 Hi, I'm Zor. Welcome to Unizor Education. Today we will solve a couple of problems related to oscillations. Now this lecture is part of the course called Physics for Teens presented on Unizor.com and I do suggest you to watch this lecture from this website because every lecture has very detailed notes with all the formulas derived much more nicely properly than whatever I'm doing here at the board in more details probably and there are certain verifications like checks which I don't really do here but I do it in the notes so you will be absolutely sure that the formulas are correct. Plus there is a prerequisite course called Mass for Teens. Mass is mandatory course for the physics and especially calculus, vector algebra and some other aspects, complex numbers. By the way today we will actually talk about complex numbers as well. Okay, so I have two problems. The first problem is related to a lecture where I was deriving a formula for oscillation of, if you have something like this, you have a spring with an object. Spring has certain characteristics like elasticity for instance, this object has mass obviously so that's basically enough to describe the oscillations outside of any influence, any force but now there is a force actually which is periodic. It's like basically you have something like a swing and you are pushing the swing with certain periodicity. So this is angular speed, sometimes it's called angular frequency and obviously its own oscillations, natural oscillations without the force as I was explaining one of the lectures was with angular speed equals to K over M square root. Now this is all derived previously and in the lecture I'm referring to right now I actually derived the oscillation of this object as a formula, as a function of time from if this is some neutral position and X of t is deviation from the neutral position and the spring can go both ways either stretching or squeezing by itself and obviously there is a force which is also involved and then I have derived this formula which I will put here, F0 divided by M omega 0 square minus omega square cosine omega t plus g cosine omega 0 t minus phi. Okay so don't get scared, I mean if you have just got to this lecture and I'm not going to explain where I got this formula from, go to the lecture which is called force oscillations I think it's force oscillations number one because there are two of them where this formula is derived properly without any kind of shortcuts. So I'm using it as is to basically describe the problem which I have. Now the F0 is this coefficient, now what are d and phi, these are defined by initial conditions because the spring can be initially stretched to certain distance or the object can be pushed to give it certain speed initially. So these initial conditions are defined. For instance in this particular case since we have this force involved we can actually start from an initial position at zero at neutral point and initial speed at zero and then the force will basically start acting and the whole system will start oscillating. Now my question is if I did not have this that's the periodic function, that's the cosine. If I don't have the force, its own oscillations are also periodic because there is a cosine here. My question is, is the sum of these, the oscillation which is based on both its own properties and the properties of the force are these oscillations periodic? That's a very important question. So that's my first problem. Are these movements represent periodic oscillations? Okay, so first of all you have to basically think yourself. You can pause the video and just think about this yourself. Now how can I basically answer this question? Well what's important is, let's forget about the details what's important is this and this. So the periodicity of this T1 is equal to 2 pi over omega. If omega is angular speed which means how many radians per second then the period is obviously supposed to be 2 pi. The period means when the full cycle actually is done. So the full cycle is 2 pi. If you divide it by distance, divide by speed you will get the time. This is the time when the whole period 2 pi will actually come into play. Because without omega the cosine T has the period 2 pi. But with omega it has 2 pi over omega. That's obvious. Now that's one period. That's the periodicity of this member. The periodicity of this member is 2 pi over omega 0. So my question is when the sum of these will be repeating itself. So this component repeats itself after this amount of time. This component repeats itself after this amount of time. Now if I will come up with certain amount of time which is multiple of this and multiple of this then the sum would actually be of this function. Then this period which is multiple of this and multiple of this will actually be the periodicity of the sum of these. Because if for instance I have certain T which is equal A times T1 and B times T2 where A and B are integer numbers so after the T time which is after A times T1 this will repeat its own value. But at the same time this is an integer number of T2 so this one will repeat its value. Which means their sum will also repeat its value. So all I have to do is to find such a period which is the multiple of this and multiple of this of this and of this. Well, this is not always possible because if you will convert this thing it will be T1 divided by T2 is equal to B over A where B and A are integer numbers which means this ratio is supposed to be a rational number the result of dividing integer by integer. If it's irrational number which is possible. I mean you don't... these omega 0 are completely independent angular speeds so they are not necessarily... their ratio is not necessarily a rational number it can be irrational. For instance this can be a pi and this can be 27. Obviously their ratio is irrational. Their ratio is irrational. In which case there will be no such period which is the multiple of this and multiple of that. So the example which I have presented in the lecture where I derived this formula I was actually using two and five. Two and five obviously their ratio to fifths is a rational number and the graph which was presented in that lecture represented actually the periodicity. It was not like a smooth sinusoidal it was something like this but it's still periodic function because if you will start from here to here this piece is repeated itself. So the answer is if the angular speed of its own oscillations of this spring with this mass is such that it's in a rational ratio with frequencies between this and this and their ratio is actually equal to ratio of angular speeds because two pi obviously is cancelling out so this is the same as omega zero divided by omega. So if this is a rational number then you can expect the periodicity of the oscillations. If not this thing will not be periodic. I mean one of the pieces maybe will be stretching then make it a little smaller and then maybe stretch to another it will be more chaotic let's put it this way it will be much more chaotic. If this ratio is an irrational number it would be much more chaotic. That's it for the first problem. Now my second problem is it's kind of a summary of whatever it was before so it's like a lecture but I present it as a problem. Now we discussed how the oscillations are happening just by themselves without any external force. Then we introduced the concept of a friction so what happens with oscillation if there is a friction which is not part of this particular problem. Then we were discussing in another lecture the concept of viscosity when there is some kind of a resistance to a movement proportional to the speed it's like you are making certain oscillations inside the water if you're moving slowly it doesn't really present much of a resistance but the faster you go the faster the oscillations are the stronger resistance of the water will be that's what viscosity actually is all about. And then we were discussing what happens if there is some kind of an external periodic force like with a swing. Now what this problem is it's a combination so what happens if you have a spring you have a viscosity of the environment where the oscillations are happening and you have the external force so it's like a combination of all the previously considered cases. So let's do it step by step. So let's say you have elasticity you have mass you need a viscosity now what is viscosity? The function which is the result of this resistance of the environment discuss resistance it's supposed to be proportional to speed and the coefficient of proportionality I call it C and the speed is the first derivative of the displacement from the neutral position so we're assuming that we are along x-axis this is at zero in initial position zero but let's just assume that the position is zero and the speed initial is also zero it's the external force which starts the oscillations but I have to put minus sign because if the speed is in one direction the resistance, the force is resisting which means it's directly opposite so that's why there is a minus sign same thing very analogously to a spring to the force of the spring the force of the spring is dependent according to the Hooke's law it depends on the displacement how far we stretch the farther we stretch the stronger the force but again if we stretch the positive direction the force will be directed negatively and the coefficient of proportionality is elasticity and it depends on the position so that's the second force which acts on the object now the third force that's this one external force called fx external so that's f0 times cosine omega t where omega is given angular speed so all these three forces are acting at the same time now my problem is derive the equation of motion and solve it now I will derive equation of motion and I will talk about how to solve it the complete solution is in the nodes because it's kind of lengthy and most likely I will make a mistake if I will start reproducing it again at the board so I would refer you after this lecture to the nodes to see exactly all these calculations in all their small detail okay, so first we have to come up with an equation of motion now it's based on Newton's second law which says that mass times acceleration or instead of acceleration usually it's what are a but I will use the second derivative of the displacement, right first derivative is speed and the second derivative is acceleration so this is supposed to be equal to the total of all the forces acting on this object which means it's equal to minus cx first derivative minus k function and plus f0 cosine omega t now what is this? well it's a differential equation of the second order now it's not a homogeneous because homogeneous is if you didn't have this member it would be homogeneous which means if you found function x of t then the function x of t multiplied by any number would also be a solution so without this if x of t is solution and we multiply x of t by some constant obviously this constant will be multiplied here and here and the function multiplied by a constant will also be a solution so how to solve these non-homogeneous equations if a simple function can be just removed from it and it will be homogeneous well I did address this before and I'll just basically repeat the whole thing here is how we will do it now this is a repetition it was in one of the lectures before first of all let's just change this to this I will put all x related things here so I put all the x related things to the left and divided by this coefficient so I have coefficient 1 at the second derivative so that's why it's divided by m so this is my differential equation now let's assume that there is a function which is a solution to this let's call it x1 and then there is an x2 which is also a solution so we have two different functions what if I will subtract one from another well the first and the second derivatives are linear so derivative of a sum is equal to sum of the derivatives or a difference between functions so if I will subtract x1 and x2 which is equal to the same thing this will cancel out and I will have x1 minus x2 second derivative then c over m the first derivative of the difference between functions and then the functions themselves that would be equal to 0 which is a homogeneous equation which means that the difference between two solutions to non-homogeneous equation the difference between two different solutions is a solution to a homogeneous equation so from this follows the following to get all the different solutions for a non-homogeneous equation we can have only one of them particular solution to non-homogeneous equation and then add all the homogeneous equation I mean all the solutions to a homogeneous equation so if this is fine I'll continue if not you can just stop here and I would refer you to the first lecture actually about forced oscillations where I explain it again and it's basically in writing as well and it's in the notes for this lecture as well so again to solve the non-homogeneous equation it's sufficiently to solve completely homogeneous equation when it's equal to 0 and all the solutions of these can be added to one particular solution of this to get all solutions to non-homogeneous equation okay now so we have two problems number one problem is solve homogeneous equation number two is it needs totality which means to get all the solutions to a homogeneous equation and the second problem is find just one particular solution to a non-homogeneous when there is this function on the right so with a 0 it's a homogeneous with a function it's non-homogeneous it will try to find all the solutions of homogeneous equation when it's equal to 0 and then try to find a particular solution to a non-homogeneous so if I have a homogeneous equation how can I solve it? well, people obviously did think about this and they have, you know come up with certain very good results but first of all this is homogeneous it's linear of the second degree because there is a second derivative here now first of all people have proven that it's enough to find two different functions and the linear combination of these functions would represent all linear combinations and would represent all the solutions to this particular equation again, we did address this before it's basically proven in mathematics you can just take it as given or you can just go into the differential equations theory and basically get it from there so if there is such an equation so all the solutions can be basically obtained two particular solutions by combining them in all the different linear combinations why two? because it's the second degree if it was a third degree we would need three independent solutions etc now when I'm saying independent it means that one solution should not be a multiple of another solution because that would be obviously only one because we are edging all the linear combinations and we will not have a multiplicity of solutions and how can I find two independent solutions? that's easy actually because I will do the following I will assume this let me see maybe this may be a solution now, why am I so smart to suggest it? for a very simple reason because the first derivative of X is gamma times e to the gamma t and the second derivative is gamma square and if I substitute it here I will have this multiplier here, here, here this is zero so I will just cancel it out and what will remain? it will remain gamma square plus c over m gamma plus k over m equals to zero right? so if I will put this, this and this into this equation my e to the gamma t would just be here, here and here I will just cancel it out because it's never equal to zero as we know e to the power of gamma t for any gamma m t this is not zero so I just cancel it out divide everything like this and now by solving this quadratic equation I will get two different roots for gamma and that would represent two different functions and combination, linear combination of these two different functions one of them would be e to the power of gamma one t where gamma one is one solution x2 of t would be e to the power of gamma two t and then any linear combination of them where a and b are any numbers would be a solution and that would actually constitute all the solutions to this quadratic to this second order differential equation so how to do this again it's just quadratic equation I'm not going into all these calculations they are presented in the notes for this lecture and again that's kind of a trivial thing which I'm not going to spend any time in front of the board so we have found a general solution to a homogeneous equation there is one little detail here you see this is not necessarily two independent solutions because sometimes you might have the same solution it's like a I don't remember how it's called when two roots of this quadratic differential equation are exactly the same for instance if you have something like x2 is equal to 4 no sorry is equal to 0 for instance then you have x0 and x0 you don't have more than one equations so if this is a full square maybe x-4 square is equal to 0 which is x2-8x plus 16 now this also has only one solution 4 so it's like a double root now in this case I will not have two independent solutions and they have to resort to something else and I did actually explain how to deal with this in the lecture related to free oscillations the first lecture about free oscillations here just for simplifying my work I would just assume that C is small enough to let the oscillations happening now why C is supposed to be small enough C is a viscosity if viscosity is very high and then I let's say push my object from the equilibrium I stretched a little bit using the force well it will go then it might actually start going back very slowly and it might not even reach back the neutral point it will just be slowly approaching and never going above it and never actually oscillating with a small C oscillations will happen with a small viscosity so let's assume that the viscosity is small enough and if it's small enough then over M square minus 4K over M which is discriminant of this would be negative and if it's negative I will have two complex roots and from the complex roots I will have sine and cosine basically and again I did explain it in more details in the notes where the exact solutions are actually presented so this is how we solve the general how we get the general solution to homogeneous equation fine that's done here it is x1 and x2 can be found from here gamma1 and gamma2 are complex numbers because this is negative but from these complex numbers we are getting only the real part again you remember probably the formula the Euler's formula about what is actually e to the power of ix that's my calculations depend on it okay that's another thing you need to know complex numbers from the math okay now let's go to the second problem the second problem is how to solve how to get one particular solution to non-homogeneous equation which is written here now to find this non-homogeneous to solve this non-homogeneous equation I also have to assume something about x of t and maybe I will be smart enough or lucky enough to get a solution to this thing alright so the idea here is basically exactly the same what we have to do is we have to find some kind of an expression some kind of a formula for x of t maybe with certain parameters and try maybe I will be able to manipulate this formula in some way with parameters to satisfy this equation and the way how I'm just thinking about approaching it this is the core sign now the derivative of sin is a core sign the derivative of a core sign is a minus sign so I think it's reasonable to attempt to search for a particular solution in a form let me try, maybe with certain a and b this function would be a solution to this equation now why well again because this is core sign and sign is the first derivative from the core sign it's a sign, from a sign it's a core sign from this it's also second derivative so it's all signs and core signs and nothing more and maybe with these coefficients I will be able to make this an identity it's reasonable so let's try so the first derivative of this function is derivative of core sign is minus sign so it's minus a and then would be derivative of internal function and that's my derivative of this derivative of this would be b derivative of a sign is a core sign but then there is an inner function so also omega core sign omega t second derivative derivative of this minus remains omega remains from the core sign it's a wait a moment this is supposed to be sign it's my mistake, sorry from the sign derivative is a core sign and another inner function so it would be square here here from a core sign it's minus so it's minus b omega sign of omega t and inner function so that would be square as you see now we have to substitute these three into this and I will have signs and core signs with different coefficients obviously if I have something with signs and core signs and it's supposed to be identity so all coefficients from this related to core sign which is this, this and this must be equal to f0 over m so a plus b omega minus a omega square that's the core sign right? if I will but this should be multiply my first derivative it's supposed to multiply by c over m so it's c divided by m here my my free member should be multiplied by k over m and my second derivative is without any multiplier so this is all coefficients at core sign and it's supposed to be equal to f0 over m now, sign there is no sign here so signs here should be equal to 0 so it's again it's b but I have to multiply it by k over m here minus a but that's c over m a omega and second derivative without so it's minus b omega square and that's supposed to be equal to 0 because there is no sign here now this is a system of two equations with linear equations with two unknown a and b very easy to solve lots of calculations but it's simple it's tedious but it's simple and that's how I can get a and b now getting a and b it means I have got one particular solution for my equations now having this particular solution and having a general solution to a homogeneous equation with oscillations without the force but in the viscose environment knowing both of these solutions I'll just add one solution which I have found this concrete one and the general solution let me just write the formulas for you and the general solution which I have found for damped oscillations in the viscose environment and I will get an interesting formula I will get x of t is equal to a particular solution which I have found a cosine omega t plus b sin omega t that's which I have just found from these system of linear two linear equations for a and b plus general solution to damped oscillations e minus ct over 2m cosine of omega 0 t minus y instead of 0 I will put homogeneous which has its own formula so this part this this part represents the oscillation in the viscose environment important thing is here so it's kind of periodic I should put internal it's kind of periodic because there is a cosine but the amplitude is diminishing as the t grows because of this exponential multiplier so the graph of this thing would be so instead of plane cosine plane cosine is this with equal lengths now this multiplied by this that would be a different picture that would be this picture it will be diminishing amplitude and this represents damped oscillations inside the viscose environment now this represents the results of oscillations on the force, external force and what you see here is that it's a sum of these and this component is getting less and less and less and this is basically periodic with a periodicity of omega well, 2 pi over omega and that remains basically a relatively periodic movement so on this periodic movement related to the force we basically superpose the additional oscillations because of the system has its own spring, its own characteristics but this part is getting smaller and smaller so if I will add to this I will add something like this I don't know how to put it let's say this the result of this would be some kind of a chaotic movement in the beginning but then less and less it would be chaotic and it would be more and more like a plain sinusoid because of this part and that's very important in the environment which has no viscosity and that was the subject of lecture force oscillations number 1 we might have the periodic movement if my ratio that's the first problem which I have presented today if my ratio between the frequencies of its own oscillations and the oscillations of the force are in rational ratio if their ratio is a rational number but in any case it will be basically all the time both components will be important even if it's not periodic but the component related to its own oscillations would participate in the movement in this pseudo chaotic movement and the periodicity of this function will also have a picture if we are dealing with a viscous environment which basically slows down its own natural oscillations but the force outside force remains the same so eventually its outside force will play the most important role and the role of its own spring would be actually less and less effective ok now let me just refer you again to notes in this particular case it's important because there were a lot of calculations I could not present all of them on the board in one lecture so I put everything in writing and I do suggest you to go to this lecture which means go to unisor.com courses called physics for teens then the part is waves and inside the waves part there are many different kinds of waves of oscillations this is the mechanical oscillations problem number 4 so the notes for this lecture are much more detailed than I have presented here just because of the lack of time and again most likely I would make a mistake but the notes are correct I checked them out so I do suggest you to read them that's it, thank you very much and good luck