Sinusoidal Cycle: Tutorial
So, hello and welcome back to another lecture on introduction to soft matter. Today what we are going to do is we are going to look at some example problems. These example problems have been designed so that we can actually end up revising some of the materials that we have been discussing. So, it will be a very good work through some of the important concepts that we have already discussed.(Refer Slide Time: 00:53)So, let us take a look at another problem number two. So, the question we have is that for a strainhistory, so the strain history is given to you as here I am just going to use s again but pleaseremember that this is not a Laplace transform variable, but this is since we are going to use timein so many places. So, this is just variable represents time.So, we have . So, we have a sinusoidally varying strain. And the question is fora given this given strain history, calculate the work done per cycle. So, now let us try to solvethis particular problem. So, this is given to us as the input and recollecting from the material thatwe have already discussed for sinusoidal strain and sinusoidal stresses is that this input willcreate an output in the strain sorry, for the stress, which is also going to vary in time.But now, the expression for this we will be given by has the same meanings that we have used in the class before.So, we are already familiar with these variables. So, now the work done on a material per unitvolume is given by W =This is basically the, if you this looks like the force into displacement form right and we areintegrating over the entire strain produced in the system. But our question is to find out the workdone per cycle. So, it means per cycle of this sinusoidal input, so we would like to be able tointegrate this in time. So, if you want to now find out the work done over a time interval, over aspecified time interval then my W will now become this integral over that time integral.So, let us say that time in the time interval is, so my work will now be, has to be integrated wherethe limits are this time interval. So, let us say this is some time t and then till one unit cycle, so There represents the time period. So, this is Now, I want to integrate this over time,but, so I would like this integral to be such that this integration can be easily performed right. So,what I will do is you can change the integral to this particular form where s represents a timevariable.And let me also clarify that where, where t is the initial time. And capital T is the time period,time period of the sinusoidal oscillation of the input. Such that so, so, T is basically equal tois the frequency of the input and without loss of generality because I am interested in integratingthis over a cycle, without this law, any loss of generality, we can set the small t to 0.Now, this particular formshould already be known to us because the input is known to us. . But we also know sigma as a function of time. This we have written itdown here, so for this integral, both these two terms are known to us.So, we can write down the expression for work as an integral, but this time I am going to set thesmall t as 0. So, this is just 0 to capital T, capital T, once again being the time period ofoscillations. And here I will have if you multiply all of these you will haveSo, we have to evaluate this particular function okay.So, let us just bring this in inside this cosine and we see that. Now, were going to perform thealgebra in a second. But you can actually already anticipate the answer by just looking at thisparticular form.Now, see here what we have is a And if you know your trigonometry, well youknow that when we integrate this over one time period, this term will actually have to vanish.But here is a term where we have , I think I have left out the omega sorry and multipliedby cosine square. Now, over a time period, the cosine square obviously, because its a square, itsalways positive and you have So, this term will not vanish, but the term associated with will actually vanish. So, in thework done, only will be contributing. And you can see that without even doing the fullalgebra. But let us go through this set of calculations.(Refer Slide Time: 10:02)Now, observe that the Now, when we integrate this term, these are all constants with respect to time. This is the onlyfunction here which is which is the only variable which is a function of time. So, my integrationis actually becomes rather straightforward, because all these other terms which are independentof time, you can simply bring out and when you integrate this you will be left with sine squareomega s and this has to be evaluated over one time period plus this particular term over here,which again I will write the constants outside and I suggest that using our trigonometricidentities, we convert this to this particular form.Now, note that this is this particular function if you if you go ahead and evaluate it over a timeperiod actually results the answer comes out to be 0, 2. And for this other functionalform if you evaluate that oneo, this is ouranswer that we are seeking right here. So, as we had when we had started out with the problem,and we had not even finished the full integration, it already said, just by using our trigonometricunderstanding of trigonometry, that this first term is going to drop out and the second term is theone which is going to contribute to the work done.And surely, that is what we see, we only have the term here and the term does notcome. Now, also notice that there is no explicit dependence on in here. But, the dependenceon omega is contained in the dependence of . The other important thing is that in the verybeginning when we had discussed sinusoidal oscillations and work done.Then we had said that the work done on a solid body is fully recoverable. That is what pureelastic or classical elastic solid is supposed to do, when you do work on it over a sinusoidal cyclelike this, it is the work done is fully recoverable. So, in a given cycle, there is no work net workdone. Whereas in a fluid, there will be work done in a cycle and that work done is eventuallygoing to be dissipated in terms of in the form of heat.And the final expression that we have found, this suggests just that the storage modulus does notcontribute to the work done in a sinusoidal cycle, but its only the with the loss modulus which isimportant, which means that that loss modulus. Now, you can see that the naming is soappropriate when we say is the loss modulus as you can see this one more motivation ofcalling it that that this work done that is done over a one time period is now going to be lost interms of heat.So, overall, this satisfied satisfies our intuitive understanding of the viscoelastic materials alsovery nicely. And also, please remember that these are again, limited to the class of the linearviscoelasticity that we had all these. So, all the assumptions that we had already used in comingup with these expressions to begin with or still apply here. So, with that, we are done with oursecond problem. And we move on to one more a third problem that we will do. I will write downan expression and the task is to figure out if the expression is correct.(Refer Slide Time: 17:02)So, evaluate whether the following relationship is possible and the expression is0. So, to figure this out whether this expression is indeed corrector not, we will go through the process we have sort of seen before. So, we have seen before asyou have seen earlier that for an input a sinusoidal inputWhen you apply that then you end up get getting an expression for the loss modulus as This is something that we have already seen before. So, in essencewe are asking whether this particular integral is the same as this particular integral over here.(Refer Slide Time: 19:16)So, to evaluate this let us go back to our basics and we will write down the expressions for thestress that we had found. And we know that the stress can be written as this particular functionform , if the stress relaxation function is known and this isequivalent to , we had assumed a particular form for the stress relaxation function. So,we will assume that G(t) sorry we will assume that decreases to 0 as time goes to infinity.(Refer Slide Time: 21:21)Now, so using this particular expression here. So, we have already said we will use thisparticular formulation when you insert this over here, the derivative will only account for thesecond term, so that the derivative of So, what is it that we are really looking for here. Now, the strain is the input. And we are tryingto figure out the expression for the stress from this given that you know the functional form forthe stress relaxation function. So, we now have to assume some form of time dependence for ourstrain here, which you can take it as either a sine omega or a cosine. But for generality, what wewill do is we will just go into the complex formulation.That just makes life easier for us. So, to check generality, we will use the complex formulationwhich means we will use . So, let us go ahead and replace that in the aboveexpression. So, that implies that my stress which is the output here becomes0 as t goes to infinity.(Refer Slide Time: 24:52)So, basically this functional form has to be such that, this is here, by the way. Now, if this functionalform is valid for the stress relaxation function, then you can make an additional observation andthat observation is that the derivative of this also goes to 0 as time goes to infinity.Now, you are also to make an additional you have to make an additional assumption that thisintegrals will converge when we take the limit of t tending to infinity, but assuming that theintegrals will converge, we have made certain assumptions about the functional form of thestress relaxation function. The reason we are making these assumptions is because we are goingto look at a system where the transients have died out and we will look at the long term timelimit, which is very similar to the formulation that we carried out in the class earlier.So, we are going to do this integration such that this 0 to t this t will be replaced by infinity. Inthe case, in that case, assuming that all the integrals do converge, so that assumption is also thereI am not writing it down explicitly. But in the long timescale with all the transients have dieddown, I should be able to write if the integrals converge.I should be able to write down the output which is the stress as. So, after transients have died out,I should be able to write my stress as some complex modulus multiplied by my input such thatthis complex modulus only depends upon omega and not on time. So, basically what we aresaying is that this G star is equivalent to this particular form over here, such that the integrationis now going to be performed over 0 to infinity.This is by the way very similar to the thing to the formulation we had performed in the class. So,But now we are going to perform it this integralintegration from 0 to infinity.So, this implies that we already know that the I do not know this particular form because this has not been given to me. So, we willleave it at this, but this is exactly what we are after. So, if you remember the original question,we had asked if this particular functional form is possible. And what we did is, we have alreadybeen familiar with one of the functional forms.But we go we went ahead and we chose a different representation for our stress. And we carriedout the exact same sequence of steps that we had carried out in the class before for getting ourevaluating our complex modulus. And when you do the same set of steps, just because you chosea different expression for sigma t to begin with, you will end up with this particular expression,which we know is equivalent to the other representation as well.So, we know that this relationship is valid and we also know the conditions under which this isvalid. So, the purpose of this particular exercise was to show you that there are different forms ofthat can be written. And it is just because if you encounter a different form,you should not get confused and think that it is incorrect. So, with that we have ended ourproblem three.
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