Equations of Simple Harmonic Motion

by Patrick Ford
207 views
3
3
Was this helpful ?
3
Hey, guys. So for this video, I'll be giving you the solid equations that will need for mass spring systems and simple harmonic motions. Let's check it out. So remember that as the spring, as as its Mass is moving back and forth along the spring thing, acceleration is always changing. And because it's never constant, we can't use the Kinnah Matic equations from early on in physics, and the equation that we've seen so far are the force and the acceleration. But both of these things depend on X. So what if we wanted a new set of equations that we could actually depend on t on time for, so I'm not gonna derive them, but they're right down here. So let's take a look at the subtle differences between them. They're all signed soil, and the beginnings have a a Omega and a Omega square. Those air the front terms. Now it's We have to be careful because the sign changes between these. We've got positive, negative negative, and we've also got changes between co signs and signs. So for those of you have taken calculus, you'll be able to relate these functions using derivatives. So there's two important things about these functions. The first is that these air functions of time. So what these questions will look like is they'll give you a specific time. And if you have a and Omega, you can plug it into these equations to figure out what the position, velocity and acceleration are. The second thing is that because we're using co signs and signs, just make sure that your calculator is in radiance mode. Okay, so these were equations of time, whereas the equation that we've been dealing with so far the old equations are functions of position. So these were asking us for the forces and acceleration at a specific position. So now what if I wanted to know what the maximum values of these things are? Well, I can really this back to my simple harmonic motion diagram. I know that at the end points the acceleration and the maximum and the forces are maximum. So that means that the for the old equations, these things are maximized when X is equal to either amplitude. So because I've got these equations on the left, Aiken figure out their maximum forms. So I got plus or minus K a and the acceleration is plus or minus K over M times A. So what about these bottom equations, then? One of those things maximum. Remember that one's position and ones? Uh, once time. Well, if we take a look here, we've got co signs and signs and co signs and signs always oscillate between positive one and negative one. And they do that forever. So that means that these things are going to be maximized when the sine or cosine graphs are equal to their endpoints, which is positive or negative one. And so a good way to remember where these things are maximum and what those are are these. They're just gonna be whatever's on the front terms here. So what that means is that X max is just gonna be plus or minus a, which we actually know. The maximum displacement is the amplitude. The maximum velocity is going to be plus or minus a omega, and the maximum acceleration is a Omega squared. Okay, so we've gotten all of the maximum terms for these. One other thing that we can dio is we've gotten to expressions for the maximum acceleration. One is a function of position in one of time. So we combine those two things together. They represent the same variable, and we could do that. We can use that to figure out what this Omega term is now. Up until now, the Omega, which the angular frequency, we've only been able to relate to the linear frequency and the period. But if you combine those two equations and solve for Omega, what we'll get is that omega is equal to the square root of K over em. This is arguably the most important variable in this whole chapter, so make sure you remember that Omega is always squared of K over em. So to see how all of this stuff works, let's go ahead and do an example. So this first example. I've got a 4 kg mass and it's at rest, and it's attached to a spring. So I've got 4 kg. I'm giving the K constant, and I'm told that it's pulled 2 m. So now I'm supposed to find out what the angular frequency is, so I'm supposed to find out what Omega is. Let's write out my equation for Omega. I've got omega equals two pi times the frequency, and I'm not giving any information about the frequency or the period so I don't wanna use is this new expression that I have square root of Kate over em. So that means that my omega is equal to the square root of 200 divided by four. And that gives me 7.7 And I know the units for that are radiance per second. So now that I've got the angular frequency now for this part B, I'm asked for when t is equal to 0. seconds, What is the velocity? So I'm given a T. And I'm asked to find out what the velocity is so I can use my velocity as a function of T. And so let's go ahead and look up there for a second. So I'm told that the velocity of as a function of T is that equation. So I've got negative a omega, and then I've got sign of omega times t so I know what my omega is. I just figured that out. I'm gonna plug in that 0.5 seconds for tea. Now all they need to do is find the amplitude. So I'm told in this case that the masses pulled 2 m. So that means that I've got an amplitude that's equal to 2 m. So it means that the velocity is a function of time is equal to negative two, and I've got 7.7 and I've got sign of 7.7 times 0.5 Now, if you do this and you make sure that your calculators in radiance mode we'll get a velocity when t is equal to 05 seconds, that's equal to 5.42 m per second. So all of this this means notice how I didn't get a negative sign. This positive sign means that the velocity at this this particular time points to the right. So let's move on to part C. Now. Part C is similar to part B, except now it's giving us a position. It's telling us the position is equal 2.5, and now we're supposed to find the acceleration. So now we're gonna use the position formulas the position functions. If I want to find out the acceleration as a function of position, then I'm just gonna use negative K over m times X. So the acceleration when the position is equal to 0.5 is gonna be equal to negative. I know that K is equal to 200. Got em is equal to four. And I'm gonna multiply this by 0.5. So I'm gonna acceleration of negative 25 m per second squared. So that is the acceleration and all that means is that at this particular instant, the acceleration vector Whoops, the acceleration vector points to the left. So now for this last one here, I'm supposed to figure out what the period of oscillation is. So for part d, I'm supposed to figure out the period, so t so let's go ahead and look up in my equations and figure out which one has t. So I've got this big omega equation in here, and this Omega has contains that variable t in there, so I'm gonna write that out. So omega is equal to two pi times. Frequency equals two pi divided by t equals and then I've got square roots of K over em. So all of these things are equal to each other and I've got I've already got with my Omega term is and I'm just trying to figure out what this T term is, so I can just go ahead and use that relationship directly. So I've got Omega equals two pi divided by t and now all Aiken Dio. What I can do is just, um, just trade places so the tea will come up and then the Omega will take its place so they'll just trade places there. So I've got t is equal to two pi over omega, so t is equal to two pi divided by 7.7 So what I get is a period of 0.89 seconds. Alright, guys. So that's it for this one. Let's keep going.