It's NOT so Simple: Physics of the Simple Pendulum - Equations of Motion & Beyond

so in today's video i'm going to talk about the simple pendulum so you may have seen that objects that are hanging from some point usually create some kind of a to and fro back and forth oscillations now if you have an object that can be approximated by a point mass like a ball that is hanging from a string that kind of a setup is known as a simple pendulum it's called a simple pendulum because that is a simplistic or simplest construction possible now the question is is the nature of the motion simple enough well we will see okay so what can you expect from today's video first of all i'm going to look at the forces that are involved and i'm going to obtain the equation of motion of the simple pendulum i'm going to do that by applying the newton's second law in polar coordinates that's the first thing second i'm going to look at the torque involved and the nature of the rotation and from there i'm going to apply obtain the equation of motion the same equation of motion and third i'm going to obtain the same equation of motion by looking at the energies by looking at the lagrangian now you may say why am i obtaining the same equation of motion of the pendulum from three different approaches well why no i mean if you can obtain the same equation of motion for the system from three different approaches i mean that is quite interesting isn't it so let us do that and once you do obtain the equation of motion which is a second order differential equation that needs to be solved but the question is is the equation even solvable in the first place for that we will have to look at some approximations especially the small angle approximation so once we do the small angle approximation now that becomes a differential equation that can can be analytically solved which we are going to do and then we are going to look at the nature of the solution what is the displacement what is the velocity what is the momentum what is the energy etc i'm also going to do one more thing i'm also going to try to solve the equation of motion without doing the small angle approximation which can be done not analytically but using some kind of a computational software that is capable of solving differential equations numerically for that i'm using scilab and once we do that we will compare the results for small angle approximations and the results without considering the small angular approximations and we will see how uh that is different from each other then we will also look at the phase space trajectory we will look at how the momentum varies for both small angles as well as for larger angles we will also look at how the kinetic energy and the potential energy varies for small angles as well as larger angles so it's going to be a little bit of an elaborate video if there are portions that you want to skip through because you're already familiar with it you can do that i'll put a timestamp for the various topics in the description so let us begin so here we have a very simple construction of this pendulum where there is some kind of a mass that can be approximated by a point mass that is attached to some kind of a massless inextensible string that is uh hooked at some kind of a pivot to a ceiling or the wall and the nature of the motion is restricted to a plane in that kind of a situation we can obtain the equation of motion by looking at the forces involved now what is the most obvious force in this kind of a situation well the most obvious force is the force of gravity which is mg acting downwards there is a force which is the force of tension the tension is existing there because the massless string is inextensible and it is because of this tension that the pendulum's motion is restricted to the arc of a circle now if i want to analyze uh the nature of its motion i will have to look at the forces that are resulting in this kind of a motion on the arc of a circle for that i can look at the components of the forces so for example if mg is directed straight downwards and i look at any random point where the pendulum is present at any random point such that the angular displacement from the vertical axis is theta then i can divide the force mg into various components one i can look at the component along the radial direction now if you look at the geometric construction here this angle is the same as the angle theta all right so the component along the radial direction is simply going to be equal to mg cos theta right and the component the force along the tangential direction is going to be simply equal to mg sine theta all right now it is clear from the diagram that the radial force is not contributing towards the motion because the radial distance is a constant of time it is only the tangential force component that is responsible for the to and fro motion along the arc of a circle so let's first obtain the equation of motion from the newtonian perspective by applying the newton's second law in polar coordinates the newton's second law says that the force acting on any kind of a point mass object leads to some kind of an acceleration a however because the nature of the motion is along the arc of a circle we can apply the polar coordinates we can write the newton's second law in polar coordinate so if you write the newton's second law in polar coordinates this simply becomes the force components along the radial direction the force components along the transverse or the tangential direction is equal to m the acceleration component along the radial direction plus m the acceleration component along the transverse or the tangential direction now if you know how to transform the acceleration terms in from x y z to polar coordinates then you would probably know then the first term here simply represents r double dot minus r theta dot square r cap and the second term is m r theta double dot plus 2 r dot theta dot theta cap don't be scared of the dots here the dots simply represent time derivative so a singular dot represents single time derivative the double dots represent second order time derivative now if you look at the forces fr which is basically equal to this particular term and if you look at f theta which is basically equal to this particular term now you must understand from the diagram itself that f r is not creating any kind of a motion because the radial distance of the pendulum is not changing l is a constant of time so the only thing that we are interested here is the relationship between f theta is equal to m r theta double dot plus 2 r dot theta dot again theta represents the angular displacement r represents the radial displacement the radial displacement is the distance from let's suppose the point o to the pendulum's point mass and theta is the angular displacement with respect to the vertical axis now in the construction of the pendulum the r dot is equal to zero why r dot simply represents time rate of change of the radial displacement the radius of a pendulum remains constant with time that is the construction of a pendulum so this is equal to 0 and r here is simply equal to l is a constant of time right this is the first thing second what is f theta f theta is the force along the transverse direction which is nothing but m g sine theta now i have to write the minus sign here because the force is acting in a direction opposite to the increase in angular displacement so the direction in which the angular displacement is increasing is in this particular manner but the force is acting in the opposite direction so the force is acting in a direction of the decrease of the angular displacement therefore i have written the negative sign here so minus mg sine theta is equal to m r theta double dot which leads to the equation minus g sine theta is equal to r theta double dot if i substitute the values here this simply becomes minus g sine theta r is nothing but the length of the pendulum and theta double dot is d2 theta upon dt 2. this becomes a very simple differential equation if i write it properly this is d2 theta upon dt2 plus g upon l sine theta is equal to zero this is quite simply put the equation of motion of the simple pendulum that you can obtain just by looking at the forces that are involved more specifically the force acting in the transverse direction written in terms of polar coordinates now i can obtain the same equation of motion but using a little bit of a different approach let's try that also now of course the nature of the motion is some form of rotation here right so if you apply the newton's second law in terms of rotation then you can write that whatever the torque is that is creating this motion is the result of i alpha where i is the moment of inertia and alpha is the angular acceleration this is quite simply put the nature of the newton's second law in terms of rotation so here the moment of inertia i is what if you look at the pendulum of course the string is massless by our assumption and the mass is a point mass at a distance of l so the moment of inertia is quite simply put m l square and the angular acceleration is simply defined as the second order derivative of the angular displacement which is d2 theta upon dt2 now what is the torque involved in this kind of a setup the torque is simply nothing but r cross f okay so if you look at the torque the torque is if i write it only the magnitude of the torque that is nothing but r cross f now what is f here as i already said the force is the force of gravity but it can be divided into two components one along the radial direction the other along the transverse direction the component along the radial direction is not causing any change in the radial displacement it is only the component along the transverse direction that is leading to the change in angular displacement therefore i have to write here r is nothing but the distance from the origin o which is nothing but l okay and since l and mg sine theta are perpendicular to each other therefore the cross product will simply involve the length of r that is l cross product sine 90 comes out to be 0 and force is m g sine theta and this is a minus sign here because it is acting in the direction opposite to increasing theta so this simply becomes an equation where minus m gl sine theta is equal to moment of inertia is ml square and alpha is t2 theta upon dt here of course the masses get cancelled out the l of course also gets cancelled out and you are left minus g upon l sine theta is equal to d theta upon dt 2 which can be written in the form of an equation d2 theta upon dt 2 plus g upon l sine theta this is exactly the same equation of motion that we just now obtained by taking a look at the forces so even by using the concept of torque and moment of inertia we can obtain the same equation of motion now let us again obtain the same equation of motion but using a little bit of a different approach some of you may have heard of it some of you may not have and this is known as the lagrangian approach which is the approach that is used in classical mechanics so in the lagrangian approach instead of dealing with forces we deal with energy specifically the kinetic and the potential energy now if you are completely not familiar with the lagrangian approach you can skip through to the next section where i am analyzing the equation of motion okay so in the lagrangian approach we basically look at the kinetic and the potential energy of the system so the kinetic energy of the system or which is written as either ke or t is simply equal to half m v square now as i already told you that the nature of the motion is restricted to the arc of a circle it's always better to write the physical quantities in terms of uh coordinate system which makes sense in that sort of a motion so the coordinate system that makes sense here is the polar coordinates so in polar coordinates the velocity is written as simply v square is v r square plus v theta square where v r is the velocity along the radial direction and v theta is the velocity component along the transverse direction now some of you may already know if you are familiar with polar coordinates that v r is nothing but r dot and v theta is nothing but r theta dot quite simple so if i apply this here the form of the kinetic energy simply becomes half m r dot square plus r theta dot square that's it this is the kinetic energy what about the potential energy where well to find out the potential energy will we will require a reference with respect to let's suppose the zero potential now i can choose the reference to be the ceiling or the point of the pivot or i can choose the reference to be also the lowest point of the pendulum so if i choose the lowest point as a reference for the pendulum in that kind of a situation what is the height let's suppose the height of the pendulum is given by h i can calculate h in terms of the length of the pendulum because as you see here the projection of the length of the pendulum along the vertical axis is simply nothing but l cos theta so if you calculate the height the height is nothing but l minus l cos theta as it is obvious from the construction here so the height is simply l 1 minus cos theta therefore the potential energy is mgh which is nothing but m g l 1 minus cos theta so if you are not familiar with the lagrangian the lagrangian is simply defined as t minus v so this is equal to half m r dot square plus half m r theta dot square sorry this is supposed to be r square okay r square theta dot square minus m gl 1 minus cos theta now what do i do with the lagrangian what we do is we plug it into what is known as the euler lagrange equation when we plug the lagrangian into the euler lagrange equation that is when we obtain what is known as the equation of motion so the euler lagrange equation is is written in this manner d upon dt of del l upon del q dot minus del l upon del q is equal to 0 where q is supposed to be the generalized coordinate so g generalized coordinate is just the coordinate with respect to which you are analyzing the motion here we are trying to analyze the motion with respect to the angular displacement so i'll assume that q here simply represents theta so the euler equation euler lagrange equation is simply d by dt del l upon del theta dot minus del l upon del theta is equal to zero so here as you can see if you look at the lagrangian i can find out the various derivatives of the lagrangian with respect to the various quantities so here del l upon del theta is simply nothing but minus m g l sine theta del l upon del theta dot is equal to m r square theta dot and therefore d upon dt of del l upon del theta dot is equal to m r square theta double dot so if i plug all of these equations back into the lagrange equation i simply therefore obtain an equation of the form m r square theta double dot minus minus m gl sine theta is equal to 0 the m is common i can take it out it simply is r square theta double dot plus g l sine theta now what is r r i have already mentioned is nothing but the radial distance of the point mass from the origin o here which is l okay so therefore i substitute here r square by l square and if i divide the equation by l square i simply get d2 theta upon dt2 plus g upon l sine theta is equal to 0 this is the equation of motion that we have obtained using the lagrangian approach which is exactly the same as the equation we have obtained using the concept of torque as well as the equation we have obtained from the concept of forces and polar coordinates so it is the same equation of motion if anybody asks you this kind of a problem in an examination you don't have to do all three approaches you just choose any one of them it doesn't matter the reason i am doing it because it's kind of interesting to see that the same equation of motion can be obtained from so many different ways so this is kind of interesting in a manner so this is the equation of motion and the next challenge is now can we solve the equation of motion if i am able to solve this equation of motion then the solution will some give me something like theta t it will give me the angular displacement with respect to time but the question is can i in fact solve this equation of motion well you see analytically that means using the ideas of mathematics and solving differential equations this kind of an equation is difficult to solve analytically it can be solved in two situations it can be solved either by some kind of a computational software or we can make some kind of an approximation which will transform this equation to something that can be analytically solved so i'm going to do both of that so first let us look into these small angle approximations so in this small angle approximation what we consider is that we assume that the pendulum is oscillating in such a manner that the angular displacement is quite small okay the angular displacement of the point mass from the vertical axis is quite small if the angular displacement is quite small in that kind of a situation we can make a particular approximation that sine theta this particular function of theta is approximately equal to theta itself we can take a look at their uh graphs so here you see i have simulated the sine theta graph as well as the theta graph now of course the theta graph with respect to theta itself is a straight line while sine theta with respect to theta is the obvious function that you already know of now if i consider a very a small region where let's suppose this is for corresponding to 90 degrees knight so if i go up to from 0 to 45 degrees so if i go from 0 to 45 degrees as you can see the red line represents theta the blue line represents sine theta so clearly as theta is decreasing sine theta and theta kind of coincide if i even go to smaller angles let's suppose from here to here now both of them are kind of indistinguishable so the sine theta varies in such a manner that for very very small angles like let's suppose 5 degrees 10 degrees the function is almost equal to the variable theta itself so coming back to our expression as i showed sine theta is near about equal to theta if angular displacement is very small in that kind of a situation how does the equation of motion look like the equation of motion simply looks like this d2 theta upon dt 2 plus g upon l theta is equal to 0 now this is an equation of motion that can be analytically solved and we can obtain that solution and we can study that solution so let us do that now there is a standard uh solution for differential equations of this form and if you are already familiar with that kind of solution you can skip forward to the time stamp that i have provided where we will analyze the results but if you're not familiar how to solve this kind of a differential equation so you can stick with me here g and l are a constant and we are going to assume that g upon l is something called omega square now what omega is i'm going to come to that in a moment but the moment i do this substitution the equation becomes d 2 theta upon d t 2 plus omega square theta is equal to 0 what is the solution of this kind of a differential equation where omega here is a constant well to do that equations of this form there are solutions theta are written in terms of the complementary function and the particular integral now of course this will require a little bit of an idea of how to solve differential equations and i am assuming that you have some kind of a idea of how to approach differential equations so differential equations of this form can be solved by looking at their what is known as auxiliary equation d square plus omega square is equal to 0 where d simply is a representation of the second order derivative that we have here if i solve this this simply becomes d square is equal to minus omega square or d is equal to plus minus i omega so this is a complex root 0 plus i omega and 0 minus i omega so for complex solutions the complementary function can be written in terms of some kind of a constant c 1 e to the power 0 plus i omega t plus some kind of a constant c 2 e to the power 0 minus i omega t which is quite simply put c 1 e to the power i omega t plus c 2 e to the power minus i omega t this is the most general expression for the complementary function where the roots of the auxiliary equation are complex since the right hand side of the differential equation is 0 the particular integral in this case also comes out to be zero therefore the solution of the differential equation here that is theta which is a sum of complementary function plus particular integral is nothing but c1 e to the power i omega t plus c2 e to the power minus i omega t i can simplify this by writing the expressions in terms of sines and cosines now you may already know that the exponential e to the power i theta can simply be written as cos theta plus i sine theta so if i write down my solution in these terms this becomes c1 cos omega t plus i sine omega t plus c2 cos omega t minus i sine omega t so this simply becomes c 1 plus c 2 cos omega t plus i c 1 minus c 2 sine omega t so as you can see here c1 plus c2 is nothing but some kind of a real constant while ic1 minus c2 is nothing but some kind of a complex constant which i can denote by let's suppose a cos omega t plus b sine omega t so a here is a constant so is b so this is the expression for the general solution of theta so the solution for the equation of motion of the simple pendulum under small angle approximations is simply given by theta is equal to a cos omega t plus b sine omega t however this is a very general form of a solution we want to obtain a particular solution for a particular given initial conditions so basically what i'm going to consider is that let's suppose the pendulum is stretched from its equilibrium to a particular initial condition that i stretch it to a particular angular displacement theta naught and i just leave it without giving it any initial velocity under that kind of an initial condition we can find out the values of the constants a and b that would give us a little bit of a better expression to study so let's do that let's apply the initial conditions so according to the initial conditions i'm saying that theta at time t is equal to 0 when we displace the pendulum from its equilibrium position is some kind of a value let's suppose theta naught some small angle theta naught may be 5 degrees maybe 10 degree something like that and theta naught which is the velocity that we impart at time t is equal to 0 is 0 that means we don't give it a push we just stretch it from its equilibrium and leave it there now what should this give us so theta at time t is equal to 0 is nothing but a cos 0 plus b sine 0 now you may already know that sine 0 is nothing but 0 while cos 0 is equal to 1 and theta at time t is equal to 0 is theta naught so this is equal to a so theta naught is nothing but a all right while on the other hand and find out the time derivative of theta the time derivative of theta can be obtained from this expression the time derivative of theta is nothing but a omega minus sine omega t plus b omega cos omega t now i have already mentioned that we do not give it any kind of an initial push but rather separate it from its equilibrium and leave it there that means we don't give it an initial velocity so the initial velocity at time t is equal to zero at time t is equal to zero is zero so this is equal to minus a omega sine 0 plus b omega cos 0 again sine 0 is 0 while cos 0 is equal to 1 that simply gives us that b is equal to 0 so by giving the initial conditions that we displace the pendulum from its equilibrium by a very small angle and then leave it there by doing that i have obtained the values of the constants a and b if i substituted here i finally get the same solution of the pendulum for our situation of initial conditions which is theta t is equal to theta naught cos omega t right i put up the values of a and i get theta t is equal to theta naught cos omega t so i simply get a cosine function or i get some kind of a sinusoidal motion because what is cos omega t cos omega t is nothing but sine pi by 2 plus omega t so what i end up getting is a simple harmonic motion i end up getting a sinusoidal tragic tree we can look at it graphically okay so here i am uh simulating the solution for a pendulum that has been initially displaced from its equilibrium by 15 degrees as you can see the pendulum is displaced by 15 degrees here the axises are written in terms of radians okay and the pendulum then comes towards the equilibrium goes to the other side reaches the extreme again comes towards the equilibrium goes to the other side reaches the extreme and this kind of a two and fro motion is happening what is interesting is that this kind of a two and fro motion is nothing but a simple sinusoidal wave it is a simple harmonic oscillation so when we solved the equation of motion for small angle approximations we obtained a result which is simply sinusoidal oscillations which is known as simple harmonic oscillations so the pendulum basically behaves as a harmonic oscillator for small angles right so the pendulum is behaving as a harmonic oscillator it is executing some kind of an shm for smaller angles now what is the frequency and various other quantities associated with this kind of a harmonic oscillator well we can obtain it from the nature of the uh the solution here so solution is nothing but theta t is equal to theta naught cos omega t the theta naught simply refers to the amplitude right what is the amplitude here that simply gives us an idea about the theta naught which is the initial displacement of the pendulum right what about omega what does the omega tell us omega tells us i told you right what is the omega the omega we initially assumed that omega is equal to root over g upon l you remember we initially assumed that when we were constructing the differential equation that omega is nothing but root over g upon l here now as you can see from the solution the omega is nothing but the angular frequency of this kind of a simple harmonic motion by looking at the nature of the motion which is the shm you can see that the omega is nothing but the angular frequency and if you know the angular frequency you can obtain the frequency you can obtain the time period so the angular frequency is simply equal to root over g upon l so you can obtain the time period of the pendulum also so if a pendulum is displaced from its equilibrium by a very small angle what is the time taken for one complete oscillation that is the time period of the pendulum which is simply given by t is equal to 2 pi upon omega which is equal to 2 pi upon root over l by g now it is very very interesting to note that when we look at a simple pendulum which is displaced from a very displaced by a very small angle from its vertical axis the time period only depends upon the length of the pendulum the acceleration due to gravity now what is interesting to note here is that the time period is not dependent on the mass it is not dependent on the angular displacement also it is only dependent on the length of the pendulum and the acceleration due to gravity so you may have a heavy object or a light object at the same distance from the pivot and then they will create the same kinds of time periods even for smaller angles as well as for a little bit larger angles let's suppose 15 degrees or 20 degrees so the time period is independent of the mass of the pendulum as well as the initial angular displacement now remember that we are only considering those classes where the initial angular displacement is small because only in those situations i have proved to you that the nature of the motion is a simple harmonic motion okay so by looking at the equation of motion for uh the simple pendulum that we have obtained and by making an approximation that for small angles we have obtained the solution to be that of a simple harmonic motion and from here we can obtain various other quantities also so for example we can also obtain what is the nature of velocity how does the velocity of the simple pendulum change with time as the pendulum is oscillating back and forth in some kind of a simple harmonic fashion how does the velocity component change well that can simply be obtained by looking at theta dot so what is the time derivative of theta so this is d by dt of theta dot cos omega t as you can see this gives us quite simply minus omega theta naught sine omega t so this is also a kind of a sinusoidal uh a function so here i have created a program to plot the shm angular displacement as well as a velocity the blue line represents the angular displacement and the red line represents the velocity now of course the amplitude of the velocity and the amplitude of the angular displacement is not necessarily same therefore one is a little bit larger than the compared to the other but both of them are simple harmonic in nature although they are not necessarily in phase the blue line decreases initially because angular displacement is decreasing but the red line increases although in the negative direction so therefore the magnitude is increasing now before we analyze other quantities like momentum i want to do something interesting let's go back okay let us go back to the equation of motion that we had obtained now the solution that we obtained was for a very specific case of small angle approximations right now what if we do not make the small angular approximation what if we do not what if we want to find out the nature of its motion if the pendulum was displaced to let's suppose a large angle if the pendulum was displaced to a large angle what would be the nature of motion there for that we will have to solve this equation right however as i just now told you this is an equation that is difficult to solve analytically but it can be solved by some kind of a computational software that is capable of solving differential equations numerically you see usually we solve differential equations analytically by looking at the the logic of mathematics and the flow of it but there are certain kinds of differential equations which cannot be solved analytically but they can be solved numerically by various numerical methods so if you know some kind of a computational software you can do use that to solve this kind of a differential equation i have used scilab so maybe you know scilab maybe you do not know scilab if you do not know scilab maybe you know something else like matlab or mathematica or some other programming language python in which you can create some kind of a program to numerically solve a second order differential equation then you can analyze this equation now i'm not going to go into the details of the program that i have created i am just going to show you the nature of the solution that i have obtained from the simulation of numerically solving a second order differential equation of this particular form so let's go to that particular program okay so here i have created a program in scilab to find out the nature of the solution of a second order differential equation no need to go into the details of the program if you are familiar with scilab you can copy it from here and try to create the program yourself otherwise just look at the results because it is the results that we are interested in if we solve that second order differential equation using some kind of a computation software then what i can do is i can look at the solutions for various angles so here in this particular line so i have taken mass to be 1 kg length of the string to be 0.5 meters and time period from zero to two pi and acceleration due to gravity is nine point eight two meter per second and here i am taking for four different situations if the pendulum is displaced by ten degrees what is the nature of its solution if the pendulum is displaced by 15 30 and 45 degrees what is the nature of its solution let us take a look so here it is so here you can see the four different solutions for theta is equal to 10 15 30 and 45 degrees respectively so as you can see here in the in the four graphs the first graph represents the motion for theta is equal to 10 degrees second graph represents the motion for theta is equal to 15 degrees the third represents motion for theta is equal to 30 degrees and the fourth one represents motion for theta is equal to 45 now there are two plots in every single graph one is a red line the other is a blue line can you see that one is a red line the other is a blue line the red line represents some kind of a sine wave some kind of a sinusoidal motion a simple harmonic motion the blue line is the solution corresponding to the differential equation where the sine theta term was present remember the original differential equation contained the term sine theta and not theta so the blue line is the actual motion of the pendulum the red line is the solution for simple harmonic motion now let us look at the first graph the first graph is for displacement 10. here if i zoom in you can see that the blue line and the red line are very near to each other they are almost merging together so what does this mean it simply means that the blue line which represents the motion of the pendulum the blue line represents the motion of the pendulum and the red line which is simply an shm are both merging together that means the pendulum creates a simple harmonic motion now if i go to the second graph for initial displacement is only 15 degrees again you can see that the blue and the red line are kind of merging but there is a certain amount of uh distance between both these two solutions so even though the pendulum is as it creating shm but you can see that the blue line in the red line is slightly separate from one another yes now if i go to initial displacement 30 degrees you can clearly see here that over time the blue line and the red line are sort of separating from each other that means the red line which represents this a simple harmonic motion and the blue line which represents solution of the pendulum are not exactly in phase as you go forward in time and this becomes more and more clearer for angle displacement theta is equal to 45 degrees so as you can see with time the pendulum is sort of going out of phase from some kind of a simple harmonic motion so let me let me bring this particular graph to my notepad so this is quite interesting you see if we make a comparison between the solution of the simple pendulum for small angle approximations with that of the solution of the equation without small angle approximation we come up with something really interesting because uh the red line here is of course as i told you is the shm motion which is a sinusoidal motion and the blue line here represents the actual motion of the pendulum right now as i showed you earlier for smaller angles theta is equal to 15 degrees 30 degrees both of them were very very close to each other but for large angle like 45 degrees both of them sort of keep going out of phase with time so as time is going forward as the pendulum is moving oscillating on and on in time the pendulum develops a certain phase difference from the shm it is no longer an shm so now you may remember that if you have performed this kind of an experiment in a lab maybe your teacher told you that please make sure that you only take small angular displacements well when trying to create oscillations well this is the reason the reason is that the pendulum stops behaving as a harmonic oscillator for larger angles so for considerably larger angles the pendulum is not behaving as a harmonic oscillator anymore the pendulum's motion can be approximated to be the same as a harmonic oscillator only for smaller angles like 10 degrees 15 degrees 5 degrees less than that so for smaller angles the solution of the nature of motion is that of an shm it behaves as a harmonic oscillator it creates sinusoidal motion but for larger angles the pendulum's amplitude lags behind a sinusoidal function so as you can see the the motion of a pendulum is not that simple i mean it is simple it looks simple for small angles but for larger angles it is a little bit tricky yeah okay so let us move ahead to the next topic that i wanted to discuss which is the phase space trajectory what is the nature of the phase space trajectory for a simple pendulum by phase-based trajectory you may already know that when we talk about phase we are basically talking about momentum so if you look at what is the nature of the momentum and how it varies with displacement now again i am going to look at or i am going to do the analysis from two perspectives one i am going to look at the solution that i have obtained for small angle oscillations and from there i am going to try to find out what is the relation between momentum and angular displacement but i am also going to do the solutions using the simulation that i have created the program that i've created where i'll obtain the momentum from the solution of the differential equation done by the numerical method and then i'm going to compare both so in a way we will obtain the phase space trajectory for small angles as well as for larger angles so let us do that also so here the displacement is clearly theta and the momentum is the momentum associated with theta which is the angular momentum i'll write it as p theta all right so how is p theta related to theta well we can figure it out by looking at the solution that we have already obtained which is theta is equal to theta not cos omega t so from there we can obtain what is the momentum now you may already know the definition for angular momentum which is nothing but r cross p where r is the uh radial distance from some kind of an origin o and p is the linear momentum so if we look at the pendulum the original pendulum construction that we made the p that is the linear motion is along the transverse direction while r is along the wire which is along the radial direction both the transverse and the radial directions are perpendicular to each other so therefore the cross product simply it looks like r is replaced by l and p is nothing but the definition of momentum which is mass times velocity l m l omega so this is equal to m l square omega so this is the angular momentum magnitude of a pendulum for shm when i say for shm i simply mean that for small angle approximations what is the relationship that exists between angular momentum magnitude and theta p theta is simply equal to m l square omega what is omega here omega is nothing but the angular velocity that is theta dot and what is theta dot theta dot is equal to omega theta naught minus sine omega t right so theta dot is nothing but m l square minus omega theta naught sine omega t so this is simply equal to minus m l square theta naught omega into root over 1 minus cos square omega t if i take a square on both sides this simply becomes p theta square is equal to m l square theta naught omega square 1 minus cos square omega t however what is cos omega t i can obtain that cos omega t from here simply is nothing but cos omega t is equal to theta upon theta naught if i use this to substitute in this equation here it simply becomes p theta square is equal to m l square theta naught omega whole square 1 minus theta upon theta naught square this i can further simplify by writing p theta square is equal to m l square omega whole square if i take theta dot square inside this becomes theta naught square minus theta square now what is m l and omega they are all constants of motion for this particular situation so i can simply write this as some kind of a constant c theta naught square minus theta square so theta naught is the maximum amplitude while theta is the angular displacement at any given point in time so this can be further simplified to write as p theta square upon c plus theta square is equal to theta naught square so what is the nature of this expression or equation p theta square upon let's suppose some constant c which can be written as some other constant let's suppose a square plus theta square upon 1 square or 1 i can just write it here as b square so this is equal to let's suppose theta naught square this is simply the equation that corresponds to an ellipse okay this is an equation of an ellipse now it doesn't matter what a b and theta are here in this uh situation because i am just trying to look at the nature of the trajectory in phase space so if you plot the momentum with respect to the angular displacement then you should get some kind of a elliptical trajectory so let us try to do that so here i have a program again in scilab where i am trying to plot what is p theta with respect to theta for some angle 10 degrees okay so 10 degrees is small enough because i have done the analysis for small angle approximations i have used that solution so for that if i come up with the graph so this is the graph so this is clearly a elliptical trajectory right so the y axis represents the angular momentum magnitude and the x axis represents the angular displacement so clearly you can see that this is some kind of a ellipse that has been formed so this is a phase space trajectory for small angles now the question is uh what if we uh look at the phase space trajectory for larger angles right now for that again i'll have to take help of some kind of a computation software which i am taking here where i solve the equation of motion using some kind of a numerical uh solution for second order differential equations and from there i obtain the momentum and from there i plot the momentum with respect to increasing theta so here what i'm going to do is i'm going to take the starting angle as 1 and the last angle is 179 degrees and i'm going to look at 10 different plots of increasing theta so let us run the program and see what we get whoa okay okay so this is an interesting diagram here so if i zoom in this is the phase space trajectory for theta is equal to 1 degrees which is clearly an ellipse right this is for a greater angle again an ellipse right this is for a greater angle again an ellipse so for smaller angles the phase space trajectory between the momentum and the y axis and angular displacement in the x axis looks like an ellipse but when angle becomes larger and larger clearly as you can see the graph short of sort sort of keeps on changing a little bit the ends sort of taper a little and this becomes like some kind of a i all right so the inner circles i have obtained using small angle approximation the outer circles have obtained from the solution of the differential equation for larger angles so this is the phase space trajectory for a pendulum starting from 1 degree of initial displacement to 179 degrees of initial displacement so it is it is quite interesting to see isn't it it looks like some kind of a human eye or something it is quite interesting yeah so again i'm going to do the analysis for both small angles as well as larger angles now if you went through my video where i was discussing the lagrangian i have already calculated what is the kinetic and the potential energy the kinetic energy here that i have obtained is simply written in terms of polar coordinates in this particular manner while the potential energy is written in terms of the ground ground here is the lowest point of the pendulum this is mgl1 minus cos theta so the kinetic energy expression was this but i already told you that since the length of the pendulum does not change with time r dot is obviously zero so the kinetic energy t is nothing but half m r is nothing but the length l square theta dot square while the potential energy v is equal to mgl 1 minus cos theta i want to look at how the potential energy varies for small angles as opposed to larger angles so that approximation i want to apply here also so if i apply the small angle approximation i can approximate cos theta okay so cos theta can be approximated for very very small angles as 1 minus theta square upon 2 if i substitute that in the potential this simply becomes mgl 1 minus 1 plus theta square upon 2 which is nothing but mgl theta square upon 2 so this can be written in the form of half ml square into g upon l theta square so this is equal to v is equal to half ml square remember it was the moment of inertia of the pendulum g upon l is omega square and theta square so clearly half i omega square is dependent on the setup and theta is the angular displacement so v is directly proportional to theta square this by looking at this you can say that the nature of the potential energy should be a parabola okay now what is what does the parabola signify the parabola simply signifies that it's a harmonic oscillator for small angles as i have already mentioned before the simple pendulum behaves like a harmonic oscillator therefore the potential energy looks like a parabola right and the kinetic energy is simply given by this particular expression which is half mv square the sum of the kinetic and the potential energy is simply nothing but the total energy which is e is equal to t plus v so what i'm going to do is i'm going to look at how potential energy and the kinetic energy varies and the total energy should remain a constant for both small angles as well as larger and okay so this is a program here let us run it and see what results we get okay so the left hand side is for smaller angles the right hand side is for larger angles so as you can see in the left hand side the blue graph represents the potential energy which represents the parabola corresponding to a harmonic oscillator the red one represents the kinetic energy now of course as potential energy decreases kinetic energy increases and as kinetic energy decreases potential energy increases and the black line here at the top is a straight line because the sum of potential and the kinetic energy is constant that's just the conservation of energy principle however if you look at the right hand side this is for larger angles from uh 1 to 179 here you can see that this is not exactly a parabola but for smaller angles even this will approximate to a parabola so for smaller angles the potential energy will approximate to a parabola corresponding to a harmonic oscillator but it is not necessarily true for larger angles so there it is for larger angles the pendulum does not behave like a harmonic oscillator and therefore the energy does not represent a parabola the potential energy does not represent a parabolic curve but for smaller angles it behaves like a harmonic oscillator therefore the potential energy behaves like a parabolic curve and the same with the nature of the motion the pendulum executes simple harmonic motion for smaller angles but it does not execute simple harmonic motion for larger angles with time the actual motion of the pendulum for larger than just lags behind some kind of a sinusoidal curve so that is all the discussion i have for you today in terms of the simple pendulum i hope you enjoyed the video you learned something from it so let me revise what we have done today first we obtain the equation of motion using the newtonian approach in polar coordinates second we obtain the same equation motion but in terms of torque third we obtained the same equation of motion in terms of the lagrangian approach and then we looked at what happens to the equation of motion so we use the small angle approximations where we reduced sine theta to theta and then we solve the differential equation to obtain the nature of the solution for a certain given initial conditions where theta at time t is equal to zero is theta naught and theta dot which is a velocity at time t is equal to zero is zero the the solution is theta is equal to theta naught cos omega t from here we can see that for small angle approximations the simple pendulum behaves in a simple harmonic fashion from here we can calculate the angular frequency we can calculate the time period it is interesting to see that the time period is independent of the mass as well as the angular displacement initially i also obtained the similar kind of a graph from solving the differential equations for larger angles using some kind of a computation software where we can see that for larger angles the actual motion the pendulum develops a phase difference it lags behind an actual shm with increasing time we also looked at the phase space trajectory where the nature of the momentum how it varies with the angular displacement we saw that it looked like an ellipse for smaller angles but for larger angles it deviated from the shape of an ellipse from the energy analysis we can see that the potential energy is a parabola for small angles but not so for larger angles so that is all for today i hope you learned something from this video i'll see you in the next one thank you very much you

Channel: For the Love of Physics

Views: 8,641

Keywords: small angle approximation pendulum, simple pendulum, simple pendulum physics, for the love of physics, physics, education, lectures, simple harmonic motion, pendulum, shm, simple harmonic, simple pendulum explained, force, torgue, lagrangian, euler lagrange equation, lagrangian analysis, simple pendulum experiment, polar coordinates, lagrangian of simple, lagrangian of simple pendulum, lagrangian of simple harmonic oscillator, small angle approximation, velocity, phase space trajectory

Id: v-yEApxAKsE

Length: 57min 33sec (3453 seconds)

Published: Fri Mar 05 2021