Lecture 11

[Music] hello and welcome to another discussion on classical motion of a single particle so in this Su lecture we will discuss uh the basic motivation of classical mechanics so if you remember we said that in classical mechanics um the kinematics is to describe the motion to know where the particle is or whether at that time in stand the particle is trying to move at all or not but Dynamic says that how this uh nature actually has been generated or I mean what is the source of the motion or the I mean explanation of the motion so now if we have by uh I mean thanks to Newton's law the equation of Dynamics then from that the main OB objective of classical mechanics is to know the position and the velocity for a constant uh for a particle with constant mass of course at any given time and we said in a previous lecture that this is only possible when we have the uh total uh set of equations of Dynamics with appropriate initial conditions so the condition or the deal is that you give me the equations of motion and you also give me the position and the velocity at any point of time then I will exactly predict you at a later instant of time okay or at some arbitrary instant it can be later it can be actually previous if it is Meaningful of course huh it depends how you define your instant time instance huh so it can actually so by the help of the equations of Dynamics and the initial conditions I can uh exactly tell you the position and the velocity of the particle at an arbitrary time okay which was not known in general the position and the velocity okay so that is the main motivation and the main objective so whatever be the process of solution if I have the Newton's law of motion I mean Newton's laws of motion or the equations of Dynamics if somehow by using any correct method huh if I can just tell what is the x of T and X do of T actually my purpose is served that is the classical mechanics main goal so uh of course I can tell you that in one dimension of course in in in two Dimension or in three dimension this is the purpose is to know r as a function of T and R dot as a function of T starting from starting from D2 r dt2 or R double dot is equal to F by m okay where f is the force now in this discussion I will actually try to to show you that uh the integrals of motion or the special constants of motion sometimes are very helpful in uh solving those equations in classical mechanics and that is actually very evident and very direct in case of onedimensional motion and I will also discuss about that okay so I am taking the very condition or the very case of a a one-dimensional simple harmonic oscillator that means you have M d2x dt2 and the force is min - KX if the force is minus KX then you all know that what is the corresponding potential the potential is so this is totally can be written as a this is a force and if that force is equal to minus du da X then that U should be half K x² okay and so of course you know in uh this case the total mechanical energy is nothing but half MX do s + half K x² is constant now before using this constants I mean before using the uh conservation of this total energy E I will first tell you that the I mean conventional method let's say what is the equation of Dynamics for a simple harmonic oscillator in 1D this is simply this one and then you can easily write d2x dt2 is equal to Min - K by m X then you know this is d2x dt2 + Omega S xal 0 where Omega s is equal to K by m now you can now have a second order ordinary differential equation and you know the trial solution from our knowledge we know that if x is equal to some a uh I think I just write like this a sin Omega t plus b cos Omega T I mean you can do the other way as well then this solution should be satisfying this equation okay and uh why is this because you know like if I do dxdt from my knowledge that will be of course a Omega cos Omega T minus B Omega sin Omega T and d2x dt2 will be equal to minus E Omega s sin Omega T minus B Omega s cos Omega T then of course minus Omega squ common you have a sin Omega t plus b cos Omega T and this is nothing but your a so you know that this solution actually satisfies this equation d2x dt2 is equal to minus Omega s x okay now the point is that if we want to do it I mean want to solve it more directly then the conservation of energy total mechanical energy actually helps us we can do this in the in this way I already said that MX dble dot is equal to to minus KX can be written as MX dot X double dot is equal to KX x dot I can write this thing as you already know I have done this trick also for the conservation of total energy huh this is nothing but DDT of half MX do s plus half k k x dot h sorry half K x² is equal to 0 and then I simply say that my M by 2x do s plus half K x² is equal to a constant and that constant I now call it as e okay so this e for example let's say I mean in a in a in a in a motion this e should not change okay if if I have this Force only and there is no other forces acting then this should not change okay so then I can easily write from this step I can easily write that x dot is equal to root 2 e/ M minus Omega s x² where Omega Square as I just said previously as well is equal to K by m if just from this equation you can easily write this equation okay now what is this this is DX DT is equal < TK 2 e by m - Omega s x² okay now I can say okay I can just take everything including this X part on one side and the DT on the other side okay DT I also multiply with this constant Omega both sides and then I say d of Omega X by root 2 e by m - Omega x² is equal to D of Omega T okay I can actually integrate now at and so you see this is nothing but an integration of the form DZ over < TK a² minus Z squ where a is nothing but < tk2 e by m okay this E I can just write like this okay and if you if you want to calculate that I am also trying to calculate that so the easiest way you just do some substitution Z is equal to let's say a sin Theta and so DZ will be equal to a COS Theta so integration DZ will be a COS Theta D Theta sorry it will be a COS Theta D Theta I'm sorry root a² minus a² sin s Theta if you do this integration so it will be actually again a s - a square cos Square Theta I mean a square - a square sin Square Theta which is a square cos Square Theta under root it will be a COS Theta so a COS Theta and a COS Theta will go away you will have integration Theta which is Theta + C now that Theta is nothing but sin inverse or R sign Z by E okay so Plus some constant I can do the same thing over here and I can actually again go to the green yeah I can actually uh calculate that and clearly write this is nothing but sin inverse Omega X by < TK 2 e by m is equal to some omega t + 5 5 is an arbitrary concept of integration okay and finally I can actually write this Omega X okay by < TK 2 e by m is equal to sin Omega t + 5 okay and then finally you can write X is equal to < TK 2 e by m Omega squ sin Omega t + 5 okay you see here you can find another another solution which is exactly of this same thing okay but now maybe you cannot easily understand the for because if I now tell you that okay this a sin Omega t + B cos Omega T is equal to some C uh I think like sin Omega t + 5 where a I can Define as some C * cos 5 and B I can Define some C * sin 5 then it will be so of course so C is nothing but root over a square + b square and F is nothing but tan inverse B by a okay so if you do if you if you choose a and b in such a way then you can actually find C and F and you you can write this expression as this one so you see this C is nothing but this constant now we root over 2 e by m Omega squ and you see this is the correct con correct constant Dynam I mean from the point of view of Dimension as well because e has a dimension of m x do s and has a dimension of K x² okay and also we are saying that Omega square is having a dimension of K by m so if I now say that this uh I mean I can actually say like this that if if I are write completely you will see that e is actually dimension of M Omega Square x² because you see that this m Omega square is nothing but K so KX squ can be written as e of M Omega Square x² so 2 e by m Omega Square will have the dimension of x² and root over of that will be the dimension of a length this has Dimension less so this is also correct so you see in the first method you don't know what c is that will be given by of course the uh the initial conditions but here actually you know what this is 2 e by m Omega squ but the point is that which one is more advantageous actually both are equally advantageous because this e is also given by the initial condition because whenever I give you the initial x and x dot or the x and x dot of any instant actually you know that this value at that instant and since that that instant is I mean since that I mean constant is a constant of motion this quantity is a constant of motion basically you can uh determine this one given an initial condition and this constant then you use throughout okay so in this way you can see that you can uh just using the invariance or the conservation of uh the total mechanical energy you can exact L solve uh one-dimensional motions I can actually give you some uh problems on some I mean on some other type of onedimensional motion where the uh motion is not simply U like harmonic oscillator type okay for that you can actually also find out how you can solve for x and x dot to to find the solution of course after having after having found X as a function of t now you can take x dot and to find the kinematics the complete description of the kinematics of the particle now before ending this lecture I would like to emphasize on a very important Point what happens for more than one dimension okay then you here you see here basically in one dimension you had two variables x and x dot and you had one equation okay and then I will use the information of the conservation of total mechanical energy then basically you actually constrain between X do and x and that's why there is one constraint and uh two variables so finally the effective variable becomes one and actually you can solve for the other one two variables are x and x dot of course okay now in 2D for example if you just go to a little bit higher Dimension let's say just two okay so there you have the unknown quantities or variables like x x dot Y and Y Dot and finally if you think that if we know that there is the total mechanical energy is again conserved just let's say for the case of uh simple harmonic oscillator huh so we actually know that half of m x do s + y do s + half of K x² + + Y2 is equal to constant so I have one constraint connecting all these four variables but I have actually two differential equations of second order one is MX d dot will be equal to minus KX another will be m y d dot equal to minus KY okay so I actually have this and I actually have but if you if you want to write as uh like in terms of just first order ordinary differential equations you will actually have four okay two for each Dimension and how many constants we have only one so this is not sufficient to just solve the whole system in more than one dimension for more than one dimension you actually need more conserved quantities or more integral result of motion that will give you further constraints between I mean among those uh variables okay so just you comment me in the Forum discussions okay how many conserved quantities do you need in 2D and 3D in order to uh completely solve a mechanical problem I mean of just an equations of Dynamics okay with that I just wrap up this thank you [Music] [Music] [Music]