2 dot at t is equal to 0 is 0 so from here i will get at time t is equal to 0 the 0 is equal to minus a omega 1 sine 0 plus b 1 no sorry this is just b not b 1 b omega 1 cos 0 minus c omega 2 sine 0 plus d omega 2 cos 0 right again the sine terms vanishes and we are left with b 1 omega 1 plus sorry it's just b not b 1 b omega 1 plus d omega 2 is equal to 0 similarly for the second equation what do we get 0 is equal to at time t is equal to 0 minus a omega 1 sine 0 plus b omega 1 cos 0 plus c omega 2 sine 0 minus d omega 2 cos 0 this here also the sign terms vanishes and i am left with b omega 1 is equal to d omega 2 so this is one equation and this is the other equation so if i apply uh this equation to the uh first equation here this simply becomes b omega 1 plus d omega 2 is equal to b omega 1 right so this is 0 or b omega 1 is equal to 0 omega 1 is clearly not equal to 0 we know that we end up getting b is equal to 0 and if b is equal to 0 then from this expression i can immediately say that d is also equal to zero so i end up getting b is equal to zero d is equal to zero and a is equal to a by two and c is equal to minus a by two if i put these back in our original expressions in these expressions the nature of the equations look something like this so finally these are the solutions of x1 and x2 when we provide them with this kind of an initial setup where we only displace the second mass by some kind of a fixed amount that's it so when we displace the second mass by some kind of a fixed value capital a then the solutions become this now as you can see this is again some kind of a linear combination of distinct frequencies omega 1 and omega 2 now before i show you what is the nature of the motion under this kind of a configuration let me sort of do some trigonometric substitutions here to one of the expressions just so that you can appreciate the results that we'll finally obtain so what i'm going to do is i'm going to substitute omega 1 with let's suppose half of 2 omega 1 which is just equal to half of omega 1 plus omega 2 plus omega 1 minus omega 2 that can simply be written as omega 1 plus omega 2 divided by 2 minus omega 2 minus omega 1 divided by 2 similarly i'm going to substitute omega 2 with half of 2 omega 2 that can simply be written as half of omega 2 plus omega 1 plus omega 2 minus omega 1 so this is equal to omega 2 plus omega 1 upon 2 ma plus omega 2 minus omega 1 upon 2 so as you can see i have written omega 1 in terms of this particular frequency which is the average of both the frequencies and this which is the average of the difference of both the frequencies and omega 2 also as a sum of the average of both the frequencies and half of the difference of uh both the frequencies here now why am i doing this you'll just discover it yourself let me substitute these two expressions in let's suppose let me do it for x1 and then it will become obvious for x2 so let me do it for x1 and then this final solution for x1 simply it transforms into cos omega 1 t can be written as cos omega 1 is nothing but omega 2 plus omega 1 upon 2 t minus omega 2 minus omega 1 upon 2 t right and minus cos omega 2 t can be written as omega 2 plus omega 1 upon 2 t plus omega 2 minus omega 1 upon 2 t now what is cos a plus b and cos a minus b well it is nothing but the expression here so according to this if i write down the entire expression then this simply becomes cos omega 2 plus omega 1 upon 2 t cos omega 2 minus omega 1 upon 2 t then i'll end up getting plus sine omega 2 plus omega 1 upon 2 t sine omega 2 minus omega 1 upon 2 t for the first expression and for the second expression i'll simply get minus cos omega 2 plus omega 1 upon 2 t cos omega 2 minus omega 1 upon 2 t and minus minus plus sine omega 2 plus omega 1 upon 2 t and sine omega 2 minus omega 1 upon 2t now you might be wondering why we are doing this i'll come to the results in just a moment okay so this is a term and this is a term that no sorry not these two terms but rather this term and this term they are the same but opposite so they get cancelled out and finally we are left with these two terms which are exactly the same okay these two terms are exactly the same so so twice of a by two simply becomes a and sine omega 2 plus omega 1 upon 2 t and sine omega 2 minus omega 1 upon 2 t is just the solution x 1 t okay this is just a solution x 1 t i can obtain a similar solution for x 2 t also but let me first show you the graphical representation of what this kind of an expression actually means okay so here we have the graphical representation of both x1 and x2 for the initial conditions that i have provided here okay for these initial conditions how does x1 x2 vary with respect to time so let us try to study this and see what sort of a peculiar behavior we can understand so x1 here is varying with respect to time which is here in the y axis okay and x 2 is varying with respect to a time all right now what sort of information can we get from these graphical representations so x1 so let us first look at x2 so if you look at the initial conditions we only provided some displacement to x2 we did not give any displacement to x1 we did not give any velocities to x1 and x2 at time g is equal to 0 we only gave some kind of initial displacement to x2 and because we gave some initial displacements to x2 you can clearly see that x2 starts oscillating but as it oscillates its amplitude of oscillation decreases so x2 starts oscillating but with time its amplitude starts decreasing and after some time its amplitude again starts increasing and after some time it again starts decreasing and after some time it again starts increasing so the system of x2 is oscillating its amplitude decreasing and after some time its amplitude increasing and after some time its amplitude decreasing and after some time its amplitude increasing and on and on and because x1 the first mass is connected to x2 what you see for the first mass is that x1 starts oscillating with increasing amplitude which decreases after some time with increasing amplitude which decreases after some time so initially the x1 is at rest but when we provide displacement to x2 because it's connected to x2 the x1 also starts oscillating it amplitude increases it reaches a maximum and the amplitude decreases becomes minimum again the amplitude increases becomes maximum again the amplitude decreases becomes minimum so the behavior of both the masses is such that when i displace x2 from its initial configuration it starts oscillating amplitude decreases at the same time x1 starts oscillating it amplitude increases and after some time x2 x1's amplitude decreases and x2 is empty starts increasing so two very interesting things are happening here the first interesting thing is that what we just now derived this frequency omega 2 plus omega 1 and omega 2 minus omega 1 these are the two frequencies that you can see here the frequency of the oscillation of x1 and x2 and the frequency of the modulation with which this wave is happening these are these two frequencies so the frequency with which the oscillations are happening that is nothing but omega 2 plus omega 1 upon 2 the frequencies at which the oscillations are happening that is omega 2 plus omega 1 upon 2 but the frequency angle of at which the modulation is happening you see the amplitudes are modulated amplitudes are slowly increasing and the amplitudes are slowly decreasing the frequency with respect to this modulation that is happening to that original oscillations that is omega 2 minus omega 1 upon 2 so a graphical representation of the general solution for our initial conditions provides these beats that you have these oscillations but these oscillations are being modulated by a further oscillation right you have these oscillations but these oscillations are being modulated by a further oscillation and the oscillations are happening with this frequency and they are being modulated by this particular frequency so as you can see this is the sign one frequency and assign another frequency term similarly the same thing is also true for the x2 case it is also oscillating and these oscillations are being modulated by a further oscillation all right so this is very interesting that when we only give one displacement to one of the masses that since that mass is connected to the first mass they both are oscillating with a particular frequency which gets modulated uh by another frequency and what's more interesting is that both of them are exchanging energy with respect to time so as you see as the first mass increases in amplitude the second mass decreases in amplitude and when the first mass reaches maximum the second mass amplitude becomes minimum as the second mass amplitude increases the first mass amplitude decreases and when the second mass amplitude becomes maximum the first mass amplitude becomes minimum after some time the first mass amplitude increases the second mass amplitude decreases so when the first system is oscillating at its maximum amplitude the second system is zero at rest and when it starts losing energy the second system starts gaining energy and it reaches the maximum amplitude and after some time the second mass starts losing energy and the first mass starts gaining energy and reaching maximum amplitude and this starts losing energy and this starts gaining energy so this keeps happening over and over again this loses energy this gains energy this gate loses energy this gains energy and both these two subsystems in the coupled oscillator are sort of oscillating by exchanging energy from one to the other from the one to the other and on and on and on so this is quite uh interesting to see all right so that is all for uh today's class i hope you enjoyed and learned something from the sort of lecture let me do a very quick revision we took the case of a coupled oscillator where there are two masses and three springs attached together and the outer springs are connected to two walls we obtained the force equations using the newtonian approach from there we obtained two differential equations that can be solved and when we did that we found out that there is a coordinate called x1 plus x2 that is executing simple harmonic oscillations or sinusoidal frequencies and there is another coordinate x1 minus x2 that is executing a simple harmonic oscillations constant frequency and i say said that these are known as normal coordinates from there i obtain the general solutions the general solutions are nothing but linear superpositions of two distinct frequencies omega one and omega two from here i talked about normal modes that if i provide certain initial configurations to both the systems they can exhibit simple harmonic oscillations of distinct frequencies so this is the first normal mode and this is the second normal mode and then we talked about when we provide one displacement to one of the masses then the nature of the motion is such that that mass will start oscillating but it will keep on losing energy over time while the other mass will start oscillating and keep on gaining energy over time and this kind of an oscillation is modulated by a further oscillation and the values are omega 2 plus omega 1 by 2 and omega 2 minus omega 1 by 2 respectively so that is all for today thank you very much

Channel: For the Love of Physics

Views: 3,388

Rating: 5 out of 5

Keywords: coupled oscillators, Coupled oscillator, coupled oscillators and normal modes, coupled oscillations, coupled oscillators explained, normal modes, normal frequencies, normal coordinates, oscillators, normal coordinates and normal frequencies of vibration, for the love of physics, physics, education, lectures, normal coordinates in classical mechanics, spring mass system, 3 spring 2 mass system, 2 mass 3 spring system, beats, force, simple harmonic motion, derivation

Id: 4WhrNjg3I_o

Length: 49min 56sec (2996 seconds)

Published: Thu Apr 08 2021