 So, this is the second session on the oscillation or you can say simple harmonic motion, okay. So can anyone summarize what we had done the last class quickly like whatever you think of whatever you remember quickly type in or you might have seen it before coming to the class quickly type in equations of SHM okay and causes of SHM all right. So when we discuss the causes of SHM what is specific thing we had discussed last class what are the things we have discussed specific things okay what else steps of problem okay what else anything else we discussed. So we had discussed a spring block system right we had seen in a spring block system how we can find out the time period of oscillation all right yes constant external force has no effect you need a force which will try to push you towards where you need a force which will try to push towards the mean position okay what is a mean position what it is mean position is what where yeah position of equilibrium position of equilibrium when that force is 0 okay that is the mean position and when we write a is equal to omega square x what is x acceleration when we write is equal to minus omega square x this is the SHM this thing equation so what is x x is the location with respect to the mean position okay with respect to the mean position it may not be extension in the spring it may not be elongation in the spring okay it can be elongation or extension also when spring is horizontal okay otherwise spring will be already extended when you put a spring vertically like that there will be extension in the spring already so add x equal to 0 itself a spring has extension okay so keep these basic things in your mind and then only you should proceed okay so I am assuming like you guys have attempted the assignments also and last statement the last statement what I said was let me write it because it is important what I said was when you write a is equal to minus omega square x and when you write v is equal to omega root over a square minus x square the x is what x is position with respect to the mean position okay so your mean position is x equal to 0 alright that is what I said okay so now you know that just like any other mechanics chapter this chapter is also full of numericals okay so it's not that we'll just discuss a theory and just run through the chapter okay so there are still many kinds of questions that we can solve alright of course I cannot take up all the questions otherwise for entire year I'll be doing only one chapter okay so let us start the session with couple of numericals based on whatever we had learned the last session itself so that it get refreshed and try doing this Pranav got something others what is the amplitude everyone amplitude is how much amplitude is 5 meters okay what is the time period formula how you write time period what is the formula for time period looking at this can you tell anything you can find out omega right 2 pi by omega is a time period omega is what omega is 5 right a sin omega t plus 5 omega is 5 so this will be equal to 2 2 seconds alright and then we need to find maximum speed the formula for the speed is written above so using that formula how you find the maximum speed this one what is the maximum speed in SHM V max is omega into a right omega is pi is 5 so 5 pi meter per second alright and also to find velocity at certain time how you find out velocity as a function of time x as a function of time is given you can differentiate it to get V as a function of time it will be 5 pi cos of pi t plus pi by 3 and you know you could have differentiated this and then looked at that and find out when this velocity will be maximum to answer the V max like that also you can do it maximum value of cost function is 1 so V will be maximum which will be equal to 5 pi from here also you can get it now when you put t equal to 0 here t equal to 1 here it will be 5 pi cos of pi plus pi by 3 what is cos of pi plus theta you know that minus of cost right so minus of 5 pi cos of pi by 3 alright now cos of pi by 3 is half so the answer is minus of 5 pi by 2 negative velocity means what what does it mean negative velocity which direction it is going left or right it is going to the left hand side okay and can you tell me the location where it is at t equal to 1 second can you find out the location where it is how you find the location at t equal to 1 second everyone how you find it put t equal to 1 second here put t equal to 1 second there that is a function of x with time right so when you put t equal to 1 second over there you get x equal to 5 of sine pi plus pi by 3 what is sine of pi plus theta you remember that sine of pi plus theta is what minus sign you know that all that right so minus of 5 root 3 by 2 so it will be here somewhere it will be here somewhere and going this way negative side like that okay this is how you solve this question anyone has any doubt you can type in do this okay Prano got something okay do you remember for a splint block system what is a formula for time period do you remember that time period is 2 pi root of m by k okay this is a time period so from here this is 3.14 is pi only so pi is equal to 2 pi root over m by k why we are doing all this because I don't know what is k I'm trying to find k why I'm trying to find k because the answer is k into maximum extension which is amplitude okay that's why so pi and pi got cancelled and 1 by 4 is equal to m by k so k is equal to 4 m m is 5 so k is 20 Newton per meter so force is 20 into 0.1 that is 2 Newton okay so you don't need to again and again find out what would be the time period as a function of m and k I mean just like you remember the time period of the pendulum formula with practice you will remember the time period of the spring block system as well okay don't worry much about it in terms of time period t you have to find out how much time it will take okay Aditya got it Gayatri got something others see you can assume a simple form a simple simple harmonic motion equation that is a sin omega t okay omega is 2 pi by capital T this into small t now I want to find out how much time it takes to go from 0 to a by 2 distance so x equals to a by 2 okay so sine of 2 pi by t into small t is equal to half so 2 pi by capital T into small t will be equal to pi by 6 okay because sine of pi by 6 becomes half first when it starts going like this so t is equal to capital T by 12 directly you can get it okay and here you can get the multiple answers because it will not just go there once it will reach there then come back to the same position go there again go back to the same position and then again go there so when you get the solution of this equation you will get multiple answers okay but I am interested in only the first answer of sine of what angle becomes half sine of pi by 6 is also half sine of 5 pi by 6 is also half sine of 2 pi plus pi by 6 is also half multiple angles for which it will be half do this can anyone tell me what is the amplitude of the oscillation for this one in terms of k m and all that what is the amplitude so see the mean position at the mean position at the mean position spring is extended by mg by k amount okay and about this it oscillates it goes up and down from this position so when it goes down it further stretches but when it goes up the stretching comes down and if at the highest point it becomes unstretched it means that amplitude of the oscillation is exactly equal to the initial extension so that as highest highest point spring becomes the unstretched position all of you understand this right everyone now what is omega omega is root over k by m okay and time period is given to us 10 times what is this 10 times per second is what time period or what it is frequency time period is 1 by 10 1 by 10 is 2 pi root over m by k okay got it so this is time period and we need to find what maximum speed of the block maximum speed is omega times a okay omega is root over k by m root over k by m and a is m by k times g so we are getting it as root over m by k times g all right so from this equation root over m by k root over m by k is 1 by 20 pi from this equation okay so when you substitute it you will get g which is 10 into 1 by 20 pi so 1 by 2 pi meter per second is the answer for the a part all of you understand this type in everybody got this type in do the next part could you explain again see whatever is given to us we are reusing it that's all time period is given to us 1 by 10 whose formula is 2 pi under root m by k so from there I'll get the value of root over m by k root over m by k is 1 by 20 pi from equation number one which we have and maximum velocity is omega into a a is m by m g by k a because it becomes unstretched top most position so all that combined will yield this result b part anyone got the b part different answer check for the silly error of yours and mine also do the b part everyone yes for now that's fine that's fine instead of writing 10 I could have written pi square over here answer is same don't worry pi square is roughly 10 only how will you do the b part everyone can you tell me what is the amplitude of the of the spring oscillation how much is the amplitude anybody got the amplitude see from equation number one again you know I'm doing the same thing root over m by k is equal to 1 by 20 pi okay so if I square it I'll get m by k is equal to 1 by 400 into pi square and then mg by k g is pi square is 1 by 400 meters so this is the amplitude okay how much it is how much it is in centimeters 1 by 4 centimeters that is 0.25 centimeters is a okay 0.25 centimeter is a now you have to use v is equal to omega root over a square minus x square formula can anyone tell me what is the value of x for the b part if amplitude is this what's the value of x will be will it be above the mean position or below the mean position everyone will it be above the mean position or below the mean position at mean position at mean position spring is already extended by 0.25 centimeter which is the amplitude also now when spring is stretched by only 0.2 it is above the mean position because at mean position itself it was extended by 0.25 so it is how much above from the mean position 0.05 centimeter okay above the mean position so you need to calculate x from the mean position so x is 0.05 don't ever put x blindly from here as 0.2 x is what distance from the mean position it is initial stretch at mean position was 0.25 final stretch is 0.2 for the spring so it is 0.05 distance away from the mean position so now you can do it yourself right omega v no a v no and x v no just take care of the meters and centimeter s i units all of you understood this question this was a good question everyone clear right no you guys are ready for these or not let us see do this all of you what is the time period of oscillation everyone 2 pi under root small m by k others okay Hariran got something others Prano got something you don't need to redo it you already know for spring block system what is the time period you need to fit in the formula for this situation okay Shreyas also got it so both the blocks are performing the shm together so 2 pi under root of m plus capital M by k this is the time period right now total mass of m plus m is performing shm there is no slipping which we are we are assuming that there is no slipping both of them are doing the shm together okay now b part this is a yeah no slipping right now no slipping so but isn't the friction force also changing yeah let friction force change it is an internal force if I tell you a single block also like this single block then also I can say that the single block can be assumed to be made up of two parts one above part one below part okay how do you know why do you care what is the force between above part of the mass and below part of the mass as long as it is internal so if they behave like a single mass it is as good as this which block has only friction force in horizontal direction which block has only friction in the horizontal direction smaller one so draw the free by diagram of the smaller one okay look at the forces on it and write down force is equal to mass m acceleration you will get the answer it is straightforward do it let me know once it done okay here I have got something let's say it is at the right hand side of the mean position so what will be the acceleration of this block how much will be the acceleration at a distance x away from the mean position everyone acceleration will be omega square x okay and there is only one friction that one force that can provide that acceleration and that is friction okay rest all of the forces are vertical in nature vertically they are balanced normal reaction and gravity okay so the friction force should be equal to m omega square x omega is what what is omega omega is root over ky m plus capital M this is omega so when you substitute it you will get m divided by m plus capital M into k times x this is the force of friction okay now do the c part throughout we are assuming no slipping okay why do you think there is a maximum force of friction why what is the reason why there is a limit because friction force has a limit it cannot be beyond a certain value right how much maximum friction will be maximum value of friction is what mu time normal reaction is the maximum friction right normal reaction is mg so mu mg okay this force should be greater than or equal to largest value of mass time acceleration which is omega square into amplitude all of you understand this whatever is the largest value that can happen m into a friction force should be sufficient to provide that acceleration this is clear to everyone right so a and a gone so amplitude should be less than or equal to mu g divided by omega square mu g by omega square is mu m plus m by k this is the maximum value of the amplitude all of you clear right any doubts quickly type in see there are such there are questions in which two masses together are doing simple harmonic motion we have only catered till now situations in which only one mass is doing shm okay so i'm not sure whether you guys have done enough practice on the shm or not so probably next class we can take up some questions in which two two masses together are doing shm all right okay so maybe end of the today's class itself if we get time we can take such questions how does that happen a very simple example is okay i'll put the question here and we'll come back and solve it later on this is the example you can see both the masses are doing shm together okay this is doing shm this is also doing shm there is no fixed wall as such over here okay so we will take this at the end right now i'll leave the space to go to the next okay write down the angular shm that question was in your assignment that's good so you guys have little bit struggled through now when i tell you you will learn quickly okay so angular shm can anyone give me one example of angular shm whatever comes in your mind pendulum pendulum okay so in case of angular shm what happens is that the oscillation with respect to angle happens it means a very simple thing that in case of linear shm we track what we track the position with respect to mean point okay in case of angular shm we track angle with respect to mean position okay so like you guys said pendulum okay a straight forward example is a simple pendulum okay so at mean position it will be here like this mass m length l at mean position when you displace it slightly it goes this way and then it goes that way okay so it swings with respect to its initial position and this angle changes angle with respect to its mean position all right now can you tell me mean position with respect to the linear shm is when net force is zero okay so i'll write down mean position for linear shm is when net force is zero what is the mean position for angular shm what do you think net torque is zero okay net torque should be zero fine so from this position where the net torque is zero you need to then find out how it is getting deviated now things are little bit similar in fact they are exactly the same compared to the linear shm mathematically they are exactly the same what i'm trying to say is that if if a is equal to minus omega square x is for the linear shm then alpha is equal to minus omega square theta is for the angular shm mathematically both are same because a is what d square x by dt square double derivative of x and alpha is double derivative of theta now over here i can say that what else you have v is equal to omega under root a square minus x square okay here what we can write everyone what we can write here if this is true for the linear shm v is equal to omega root over a square minus x square what we can write here instead of v what it will be over here v is d theta by dt don't say v corresponds to omega because omega is angular frequency already taken up don't write omega here d theta by dt is omega root over amplitude is maximum angle minus theta square yes this is angular velocity this is angular velocity this is angular frequency omega is angular frequency d theta by dt is angular velocity okay then you have x is equal to a sign of omega t plus phi and over here what should i write everyone what should i write here right theta is equal to theta not sign of omega t plus phi okay rest all things you can differentiate it and get it from here onwards all of you are able to understand it kindly type in quickly there is this correlation between linear and angular shm how is d theta by dt different from omega what is omega omega is constant of proportionality between a and x or alpha and theta how is omega d theta by dt why do you think it is that just because we are used to seeing omega as angular velocity we tend to think that omega should be equal to d theta by dt okay what is the time period time period is same in both the cases time period is 2 by by omega only that doesn't change because do you think for mathematics these two equations they are different or same for mathematics mathematics will look at these two equation and will treat in exact same way or not these two instead of a i have written alpha instead of x i have written theta so mathematics doesn't care whether it is alpha or a or theta or x just like a is d square x by dt square similar way alpha is d square theta by dt square okay previous slide somebody has to see that's all i move forward okay fine so now the question comes how do you find out the time period of oscillation for the angular shm okay all the steps are exactly the same okay all the steps are exactly the same i will not going i'm not going to write down the same steps again just that instead of displacing by a small distance x from its mean position what do you have to do can you tell me previously when we wrote steps of solving problems for linear shm we displaced it by a small and a small distance x now in this case what will happen displaced by small angle theta okay here what you will do is that displace by small angle theta then what i have to see in earlier case i had seen that there is a force a restoring force that get developed that again tries to bring the mass towards the mean position here what will get developed correct here a restoring torque restoring torque gets developed so what you have to do is write down torque is equal to i alpha equation all right then get the alpha is equal to some constant times theta and alpha should be in opposite direction of theta okay and then you can compare it with alpha is equal to minus omega square theta you'll get omega straightforward thing okay now if theta is very small if theta is very small we are going to use this approximation again and again that sin theta is roughly equals to tan theta and is roughly equals to theta itself remember these approximation anyone has any doubts sin theta if there is a right angle triangle just taking an analogy here is perpendicular upon hyperton is this is p by b so when angle is very small hyperton is and the base becomes almost equal if the angle is very small this hyperton is and this base becomes equal almost so that is what it is and moreover it looks like an arc length which is perpendicular divided by base so that is the definition of theta itself okay so that is why is equal to theta also okay so we are going to use this approximation multiple times okay let's now again go back to the simple pendulum and see how to get the time period is this chapter done in your school by any chance not in dps not in ypr no not in hsr also okay so what i'm going to do is i'm going to i'm already displacing the mass m by a small angle theta this is the length l okay now i want you to get the answer yourself try doing it okay aditya got something there'll be mg force acting on it and there'll be tension force acting on this t so how much is the torque about this point oh everyone what is a torque of mg torque due to mg is what mg into l sin theta l sin theta not cos theta this is the perpendicular distance right this is the line of force i drop a perpendicular on it mg into l sin theta is the torque okay this is the torque not torque is equal to i alpha what is i moment of inertia about this point everyone how much it is for this mg what is the moment of inertia for mass m it's a point mass ml square so alpha is equal to mg l divided by ml square into sin theta okay m is gone one of this is gone now theta is very small so sin theta is theta only now is theta and alpha they are in opposite direction or in the same direction theta and alpha theta goes this way alpha tries to bring it back okay alpha is in this direction theta is in that direction clockwise anti clockwise so minus so now when you compare it with alpha is equal to minus omega square theta you'll get omega is equal to root or g by l so time period is 2 pi under root l by g this is the time period okay now why it is called simple pendulum it is called simple pendulum because it is a simplest kind of pendulum can you think of any other kind of pendulum okay think of any other kind of pendulum correct there can be there can be a rigid body okay like this pen itself it can have an shm this bottle itself if I displaced a bit it can perform some oscillation like this okay but then this is not a simple pendulum this is a this is a different kind of pendulum and the name for this is physical pendulum so write down physical pendulum can both angular and linear shm what do you mean didn't get okay physical pendulum let's say we'll take an example to discuss that okay let's say you have a rod like this okay this rod is hinged over here total mass of the rod is m length is l this distance the distance of the topmost part from the hinge is a okay I want you to find out the time period of oscillation for this rod at the mean position it is vertically straight like this isn't it net torque is zero right now then what you do you displace it both angular and linear shm yes they can happen together it's can be like this a rod then there is a spring mass like this so when rod is oscillating this block also oscillates but then you guys have to do a lot of problem practice your own the basic level okay so Pranav got something others anybody else got anything nobody else okay so at this position gravity force will act from where gravity will act from center of mass so this is mg can you tell me what is a torque due to mg about this position where it is hand what is this perpendicular distance this perpendicular distance is how much this is theta this angle is theta okay so this angle is also theta this angle is theta okay from here to there what is the distance this distance is what l by 2 minus a see I intensely made it little geometrically like this so that it should not be a straightforward thing right so this is l by 2 minus a so the torque is mg l by 2 minus a times sin theta all of you understand this type in this is a torque we are trying to use torque is equal to i alpha okay now can you tell me what is moment of inertia i which i should be using there if it is icm it is not ml square by 3 plus ma square if it was icm plus m into distance from center of mass square icm is what ml square by 12 plus m into distance from center of mass square that is square okay now I think let's call this as this bracket the white bracket let's say c so alpha is equal to minus c times theta is very small this okay so omega is root over c where c is this okay and time period is 2 pi root c is it clear to everyone all of you anyone has any doubt quickly type in let's proceed they are now you guys have learned almost everything about the problem solving it's all about practice now okay so we'll do one more question because you guys will see these kind of things quite often so hence you should be so I think the arrangement you can make of this is k1 spring k2 spring constant mass m length l this rod can freely rotate about one of its ends which is topmost point it can rotate so you need to find out the time period of its oscillation yes k2 is a midpoint sorry anybody close to the answer should I wait or solve okay one or two minutes take all right so nobody got it so let us say the extension on the bottom most spring is x1 so it'll be k1 x1 force over here the compression on this is spring this spring will be compressed the bottom spring will be extended but the good thing is that both will create the force in the same direction this will be k2 x2 okay and that forces gravity mg okay now the torque about point o is mg into l by 2 sin theta plus k2 x2 into l by 2 plus k1 x1 into l do you all understand this torque type in quickly angle is very small so the force into perpendicular sense remains that much only all of you type in everyone is it clear yeah yeah this is l now how how is x1 and x2 they are related is there a relation everyone type in is the relation between x1 and x2 okay can I write x1 x2 in terms of theta can I write x1 and x2 in terms of theta yes directional spring force will change and it'll always try to bring the rod in its main position that's how the force should be which is helping you to do the shm okay so x1 all of you agree is l into theta if theta is a small and x2 is equal to l by 2 into theta and sin of theta is roughly equal to theta okay so we can write down torque about o to be equal to mg into l by 2 theta plus k2 l square by 4 into theta plus k1 l square into theta okay this is the torque this is equal to i alpha that is ml square by 3 into alpha so alpha comes out to be um mg l by 2 okay we can cancel out one of the l l is gone how it is l square by 4 okay x2 is l by 2 theta substitute here one of the l is gone so alpha is mg by 2 plus k2 l by 4 plus k1 l see I just made up a question so that's why the values are not looking so good but there are many questions like this when you practice your own that divided by ml by 3 this into theta now the problem is done this bracket term this term is your omega square and time period is 2 by by omega okay so like this you solve this particular question okay so let's move forward there are many questions you know so let me just move forward and do a little bit of more theory then we'll again come back to the numericals okay so write down the energy simple harmonic motion so basically our main interest is spring block system it's a relatively small topic because there is nothing new as such whatever you know about the springs energy and the blocks energy same thing it's a repetitive stuff but slight bit of specific details are there so if a block is doing shm with a spring or spring constant k okay then the value of omega is root over k by m okay the amplitude is let's say amplitude is a so can you tell me maximum potential energy how much it will be and maximum kinetic energy how much it will be everyone maximum potential energy and maximum kinetic energy how much it will be maximum kinetic energy maximum sorry potential energy is half k into x square what is the maximum amount of x a only so half k a square maximum kinetic energy is what half m into v max square v max is what omega into a whole square omega is root over k by m so you will get half k a square here also okay got it now in terms of x at a particular moment in terms of x the potential energy is half k x square and we know that in an shm v is equal to root over a square minus x square so kinetic energy is half m into v square that is omega square a square minus x square now omega is root over k by m so this can be written as half k times a square minus x square so can you see that the sum of kinetic energy and potential energy is a constant can you see that it doesn't depend on x at whatever location you go to when you add potential energy and kinetic energy it is equal to half k a square only all the time okay so why don't you do one thing draw a plot let's say this is the this is the zero energy zero energy and there's a maximum energy which is half k a square okay can you plot the potential energy and kinetic energy on it this center line is x equal to zero this is x equal to zero this is not zero left hand side is x equal to minus a right hand side is x equal to plus a i want you to plot the potential energy on it everyone it doesn't matter from where it starts you have to plot this on the graph let me know one step okay done so potential energy will be maximum at x equal to minus a and x equal to plus a and it is a parabola like this so i hope all of you have got this okay this is from your n crt whatever i'm discussing okay now i want you to plot the kinetic energy on the same graph which are different color can you plot the kinetic energy this is your potential energy you plot the kinetic energy energy is this let me know one step look at the critical points at x equal to zero what will be the kinetic energy at x equal to plus minus a what it will be and then join it with a curve what will be the kinetic energy at x equal to minus a and x equal to plus a look at the formula zero and at x equal to zero it will be half k a square it's a downward opening curve right to plot it it'll be like this this is your kinetic energy i hope it is clear to everyone nobody has any doubt so if the shm is a system which is non-desipative non-desipative as in in which friction is absent and there is no let's say drag force or anything like that the mechanical energy will be constant okay so there are ways to solve a problem by using the mechanical energy conservation you can get the value of omega okay so if you want i can do one question where you can do the conservation of energy and get the answer of finding time period and things like that i'll take a simple question only let's say you have a spring block system mass m spring constant k okay then there is no friction anywhere whatsoever so at any time at any time kinetic energy plus potential energy will be a constant right kinetic energy is half m v square potential energy is half k x square this is a constant if it is a constant what will be the derivative of this if i differentiate it if it is a constant what should it become z will become equal to zero it's a constant so when you differentiate it you'll get m v dv by dt plus k x dx by dt is equal to zero now dx by dt is your velocity also goes away okay from here you'll get dv by dt is equal to minus of k by m times x d by it is acceleration so a is equal to minus k by m times x so you got the shm equation by using the conservation of energy also okay this kind of method is very useful and easy but it requires a lot of practice okay so i'll send you a worksheet in which you can practice something like this but right now you can focus on the force and the torque equation and if you have time right now after the class only then you can do the same question like this all you have to do at any point in time write down kinetic energy plus potential energy it should be a constant and you will get the answer okay in case of angular shm also you can write down the kinetic energy half i omega square plus mg h and differentiate it okay when you differentiate mg h you get velocity of center of mass which is equal to omega into l by 2 so there are some certain things that you need to take care of but then you'll get the answer okay yeah so let's take a question i have now i think i am um more or less done with the basics of the shm i can take few questions and then there is a third part of the chapter that i can discuss do this your concepts of rigid body motion will also be used in shm okay pranav got something others okay so moment of inertia is not coming in the equation i is not coming are you sure pi t i what is that 2 pi r t i root i by k okay root i by k okay so all of you try it right so now let us move this down by a distance of x who is doing shm the pulley is doing shm okay so you just look at the movement of the pulley that's all so movement of the pulley is such that it goes down by a distance of x from its mean position it goes down by a distance of x from its mean position how much spring will be extended extension the spring will be how much this length will be extended by x even that will be extended by x isn't it if the pulley has to go down by x it increases excellent this side excellent that side from where it comes from the spring a spring get extended by 2x are you getting it everyone quickly is it clear to everyone why 2x because if cylinder has to go down by x the length of this portion has to change by 2x okay yeah let it rotate pulley is rotating so what but the length has to change you know this length and this length both are x from where it will come from from the spring only 2x length will come then only cylinder will go x down now draw the freeway diagram this is 2kx we don't know what is the tension over here this is mg okay fine so the there was an initial extension on the also on the spring let us do that as well so that it had a complete picture when initially it was extended kx not this side is t this is your mg everything was at rest so pulley was not rotating so this tension was also equal to kx not only so initially kx not plus kx not is equal to mg now I've displaced it by a distance of x so the spring force has become k times 2x plus x not tension has changed from the earlier tension which was kx not plus let's say delta t okay so the net torque about the center is what k times 2x plus x not times r minus k times x not plus delta t times r this is equal to i alpha do you understand this equation all of you type in kx not r minus delta t r is equal to 0 okay minus sorry kx not r and kx not r will go away so you'll get k times 2x times r minus delta t times r is equal to i alpha okay now we also know that net force is equal to mass acceleration okay earlier forces were all balanced as in kx not kx not and mg they were balanced so if you write kx not plus delta t plus k times 2x plus x not this is the upward force minus mg is equal to m into x version of center of mass so kx not kx not and mg they get cancelled over because 2 kx not is equal to mg so we have delta t plus k times 2x is equal to m into a sin center of mass now what to do everyone these are the two equations now what to do i have to write alpha i have to write alpha in terms of x version of center of mass a center of mass divided by r okay pure rolling condition so substitute it over there using these two equations remove delta t remove delta t using these two equations then write down x version of center of mass to be equal to something into x and this something will be equal to omega square so like this you solve this particular question all of you understand this just go through it once there is no hurry you can spend one minute you don't need to hurry up it can be in terms of r see r will be there r is a constant it should be in terms of x so that you can compare a is equal to minus omega square x this equation when you compare then only you'll get omega right inside this bracket there can be r there can be m there can be i first equation again see first equation is torque about center of mass is equal to i alpha earlier kx not kx not and mg were acting okay so 2kx not was equal to mg at the mean position everything was at rest then i have displaced the mass by distance of x when i'm displacing by distance of x the left hand side the string length increases by x the right hand side also the length increases by x the total increase in this overall length of the string is 2x from where this 2x comes from the spring only so spring generates 2x into k force because the block gets pulled down by x so k into 2x the new extension plus the earlier extension here tension also changes when you leave it let's say tension earlier it was kx not now it is kx not plus extra tension then i'm writing torque equation torque due to this force is that force into the radius perpendicular distance torque due to this force is force into radius perpendicular distance this is equal to i alpha and this is what the equation is clear any other doubt normal reaction where is a normal reaction there is no normal reaction from where you got it any other doubt everyone okay so now we have seen many spring and block systems let's take a situation in which there is no spring block but still it is an SHM only have you used have you done this question in which there's a beaker and water have you done this no right so let's do this you have a beaker full of a liquid this is from your ncid textbook the last question of your unsolved x problem don't look at the final answer do it yourself this has density of row okay the beaker is full of a density of liquid row and there is a cube that is floating on it there is this cube whose edge length is a and it is floating over it it has a density of sigma sigma clearly is lesser than row that is why it is floating okay you need to find out time period of oscillation if what I do is that if I push it slightly down and release it if I push it slightly and release it then with what time period will oscillate all of you do this what length is it submerged to start the oscillation doesn't matter it doesn't matter you can find out initial submerged length also if you want but it doesn't matter okay Pranav got something Hariharan others when you push it down further what exactly happens what are the changes that will happen everyone when you push it down by a distance of x what is the change that will happen all of you type in what do you think buoyant force increases very good buoyant force increases so earlier we had earlier the buoyant force balancing the gravity force earlier now extra buoyant force is created how much extra buoyant force is getting created everyone if I push it down by a distance of x what is the extra buoyant force that gets created so the net force is the earlier buoyant force plus extra buoyant force minus mg okay the earlier buoyant force was equal to mg so the net force is the extra buoyant force only which is rho dv what is dv Hariharan what is that don't include another variable already we have so many variables how much volume gets submerged further a square into x so extra displaced liquid weight is rho a square x this is mass into g okay this is the net force this is equal to mass and acceleration sigma a cube is the mass times acceleration let's say capital a is acceleration small a we have already used actually so capital a is rho g by sigma a times x and clearly a and x are in opposite direction so minus sign this bracket term is your omega square all of you got this everyone any doubts this is the time period of oscillation clear to everyone type in can you explain the last two steps last two steps though nothing it is you you want to see this one this is the equation of shm a is equal to minus omega square x if acceleration is minus omega square x then it is shm so I am comparing whatever I have got with a standard shm equation and when I compare I get omega as under root of this bracket term and if you get omega time period is 2 pi by omega okay any doubt anyone has clear to everyone so we'll meet after the break now okay fine so let's take a break now we will meet at 6 22 p.m