 Good morning to all of you and this is the ninth day of this workshop. So as you will see actually that we are going to complete our syllabus in this morning session. So what we have kept is mostly mechanical vibrations that is what we are going to look into. Now remember we have also reviewed you know many syllabus of different colleges. Again this topic is not covered and there could be reason because as such you know this is more applied towards mechanical engineering and in civil engineering we tend to actually even in IIT we do not have a you know course undergraduate course on mechanical vibrations. Only thing is that if student requires then they actually tend to you know take a graduate course as an elective that I offer. So now what is mechanical vibrations as you see that in a simple term it is basically oscillation of a body about its equilibrium configuration. Now how does that happen? It can happen in variety of reasons as you see that for example in mechanical systems if you look at it and if we apply a force right. So that force should be again now bounded by time that means it depends on time. So usually we try to take examples of you know periodic loading and this periodic loading you know that is the function of sine or cosine. So if you induce that kind of force onto the system then the structure vibrates. The other could be just the free vibration. So free vibration is basically suppose let us say I have a spring attached to a mass and I pull the spring right. So I pull the mass and then release it. So what will happen basically that mass is going to oscillate about the equilibrium configuration. So we are really going to look at the oscillation of a system however remember that what is most important to understand that entire thing is basically characterized by 3 important quantities. The first one is the stiffness, second one is the mass and third one is of course the forcing function that we have ok. Now remember if we are talking about free vibrations then the forcing function is not coming into play. So the motion will be determined by the initial conditions that means what kind of displacement and velocity is being given let us say at t equals to 0 time equals to 0 ok. So as you all know that why it is so important firstly we see that you know in cars we have a suspension system that is coming through the you know spring and dashpot we call it a damper right which will reduce the unnecessary vibration that is coming. Similarly you know spring mass systems also used as different types of absorbers and even you know supporting the structures also you tend to use the spring mass type of system. So what we are basically trying to you know try to do in this session is try to derive the equations of motion for spring mass system and try to understand that what is the natural frequency of vibration. Remember the natural frequency of vibration is an important aspect which is totally controlled by the geometrical properties as well as the material properties that means spring mass that is important for rigid bodies and if you are looking at the continuous system then also we see that you know again we have the stiffness mass that is controlled by the geometry also ok. Now we all know that we know the examples you know very old example is there that you know I think may be hundred years back something around 1830 we have seen that there is a bridge in England I think it is called brofton bridge and if you look at look around the wave to pull up that example what happened because soldiers were marching so soldiers were marching on that bridge and what happened that bridge experienced a large displacement. Now why that happened because what happened now when the soldiers are marching they have some marching frequency right that is we call forcing frequency of the system that is the forced vibration. Now we all know when the forced vibration you know frequency matches with the natural frequency of the vibration of the system then what happens is the resonance we call it a resonance that means amplification of the displacement. So this is very critical parameter therefore this how to get natural frequency of variety of systems that is what we are going to study and we are only going to restrict ourselves to study of single degree of freedom systems ok. So we will try to look at single degree of freedom system what is the natural frequency then we are going to study if it is a free vibration problem let us say then given an displacement or velocity what kind of you know due to this initial conditions what are my maximum amplitude maximum displacement maximum velocity maximum acceleration to the system and we are going to look at both the undamped case that means no damping as well as damped case. So with that I will just you know try to now look at how to really get the equations of motion. So far as we all know that for a rigid body under the action of external forces if it is under motion then we know that from the Newton's second law if we apply the resultant force along x direction must be equals to mass times acceleration. The resultant force along the y direction again mass times acceleration in that direction and the moment about any point g which could be let us say in this case the mass center of the body that is equals to rate of change of momentum angular momentum right. So what it means that is our J alpha so that is the mass moment of inertia multiplied by angular acceleration. So this has already been done now ultimately what we see yes if the body is under motion and it is accelerating system or decelerating system yes then we can use the Newton's second law right to establish the equations of motion. Now this you have already studied now what we are going to do we are going to little bit deviate from that and what we will say that okay we will do one thing is to understand that each and every instant of time I will represent the body through equilibrium that means again I will say that each and every instant of time the body is in equilibrium and therefore the resultant force should be disappear that means resultant force should be equals to 0. Now how do I do that so there is a principle which is called the D'Alembert's principle now through the D'Alembert's principle remember the explanation given is like this think of a particle that particle is under acceleration as it is shown in this direction due to the resultant force acting on that direction therefore if we take a fixed set of reference axis that means x y and z if this is my reference axis right then we can simply apply that sum of force must be equals to mass times acceleration. Now suppose I observe the particle from a moving system x y z which is attached to the particle so small x y z is now attached to the particle then what is happening as an observer I will see that particle is necessarily at the rest okay so it is at a rest condition or in equilibrium now how that is possible then I have to imagine that if I am in this frame okay which is small x y z then I can simply experience a force which is fictitious in nature and I say that is negative mass times acceleration so that force is actually in terms of D'Alembert's principle we call it a inertia force now the inertia force is coming into play which is trying to oppose the motion but remember this entire thing is now applied at a moving frame such that the net resultant force is equals to 0 okay so if we try to look at this situation or if we try to now convert the problem like this what will happen that one can transform an accelerating rigid body into an equivalent static system that means what is happening now I can solve this problem at an instant of time T by simply saying that I have an equivalent static system now on this equivalent static system how I have taken the effect of the inertia forces so I have introduced this inertia forces as mass times acceleration so remember that is opposing the motion so acceleration whose actual acceleration of the body is this way so that is mass times acceleration is opposing the motion similarly if the body is rotating counter clockwise then I have a inertia force that is my rotational inertia force it is clockwise okay therefore the body is assumed to be at rest so now I am really going back to static so through the static now I will be able to solve this class of problems however as I mentioned that this is just one approach that is to use D'Alembert's principle you may want to also use the Newton's second law to derive the equations of motion but you will see that D'Alembert's principle will effectively what will happen if we try to realize the system as an static system then some of the unknowns that are coming from the support reactions we need not to solve for those okay because when we take the moment about let us say support where the reactions are we can simply eliminate the effect of the reactions okay but remember one important thing is that the translation inertia that must act through the center of mass and the rotational inertia can act anywhere because remember rotational inertia will act as a couple moment so it can be added anywhere but remember that rotational inertia we are going to calculate about the mass center all the time and remember we have we already know what is the mass moment of inertia of various planar bodies right about its mass center we have already studied that so therefore ultimately what the problem boils down to that I have to correctly and identify the translational inertia and the rotational inertia that are acting at the mass center of the body and I must take this translational inertia and rotational inertia in opposite sense right of the motion okay so with that now we can start you know the working on it so we will start again by simple spring mass system so let us say a mass is connected by a spring okay and the mass is resting on a frictionless surface and let us say I also apply a force P right which is function of time now I am interested in finding out what is the displacement at instant of time t okay so now how do I derive the equations of motion now you can do it in two different ways as I said you can either adopt the Newton's second law okay or you can use the D'Alembert's principle now remember if I am adopting Newton's second law then there is a way to write this equation of motion what is the way to write it since the body is displayed along the positive x direction therefore I have to find the resultant force along the positive x direction so therefore what is the resultant force remember I have applied loading that is Pt and I have the spring force that is Kx but that Kx is the force that is given by the spring to the body so it should act in the negative x direction okay remember spring will be on the tension so as per Newton's third law we can also see the direction of this force on the body should be in this direction negative x direction so I have Pt-Kx that should be equals to Mx double dot that is the Newton's second law mass times acceleration remember acceleration will be given d2x dt2 we will always denote like x double dot and if it is a velocity it will be denoted by x dot so x dot would be the velocity that is dx by dt okay derivative of the displacement with respect to time so now if we look at the D'Alembert's principle we have to represent it by an equivalent static system and you can clearly see that what I have done here I have taken the inertia force that inertia force is Mx double dot that is mass times acceleration this is my translational inertia that is on the negative direction so when I am looking at D'Alembert's principle I will simply write it Mx double dot plus Kx negative Pt equals to 0 now you can also write Pt-Kx-Mx double dot equals to 0 so I am basically using the equivalent static right sum of force must be equals to 0 okay so now that will make you know things to be much simpler and we will have we will be in you know less prone to error actually as long as I identify the inertia force on the body correctly and the problem becomes a static problem at any instant of time t now let us try to study when the mass is under the action of gravity so previous problem is just inverted now okay now what is happening first thing to understand here is that we have the spring unstressed configuration right so that is basically the original spring length and as we attach the mass it is going to get the displacement now that displacement is going to be static displacement that is the stretching of the spring okay so if you look at statically then there is a tension in this spring that tension equals to K delta st right so on the body therefore it will be upward and I have the weight that is downward so we can clearly see that I have a static equilibrium position where by K delta st equals to mg so that is the force balance and through that I can find out what is the static displacement so now I am going to call this static equilibrium configuration okay remember now once we are in this static equilibrium configuration now what I am going to do I am simply going to displace the spring by a distance x0 downward and also I am going to give it a velocity v0 so now I have an initial condition let us say I pull this spring okay and I give it a velocity as well as it is now displaced by an amount x0 from its equilibrium configuration okay and I am going to release it so what will happen suddenly if there is no damping let us say we ignored the damping effect there is no damper installed here so it is going to oscillate with respect to its equilibrium configuration okay so what is the governing equation of motion now when we look at the governing equation of motion how do we do that again it boils down to drawing the free body diagram correctly so what is the free body diagram so I have the at an displacement x that means at an instant of time t I have the displacement x from the equilibrium configuration then what is the spring force the spring force is equals to k delta st plus x because that is the net stretching of the spring delta st plus x okay and then remember we have to add the inertia force inertia force will try to oppose the motion so since I have assumed this way particle is displaced right mass is displaced in the you know this direction so therefore we have the inertia force that is opposing the motion which is in the upward direction so what we essentially have if we really take the equilibrium mg minus k times delta xt plus x negative mx double dot equals to 0 so again from this free body it became a static okay so ultimately what you get remember the k delta st equals to mg so therefore that going to go away and we have mx double dot plus kx equals to 0 so x double dot plus k over mx equals to 0 that is the equation of motion remember this equation of motion is measured or derived from the equilibrium configuration which we call static equilibrium configuration so therefore what we have learnt is that when we start measuring the vibration from the static equilibrium configuration for this particular system then there is no effect of gravity coming into the equation of motion and that is very unique so in this problem the gravity effect is not coming into the equation of motion as long as I derive the equation of motion from the static equilibrium configuration so that means my x has to start from the static equilibrium configuration okay now remember we can now this in k over m is a constant so we will just try to write it in terms of omega n square equals to k over m and we will later tell this omega n square right this quantity that means this omega n that is circular natural frequency now why it is circular natural frequency that we are going to come to a minute okay but what is important also to understand why do we call it a harmonic motion why it is harmonic because remember force the spring force that we have that is proportional to the displacement that is force equals to k delta so that implies that you are always going to get a harmonic motion okay so it is going to simply oscillate in its natural frequency about the static equilibrium configuration now if let us say f is not equals to k times x but let us say k is non-linear either that means k is now you know some function of x then you are not going to get an harmonic solution to it that means you are not going to get a oscillation about this equilibrium configuration you may have some other harmonics that are going to occupy your solution that means higher order harmonics are going to come into play so that requires more advanced study but remember the term harmonic has come from the fact that force is proportional to the displacement okay so now we will try to see okay see the spring now how do we interpret the motion of the spring therefore remember the spring was given a displacement that is x0 and also a velocity v0 at time t equals to 0 which is somewhere right because first I pull it and then I am releasing it okay so there was an initial condition therefore this equation of motion can be solved let us say what is the general you know solution to this equation of motion we all know it is going to be periodic function so we are going to have c1 sin omega nt plus c2 cosine omega nt because the root of that equation will be determined by plus and minus i omega n these are the two roots plus and minus i omega n and that will give me the solution general solution real solution in terms of c1 sin omega nt plus c2 cosine omega nt okay now how do I get c1 and c2 from the initial condition so at time t equals to 0 x equals to x0 right so therefore c2 equals to x0 and time t equals to 0 I have velocity equals to v0 so that will give me c1 equals to v0 omega v0 divided by omega n so now I have the complete solution subject to these two initial conditions now suppose I do not have a velocity that means let us say at time t equals to 0 what I am doing I am just giving it a displacement x0 and releasing it so now I do not have the velocity let us say then you can see the solution boils down to simply x0 cosine omega nt okay now let us try to see that how do I interpret the motion of the mass right along the x axis so ultimately what we have just look at it my solution is xt equals to v0 by omega n sin omega nt plus x0 cosine omega nt right now suppose I have two vectors what I am going to do let us say I have two vectors c1 and c2 what is my c1 c1 is v0 by omega n and c2 is equals to x0 and remember what is happening they have a resultant that resultant is simply given by xm which will be the maximum amplitude of the motion and given by this now what is happening I can always say this c1 and c2 if I interpret these two as vector they are rotating at an angular speed constant angular speed which is given by omega nt so that angular speed if it is omega n let us say right then what we can say that now remember the amplitude xm of the motion right the projection of this xm on to the x axis now remember this x axis is this way that is the vertical right so projection of this xm on to this line vertical line is what the spring motion is in other words if I think that there is a particle right here that particle is actually being rotated at an angular speed omega n right then the motion determine that by that particle along the x axis that means this axis is actually the motion of the spring okay because spring motion is always going you know positive x to negative x so ultimately what happens we can clearly see now if you look at this way so now this way is the time and this way I have the positive x okay now how the particle if you look at it so I have the spring motion that is interpreted by this so which is this function right here x equals to xm sin omega nt plus phi so that is the motion now each and every instant of time what is happening I can actually determine the position of the spring is also controlled by the position of this particle on the circumference of the circle okay as we can clearly see that at an instant of time t I have the displacement x so the particle position on the circle is at q now remember therefore what is the projection of that on to this vertical axis that is OP so what is OP OP is nothing but sin omega nt plus phi now remember what is happening in this case there is a phase lag there is a phase angle why this phase angle is coming into play we can clearly see it is because we have a initial velocity on to the system since we have a initial velocity on to the system what is happening is that now you have both c1 and c2 okay now if phi equals to 0 right then I can set this equals to easily to equals to 0 right so ultimately what you see it is going to be controlled by just c2 not c1 okay so ultimately what we see now few important parameters here that means how do I define the omega n that means to make a full circle right the particle starts from here somewhere that is my at t equals to 0 I have x0 and the v0 is what v0 is going to be the slope of this curve displacement time curve right that is going to be the slope and we have also x0 so ultimately particle will start some from somewhere here and it will make a complete loop right in circle and come back so that is going to be determined the time period okay so that means to make one complete cycle how much time it takes that is going to be determined by the time period so you can also look at from two peaks let us say first peak is this second peak is this that means what is happening particle is started starting from right here so from here it is going to make a full circle and come back so you are going to land up getting two position one is this one and another is that one so that means to complete one full oscillation the time taken is given by time in a time period now this time period in some text book is also denoted by capital T do not forget but here we do not want to say all the time let us say capital T so we are going to use this symbol that is tau okay so now if time period is defined then what is my frequency natural frequency that is simply going to be 1 over T 1 over tau right that means what is natural frequency natural frequency is nothing but how many cycles will be completed in one second so let us say I complete 10 cycles in one second that means my natural frequency is 10 hertz okay so it is unit will be hertz now what is the time period then that means the time period will be simply 1 over 10 that is 0.1 second so in 0.1 second I will be able to make a full oscillation complete oscillation that is the time period okay so with that now what we can see clearly remember once I have the displacement time curve okay then I can also construct the velocity time curve and also acceleration time curve so let us look at the next so mathematically we can say okay this is the displacement at instant of time T which is defined by xm sin omega nt plus phi then my velocity will be x dot and acceleration will be x double dot so derivative of this first derivative of this is velocity and second derivative of the displacement is going to be acceleration so ultimately what you see now velocity is equals to xm omega n cosine omega nt plus phi now you see this is a difference with the there is a difference in phase okay so how much the phase is changing phase is changing by pi over 2 so that means velocity and displacement is always going to be we call it out of phase the phase difference is 90 degree whereas if you look at the acceleration is going to get the same pattern however with a negative sign so we say that acceleration and the displacement they are in phase but with a negative sign so therefore actual phase is actually pi so these are all happening because we have a simple harmonic motion now again if you look at the particle right the one we discussed on this circle now I have the acceleration this way and the velocity this way right look at what is the actual so projection of this as a component of acceleration along the x direction so this is the actually happening in the mass spring system that is the acceleration which is negative to the displacement right so if you look at it so we have the acceleration which is simply you know projection of this onto the x axis or vertical component of the acceleration given by the particle which is nothing but sign of omega nt plus phi but with a negative sign okay similarly for the velocity what is the vertical component it is going to be cosine of omega nt plus phi okay so therefore now what I will do I will try to again try to explain this let us say you know in terms of when I have v0 equals to 0 so that will make it very simple if I have v0 equals to 0 then how do I actually plot all of these curves so I will just try to show it again what do I mean by phase and out of phase suppose you have this spring and mass and this is the static equilibrium configuration so now remember now what the problem here is I will simply trying to pull this by x0 okay so this is going to be my x0 and this way I am always measuring the positive x right and let us say v0 equals to 0 so I displaced the system through x0 that is my maximum amplitude now and release it so remember therefore what we have xt the solution should be x0 cosine of omega nt okay now how do I draw this then so now let us say I have xt this way now to make this function so I must say I have 0 here pi by 2 here pi here so at t equals to 0 I have the solution x0 right and then how it moves it is basically going to okay so that is the property of the cosine function right so that means this is going to be my full cycle full cycle of you know oscillation so that is my t so now this is the displacement now how do I draw the velocity on top of this if I want to do the velocity remember I say velocity is out of phase what does that mean that means that xt is now sorry vt so velocity vt that is equals to x dot that is going to be equals to what that is going to be equals to negative of x0 omega n sin omega nt okay so ultimately you can write it x0 omega n cosine of omega nt plus pi over 2 so what do I mean by that means the velocity will be drawn how it will have its own amplitude so do not worry about that let us say okay so that is my velocity remember velocity will have maximum peak right here and you can clearly see now I do have a phase lag that phase lag is determined by pi by 2 so if this is the velocity curve this is the displacement curve therefore it is differing by pi by 2 okay similarly what is the acceleration acceleration is at so that is x double dot now remember in x double dot if I do then again it is going to be negative omega n square x0 sin a cosine of omega nt okay so ultimately now we can say that it is equals to omega n square x0 cosine of omega nt plus pi okay so that is plus pi so now do you have again a phase lag so how do I draw that now now remember what will happen in case of the acceleration it will start from here right then how it is going to go okay so that will be your acceleration curve so as if you can see that this is shifted by just pi amount of pi okay so that is how we determine that what is you know so basically you see that acceleration is going to be negative of that of the displacement that we see okay and velocity is always going to be out of phase by pi over 2 okay so there is a you know pi gap when it is coming when you look at the zero point to the zero point on the time that will always differ by pi so in other way to you know look at these things is to extend this on beyond this line and just complete this circle so what is happening in this case like this okay so now we can go up to you know this point right here so this point from here okay as if this distance is how much pi so if you take the entire acceleration plot let us say you take the acceleration plot and shift by pi you are going to get the displacement is in that okay similar thing we could do for the velocity so velocity is going to come somewhere here okay so this is let us say negative pi over 2 this is let us say negative pi okay so if this point is negative pi over 2 again look at the velocity so velocity wise what is happening you can again you know take this and simply shift it shift it by plus pi over 2 so if you take the velocity if you take the velocity curve and simply shift it by plus pi over 2 you are simply going to map the displacement curve is in that okay so therefore there is a lag of pi over 2 between the displacement and velocity and between the displacement and acceleration I have a lag of pi okay so that is how we define the phase lag but that is for a simple system okay now we are going to go back a little bit and let us try to study a few problems that how to derive the equations of motion for let us say various types of springs and mass systems and we can see there is some you know experimental demonstration so this was taken from one of the you know MIT course of web so you can see that spring and mass is attached and ultimately once I pull it and release it then we are going to look at the equation of motion of this first of all what is the natural frequency what is the maximum displacement maximum acceleration and so on so forth so ultimately let us try to say that we have two different systems we are talking about in first case I have two springs like this in the other case I have two springs in series so in this case let us say they are in parallel and in this case they are in series okay so now what we are going to do in part a basically pull the block 40 millimeter down from its equilibrium position and release it so for all of these cases we have to find out the period of vibration and the maximum velocity of the block and maximum acceleration of the block okay so let us try to look at how do we you know solve this problem although from static you can easily figure out that what is the equivalent spring constant that I am looking at for these two systems okay so first part will be really the you know static problem so let us do one by one let us say when my springs are in parallel so I have k1 and k22 springs are there okay now what happens remember if I give it a displacement that means at an instant of time t if I try to draw the free body diagram of this mass then what happens in this case displacement will remain same for both the springs right so what will happen that I have spring force equals to k1x here spring force equals to k2x okay so ultimately now if we look at the free body diagram I have k1x k2x and the inertia force which is mx double dot so mx double dot plus k1x plus k2x equals to 0 so I get the k1 plus k2 as my equivalent spring constant okay so when it is in parallel arrangement then I know k equivalent equals to k1 plus k2 okay the reason is very simple because displacement has to be same on both of these okay so therefore what is the natural frequency of vibration I can get that what is the maximum velocity maximum velocity is going to be equals to xm omega n where xm was given xm was given by 0.04 meter right so 40 millimeter was the you know pull of this mass downward so that is the maximum velocity and maximum acceleration you know amplitude we can also calculate through this xm omega n square so it is a very simple problem next problem would be when they are in actually in series firstly why it is a single degree of freedom system because I have only one mass remember there is no mass right here so it does not have an inertia okay so I have only one mass that is going to take the inertia force but remember the logic will be very different than the previous problem so what will happen if I give it a displacement x that displacement is now going to be shared by spring 1 and spring 2 so you can write that x1 plus x2 must be equals to x right such that x2 is nothing but x minus x1 so therefore if you look at from the equilibrium perspective of force your force remain constant here see this is going to see the same force force is going to be same in both the springs so ultimately if you compare with an equivalent system which I am trying to find out what is the spring constant of this equivalent system then k equivalent right that is the spring force mx double dot that is the inertia force right now look at the static condition here my force must be same on the both springs so what we have I can easily write this equation k2 x minus x1 should be k1 x1 as I said x1 plus x2 should be equals to total x okay so therefore I can actually solve x1 in terms of the total x right now remember the force is equals to k equivalent multiplied by the x so in this system right so therefore what I have is I can now write p equals to k1 k2 k1 plus k2 x so therefore this is actually k equivalent k1 k2 divided by k1 plus k2 that is my k equivalent so I can write it in this form that 1 over k equivalent equals to 1 over k1 plus 1 over k2 okay so ultimately what is the natural frequency of vibration that is going to be k equivalent divided by m which we calculate from this then we have the velocity and then we have the acceleration okay so let us say we have completed the spring mass system now so far whatever we have studied if there is any discussion we can do and we can simply do a very small problem right now at this stage okay and then we can go to you know another type of single degree of freedom system that we call generalized single degree of freedom system and we are going to study that from now onwards so let us say if we have to discuss you know just couple of discussions we can do good morning your question is what is the maximum value of theta right in simple remember simple pendulum if you are talking about that problem we will come to that next do not worry simple problem pendulum here our consideration is always small oscillations okay so whatever we are studying now these formulations are valid for small oscillations pendulum can also have large oscillations right now when it is large oscillations then your natural frequency is going to deviate then what you solve based on the small oscillation consideration remember you can very easily show that for the simple pendulum problem let us say you have the length L and the pendulum is attached here okay so the form would be you know their gravity will come into play and we can show that it is actually square root of G over L right so that is coming into play okay so that is the form now that is only from the small oscillation if you go to the large oscillation then you have to take the because weight weight is going to balance your inertia in this case component of the weight is going to balance the inertia translator inertia translator inertia will be balanced by the component of the weight now when you are taking the component of the weight we have to be extremely careful because there is w sin theta involve w sin theta you are going to equate with the translational inertia force that is how you get the equation of motion isn't that right so now question is what is that theta so for small oscillation sin theta equals to theta but for large oscillation it is going to deviate okay you can look at any standard text boot it is not a you know any fancy problem or what so ever okay is that clear so we are really talking about small oscillations here so sin theta tends to theta that is the assumption we are making okay okay yes sir next question sir hello hello yes sir in in in this question why only consider simple harmonic motion and other motion when we are considering there is nothing simple harmonic simple harmonic because as I said force is proportional to displacement as long as you have force is proportional to displacement you remember the spring force right kx why are you saying kx at the beginning just answer that why it is k times x as long as force is proportional to the displacement it is always going to take the form of that force so that force no is actually you know when you are looking at the oscillation right force is changing the direction right and it is able to make the system oscillate from some positive value to the negative value okay but as soon as if you change the k right if I say now k is dependent on x that means you do have a non-linear spring then it no longer will be able to follow that routine thing that is plus to minus that is not going to happen so your solution will be more complex then you are going to have non-linear effect in your solution so you are going to give higher order harmonics okay and that is a complete non-linear problem even in contact mechanics when you look at the vibration with contact problem they are highly non-linear in nature and you are going to generate higher order harmonics okay it is not just f it is going to be 2f, 3f, 4f so on so forth it is not the first frequency but other frequencies are going to be generated okay so concept the harmonic simple harmonic motion that harmonic term is coming because f is equals to kx that is a force is proportional to displacement and k remains constant do not forget k is not varying with x okay clear. Yes sir thanks. Please explain by giving some practical examples regarding critical damping over damping and under damping. We are going to go to that just hold on which will be on the next session so your question is under damped critically damped and over damped system and if you look at the slides which are already in posted in the module the last part will be that one okay and you can if you are asking me to example very quick example I do not know if you are sitting in a room which has a door and that door has a automatic door lock are you in a room which has a automatic door lock just look at that that is a critically damped system or over damped system okay because your door will never be able to you know go beyond its equilibrium configuration okay so we will come to that just hold on to that that is the discussion on the next session okay.