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>> Hello class, Professor Anderson here. Let's talk about the difference between a horizontal spring and a vertical spring. So, let's say we do the following experiment. Let's say we have a spring with a mass attached to it, and we stretch it out, and we let it go and there's no friction or anything to worry about. And, let's stretch it out to a distance, A. We'll let it go back through equilibrium. We know that it's going to go compress the spring to a distance minus A, okay? So, where is the velocity going to be a maximum? Well, the velocity is a maximum when it goes back through the equilibrium position. And, so if we think about conservation of energy here, what can we say? Ei equals Ef. When we start, we are all spring potential energy one-half KA squared. When we go back through equilibrium we are all kinetic energy; one-half mV max squared. You can quickly solve this for Vmax, right? Multiply by two, divide by the m, we get Vmax equals square root K over m times A. But, we also remember that square root of K over m is our good old angular frequency omega. Now, how does that differ if we take that system and instead we hang it from the roof? So, let's draw the system hanging from the roof. First off, let's just hang the spring, okay? It's hanging at its equilibrium length. And, now let's attach the box. When we attach the box, it stretches, a little bit. It stretches a distance L. And, now let's stretch it by hand a distance A [drawing], okay, from the equilibrium spot down to there. We'll let it go. What's it going to do? Well, we know it's going to oscillate up and down, and in fact, we know it's going to go to a maximum height of A above that equilibrium spot. But, how does the maximum speed compare to our horizontal table? Let's define this equilibrium position as gravitational potential energy there equal to zero, and now let's look at this from a conservation of energy standpoint. So, let's call this lowest position here, position one. We will say that when it comes back up through the equilibrium position that is two, and when it gets up to its maximum height above it that is position three, and let's see what we get for conservation of energy. Before we do that, let's see if we can figure out what this new equilibrium position is. When you hang the mass from the spring there is gravity acting on that mass, but there is the spring force holding it up. And, if it stretches a distance L, then the spring force is equal to kL, and so now look what we have. Those forces have to be zero when it's just hanging there, and so we get kL minus mg equals zero, or kL equals mg. Now, let's do one more little trick which we will use later. Let's multiply both sides of this equation by A. kLA equals mgA, and we'll see why we did that in a second. Okay, so now let's go to our positions one, two, and three and let's use conservation of energy. Okay, so what's the energy at position one? The energy at position one is, well there is no kinetic energy, because it's not moving at the bottom, so that's zero. We do have some spring energy, and how much has that spring stretched out from its original equilibrium? Length, it has stretched out A plus L. And, if we set position two to have zero gravitational potential energy, then we have to subtract mgA from this expression. Okay? And, so that's where we end up with for E one. Alright, what about E two? E two is when the block is moving back upwards. It has kinetic energy at that point. We suspect that's going to be the maximum speed, but we're not totally sure yet. It also has a little bit of spring potential energy, because even when it's at position that spring is still stretched a little bit. It stretched by L, and we have zero potential energy due to gravity, because we said that's where it equals zero. Alright, what about E three? E three is when the block has moved all the way up to the top of its motion, and let's see what we have for those terms. We don't have any kinetic energy. It's come to rest up there, so that one's zero. We might have a little bit of stretch in that spring still, and the amount of stretch is going to be the difference between A and L. So, we have to subtract those two. And, we have some gravitational potential energy, because it's up at a height A, and so we have to add mgA. Okay? And, now we're going to take advantage of this little last statement here, kLA equals mgA. Okay, so let's put in these terms here. So, we have one-half k, A plus L, quantity squared, minus mgA, but mgA we could write as this, kLA. E two, we'll leave that alone for a second. E three becomes what? One-half k, A minus L, quantity squared and then we have plus mgA which is plus kLA. So, now we have all of those energies in terms of the spring constant k, the rest length L, and how far we stretched it, A. And, now let's see if we can solve this thing for velocity. Alright? So, to do that we just set these energies equal, and why don't we take equation one and set it to equation two. So, if we say E one equals E two, what do we have? One-half mV squared plus one-half kLA squared equals E two. E two is right here, and let's write that down here, but let's multiply it out. So, we have one-half kA squared, plus we've got to do the foil on this, so now we have two AL in there, and that two is going to cancel with the one-half out in front. So, we have kAL, and then we have one-half kL squared. And, then we're still subtracting this last term kLA. And, now look what happens. kAL and minus kLA we can get rid of those. And, one-half kL squared is over there as well, so we can get rid of those, and that looks very familiar. That looks like our old equation, and in fact, V is equal to what? Well, we multiply by two, we divide by m, we get V max equals square root of k over m times A which is omega times A which is exactly the same as the horizontal case. So, a vertical spring behaves exactly the same as a horizontal spring, as long as it's all relative to a new equilibrium position, and that's the whole point of this exercise. Alright, hopefully that's clear. If not, come see me in my office. Cheers.

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