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Current time:0:00Total duration:14:27

let's say you've got a mass connected to a spring and the mass is sitting on a frictionless surface if the mass is sitting at a point where the spring is just at the springs natural length the mass isn't going to go anywhere because when the spring is at its natural length it is content with its place in the universe it neither pushes nor pulls it has no spring energy this is like me most days in the summer so we call this point where the spring neither pushes nor pulls the springs natural length and for a mass on a horizontal spring this is going to be also the equilibrium position and what we mean by equilibrium position is the point where the net force on the mass is zero so for a mass on a horizontal spring the equilibrium position is at the point where the spring is at its natural length because the spring wouldn't be pushing to the right or the left and if you just put the mass there at that point it would just stay there forever at rest that'd be boring so let's say we pull the mass to the right a distance D if we do this we give this spring spring potential energy and if we release the mass from rest while this spring has spring potential energy the spring is going to pull the mass back to the left and that mass is going to move through the equilibrium position with some speed and we could figure out what that speed is just by using conservation of energy and it's not that hard the potential energy the spring would start with would be one-half K the spring constant times D the amount the springs been stretched squared and there'd be no kinetic energy to start because we released the mass from rest and as this mass flies to the left it would start gaining kinetic energy the spring energy would start turning into kinetic energy and when the mass gets to the equilibrium position the D would be zero so at that point there would be no spring energy and all of this spring energy will have turned into kinetic energy and you get this simple relationship that says all the spring energy equals all the kinetic energy at the equilibrium position so if we solve for V we can get that the V of this mass at the equilibrium position would be the square root of K over m times d squared you could pull this D out because it's d squared in a square root but this is the idea this is the speed you would get of the mask passing through the equilibrium position let me ask you this what if it was a vertical spring and this mass is sitting here we come find a vertical spring with a mass hanging on it we're like hey I want to pull this down to distance D if I pull it down a distance D when this mass reaches the equilibrium position again will it also be going route K over m times d squared or will it be going at some different speed because now it's hanging vertically well it turns out the spring constants the same and you pulled it down from the point where the mass is hanging this exact same procedure is going to hold over here and you can find the speed in the exact same way and that should be surprising that should not be obvious because when the mass is hanging over here you don't just have a spring force you've got a gravitational force and you don't just have spring energy and kinetic energy think about it this mass is moving up and down you've got changes in gravitational potential energy so why don't we have to take into account the gravitational potential energy when we're doing conservation of energy in this equation well that's what I want to prove to you in the rest of this video if all you wanted was the result of all you if you're good if you're like man REI can do the same thing in both cases don't even tell me anything else you're good but I suggest you watch the rest of it because knowing why you can ignore the MGH in this calculation gives you better insight into what we really mean by this D and this H and this V as well as what we really mean by the equilibrium position and that will conceptually aid you if you get a problem that's more challenging so let's prove this and figure out why we can get away with ignoring this MGH right here let me get rid of all this let's start fresh let's just say we had a spring hanging from the ceiling of spring constant K and let's say this spring is light if it was heavy it might pull itself down by its own weight I'm assume this is a very light spring and it's hanging right here there's no mass connected to it initially so it's just hanging at its natural length knot it's neither pulling up nor pushing down as you see it right here because it's at its natural length but we connect a mass M to it and when we do that we lower the mass with our hand we don't just let it fall and start oscillating we first lower the mass we connect it and lower it we find the point where the mass is just going to stay at rest that would be the new equilibrium position so this right here is essentially our new equal in position in other words that's the point where the net force on the mass would be zero but this time that's not at the springs natural length the way it was in the horizontal spring this time the equilibrium position is displaced the distance a away from the springs natural length because right now it's battling the force of gravity in other words the spring force exerted upward KX minus the gravitational force which is M times G has got to equal zero in order for this mass to be in equilibrium so we can actually figure out what this distance a would have to be in terms of given variables since at the equilibrium position X the distance the spring has been stretched is just going to have to equal mg divided by the spring constant K this is what a is going to equal so the distance the mass hangs down at the equilibrium position from the natural length of the spring is just going to be mg over K this is a in this diagram and this is going to be key so we're going to hold on to this result right here but let's do this let's ask if we take this mass and pull it down an extra amount B from the new equilibrium position well at that point the forces won't be equal this spring is going to be stretched extra it's going to be pulling up with more force than the force of gravity so this mass is going to accelerate upward it's going to reach this equilibrium position with some speed it's going to shoot past that now the spring force is less than the force of gravity and so gravity wins in that case and it keeps going back and forth and we can ask ourselves the same question we did before if we pull this down a distance B what is the speed of the mass when it passes through the equilibrium position and again we're going to use conservation of energy to answer this so we're going to say that the initial energy in our system is going to equal the final energy in our system we're going to choose two points let's choose initially this point down here we release the mass from rest when it's pulled down a distance B below the new equilibrium position and then our final point is going to be right here at the equilibrium position because that's where we want to know the speed of the mass so let's try to figure out how much energy there is in the system initially if I pull this mass down and let it go well initially if I'm just letting this mass go then asses starting from rest if the math starts from rest that's got no speed and if it's got no speed it's got no kinetic energy so there's no kinetic energy to start with if this mass is starting from rest but there is going to be spring potential energy and there's going to be a lot of spring potential energy because think about it not only is this mass stretching the spring past the new equilibrium position by an amount B but the new equilibrium position itself is stretched from the springs natural length a and in this formula when you have 1/2 the spring constant times the length the spring has been stretched that's the total amount the springs been stretched so the total amount the spring has been stretched from its natural length is going to be a plus B and I've got to square that whole term this is how much spring potential energy there's going to be in the system initially so how much gravitational potential energy are we going to start with well that's kind of up to us because you can always choose where you want your H equals zero reference line to be in other words I'm going to choose this lowest point here because that's often convenient I'm going to choose this to be the H equals zero reference line will measure all Heights from that point and this is allowed because it's only really differences in gravitational potential energy that matter so you can do this you just have to be consistent with your choice but with that choice where this is H equals zero the height my mass has at the initial position is going to be zero so that means the gravitational potential energy which is given by MGH is also going to be zero at that initial point so in terms of initial energies this is all I've got this is my total initial energy just the energy from the spring and now we can set that equal to our final energies so are we going to have any spring potential energy here at our final position you might think no because the final position is at the new equilibrium position but remember this new equilibrium position is still displaced from the springs natural length so I am I'm going to have 1/2 K times the amount the spring has been stretched from its natural position and at this new equilibrium position the amount has been stretched is just a so I'm going to write a here because that's how far the spring is stretched at this new equilibrium position and I have to square that because it's one-half KX squared and we're going to have kinetic energy this mass is going to gain speed as it flies upward and it's going to be moving with some speed when it gets up to that point so the kinetic energy is going to be one-half times the mass times the speed the mass has at the equilibrium position squared that's what we want to determine but there's also going to be gravitational potential energy we said H is zero down here so if the mass is not there it's going to have potential energy due to gravity and if it's the above this point look at if we pulled it down B so when it gets back to the equilibrium position if this is H equals zero that's going to be H equals B above where it started so I have to put in M times G times the B value this length right here since in moving up this mass gained mg times B of gravitational potential energy so how do we make progress here I want to solve for the speed V but I've got this mess over here like that I've got a plus B squared so I better handle that first so let's say we do the one-half K and then we square this out remember we do foil so its first outer inner last I'm going to get a quantity of a squared and then this cross term I'm going to get a plus two times a times B and then plus B squared that's what happens if I square this whole term right here and this is starting to look really bad but don't despair something something great something wonderful is about to happen because I'm going to set this equal to the right hand side so if we multiply out on the left hand side we're going to get one-half times K times a squared plus a 1/2 times K times this to a B term plus I'm going to get another one one-half K times this B squared term and we can say that that's supposed to equal everything on the right hand side so we can already see some things that we can cancel I've got a 1/2 K a squared on each side so if we subtract that from both sides I can get rid of that and this 1/2 here is going to cancel this 2 and I'm left with K a B on the left hand side plus 1/2 K B squared but what is K a if you're clever and you look up here you're like wait a minute I remember what ka was K times a if we just multiply both sides by K here has got to equal mg because that was just the statement of equilibrium that at the equilibrium position K times a has to equal M times G so I could replace K ty a over on this left-hand side with M times G and you might be like why would I ever want to do that why do I want to replace K times a and I still multiply by this B why would I replace ka with mg because now that term is going to cancel with the other mg B on the other side of this equation one-half MV squared plus mg B but I've got mg B on the left and mg B on the right now that's why I replace this ka with mg I can subtract that from both sides and magically I just get the exact same relationship we had for the horizontal spring measured from the new equilibrium position and this is important so let me restate this you can either when solving these vertical spring problems measure your spring displacement from the natural unstretched length of the spring like we did right here we had to add a plus B because that was the distance from the natural unstretched length of the spring all the way to where the mass was you can do that and include gravitational potential energy and get the right answer but what we just saw is that these terms always cancel so the alternative is that you can just measure the spring extension the spring displacement from the new equilibrium position and if you do that you just leave off all mention of gravitational potential energy and you'll get the same answer you can think of gravity simply as shifting the equilibrium position down a bit and then the mass and spring behaving just as they would on a horizontal surface as long as you only think about spring displacements from that new equilibrium position so in other words if you were given a problem let me get rid of all this if you're just given a problem and you were told a three kilogram mass is hanging from a vertical spring of spring constant 50 Newtons per meter and this line here represents the equilibrium position it's just hanging out right there at rest you pull this mass down from the point where it was hanging at rest point three meters and let go and you want to figure out what speed will it be going when it reaches equilibrium position you can just do this you can say all right down here it's got spring potential energy one-half K is 50 Newtons per meter times the amount the spring has been stretched but I'm just going to consider stretches from the new equilibrium position so I'm just going to do 0.3 I'm not going to worry about the fact that the spring has actually already been stretched to get to this equilibrium position I'm just going to put this stretching from equilibrium I'm going to set that equal to the kinetic energy the mass has at the equilibrium position and we know that M that M is 3 kilograms times V squared and I'm not going to include information about the gravitational potential energy at all because I only measured the displacements from the new equilibrium position so at this point I can just solve for my V and if you solve that algebraically for V you cancel out the twos you divide both sides by 3 and take a square root you get the speed of the mass at equilibrium is going to be 1.2 meters per second so to recap even though it seems initially like vertical Springs would be much harder than horizontal Springs because you've got gravitational forces and gravitational potential energy to worry about if you measure the spring displacement from the new equilibrium position as opposed to the natural spring length you can simply use conservation of energy without mention of gravitational potential energy at all