17. Periodic Motion

Simple Harmonic Motion of Vertical Springs

# Simple Harmonic Motion of Vertical Springs

Patrick Ford

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But we're gonna see that vertical systems are very very similar to horizontal and the only difference has to do with the equilibrium position. But rather than tell you, I just want to show you using this example. So let's take a look. We've got a 0.5 m spring and it's hanging from the ceiling. So I've got that distance 0.5 m. And that's gonna be the original length of the spring and then it stretches down by some distance after I attach a mass to it. Now, why does it stretch down and stop? Well, we've got an extra force to consider because now we have the objects weight that pulls it down. Whereas originally we didn't have that before in horizontal systems. So as the thing is pulling the spring down, the restoring force gets higher and higher upwards. And so what ends up happening is that the restoring force ends up canceling out with the gravity. And so it reaches a new equilibrium position. So in horizontal mass spring systems we said that X equals zero is where like no forces are acting on it. But in vertical mass spring systems, the equilibrium is where these forces will cancel out. That's the important part. And so we call that distance that hanging distance until it reaches equilibrium DELTA L. So that in this case is equal to 0.2. So again that Delta L. Really represents the the spring systems hanging deformation or like hanging distance. It's the amount that you need to stretch the spring so that you reach equilibrium. And equilibrium condition is where these two forces balance out. Well the upward force is gonna be K times delta L. And the downward force is going to be M. G. So that means at equilibrium we've got K. Delta L equals MG. So let's look at the second part of this problem. So after all of this stuff happens, we're gonna pull this spring system an additional 0.3 m downwards. So now just like in horizontal spring systems, your additional push or pull was the amplitude. So that means that once you pull this thing downwards 0.3, now this thing is just going to go up and down between these two amplitudes. So this is gonna be the positive and that's gonna be the negative amplitude. So it's just gonna oscillate up and down like that. And so it's important to remember that this amplitude is the additional push or pull and it's not the delta L. So it's not the natural hanging distance or defamation, it's the additional pusher pool. So in problems, what you'll see is that like you're gonna attach a mass and it's gonna stretch by some distance. What that represents is delta L. And then you're gonna pull it down an additional something and that is going to be the amplitude. So don't confuse those two. Alright, so now basically we have everything we need to solve the problem. This first part is asking us for the force constant. So it's asking us for K. So we're just gonna use this new equilibrium equation that we have. So we've got K times delta L. Is equal to MG. Now I know what MG and delta L. R. I just have to figure out what this K constant is. So I've got K once I just move this to the other side, I've got K is equal to five, which is the mass. And I've got, I want to use 10 for gravity. And then the delta L. Is what will the delta L. We said that natural stretching distance was 0.2 m. So that means I gotta K constant of 250 newtons per meter. And that's it. So what about the second part here? The second part is now asking us at its maximum height. So it's oscillating at maximum height. How far is that ceiling from the block? Great. So let's check it out. So we've got this this thing here, uh and so I'm going to sort of represent this whole entire line here as this motion. The original distance, the original sort of length of the spring, was that black dot. And then you hang it down and reaches some equilibrium position and then you're gonna pull it down an additional 0.3 m. So that's the amplitude. And it's just going to oscillate between these two points. Okay, so we're being asked for basically this distance right here, what is the distance between the ceiling and its maximum heights? That's where it reaches that maximum amplitude. And so that distance here is going to be the letter D. Well, we're told that this thing has an original length of 0.5 m. And the equilibrium position is when it stretches an additional two. So that means that the Length of equilibrium I'll call this Q is equal to 0.7 m. So now what happens is it goes 0.3 down and then 0.3 up. So at this bottom part here, this length here is 1.0 and at the top point right here it's 0.7 -0.3. And so we say that at maximum height, so at the max height, The distance d away from the ceiling is equal to 0.4 m. And that's it. Alright guys, that's it for vertical oscillations.

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