 All right, my name is Brad Langdell and it's it's the dead of winter But I want to talk to you about spring and not the kind of spring that we're all hoping is gonna come soon We're gonna talk about springs that have masses attached to them which are probably less fun and Less interesting to look at but that's what we're gonna look at anyways under the context of simple harmonic motion Which is motion which is a repetitive back-and-forth movement fueled by a restoring force And we're gonna look at three things here actually for I guess of the system We're gonna calculate acceleration. We're gonna calculate force. We're gonna calculate velocity, and why not when we're at it? Let's figure out what period is to so here's an example of a question From this unit of study in the physics 20 curriculum a mass of three kilograms is attached to a spring spring constants 1.4 newtons per meter meter the spring is stretched to 0.5 meters past equilibrium and undergoes simple harmonic motion is going to vibrate back and forth back and forth Back and forth kind of like this. I got my little ghost block there to show that Determine the maximum acceleration maximum force maximum velocity and the period of the mass spring system All right, so why don't we start off here by just talking about when we're gonna get these different things to find maximum? Acceleration that is going to occur when that mass is not of the equilibrium point where it would normally be if you just left This thing alone for a day and came back to check at the next day This is gonna be when you pull this back to the maximum displacement to the amplitude That's when you get maximum Acceleration it would also be once it swings back to this side the the acceleration is at a maximum Also here where the amplitude is at a maximum on the other side So I'm gonna start it from here and let's think about the forces taking place at that point Well, the only force really acting at this moment is the force of the spring and that force The spring is going to be what we're gonna use to find acceleration Any acceleration or any force I should say can be equal to mass times acceleration According to Newton's laws of motion So the force of the spring which is negative kx Is just equal to ma and and we can go through and solve it from there That's all you really have to do from there It's just thrown in the numbers in the right spot and being careful with things like significant digits So we've got our K we've got our x 0.5 meters We got our mass three kilograms and we're gonna solve for a So we'll grab out the handy-dandy calculator and see what we're going to get here Now one thing that's interesting to note is we're going to get a negative Acceleration and hopefully that makes sense according to the diagram because we pulled this thing back to the right It pulls something back to the right The acceleration is the opposite direction is going to be to the left So as it turns out the acceleration we get here is negative 0.23 repeating 0.23 It's a negative meters per second squared. So there's my acceleration It's not bad to do actually pretty simple Now let's find the force. Well, you know what we've got a formula here for force. We can just use that Force in the spring is the same as the force that the object will experience the mass will experience And again, this is at this point when you pull the mass back the farthest it'll go negative kx All right, so negative 1.4 newtons per meter times the bi 0.5 meters I think I can do that with my head negative 0.70 newtons All right, so there's our force and our acceleration done Okay, now we're going to find the maximum velocity in the period. So first of all our maximum velocity It's not going to occur at the point where we had our maximum acceleration and our maximum displacement Put our line back Our maximum velocity is going to occur here at the equilibrium point At that moment when all of the energy stored in the spring has been converted to kinetic energy of the block So this is where our maximum Velocity is going to be and there's two ways you can find this The easiest way you can probably look at doing it is using the conservation of energy So we're going to go through and do that. We're going to say that the energy stored in the spring Potential spring energy or elastic potential energy it will be equal at that point to the kinetic energy So the one half kx squared equals one half mv squared And at that point you can cancel the halves and that's all you need to do You just need to plug and chug from there again not too bad with the theory and the formulas don't appear in your formula sheet Made equal to each other, but that's okay. I mean, hopefully you understand Where the equality comes from it comes from the fact that all that energy that was stored in the spring is now been transferred to the mass And of course, don't forget to square root in this equation. You know people love to forget to square root things It's just one of those things that are kind of tricky to do till you practice a lot and even after your practice You'll probably forget sometimes So we'll go through remember to square there. That's good divided by three And then i'm going to square root in the end to find what that velocity is So the velocity is going to be 0.34 meters per second And the last thing you might want to do because you did square root that velocity. It could be a positive or a negative number um Well, it depends on which way you're pulling it my diagram because I pulled it back to the right The first velocity would be going to the left. So maybe I'll stick that negative sign in there Now for period. I think period's the most interesting one to do There's one way of finding period that's really straightforward the formula sheet has a formula for period Of a mass spring system. Okay, that's pretty simple throw the numbers in the right spot. There's your mass There's your spring constant plug it through and that's easy enough But there's another way for finding period two because a mass spring system is so darn similar Mathematically anyways to a circle in uniform circular motion. We could also do it with uniform circular motion Did you know that look at this? We already have a lot of this info. We know what the speed is 0.34 meters per second Now this of course doesn't make a circle it goes back and forth back and forth back and forth So it doesn't have a radius, but if we let the radius Equal the displacement the amplitude if we put in place of that 0.5 meters And solve for period we're going to get the same thing we would get out of the period formula Which is kind of neat that the two are related even though they probably don't seem all that related All right, so let's just see what we get here two pi Times 0.5 divided by 0.34 I think that looks pretty good is going to give us a period of 9.2 seconds And just to prove it to you we'll run that other calculation as well if I went two pi Times the square root of we're going to go m, which is three kilograms divided by k, which is 1.4 Let's see what we get Isn't that cool 9.2 seconds to get the same thing So two ways to find period there at the end either one is fine One just using the formula off the formula sheet one using the principles of uniform circular motion So I hope that helps for more information about mass spring systems. 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