 So, now let us discuss about the cause of SHM, what is causing the SHM? Can you tell me a basic requirement for the SHM, basic requirement for any SHM, a force is required, what kind of force is required, that is what I am asking, see here what happens is that force is 0 here, no doubt about it, but when object is going this way, the force should be in that direction, when object is going this way, the force should be in that direction. So, what happens is, when force becomes 0 at the mean position, the velocity does not become 0, acceleration becomes 0, because velocity is not 0, it continue to move in this direction and as soon as it starts moving this way, force acts in the opposite direction and it tries to push it towards the mean position, every time the object goes away from the mean position, a force is there to push it back to the mean position, so the basic requirement for any SHM to happen is restoring force, that we have discussed last class, so a restoring force is required for oscillation to happen, what is the restoring force, a restoring force, restoring force always tries to push towards mean position, it tries to push it towards the mean position, but unfortunately what happens is, it is like there is an invisible hand which is trying to bring an object closer to the mean position like that by applying force from this way and from that way, I am trying to bring an object at the center, but unfortunately, but even though force is 0 at mean position due to inertia and velocity object moves away from mean position, away from mean position and when it goes away, again another force tries to bring it towards the mean position, again its velocity makes it go away from the mean position on the other side, so again force comes it and make it come towards the mean position, so like this it keeps on happening, so what is required, a restoring force is required then this oscillation will happen, a force should be such always it will make object go towards the mean position as in accelerate towards the mean position, so if that is the case then your force should be proportional to negative times x, when x is positive your force should be negative, when x is negative your force should be positive, force is this way when you are here, force is that way when you are here which is perfectly in sync with what is acceleration in the SHM, for acceleration A is equal to minus of omega square x, we know it already, so force is equal to minus of m omega square x, now can you tell me a scenario in which you can imagine an SHM other than pendulum, can you imagine a scenario in which simple harmonic motion can exist to and fro, correct, so a spring block system, I will just analyze one of them to you, did I give any assignment last week, did I give, is that the reason why you guys are completely unprepared today, should have given assignment, here is a block of mass m, okay, now at present the spring is in its natural length, spring constant is k, right, spring in its natural length, now what you do is that you are moving the mass by a distance of x, let's say, by some distance x you have moved it and then you have released it, can you imagine that SHM is happening like this, can you imagine that, all of you, once you stretch it and leave it, it will do SHM, oscillation it will do, are you able to imagine it, if you are not able to imagine it, then leave the chapter, you need to understand physics is best done by visualizing it, you need to feel that yes, there is a spring, there is a block, if I stretch it, leave it, it will start doing oscillation, are you able to feel it or not, okay, shittage, auro, pradyum, others, okay, great, all right, now if you have a block over here, you have a block over here, then after you release it, spring is stretched, after you release it, can you tell me what is the acceleration of this block, once you have stretched it by x and then you are about to release it, what is its acceleration, all of you, half kx square, half kx square is its potential energy, okay, so will the acceleration, as I forget about acceleration, the force, the force is k times x horizontally, right, spring is stretched by a distance of x, so acceleration is kx by m, this is the acceleration, now tell me is acceleration and x, they are in opposite direction or in the same direction, everyone, a and x, they are in opposite direction or in the same direction, x is like this, this is x and a is like that, right, they are in opposite direction, so if you want to write it in vector form, there will be a minus sign over here, okay, now clearly you can compare it with SHM equation, a is equal to minus omega square x, right, so when you compare it, you get omega is equal to root over k by m, right, so what is the time period of SHM, everyone, time period of SHM is what, 2 pi by omega, 2 pi by omega, right, so it will become 2 pi root over m by k, this is how you do this, alright, so let's take few more spring block system because spring block system is a very common system which can perform simple harmonic motion, in fact in your NCIT book, most of the examples are spring block systems only, I forgot to ask what is happening in the school, still SHM waves, waves, waves, so you have two batches in Kormangalam, different classes, okay, okay, so see we can, what we can do is that, we can take few questions right now or I can talk about steps of solving problems which is common for all the questions and then we can take multiple questions, okay, we will solve every question step by step like this, write down, these steps of problem solving, they are not meant to constrain your thinking, you should not force yourself to do it like this only, these steps will help you to align your thought process properly, that's all, okay, after maybe doing 5-10 questions, you can forget about these steps of solving problem, so we had done just, just now we had done one problem wherein we need to find the time period of a system, time period of oscillation of a system, so the first thing, first step is to identify, identify the object performing SHM, which you do it anyways, but sometimes even that is tricky, okay, so we will see that later on. Second is, second is what we need to identify, identify the path of SHM, third, identify what, can you guess what we need to identify next? Yes, identify the mean position, how you identify it? At the mean position, at the mean position, net force is 0 along the path of SHM, okay, next, displace the object by some distance x away from mean position along the path of SHM, okay, so when you displace it and you leave it, your x is amplitude, x is your amplitude, look at it, previous question, when you leave the mass from here, it will try to go back, so this point automatically becomes mean position, sorry extreme position, all right, so x is amplitude, that is the maximum distance from the mean position because after you leave it, it starts going towards the mean position, fifth, identify net balanced force along the path, what path, SHM path, now you know, there is a trick here, which I should tell you right away here that the net force, net force at mean position is what? At mean position, net force is 0, okay, does it mean that no forces are applied there? Does it mean no forces are there or total force is 0? What does it mean? Total force is 0, okay, there might be, no, having no force and total force 0 are two different things, okay, total force is 0 there, let me get total, sum of all the forces, so if sum of all the forces are 0, it doesn't mean that there are no forces, there might be multiple forces, they all add up to 0 at the mean position, okay, so when you displace it, when you displace the object little bit, you need to only focus on those forces which are coming because of the displacement, because whatever was a force before the displacement, they were all balanced and they were equal to 0, so when you displace it, whatever extra force comes because of displacement, that is the net force, okay, you will understand what I'm talking about when we take a question after writing all of this, so the force due to displacement is the net force, due to displacement is net force, then once you get the force acceleration is, write it like this, equation is equal to omega square, okay, the point of question here is that the net force must be towards mean position, otherwise it will not do SHM, okay, when you displace it this side, the force should appear on the other side, okay, so then the last step is compare it with this equation, omega square x and you can then write down time period as 2 pi by omega, anyone has any doubts in these steps? Now whatever problem we do going forward, two, three questions, we'll be doing every step like this only, all one by one, anyone has any doubts in these steps of solving problems? Fifth step, what is your doubt? Doubt is what? No doubts, is it clear to everyone, type in, all right, step number five will be clear when we take examples, I don't want to tell stories without taking an example, so let's take an example, I'll relate to that, okay, so all of you all set, here is the next example, vertical spring is there, like this, there is a mass M and spring constant K, earlier case spring was horizontal, now the spring is vertical, same scenario, you need to find that time period here also, okay, we have already done for the time period of the horizontal spring, can I do it for this, how much it is, G is vertically down, in the previous case at the mean position, the spring was in its natural length, here is the spring as its natural length, when it is at the mean position, path of SHM very clear, path of SHM is this, vertical, okay, along this path you do the SHM, okay, so at the mean position, can you tell me what is the extension in the spring at the mean position, extension in the spring is how much, condition for the mean position is what, net force which is MG minus KX naught, initial extension KX naught plus zero, upside the force is KX naught and down the force is MG, okay, so extension in the spring when it is at the mean position is this, okay, this is extension, all right, now when you displace the object, you displace from this location, this is already there, okay, so now you are moving down let's say by a distance of X, okay, by distance of X you have moved down, can you tell me what is the net force along the path of SHM, do it yourself, this is probably the simplest of all the questions to come, you need to understand this, doing yourself is the best, not getting it, when I am displace in the object by a further distance of X, further if I displace it by X, total spring force will be how much on it, total spring force will be how much, everyone, X naught was anyway there, further you have extended by X, total spring force will be how much, KX plus X naught, KX plus X naught above and MG down, so the total force is KX plus X naught minus MG, okay, this is equal to KX plus KX naught minus MG, what is KX naught minus MG, this is what, the bracket term 0, so the total force is KX, all right, getting it, now you could have got KX directly, you don't need to do this circles by writing KX plus X naught minus MG, why, because earlier all the forces were balanced at the main position, whatever extra force comes, because of the further displacement from the main position, that is the total force, so from the main position, if I displace by X, only K into X force extra will come, so net was anyway 0, whatever extra comes will be the total force, so that is why it is KX, so this KX should be equal to mass time acceleration, and you can see that the equation looks exactly the same, from here omega is root over K by M, and time period is 2 pi root M by K, all of you clear about this, type N is it clear, this is your school level questions useful for your UTs, this kind of question was done in your school or not, nobody wants to comment, this is theta, you have a block of mass N, kept like this, okay, this surface is smooth, again find out the time period of SHM, everyone, exact same way you have to do the way we had done the previous question, you will get the answer, what is the initial extension in the spring, anyone got it, correct, path of SHM is it clear, along the incline it is MG sin theta, this is normal reaction that is MG cos theta, and initial extension is X naught, so this is let us say KX naught, so when MG sin theta and the other forces in the direction of SHM when they add up to 0, that is when you will reach the mean position, so at the mean position extension in the spring is this much, okay, this is the extension, now what you do, what is the next step after you get the mean position, what is the next step everyone, displace an object, displace that object by distance of X, let us say you have displaced this way, you can displace either ways, you could have displaced upward also, does not matter, this is X, so can you tell me the net force in the path of SHM, everyone, net force, total force, after you have displaced it by distance of X, what is the net force, can I directly write net forces KX, can I directly write it, before displacement, before you are displacing from the mean position, add the mean position, all the forces were anyway added up to 0, okay, so whatever extra force that comes due to the displacement that is a net force, clear, right, so when you are displacing it by X, some extra force of KX comes in, that is a net force, are you getting it, all of you, you do not need to write again that KX plus X naught minus mg sin theta, you can write this, then also get KX only, no problem, but why you are lending the process, so acceleration is K by M times X in a vector form negative because they are in opposite directions, so omega is root of K by M and yes, surprise that the time period does not depend on how the spring and block are kept, it is always 2 and a root M by K only, okay, see my logic is very simple, the logic is if you write 8 minus 8 is 0, okay, you anyway know 8 minus 8 is 0, so when you add 4 to 8, you do not need to do 8 minus 8 again, whatever extra you are adding is your total, same thing here, at mean position, anyway net force is adding up to 0, so whatever extra force comes in, is your net force, is it clear now, keep it simple, now tell me, do you remember in word pioneering chapter, we had done multiple springs together, spring in series, spring in parallel, you remember that, do not remember, okay, so I am not going to derive it again, derivation you can look at your notes, so if springs are connected like this, one spring like this, another spring like that, then this is the parallel connection, so equivalent spring constant is what, in this case is K1, K2, I can say that this is equivalent to a single spring and a block where spring constant K is sum of 1 and 2s, okay, and if spring is connected like that, then if I tell that it is equivalent to a single spring, then what, how will you find out the equivalent spring constant in this case, do you remember that, correct, Gurman got it, 1 by K is equal to 1 by K1 plus 1 by K2, okay, this is the parallel connection and this is the series one, okay, remember this will help you in questions in which multiple springs are there and yes, that way it will be useful. Now let us take up a few more questions, mechanics is all about problem solving, theory does not help in any ways, you have to do lot of questions, then only it will be clear, there is a spring of spring constant K, there is a block here which has mass m, this mass is projected with a velocity of v, okay, this distance is L, all collisions, sorry I will write it, collision between wall and mass is elastic, okay, and mass after hitting massless spring compresses it, starts compressing, so can you describe the movement here, the mass will go hit the spring compresses the spring, then what will happen, spring will come back and then what will happen, all of you the mass goes this way, then it will come back, what will happen to the spring, will mass leave the spring, everyone, after this cycle is completed, till the spring is getting compressed, spring is with the mass, okay, as soon as the spring reaches a natural length, will spring be in contact with the mass or not, no, right, no, it will leave the spring and what will happen, the mass will go hit the wall, again come back, again compresses the spring, so this will keep on happening, all right, you need to tell me time period of oscillation for the mass, time period of oscillation for mass, do it, in this zone, in this zone when it is not in contact with the spring, can I say that velocity will be constant, can I say that, when it is moving without any contact with the spring, is the velocity fixed, see kinetic energy gets converted into potential to the spring, then again it gains the kinetic energy, velocity reverses, comes back, hits the wall, with water velocity, it hits the wall, it comes back, so in this zone velocity always remains v and the spring, when it hits the spring, it completes half the oscillation every time, half the oscillation is with the spring and other half of the oscillation is with the this non-spring thing, so it is covering to a length, to a length is covered with a velocity of v and you have half of time period for spring, so total time period is 2L by v plus, you remember time period for the spring block system is 2 pi root m by k, so half it because in every oscillation only half of the oscillation for the spring is taking part, so this is how it will be in J advanced, you should be aware that's why I told you, all right so let's take a break now, I think now we can take a break and after the break we can continue with same thing, we will continue taking couple of more questions, then I will talk about the energy related to the simple harmonic motion, okay fine, so we will meet at 6.16 pm, let me take up questions directly from here, this one, there are many these kind of questions which you will see, it's a simple question only, don't get worried about too much of text, there are multiple parts of the same question, try doing it, time period is what, first tell me that, time period is what, both the masses are moving together, so time period is, look at your notes, just now we discussed spring block system, okay, 2 pi under root, do you all agree, it will be this, 2 pi under root m plus capital M by k, do you all understand this, how it comes, type in, so answer it, okay, otherwise I'll feel that you are not able to do it, then I have to do this thing again, right, this is part A, now part B, Pradun where you went in middle of a class and then rejoin, U, Ansh and I think RA also, where you three went, virtually, okay, okay, all right, do the B part, all of you, how you do the B part, draw the free body diagram, you'll get the answer easily, free body diagram of small m, anyone close to the answer or should I have to, have you drawn the free body diagram, everyone, so what is stopping you to get the answer, tell me, this is normal reaction, this is small mg, will there be an acceleration, will there be an acceleration, all of you, the displacement from the mean position is x, so is there an acceleration for small m, how much, how much, if x is in this direction, this block is here right now, so how much and in which direction is the acceleration, tell me direction first, left or right, acceleration, left, how much omega square x, omega is what, for this system what is omega, under root k by m is omega, so k divided by m plus capital M is omega, so that is square x, so kx divided by m plus capital M, this is your acceleration, all of you able to understand this, type in, is it clear, now tell me friction will be in which direction, left or right, friction will be in the left hand side or right hand side, left or right, friction on small m, it will be of course on the left hand side, on the left hand side, why acceleration is on the left hand side, there has to be a force to create that acceleration and only force available in the direction of acceleration is friction only, okay, see usually what happens is, when you learn rigid body motion, you're like oh can I use the laws of motion concept or Vagpa energy chapter, then similarly you, when you learn the simple harmonic motion, you hesitate to use laws of motion concept in simple harmonic motion, you should not, everything that you have learned till now is science, science won't change chapter to chapter, you can use it, okay, equation of motion you cannot use in SHM, why, because acceleration is not constant, there's a reason to it, but you can use V is equal to dx by dt and A is equal to dv by dt which is valid everywhere, similarly Newton's second law is valid everywhere, which you can use here also, so there has to be a force to create the acceleration, so that friction should be equal to small m into acceleration m kx divided by small m plus capital M, this is the force of friction, now do the C part, it's an easier part to do, we'll get it easily, all of you attempted, what can be the maximum amplitude, if amplitude is maximum, what else will be maximum, what else will be maximum, everyone, if amplitude is maximum, what else will be maximum, x is maximum, correct, what else, acceleration is maximum, right, acceleration is maximum, correct, and because acceleration is maximum can I say friction has to be maximum because acceleration is maximum because if I have to get maximum acceleration I need to apply maximum force yes or no and what is the maximum value of friction what is the maximum value of it mu times normal reaction in this case normal reaction is mg I can write it as directly this should be equal to omega square k divided by m plus capital M into small m into a basically I am writing mass m acceleration here omega square a is acceleration into mass so maximum under force should be equal to this so from here amplitude of the oscillation maximum possible amplitude so that it does not slip over is mu g small m plus capital M divided by k a ok sorry that is it has everyone understood this type in is this clear similarly there are many such questions ok chapter is very small but because mechanic get mixed into it there is there are so many numericals you know which you can solve I will send you more questions in the assignment let us continue and let me do whatever I have to do today later on if whatever time is left may be 15-20 minutes then we can again come back to the problems ok so let us proceed further we will now talk about angular SHM all right write down what comes in your mind when you hear this term angular SHM circular motion circular motion is SHM is it SHM what comes in your mind angular SHM anything nothing nothing it is strange so pendulum pendulum will come in our mind we are basically studying the pendulum there are two kinds of pendulum simple simple pendulum and physical pendulum ok so we will discuss it one by one first let us talk about the simple pendulum ok write down simple pendulum whatever simple pendulum there is a string which is coming from a fixed point like this and a mass is attached here as a bob a small mass like this is attached mass m ok we need to find out time period of SHM here time period of SHM but the SHM that is happening here it is not SHM in a straight line you need to understand that the SHM which is happening here which variable we are taking into account when it was linear SHM your variable was x here what is a variable here the variable is theta ok and the SHM is following this curve path all right do you understand what I am talking about here everyone so what I am saying here is that if acceleration which is equal to d square x by dt square is equal to minus omega square x if this gives me simple harmonic equation as x equal to a sin omega t plus phi then then alpha which is equal to d square theta by dt square when this is equal to minus omega square theta will this give me theta equals to any maximum possible angle theta naught into sin of omega t plus phi does it make sense if this will give me this SHM SHM of x this equation will give me SHM of theta because mathematics does not care whether you write a in terms of alpha or x in terms of theta you will get the similar looking solution now is it clear to everyone now is it clear this is the angle SHM because theta is doing SHM theta is increasing goes to maximum comes back becomes 0 then again maximum on the other side simple harmonic motion which was linear it was happening on a straight line like this this one we are tracking angle angle increases decreases then again increases on the other side so that is how it is is it clear to everyone type it quick is a doubt if not clear okay you want me to take your names all right aria what are you doing there's no doubt from you it has been more than two and a half hours now focus is somewhere else I'm asking is it clear how much time and effort it takes for you to reply everyone Gourman what's going on everything is fine Gourman it appears to be you know at the start you were very proactive you were answering questions now something something has happened okay is it clear to everyone now I think only two or three people are left there are there they want me to take their names so that they can make huge huge amount of effort to type that yes it is clear so if I get an equation like this alpha in terms of theta if I get alpha as minus omega square theta even this will give me SHM for theta okay this was giving me SHM for x all right so let us try to see if we can convert this simple pendulum thing equation like that all right then we can get that time period of pendulum also okay so let's say you have a pendulum like that like this there is a pendulum mass m and length l okay you have to find time period now you remember when it was SHM on a straight line what you do to find time period you displace by small distance x here what do you have to do any guesses if you have to find out the time period here what you have to do displace by small angle theta okay here you have to do that way displace by small angle theta and then in the case of in the case of linear SHM we got acceleration once you got force from force you got acceleration here you need to get angular acceleration okay angular acceleration you have to get to get equation like this so how you get the angular acceleration in the expression from where this comes this comes from Newton's second law force equal to mass m acceleration how alpha will come in the equation how alpha will come any guesses in the school have you done simple pendulum also without taking alpha only without taking alpha you calculated the time period in the school no alpha oh so they have taken mg sin theta and okay taken components and all that okay but anyways we'll first do with alpha we'll first do with the alpha can you tell me one equation in which alpha comes just like force equal to mass m acceleration there is an equation in which alpha comes do you remember torque is equal to i alpha about the fixed axis torque equation right so there is this object is experiencing torque about this axis about that axis okay so after displacing it by theta can you tell me what is the torque about this axis fixed axis what is the torque tell me in the free by diagram there will be tension there will be mg what is the torque due to the tension everyone torque due to tension is how much type in tension torque is 0 because tension force passes through the axis all right what is the torque due to mg torque due to mg is what mg l this is the force what is its perpendicular distance from the axis this line of force perpendicular distance you would take this distance right perpendicular distance mg l sin theta okay mg l sin theta do you all agree this is the net torque mg l sin theta this should be equal to i alpha what is i moment of inertia about this axis how much this is not a rod just a point mass how much ml square right ml square it is so you can get i is ml square into alpha is mg l sin theta okay m is gone one of the l is gone alpha is equal to g by l now theta is very small so sin theta is roughly theta okay theta is small in a vector form alpha and theta they are in opposite direction so when you compare it with this equation you will get omega to be equal to root over g by l so time period is how much 2 by root l by g see why I have done like this because there are many different kinds of pendulum that the force method which you have done in the school is not useful to those cases so you must know that using torque equation also if you get alpha as minus constant times theta that constant is omega square why isn't it sin theta because theta is very small it will do s h m only when theta is small okay if theta is large it won't do simple harmonic motion theta is small so sin theta is roughly theta in radiance I hope this is clear why negative sign came negative sign came because theta is theta is what anticlockwise and torque is clockwise torque is trying to rotate this way so if you want to write in vector form alpha is in opposite direction of theta clear or any other doubt see this is what I want you ask lot of doubts alright so now let us do the way you have done in school also because school exams are also very important so let us say this is the pendulum have displaced by small angle theta this is theta okay theta is very small so this is mg the acceleration at the start acceleration will be perpendicular to the perpendicular to the string or not everyone can there be x acceleration will be in this direction only or not along the string acceleration is 0 along the string along the string acceleration is 0 so because string is of fixed length it can't move that is different if omega comes in there will be centripetal acceleration but right now omega is 0 we are just releasing it so if this is theta then this angle is theta so along sorry the perpendicular to the string the net force is mg sin theta this is a net force net force is equal to mass and acceleration alright so acceleration is equal to g sin theta now theta is very small so I can say that you know we are assuming when we do like this we are assuming that the bob is moving in a straight line and if this is x sin theta is equal to x by L because theta is very small because theta is very small sin theta you can write like this and you can assume that bob has moved along this straight line and then of course a and x are in opposite directions of this so when you compare it with this equation you will get omega to be root of g by L okay so time period is 2 pi root L by g okay why can't we use it for other cases because other cases rigid bodies are involved and in rigid bodies the best way to analyze it by using torque equation you have seen it in a rigid body right so this is the way it is done in your school so if it is a school exam I request you to do this this is for school alright don't use the torque equation for the school thing now I will tell you the physical pendulum what it is something very important so this is let's say a rod a rod is this and there is a hole here sorry there is a hinge here and this rod can swing like that so in physical pendulum basically simply I can say that in the case of physical pendulum rigid body is oscillating rigid body oscillates it's like if I take a rigid body like this some random rigid body I can take and if I move it like this it starts oscillating okay this is definitely not it is let's say it is oscillating like that okay this is definitely not a simple pendulum so like this I can take any object display it little bit and then it will do the shm okay I better example be bottle I can put a hinge over here display it like that and it will start oscillating like that okay this is not a simple pendulum okay so here is an example for you there's a rod of mass m and length l a uniform rod of mass m and length l you need to find out the time period of oscillation for this it is hinged at one of its ends over here do the way we have done the torque method you'll get the answer it's very simple make some effort from your side it is required trust me you can't just look at the screen and copy it you'll forget in some time then the normal forces on hinge yes there are hinge forces but what is a torque due to the hinge forces how much is a torque about that axis hinge axis it is zero okay so that is why it is better to use the torque method that is another reason okay prattin got prattin it should be in terms of theta alpha should be in terms of theta you've taken theta equal to something is a theta is very small okay anusha got something others okay should I attempt it okay good to see that many of you have attempted fine should I do it now the rod need to get displaced by some angle theta like this if you you know you don't need to draw the hinge forces but then okay why you need to don't need to do it because the hinge forces torque is zero about that there will be mg from center of mass okay so mg is this any of the force no other force right this is mg so what is a torque due to mg everyone torque due to mg about the fixed axis how much it is this distance from here to here that distance is l by 2 this is l by 2 so its perpendicular distance is this this yellow line let me draw a red line this horizontal red line is the perpendicular distance right so that perpendicular distance is l by 2 sin theta so torque is mg l by 2 sin theta this is the torque moment of inertia what you'll take m l square by what is the moment of inertia for this axis l square by 3 that into alpha so m goes away alpha will be equal to 3g by 2l times theta theta is very small sin theta become theta in a vector form theta is anticlockwise alpha is clockwise so that is a minus sign in a vector form so when you compare it with shm equation you will get omega to be root over 3g by 2l how do you know it's very small it is my choice this is valid only for very small angle theta okay is it clear to everyone this is a physical pendulum fine so let's take up this let's get the time period of oscillation for any random rigid body okay let's say this is a rigid body which you are you can just create a random object like this you are hanging from here can you tell me can you tell me anything about where the center of mass of this rigid body should be the rigid body is rigid body can rotate about this axis can you tell me anything about the center of mass location related to that anything right now it is at rest the question is or the question is can you tell me anything with respect to center of mass location like you can tell me that it should be right hand side it should be left hand side it should be horizontal it should be what where it should be anything related to the location can you say it has to be correct can it be anywhere other than this vertical line is it possible to have center of mass other than this straight line passing through the inch vertical center of mass can it be here this side can it be there if it is there then what will happen there's a torque it will rotate and it will not stop till center of mass is along this line so if center of mass is along this line then the net torque due to gravity also becomes zero because the gravity force is vertical right so the hinge forces torque is also zero and the gravity force torque is also zero so this is an equilibrium right now okay now if I tell you that this distance this distance is d okay and moment of inertia about the fixed axis is i can you get the time period for this all of you time period of oscillation same way you have to do everything is repetitive yes torque about fixed axis is zero you have to what is the process of finding time period you have to displace it by some angle theta and then look at the net torque and that you get to i alpha okay shittage simha got something others one second okay given one second when you rotate the rigid body by a small angle theta your center of mass reaches here so this is mg what is the torque due to gravity everyone this is d torque due to gravity is what when i'm not redrawing the rigid body when it is rotated because the random shape i'm just doing a green line green line represents what happens to that red line after you rotated entire rigid body will get rotated mg d sin theta right mg d sin theta this should be equal to i alpha okay so alpha is equal to mg d by i times sin theta theta is very small sin theta is theta so alpha and theta in opposite direction so minus and you can compare it with equation of the angular shm so omega will be equal to root over now you guys can see that the things are getting repetitive now once you do enough questions it'll become repetitive for you and you start looking at every question as if it is similar to each other okay so this is the physical pendulum for all of you you will get some questions on this in the assignment please try that it is very important that you go slightly beyond nct if you restrict yourself with nct there are many students who do that you are among them so if you want to improve your ranks in any exam which you are aiming at you have to go beyond it not too much but at least you should be aware of these things okay so we will now talk about the energy consideration energy in shm so for a spring block system let's take an example of spring block system at any moment let's say this is a scenario this is the scenario this is the mass at any moment let's say all of you agree that at the mean position the spring will be in its natural length right because it is horizontal right now so if you if we start the oscillation in this case okay then the total energy total energy will be equal to sum of potential energy plus kinetic energy right there is no other energy and friction is zero it is smooth so is there any way sum of potential g plus kinetic energy can decrease is there any way that will decrease or potential g plus kinetic energy will always remain constant there is no external work done there is no external work done so you know that work energy theorem this is equal to final kinetic energy plus potential energy minus initial kinetic energy plus potential energy so if work done is zero the sum of kinetic energy plus potential energy won't change this minus that is zero so final will be equal to initial so mechanical energy will be conserved okay so let's see how much is the mechanical energy in the case of simple harmonic motion okay so kinetic energy in this case can be simply written as half m v square what I can write potential energy as everyone u is what kx square half kx square is u now is there any relation between v and x because it is shm yes there is a relation wherein v is equal to omega under root a square minus x square so can you prove that kinetic energy plus potential g is a constant all of you try doing it try doing it and tell me how the sum of them is a constant it should not depend on x let me know once you are done anyone okay others nobody else one else okay c energy is equal to half m v square so v square will become omega square a square minus x square plus kinetic energy is half kx square can you tell me what is omega in the case of spring block system like this we have done it previously tell me what is omega look at your notes omega is how much root over k by m k by m right so when you put omega as root over k by m over here you will get half m omega square is k by m a square minus x square plus half kx square so you can see that the total energy comes out to be half k into amplitude square and this doesn't depend upon x clear to everyone so in a spring block system I can write down potential energy as half kx square and kinetic energy as half k is okay this k and that k look similar okay here there is no confusion here I will write k e so that the spring constant and kinetic energy symbol should not remain clean kinetic energy is half k a square minus x square okay so you can get the kinetic energy as a function of x here you don't need to write half m v square so graphically how does it look like if you have to plot the kinetic energy and potential energy for this system how does it look like let us see that all of you draw this box this length on the y axis you have energy okay this length from here to here this is half k a square okay that center line this is x equal to 0 over here this is x equal to 0 this is x equal to a amplitude and this is x equal to minus a all right now I want you to plot the potential energy here all of you plot the potential energy when the particle is at a 0 and minus a entire plot you have to do all of you will it be 0 anywhere if yes where all of you will it be 0 anywhere if yes where at mean position see at the good thing is that in this system at mean position the compression or extension the spring is 0 so potential energy is 0 at the mean position over here okay and where it is maximum potential is maximum where at extreme so can you plot it can you plot it will be a parabola y is equal to half k x square kind of curve like this are you getting it like that this is the potential energy okay this is the potential energy you now can you plot the kinetic energy use some other color maybe red color find out the kinetic energy what is the kinetic energy at extreme points kinetic energy at extreme points see potential energy is 0 at the mean position right when you say potential energy is 0 at the mean position is it true for all the shm or only in this case where spring extension is 0 at the mean position potential energy is 0 only in this case because luckily extension the spring is 0 at the mean position okay but when you talk about kinetic energy at extreme position kinetic energy will be 0 for all shms okay for all shms yes that potential energy is gravitational potential energy when you talk about potential when you talk about pendulum potential is gravitational here when we talk about the spring block potential is spring potential energy okay same kind of thing will be there also I will tell you after this all of you done with the kinetic energy curve will it be downward opening are you getting like this this curve is the kinetic energy getting it right now can you prove graphically that some of kinetic energy and potential energy is a constant both the curves are symmetrical it is like you have inverted the yellow curve and made it red it got rotated by 180 degree and it became red color so graphically without any equations can you just look at the graph and say that kinetic energy plus potential energy is a constant at any location over here let's say this is what from here to there this is the kinetic energy okay and let me use white color from here to there that is the potential energy so you have to prove that wherever you go the white line plus blue line is a constant can you do that all of you I can go let's say here this plus this is same it is same as this blue this white when you add it up should be equal to this blue and this white can you prove it using graphs and you don't need to do any construction as such it's all of you listen here now this length when you add blue and white this length is counted twice or not all of you that length is counted twice from here to here this length is counted twice okay so what you do you take one of those one of those you take and fix it here so it becomes total length from here to there take this blue line from bottom take that blue line from the bottom it will exactly fit here isn't it similarly take this blue line from here it'll exactly fit there because of symmetry are you getting what I'm trying to say every time when you add kandikeji and potential energy it will become half k a square graphically also how can you do that okay I am trying to prove here the fact that this length plus this length is a is equal to this total length forget the fact that the length represents energy you have two lengths I'm saying these two lengths when you add it will become half k a square wherever you go that is what I'm trying to say so when you add these two lengths you're counting the common distance twice so take part of it and fix it on top of each other like that so it becomes half k a square okay so that is what we have done you take the blue part and put it over here total energy will be constant here also take the blue part put it above all of you understood type in is it clear okay so this way you can prove graphically also that kandikeji plus potential energy is a constant okay now somebody asked about the pendulum how to do the similar thing okay so let us see that for angular SHMs in fact let me draw the full fledged this thing this is not given in any book as such but then since you asked one of you asked this is the simple uh this is linear SHM linear simple harmonic motion this is the angular SHM okay I'll direct in and out comparison I will draw a few lines like that so linear SHM is a is equal to minus omega square x angular SHM is alpha equals to minus omega square theta here you have to use force equation here you have to use torque equation solution for this equation is x is equal to a sign k x minus sorry sorry that's a wave not oscillation a sign omega t plus phi solution for that is theta equals to theta not theta not is maximum angle omega t plus phi okay so if you differentiate you get velocity is equal to a omega sign of sorry cos of omega t plus phi now tell me when you differentiate theta what you get d theta by dt is what is it this omega is it that omega only is d theta by dt is this omega only no what is omega what is the name for omega omega is what now in this case omega is angular frequency don't call it angular velocity angular frequency is the word all right don't call it angular velocity but when you differentiate theta you get angular velocity all right but unfortunately you can't write d theta by dt as omega omega is angular frequency of the SHM theta dot is theta dot is d theta by dt in fact I should write d theta by dt d theta by dt is theta not omega cos of omega t plus phi now we also know that v is equal to omega under root of a square minus x square v is the linear velocity similarly here angular velocity d theta by dt can be written as omega under root of theta not square which is angular amplitude minus theta square at whatever theta you can find out the angular velocity like this okay now what are the equation you get acceleration as minus of a omega square sign of omega t plus phi and here angular acceleration you can use alpha now okay because alpha is not here that is not confusing alpha is equal to minus of theta not omega square sign of omega t plus phi all right is it clear to everyone this comparison so there can be questions on kinematics of the angular SHM you remember we have taken a some numericals on the linear SHM kinematics exactly same kind of questions on angular SHM instead of x we'll talk about theta now but equations are exactly same all right so now I think this this should clear it shall we do one question we have I think yes you might not remember we had done a question sometime back which you guys took a lot of time and in and you guys felt that it is an advanced level question I am going to show you the same question can be done in just two steps by using simple harmonic motion concepts so suppose this is small n and on top of it are you able to recognize this scenario capital M no okay this is spring constant k right you need to find out minimum distance capital M to be compressed so that small m m jumps off the ground try doing it by using SHM principles of course it'll not be straightforward but that is how some questions are right you need to understand the difficult ones to gain over above others can small m jump off while spring is compressed can it jump off while spring is in compression no clearly no what is the condition for small m to jump off what is the condition normalization becomes zero all of you I think are smart enough to understand that and this when spring is extended this kx and mg let's say normal reaction towards when normalization becomes zero when n plus kx minus mg this is equal to zero when it is about to jump off normalization becomes zero and x should be equal to mg by k small mg by k do you all understand whatever I am doing here I am telling you that this much extension in the spring should be there for this small m to jump off the ground is it clear to everyone right now right now right now spring is compressed spring is compressed by how much how much spring is compressed right now from its natural length all of you answer it can you see it is mg by k the upper block gravity force is balanced by the spring mg is down and initial compression is x not kx not above okay this is the scenario right now now this is the mean position where spring is compressed okay spring is compressed in the mean position right now you need to make sure when the block when the block goes to the other extreme when what will happen is that when you further compress it down let's say you reach here after compressing it you reach here this much is the compression you have compressed it by a distance of let's say d so will it go above also d from the mean position while doing shm right so from the compressy I am writing this much because I am explaining we can do it in just one step from the compressed state of mg by k it should go to extended state of small mg by k for this small m to jump off okay getting it so how much is d d is how much when it goes to this position when you go to this position the extension the spring should be this so what is the value of d anyone think of it now nothing left actually how much should be the d will d be equal to see suppose I I have d equals to this I only compress the spring by this much so when it goes to the upper extreme how much is the extension the spring tell me suppose this is d it goes to the upper extreme the amplitude of the oscillation is however much I am compressing it so when you go to the upper extreme what is the compression in the spring compression of extension is how much from the main position you have compressed it by this much so from the main position it will go above by mg by k it was already compressed by mg by k so it will go back to the natural length at the end are you getting it at this position other extreme the spring will become natural length but I don't want that to happen I want it to get extended so d should be equal to mg by k this will ensure that I reach to the natural length at the other extreme plus small mg by k so my amplitude should be this much which will ensure that the spring is an extended state at the top extreme when it is at the bottom extreme when it is here at point number one the spring is compressed by mg by k was anyway there plus whatever is your compression this much is the compression in the spring initially when you are pushing it down then same amount it will go up from the main position because amplitude from the main position is the same distance the extremes all right so this is the answer fine so that's it from my side next week we will continue this chapter next week I have to cover something related to the how you can correlate the circular motion with the simple harmonic motion once you understand that simple harmonic motion analysis becomes even more simpler and then we will talk about the non-ideal scenarios wherein forced oscillation is there damping effect is there resonance is there and it is the most ignored part of the chapter okay so that is why I have kept entire class just to discuss that so that you can understand the importance of it all right so that's it from my side I'll send you assignment make sure you complete it because it requires a lot of practice to become comfortable with these topics okay so make sure you do that and submit the assignment in time bye