8.01x - Module 15.03 - Two springs with an object (m) between them.

okay then to d3 a frictionless horizontal so the plane of paper is my horizontal surface I have here a spring with constant k1 here is a mass M and here I have a spring with spring constant k2 this object is in equilibrium lying still on the table I call this x equals zero and I am going to place this in the minus X direction over the distance a a in the minus X direction I call this the plus X direction let us first make the assumption that in this particular configuration the way I have drawn it both planes are relaxed that means they are not pushed in and they are not stretched therefore they do not exert any force on the object M they are totally relaxed now I'm going to put the object here that means this spring is pushed in and so if it's pushed in it will push outwards to the right it will push this back to equilibrium and the force will be K 1 a in positive direction but of course this spring is longer than it wants to be so it will pull inwards so there will also be a force on M in the positive x direction which is K 2 times a and so the restoring force F which I write down as a vector equation equals K 1 plus K 2 times X and to remind you that it is in the positive x direction I will introduce this X roof and X roof simply represents the unit vector in the plus X direction I do not like this equation I hate equations like this because I'm used to seeing here a minus sign and the only reason why you don't see this minus sign here is that I have moved this object to the minus direction and so I'm going to change it I'm going to change it in a more general equation that doesn't mean this one is wrong but I'm going to change it in a more general vector equation minus K 1 plus K 2 times X it's a one-dimensional vector equation if X is positive the force is driving back to equilibrium is in the negative direction if X is negative as it's the case here the force will be positive as you see here it's driving it back to equilibrium in the positive x-direction this represents the equation of a simple harmonic oscillator and the period of the simple harmonic oscillator T this is the period this has nothing to do with tension equals 2pi times the square root of the mass of the object divided by K 1 plus K 2 the springs acts together that's why you see here K 1 and K 2 the net force on em is larger due to both Springs than it would be if there were only one spring and therefore on a given mass having a larger force it's obvious that everything will go faster and so the period will be shorter and that's exactly what you see the larger this value is the smaller will be the period and this is very intuitive now it is not so intuitive that this period T is independent of the displacement a a we call the amplitude I take the object and I displace it to one side and then I release it and it's starting to oscillate and the period of oscillation which you have here I will get back to that this period is independent of the amplitude and the amplitude is the maximum displacement from equilibrium in this particular case we had a that is not so intuitive that this period is independent but I will return to this let's first address the issue what would be the case if the two Springs will not relaxed when I start so suppose I have here again this frictionless table that's the plane of my paper and I stretch this one it's now much longer than in the relaxed position and I also stretch this one it's also much longer this is spring with spring constant K 2 and here is the mass M it would mean then that in this position in this equilibrium equilibrium position x equals 0 that is not being accelerated it's sitting still this is equilibrium it means then that in this situation which is now my initial condition there is a force to the right because this spring is longer than it wants to be so it wants to contract so it will pull through the right this string is also longer than it wants to be so it also wants to contract so there will also be a force to the left I will call that F 1 but the two forces obviously must be the same in magnitude otherwise this object could never be in equilibrium so this is now my starting condition and if now I move this object to the side for instance to the negative x-direction eh then I want you to convince yourself that everything we've done before remains the same in other words that the restoring force to equilibrium will be exactly as it was before makes no difference that now these Springs are stretched makes no difference and therefore that also the object will again be starting oscillating simple harmonic oscillation and that the period is precisely as we just stated now I will derive these periods for you and I do that through a differential equation so I write down a one dimensional differential equation F equals MA Newton's first law this is a vector this is a vector now AE is the second derivative of X so I can write down for a MX double dot whereby X double dot equals D 2 X DT squared and that now equals minus capital K which is the equivalent spring constant in this case k1 plus k2 times X why don't I put arrows over here well because this is a one-dimensional vector equation if X is positive this minus sign will tell you that the force is in the minus direction if X is negative the minus sign will tell you that the force is in the positive direction so I really don't have to put the arrows over here now I'm going to combine these two and I get the famous equation that you will see a zillion times before this course is over MX double dot plus KX equals 0 this is the famous equation of a simple harmonic oscillator the solution to this equation I will give you four now that X as a function of any moment in time X is the position of this object M equals some amplitude x0 which is the maximum displacement from equilibrium times the cosine of Omega T plus alpha or it could be the sine of Omega T plus alpha Omega we call the angular frequency which is 2 pi divided the period t this has nothing to do with tension this is the number of seconds for one complete oscillation and that is the square root of K over m the frequency in terms of how many oscillations per second equals 1 divided by T so there is the number of oscillations per second which we sometimes often call Hertz and Omega is often expressed in radians per second x0 is the amplitude the maximum displacement from 0 and alpha is a phase angle which is entirely dictated by the initial conditions there is really not that much physics in alpha I can make alpha 0 if I want that I can release the mass m at T equals 0 and I can do that with zero speed and if I do that from a location x equals plus a x equals a then X as a function of T would be X zero times the cosine of Omega T you can check for yourself that by stating the initial conditions the way I did I have made alpha zero the velocity as a function of time is the derivative of this so that gives me a minus Omega oh I was going to make this in a Omega times a times the sine of Omega T notice that at T equals zero the velocity is indeed zero that was my initial condition at T equals zero is zero speed and also notice that when T is just a hair larger than zero the velocity is negative that's clear because when I release the object from plus a in positive x direction it wants to go back to equilibrium and so the velocity is in this direction and therefore the velocity is negative if the velocity were in this direction the velocity were positive so that's all beautifully taken care of by algebra and now we get X double dot T which is a as a function of T that equals minus Omega squared times a times cosine Omega T which is also minus Omega square times X of T so now what you can do you can substitute this X in here oh where is my differential equation no I have to substitute it in the differential equation I'm sorry I have to substitute this X in here and I have to substitute this X double dot in here and what do I find then that this equation this differential equation is only and only satisfied if omega equals the square root of K over m which is what I stated earlier verbatim without proof here you have it square root of K over m and therefore that the period of one oscillation equals 2pi times the square root of M over K and I repeat myself the frequency how many oscillations per second is one / the periods and you notice which I stated earlier that this period is indeed independent of a which now I proved and that is by no means so intuitive I find it always amazing but it's very characteristic for a simple harmonic oscillation that the period of oscillation is independent of how far you displace it from its equilibrium point if you look at this equation in more detail you see that the larger K is the smaller the period I discussed that earlier that is rather intuitive the larger the mass M is the lower the acceleration will be for a given force on the mass if I make em larger the acceleration will be slower therefore the whole motion will be slower and if the motion will be slower obviously it will take longer to complete one oscillation so T will be higher so you see indeed if M is higher then also T is higher and that of course is intuitively very pleasing in general the phase angle alpha will not be zero but I have made it zero by choosing my initial conditions very appropriately and you can very often do that so now we get to the last part of this problem set let me do a time check boy I'm 15 seconds ahead of time that's a real treat the last part is asking you the following I now take this same system K 1 with mass M K 2 and it's now vertical and there is gravity let's first take the situation that there is no gravity that's the situation here and here I have equilibrium now all of a sudden I turn gravity on let this be the floor and this object will sag because of gravity it will sag over a distance s S stands for it will sag and so the new equilibrium position will now be here this spring will be longer than it would prefer to be and this spring is shorter than it prefers to be so if I take this object m and I look at the forces now in its new equilibrium then there is of course mg that's clear but now this spring is shorter than it wants to be so it's going to push up so it will push up with a force F k1 which equals k1 times s that's the magnitude this one is longer than it wants to be so it wants to contract so it will also push up with a force F k2 and the magnitude of that force in upward Direction equals k2 times s there has to be equilibrium the sum of all forces in this direction will have to be 0 in other words mg must be equal to k1 plus k2 times s and this if you want to allows you to calculate what s is how far this object is saggin now comes an interesting portion an interesting part of this problem suppose now from this new equilibrium position I move the object either down over a certain distance or I move it up over a certain distance and I let it do its thing and the displacement from this new equilibrium position I call this now Y so that we don't have any confusion with mine before I want you to demonstrate now that you will get now f equals ma equals M y double dot equals minus capital K times y where capital K equals k1 plus k2 and this again is the same simple harmonic oscillation that we have before again we find that Omega equals the square root of capital K over m the period is unchanged and the frequency is unchanged and both are independent of the amplitude by which I displace it from zero and so even though I've left you with this a little bit the bottom line is that even when I when I put this whole system in gravity and so when there is the SAG given the new equilibrium when I displace it away from the new equilibrium the object will oscillate was exactly the same period exactly the same frequency as it did before when we had the system on a horizontal frictionless plane you may think that's so obvious I find it always a little bit surprising but yes when you come to think about it maybe it is obvious

Channel: Lectures by Walter Lewin. They will make you ♥ Physics.

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Keywords: Walter Lewin, Physics, 2 springs, 2 springs in series, k1 and k2, simple harm oscillation, F = - (k1 + k2)x, Period T, T=2pi*sqrt[m/(k1+k2)], differential equation, Differential Equation, angular frequency, 2pi/T

Id: SGjzWPS7rOo

Length: 19min 22sec (1162 seconds)

Published: Mon Feb 16 2015