Damped Harmonic Oscillations | Equations of Motion | Solutions

if you have a spring mass system that you place in an external medium that can provide some kind of a friction or a drag in that case if you displace the spring mass system from its equilibrium then the kind of motion that you see one of the possibilities is that you will get oscillations that will slowly gap up over a period of time the amplitude of these oscillations decreases over a period of time this is happening because in the presence of the external medium the drag force is causing dissipation of energy of the system now the spring mass system is a very common example of a harmonic oscillator in ideal situations the harmonic oscillators execute you probably already know it simple harmonic oscillations but when we place a simple harmonic oscillator in the presence of some kind of a dissipating medium where there is friction or drag involved then you end up getting a damped harmonic oscillator and one of the possibilities of motion is right in front of you so in this video i want to take a look at the nature of the forces of a damned harmonic oscillator and i want to obtain the equations of motion solve them and see what possible motions are there so i'm going to talk about systems without any damping under damping critical damping and over damping so let us start so the sprigma system is not the only example of a harmonic oscillator you can have other systems like a pendulum executing oscillations with the small angle approximation or maybe something else but i'm going to use the example of a spring mass system placed in some kind of external medium to represent the damp harmonic oscillator that i am talking about so let's suppose that you have some kind of a point mass m ah that is attached to some kind of a spring system having a spring constant k and this entire system is placed in external medium that involves some kind of a drag so whenever an object is moving in this medium it experiences a drag force now the motion of this system mass m is along one dimension so i'm representing this by let's suppose the x axis now if i displace this mass m by some kind of an amount let's suppose x from the equilibrium configuration then in that kind of case the forces that are going to act on this mass m are two forces the first force is of course the restoring force due to the spring so the restoring force due to the spring that acts on the mass m when it is displaced from the equilibrium is directly proportional to the displacement and the constant of proportionality is the spring constant it acts in a direction opposite to the displacement now this is a harmonic oscillator but if i place this setup in external medium which can cause drag whenever motion is happening then usually majority of these circumstances the drag force that acts upon the mass m is seen to be directly proportional to the instantaneous velocity of the object whatever be the instantaneous velocity the amount of drag force experienced by the mass due to this medium is directly proportional to the velocity and let's suppose the constant of proportionality is some kind of a positive constant b where b can be let's suppose the drag coefficient or something like that so in a sense these are the two main forces involved in this kind of attempt harmonic oscillator one is the restoring force due to the spring system and other is the drag force due to the medium present now if i represent all of this into the force equation applying the newton's second law it simply leads to the equation that whatever the net forces are acting on the system is going to create some kind of an acceleration and since the acceleration is happening along the x axis i can represent this as mx double dot what is x double dot it is nothing but the second order time derivative that is the acceleration along the x axis now what is the net force first i said the net force is the restoring force kx acting against the displacement so there's a negative sign here and then there is a drag force b v acting against the motion so there is a negative sign here so if i write down this equation where i bring all the terms together this simply becomes mx double dot plus now b v what is v v is nothing but x dot the time derivative of x right so this can be written as b x dot plus k of x is equal to 0 if i divide the entire equation by m this simply becomes x double dot plus b upon m x dot plus k upon m x is equal to zero now to make my calculations easier i'm going to uh substitute these values with certain terms so what is b here b is nothing but the drag coefficient the constant of proportionality in the drag force m is the mass of the object let's suppose that b upon m is equal to 2 beta a beta is some kind of a constant which is equal to b upon 2 m why i am writing 2 beta well it will simplify our calculations later on as you will see and let's suppose k upon m is equal to some constant omega naught square what is this constant omega naught i'm sure you can guess but we will derive it uh in the due course so using these substitutions let's suppose this is a substitution number one and substitution number two the equation of motion therefore becomes the second order derivative of x being the displacement plus two beta x dot is dx upon dt plus omega naught square x is equal to 0 this is the equation of motion of this kind of a damped harmonic oscillator system uh and by solving this we can get some idea that what sort of motions are possible now if you are familiar with differential equations then you may know that equations that look like this where the coefficients 2 beta omega naught these are all constants they depend upon the system equations that look like this have solutions of the form e to the power some parameter alpha t multiplied by some kind of a constant a let's suppose so by guessing this kind of a solution if i substitute this in let's suppose this is equation number 3 then what shall i get let's try that so if i substitute this guess of a solution into the equation number three d upon so d2 upon dt2 a to the power e alpha t where alpha is some kind of a parameter whose value i am interested in finding out plus 2 beta d upon dt a e to the power alpha t plus omega naught square a e to the power alpha t again is equal to 0 then a is a constant it comes out i'll get alpha square e to the power l party here i will get in the second term beta multiplied by a alpha e to the power alpha t plus omega naught square a e to the power alpha t is equal to 0 now if i take the common terms out so what are the common terms here well a and then e to the power alpha t is a common so if i take this out in the bracket i am left with alpha square plus 2 beta alpha plus omega naught square is equal to 0 now a here is a constant and e to the power alpha t is not necessarily equal to 0 which brings us to the condition that the potential solution will come from alpha square plus 2 beta alpha plus omega naught square is equal to 0. so what is this if you take a look this is just some kind of a quadratic equation and i'm sure that you know how to find out solutions of quadratic equations right so the solution of the quadratic equation uh alpha is nothing but it comes from the quadratic formula minus b which is 2 beta here plus minus root over b square which is 2 beta square which is 4 beta square now i think it's going to become clearer why i took 2 beta in the substitution earlier so 4 beta square minus 4 ac where a is 1 and c is omega naught square this is 4 omega naught square and in the denominator i have 2 into 1 just 2. so this becomes alpha is equal to so 2 to cancel you have minus beta plus minus if i take the 4 out i'll get 2 which gets cancelled and i'm left with beta square minus omega naught square so the alpha parameter that i was talking about in the solution that i assumed have two possibilities right so therefore from what i can do is i can say that the solution of the equation number 3 can be written as x t is equal to some constant a 1 e to the power alpha 1 t plus some constant a to e to the power alpha 2 t where alpha 1 is equal to minus beta plus let's suppose root over beta square minus omega naught square and alpha 2 is minus beta minus root over beta square minus omega naught square so if i substitute these terms this simply becomes x t is equal to some constant a 1 e to the power minus beta plus root over beta square minus omega naught square t plus a 2 e to the power minus beta minus root over beta square minus omega naught square t the exponential to the power minus beta t term i can take out as a common term so this simply becomes x t is equal to e to the power minus beta t and in bracket i have a 1 e to the power root over beta square minus omega naught square t plus a 2 e to the power root over there's a minus sign here beta square minus omega naught square t so this is the nature of the general solution of the equation of motion that i was talking about so whatever equation of motion we obtain for the case of the damped harmonic oscillator this one basically equation number three its general solution is this point number four now before we talk about what kind of motion this uh point number four reveals to us uh there are certain situations possible here so there are four possible scenarios that can arise out of this solution the first scenario is if beta is equal to 0 now if you remember beta is nothing but b upon m b upon 2 m rather where b is the drag coefficient and m is the mass so in a sense beta has something to do with the nature of the drag that is happening it is basically giving us an idea what how strong the drag force is this is actually known as the damping parameter it gives us an idea about how much damping is going to happen because it not only depends on the drag force but it also has a inverse proportionality with the mass so if there is no damping if we have an ideal setup where there is no drag force no dissipation no friction involved in that kind of situation we have an ideal harmonic oscillator so the nature of the solutions are kind of obvious i am going to derive it of course but you will end up getting simple harmonic oscillations now the next situation arises when if you look carefully there is a relationship between beta and omega naught so let's suppose beta square is less than omega naught square in that kind of a situation let me give you a hint i'm going to derive these results later on you will get oscillatory motion but it will correspond to not shm but it will correspond to what is known as under damping when the amplitude dies out over a period of time okay and the third situation corresponds to when beta square is equal to omega naught square this is the situation where the oscillations will actually die out you will not even get a single oscillation so no oscillations and this is known as critical damping and the final possible scenario is when beta square is greater than omega naught square in this kind of a setup you will not get any oscillations so no oscillations however the system is going to take a little bit longer amount of time to come from a displaced configuration to its equilibrium configuration this is known as over damping now these possible scenarios can be written by a comparison of the beta term here and the omega naught term but i can also write them down in terms of the terms uh b k and m okay so depending upon which textbook you are reading sometimes somewhere is related in terms of the comparison between beta and omega naught in others it can be written as a comparison between b and let's suppose k and m now what is beta beta is equal to small b upon 2 m what is omega naught square it is equal to k upon m so let's suppose in the third setup beta square is equal to b square upon 4 m square which is equal to omega naught square which is equal to k upon m so this simply means that b square is equal to 4 k m that's it so beta square is equal to omega naught square corresponds to the drag coefficient b square is equal to 4 km where k is the spring constant and m is the mass similarly i can show that the second condition reflects b square is less than 4k m the first condition is of course where b is equal to 0 and the last condition is where b square is greater than 4 k m k being the spring constant m being the mass and b being the drag coefficient while beta being the damping parameter so under these four possible scenarios the general solution will reveal to us different kinds of motion so let us explore these different kinds of motion of the oscillating system that we have taken here for these possible scenarios so let's first look at the possible scenario number one where beta is equal to 0 so if beta is equal to 0 then equation number 4 simply becomes let me write it down here it simply becomes x t is equal to e to the power minus 0 which is just 1 and you will have a 1 here e to the power root over minus omega naught square so that involves an imaginary number so e to the power i omega naught t right and the second term involves plus a to e to the power minus again beta is 0 so root over minus omega naught square becomes i omega naught t now this is something actually we can simplify it by considering the fact that e to the power i theta can be written as cos theta plus i sine theta right if i use this this simply becomes a 1 cos omega naught t plus i a 1 sine omega naught t plus a 2 cos omega naught t plus i a 2 sine omega naught t if i combine the cos and the sine term separately i end up getting a 1 plus a 2 cos omega naught t plus i a 1 here should be a negative sign of course a 1 minus a 2 sine omega naught t now what is a 1 plus a 2 is just some constant that depends upon initial configuration a 1 minus a 2 is the same so i can write this as let's suppose the c cos omega naught 2 i see some real constant and let's suppose d is some complex constant sine omega naught t now of course we are interested in the real solution but even if i'm interested in illustration i can write the expression in this particular manner what is this it is actually a sinusoidal variation i can in fact show that the real terms can be simply written as some other constant a let's suppose cos omega not t plus some constant phi i can show that both these two expressions are actually equivalent to each other this result x t that we have obtained for the scenario where beta is equal to 0 that means there is no damping is the obvious result of a sinusoidal variation which is a simple harmonic oscillations so when you have an ideal harmonic oscillator setup no damping no dissipation of energy happening you end up getting a sinusoidal variation so you'll have some kind of a function that looks like this so here i have the solution for the oscillating system when it is displaced from its mean or equilibrium by some kind of an amount let's suppose a okay so the y axis here represents the displacement with respect to time x and this is the time axis so this is a sinusoidal variation all right this is the result that you get so when there is no dumping then the obvious result is simple harmonic oscillations now this was obvious enough this was quite obvious what about the second case scenario so let us explore the second case scenario where beta square is less than omega naught square so in the second case scenario when beta square is less than omega naught square or as i said this is equivalent to b square being less than 4 k m this represents a situation where i can rewrite the general solution point number four in a manner where i'm going to make an assumption here i'm going to assume that let's suppose omega naught square minus beta square can be written as a further term let's suppose omega 1 square then i can substitute these terms here what will i get if i substitute these terms with omega 1 i will get minus omega 1 in instead of beta square minus omega naught square so let us write down the solution then so for this kind of a scenario we will get x t of course the first term is e to the power minus beta t the further terms involve a 1 e to the power as i said beta square minus omega not square is equal to minus omega 1 square so from the minus term because there's a root over sign here i'll get an imaginary i omega 1 t a 2 e to the power minus i omega 1 t just take a look here i'll just get minus and then i omega 1 t so the general solution for this kind of a scenario is this now before i look at the um variation of this kind of a function i can simplify it further by making some assumptions you see a1 what is it is just a constant that depends upon the initial conditions right so let me suppose that this is equal to some other constant a e to the power let's suppose i uh some other constant theta and a 2 is equal to half of a let's suppose e to the power minus i theta if i make these assumptions then the form of this solution becomes e to the power minus beta t and then you will have half of a this will be common and then you will have e to the power i omega 1 t and then here you will have e to the power i theta plus e to the power minus i omega 1 t and you will have e to the minus i theta okay so let me write this in a simplified fashion so a e to the power minus beta t half all right and then you have e to the power i omega 1 t plus theta plus e to the power minus i omega 1 t plus theta right so if you have this kind of an expression you can make use of the identity let's suppose you have e to the power i x plus e to the power minus i x divided by 2 then some of you may already know that this is what this is cosine of x exponential to the power minus x plus exponential derivative minus i x divided by 2 is cosine of x so if i make that substitution here it simply becomes a e to the power minus beta t into cosine of omega 1 t plus theta this is x t so the displacement of the harmonic oscillator when this second case scenario is true the parameters are such that beta square is less than omega naught square is this particular solution now what is the mathematical meaning of the particular solution because it's a product of two functions you see a of course it depends upon the initial conditions but what about let's suppose this term this term is nothing but some kind of an exponential decay right but what about the cosine term cosine term is an oscillatory term right so what you end up getting is you get this oscillatory term which is further modulated by an exponential dk term you get an oscillatory term which is further modulated by an exponential dk term the graphical representation looks quite interesting so let me show you okay so if i displace the system by some kind of an amount at time t is equal to 0 let's suppose i displace it by a y axis here simply represents the displacement x with respect to time and i do not provide it with any kind of initial velocity in that kind of a situation these are the oscillations you get you see the cosine term is creating these oscillations the system is executing oscillations but the exponential decay is causing the rapid decay of the amplitude over a period of time these amplitudes are decaying or decreasing over a period of time in an exponentially decaying manner now in this kind of a setup since i gave the initial conditions of some displacement but no velocity it automatically turns out if you try to calculate it theta comes out to be 0 here so the oscillations are actually this oscillation is nothing but cos omega 1 t while the modulation that is happening this modulation is a e to the power minus beta t and here this is of course in the negative axis minus a e to the power minus beta t so you have these oscillations that are being modulated by an exponential decay so when you displace the entire system from its equilibrium when damping is there then the system starts oscillating but after some time its amplitude slowly decreases and it exponentially decays and it slowly slowly slowly decreases and with time the exponential decay term the system comes to rest this is the classic scenario of what is known as for the condition beta square less than omega not square under damping it is called under damping because you still have oscillations okay it is the damping is not so high that oscillations have vanished the oscillations are there but the oscillations have exponentially decreasing amplitude relationship with time so whenever the beta square term is less than omega not square term or the b square term is less than 4 km you end up getting a system that executes oscillations that die out over a period of time now it is interesting to note here that the oscillations are happening with the frequency omega 1. now if you compare this with the ideal harmonic oscillator where the frequency of oscillations was omega naught what was omega naught omega naught was nothing but if you remember point number 2 omega naught was just root over k upon m right omega naught was root over k upon m which is just the spring constant k upon the mass square root this was a frequency of oscillations the angular frequency of oscillations for the ideal harmonic oscillator but for the damped harmonic oscillator in the case of under damping the frequency of oscillations is omega 1 where omega 1 is what omega 1 is omega naught square minus beta square so omega 1 is actually root over omega naught square minus beta square which simply means that omega 1 is actually less than omega naught so the ideal harmonic oscillator has a angular frequency of oscillations omega naught and the damped oscillator has a frequency which is less than that of the ideal harmonic oscillator setup now the reason that these amplitudes are decreasing over a period of time is that the system is losing energy because of the friction involved or rather you can say because of the drag involved due to its interaction with the medium the system is losing energy the system is dissipating energy and the nature of this dissipation is also exponential because you can calculate the energy from let's suppose the amplitude square if you do an energy analysis you will find out that whatever the energy was there in the setup at the beginning let's suppose the energy is e0 then whatever the energy is there after a certain interval of time t if we do a calculation you can see that the this has a relationship of exponential to the power minus twice beta t so the graphical representation is just an exponential dk for what for the energy so energy if it was initially e naught it slowly ah decreases with an exponential curve as the system loses energy due to its motion now let us come back to the scenario number 3 where beta square is actually equal to omega naught square what is going to happen in scenario number three now in this setup where beta square is actually equal to omega naught square if you go back to the general solution then the terms will probably involve you know beta square and omega naught will cancel each other out and we'll end up getting e 2 0 now that will probably not give me the right answer so what i'm going to do is i'm going to solve the equation of motion again the point number 3 for this setup so let us use point number three here in scenario number three so the point number three is just equation of motion that is d2x upon dt2 plus 2 beta dx upon dt plus omega not square was there but since omega naught square is equal to beta square here i can write beta square x is equal to 0 this can be further simplified by writing d 2 by dt 2 plus 2 beta plus beta square x is equal to 0. now what i can do with the term in the bracket which involves this operator kind of an expression d2 upon dt2 i can simplify it into two individual operators so i can write it as d upon dt plus beta multiplied by d upon dt plus beta if you take a close look these two expressions operated one after another is actually overall this expression just just take a look at it it's actually the same so this equation can be written in this manner now if i make an assumption that d upon dt plus beta acting on x is equal to let's suppose y in that kind of a situation i will simply get an equation that is d upon dt plus beta y is equal to zero right because this is nothing but y so let me solve this first so this is d y upon dt is equal to minus beta y which can be written as d y upon y is equal to minus beta dt if i integrate both sides i will end up get ln y is equal to minus beta t plus some kind of a constant let's suppose c and left hand side if i just write y right hand side i'll have to take an exponential so exponential is minus beta t plus c the c can be further simplified by adding a constant here let's suppose i had a constant here now what is y why is this this is y y is equal to this expression so if i write this expression here i simply end up getting d upon dt plus beta x is equal to a e to the power minus beta t right this is the y that i assumed so what is the left hand side this is nothing but dx upon dt plus beta x is equal to a e to the power minus beta t right now if i take the exponential term to the left hand side what do i get dx upon dt e to the power beta t plus beta x e to the power beta t is equal to a now if you take a look at the this particular left hand side i can write this left hand side as the time derivative of x e to the power beta t right so if i differentiate this by time then these are the two terms that i will actually get if you take a look at it so therefore i can substitute this entire term in this manner so this is equal to a now if i do a integration with respect to time on both hand sides this simply gives us x e to the power beta t is equal to t plus some kind of a constant finally we get x which is a solution in the scenario number three is equal to 80 plus b e to the power minus beta t this is the solution that we end up getting for our case of what we had beta square is equal to omega naught square so for the case of beta square is equal to omega naught square this is what we end up getting now if you take a look at it first of all this is one function that you have here which is like some kind of a straight line and there is this further function here which is the exponential decay now of course an exponential decay is going to overcome the first function so what you actually end up getting as time tends to infinity after a long period of time the displacement of the system actually tends towards zero that means the configuration tends towards the equilibrium there are no oscillations here if you notice no oscillations here the system simply moves towards rest towards the equilibrium without oscillating in any manner so this is what's known as critical damping it's called critical damping because this lies at what is the boundary between over damping and under damping you see in under damping you have oscillations you have in over damping you do not have oscillations so in between you have critical damping where there are no oscillations but the system comes to rest in the quickest manner possible it takes the least amount of time for the system to come towards rest so having said that if you look at the let's suppose a graphical representation of this function then again here i give the system some kind of an initial displacement all right then with time the whatever the angular or rather the displacement from the equilibrium is simply comes towards this equilibrium and goes to rest this corresponds to critical damping all right now let's explore the last possible scenario which is scenario number four where beta square is greater than omega naught square so when beta square is greater than omega naught square it also corresponds to b square being greater than 4 k m as i already demonstrated under this situation what happens well first of all let me make another substitution there is omega 2 is equal to let's suppose or rather omega 2 square is let's suppose beta square minus omega naught square if i make this substitution let's go back to the general solution i was talking about in point number 4. so here i can substitute beta square minus omega not square with omega 2 if i substitute beta square minus omega not square with omega 2 then quite simply put the solution is going to look like x t is equal to e to the power minus beta t and then you have a 1 e to the power omega 2 t plus a 2 e to the power omega 2 t so combining the bracket as well as the exponential outside it i'll get a 1 e to the power minus beta minus omega 2 t plus a 2 e to the power minus beta here should be a minus sign here plus omega 2 t so the solution for this kind of a situation is this particular expression now if you take a look at this this is just the sum of two exponentially decaying functions this is an exponential dk this is an exponential decay although the decays are happening at different rates so this is decaying with b minus omega 2 in its power so if you take a look at this of course this is going to decay faster right the second term b plus omega 2 is greater than b minus omega 2 so the second term is going to decay faster so the second term is going to approach 0 quickly so it is the first term that is going to dominate so quite simply put over a period of time as t tends to infinity the uh term x is basically going to be approximated by a 1 e to the power minus beta minus omega 2 t clearly there are no oscillations happening and it is more or less an exponential decay that is happening this situation is known as over damping now it's called over damping because it is a similar to the critical damping situation it's just that it takes a little bit of longer amount of time compared to the critical damping so the mathematical or the graphical representation would be when i displace the system by some amount the system without oscillating comes towards its equilibrium configuration over a period of time so the critical damping is different from over damping that in the sense that in critical damping the amount of time taken is the minimum possible in over damping it can take a large amount of time so let us do a quick revision of all four possible scenarios with the graphs so here i have uh plotted all four graphs uh together all right so if we do a final analysis in the harmonic oscillator setup which was placed in the presence of an external medium when there was no damping then that correspond to the let's suppose a drag coefficient being equal to 0 or beta being equal to 0 it led to uh what we can see here is simple harmonic motion you see this simple harmonic oscillations are happening over a period of time so when there is no damping the harmonic oscillator provides simple harmonic oscillations however now that damping is happening in the red line this corresponds to under damping case so the under damping case happens when beta square is less than omega naught square or b square is less than 4 k m as you can see in the underdamping case oscillations are happening but the amplitudes are decaying in an exponential fashion okay the amplitudes are decaying in an exponential fashion and these oscillations are happening with respect to some frequency omega 1 which is less than omega naught all right this is the case where damping is there but it's not that high so the oscillations are still happening and it dies out over a period of time however what if damping is large enough if damping is large enough we reach a particular situation of what is known as critical damping so in this kind of a case where critical damping corresponds to beta square being equal to omega naught square or b square being equal to 4 km this is a situation where there is a rapid decrease of the displacement but there is no oscillations there are no oscillations in the case of critical damping the system takes the minimum amount of time to come from its displaced configuration to its equilibrium configuration for further situations like this blue line here where beta square is greater than omega naught square or b square is greater than 4 km these correspond to what is known as over damping so in the cases of over damping the system slowly comes to rest without oscillations but it takes its own time so a harmonic oscillator when placed in external medium that can provide some kind of dissipation of its energy that can provide some kind of a resistance can show can demonstrate these possible scenarios without damping you have sinusoidal oscillations with slight damping you get oscillations where the amplitudes decrease exponentially with further damping uh the system comes to rest with further very high over damping the system comes to rest but very very slowly so these are the different interesting possibilities of a damned harmonic oscillator all right so i'll see you in the next video that's it for today thank you very [Music] much

Channel: For the Love of Physics

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Keywords: damped harmonic oscillator, for the love of physics, physics, education, lectures, waves and oscillations physics, oscillation, oscillations, simple harmonic motion, simple harmonic motion physics, damping, damped oscillation, damped harmonic motion, under damping motion, underdamped equation, underdamped overdamped critically damped, critical damping, spring mass sytem, drag, dissipation, damped shm explained, damped oscillations, Harmonic oscillator, Damped simple harmonic motion

Id: 8AHQ6gBSxAU

Length: 40min 23sec (2423 seconds)

Published: Sun Apr 11 2021