Intro to Simple Harmonic Motion (Horizontal Springs)

Hey, guys. So for this video, I'm gonna introduce the motion of simple harmonic motion, the position, velocity and acceleration. And I'm also gonna give you guys some variables and definitions of useful for the rest of the chapter. So let's get to it. So the most common type of simple harmonic motion that will study in this chapter, or sometimes you'll see this oscillation is the mass spring system. So that's where you attach some fast to some spring. So if I pull this thing all the way back here, right, So I pull this on a spring and I let it go, The only force that's gonna be acting on that thing is the Spring Force, and that's equal to K X, Right? So I'm gonna pull this thing back. I'm gonna let it go, and basically this thing is gonna go all the way back to the equilibrium position. But it's gonna overshoot it and then land on the other side. So it's gonna go like something like that. And so basically, after that, it's just gonna oscillate between those two points forever, right? Those maximum displacements here, So the displacement is maximum on the left side, here and on the right side, right here. And so that the equilibrium position, We know that the X is equal to zero. So we call these two points on the outsides, the amplitude. So we've got plus a and minus a right here. And the amplitude is just the maximum displacement that this object has. It's always the initial push or pull that you apply against this object. So if you push it out to some distance and release it, that's the amplitude. So what is the velocity look like? When you do this, we're gonna pull this thing all the way back, and then you're just gonna let it go from rest. So that means that at this point, right here at the amplitude, the velocity is equal to zero. So then what happens is this thing starts speeding up faster and faster and faster goes through the equilibrium position. And then when it gets to the other side right here, it has to stop and then turn around. So that means that the velocity at this other amplitude is also equal to zero. And then what happens is as it's going back through here, the velocity is maximum here, and then it just keeps on doing that over and over again. So we have maximum velocity at the center, and so does this whole entire cycle over and over again. So the two variables that are related to that are the period and the frequency. The period, which is the letter T is the time that it takes for you to complete one complete cycle for you. Finish one complete cycle, and what's related to that variable is the frequency. So these things are just related by in verses of each other. So you see that the frequency is just one divided by T, So if you ever have one of them, you can get to the other one. So, for example, if I have a T that's equals, like two seconds on the frequency is just gonna be equal to one half of a cycle per second. And the units for that are gonna be in hurts now. Another variable that's related to the frequency. But not the same thing is the angular frequency. That's the letter. Omega. It's the units are rats, Radiance per second and all Omega is is just the frequency times two pi. We could also write it in a different way because the period and the frequency are in verses of each other. Okay, so what are the forces look like? A these three points. So when I pull it all the way back, the spring wants to push harder and harder, harder against me. So at this maximum displacement here that that means the force is also going to be maximum, Right? So take a look at K X if X is maximum that that means that F s also has to be maximum. And so we know that the maximum forces are gonna be here a the end points. And because of the equilibrium position, the X is equal to zero. Then that means that the force is equal to zero. Now, what about the acceleration? Will the acceleration and the force are related just by f equals m A. So if f is Max here, then that means the acceleration is Max here and here. And if f equals zero, then that means a is equal 20 So it looks like X f and A are all related to each other, and they're all in sync. So we've got X is all is zero F zero and a zero in the middle, whereas the weird one is this velocity here. So that's maximum when the other ones are zero. So let's take a look at an example here and see what we can find out.

Hey, guys, let's take a look at this. This example here. We've got a mass on a spring and it's pulled 1 m away from its equilibrium position. So we've got 1 m and we're just gonna draw a little box here, and then you release it from rest. But this is a mass spring system, right? So it's just gonna go and reach the other side where now the distance here, the displacement is gonna be negative 1 m, and that's just gonna go back and forth forever. So here in this first part were asked to calculate the amplitude. But we know that the amplitude is just the maximum displacement on either side. So that means that the amplitude of this is just 1 m. So the second question now were asked to find the period. So that is going to be that letter t. So what is the period look like? What we're told that the mass takes two seconds to reach the maximum displacement on the other side. So if you release it over here, then this time that it takes for you to go all the way over to the other side was equal to two seconds. Now the question is is that the whole period? No, it's not because we said that the period is equal to the time that it takes you to complete one whole entire cycle here. So this two seconds really only represents Ah, half period. This thing has to go all the way back to the other side for another two seconds. And that's gonna be another half period. Which means that these points in between here are actually quarter periods. Right, So this is a quarter, This is a quarter. These are all quarter periods, and this is the smallest sort of division that you could make. So you've got half periods and quarter periods, so we're told that it's two seconds to get to the other side, which means that the full period of this motion is going to be four seconds. Now we've got in this third part. Here is the angular frequency off the motion. So how do we relate angular frequency to the period? Well, we've got an equation appeared that could do that. So if we're looking for the angular frequency Omega we can use is either two pi times the frequency or we can use to pie divided by the period which we actually know. So I'm just gonna go ahead and use that I've got omega equals two pi divided by the full period of cycle, which is for and what you should get is 1 57. And that's gonna be radiance per second. So that's it for this one. Let's keep going with some more examples.

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.

Hey, guys, let's take a look at this example. So we're told that a 4 kg mass is on a spring. It's oscillating at two hertz, and it's moving at 10 m per second once it crosses the equilibrium position. So on a mass spring system what? It's oscillating back and forth as it crosses the equilibrium position right here. We know that the speed is maximum. So that's what they're telling us. They're telling us that V Max is equal to 10 m per second, so I'm gonna start writing everything out. I told that the mass is equal to four. I've got the frequencies equals two and I've got the max is equal to 10. And what I'm supposed to find is I'm supposed to find the time it takes to get from equilibrium all the way out to its max distance. So, in other words, how long does it take to get from equilibrium all the way out to its max distance? We know that that is one quarter of the period. One full period is the whole entire cycle. So we're just looking for that quarter. So if we're looking for tea over four, we might as well. Just find what t is. So let's just use our equations to find out what time is. Well, I've got the big omega equation down here, but I also know that t and F so t and the frequency are in verses of each other. And I have what the frequency is. So that means that the period is just one half of a second. But that's not what I'm looking for. I'm looking for a quarter. So if I got t over four, that's just gonna be one quarter off one half of a second. And so that equals 1/8 of a second. So that's the answer to part A. So what is part B? Ask us Party asked us to find out what the amplitude is. Someone write that down here. So we're supposed to figure out what a is. Let's look at all of our equations and figure out where is well, a is kind of president, all of them. So let's rule out the ones that we can't. We don't know anything about the mass. We don't anything about the maximum acceleration. We don't know anything about times. We can't use these guys eh? So basically, I'm just gonna have to use thes equation all my max equations. So I've got I don't know what X Max is, because otherwise I would be the amplitude, and I don't know what the acceleration Max is either, but I do know what the V Max is. So let me go ahead and use that equation for V Max, because that's the one that I know most about. So I've got V. Max is equal to a times omega. So if I rearrange for this, I've got that A is equal to v Max, divided by omega. So now I have what V Max is. I don't know what Omega is, So let me go ahead and find that. So let me check that I've got the max. Now. I just have to go over here and find out what Omega is. So let's use my big omega equation. Omega is equal to two pi frequency. Do I know the frequency? Yes, Ideo. So that means to omega equals two pi times f. So omega F is just equal to two frequencies to hurts. So the omega is equal to four pi, so I'm just gonna stick that right back in there. So that means that the amplitude is just 10 m per second divided by right. That's the V max. And then I've got four pi, so I can just go ahead and simplify that and say that it's 5/2 pi, and I'm gonna go ahead and highlight that box it so you guys can see it. So I got five or two pi. Now, this last one is asking me to find a maximum acceleration. So now we're actually gonna go ahead and solve for a max. So which one are we gonna use? We've got this a max over here, but I need to know, okay? And I don't know how the spring constant. So instead, I'm gonna use not this a max. I'm gonna use this a max. So I'm gonna use the a Omega squared. So a max is the amplitude, which is 5/2 pi. And then I've got omega, which is four pi. So if I square this guy, it's just gonna be 16 times pi squared, right? So this is four pi, So omega squared equals 16. Pi squared. So now what happens is I've got a pie on the bottom. The nominator got a pie in the numerator, so they cancel. And then I've got a 16 in the numerator and a two and the denominator. So all of that stuff simplifies. So I've got a max is equal to five times eight times pi. So that means a max is equal to 40 times pi. And that's the answer. So you should definitely become familiar with all these pies popping up all over the place is, I guess, let me know if you have any questions. Let's keep moving on.

Hey, guys, let's take a look at this example. So we're told that the velocity of a particle is given by this equation, and that's the only thing that were given were supposed to find out what the frequency of motion is. So what they're asking us is to figure out what f is. So let's take a look at my equation list. The Onley equation that involves frequency is the big Omega equation right here. The problem is, I don't have any of those other variables. I don't have k over em. I don't have the period. And I don't know what the Omega is either. The only thing I am given is just the velocity equation. So let's take a look at the actual velocity equation with all the variables involved. So I've got the velocity is equal to negative six times the sign of three pi times T. But the form of the equation is I have negative a omega, Then I've got sign of omega times t. So if you just look at this, what is this telling us? Well, we've got this negative six out here and I've got a negative a omega, so that means that these things are equal to each other. So that means I have that a omega is equal to six. And if you look inside the parentheses, I've got this omega over here that's equal to this three pie. So those two things are the same. So that means that Omega, what they're telling me is equal to three pie. So now let's take a look at my frequency equation. I was I don't have the period and I don't have the square of Cavor em. But now I do have the Omega. So let's relate that equation eso I've got Omega is equal to two pi times the frequency. But omega, I just figured out was three pie. So I'm gonna set these two things equal to each other. So two pi times the frequency equals three pie. And now the frequency is just three pi over two pi. What will happen is the pies will cancel. And so I get frequency is equal to three halves. And that's the answer. That is the frequency of oscillation. So you just have to look at the equation and compare those variables to, like the actual form of the equation. to figure out what those things are. So let's look at part B. Part B is now asking us for the amplitude so great. So how do we find out the amplitude? We've got a bunch of equations that involved the amplitude Eso Let's take a look. I'm not told the X Max V Max or a Max. I'm never told any of those things and I don't know what the IMAX or the restoring forces, so I can't use that. So I'm gonna have to again look at this equation right here and from this equation, I figured out that a Omega was equal to six. So let's write that out. If a Omega is equal to six, then a is just going to be six divided by omega. But I figured out what Omega was. Well, remember that Omega was just equal to three times pi right over here. So that means that the amplitude is just too divided by pi once you go ahead and sold that. So I've got that's the amplitude to over pie. Now, for this last one, I'm supposed to find out what the velocity is at a specific time. So this is a question where they're giving me tea and I've asked for V, so I'm just gonna basically plugged in into my formula. So the velocity when t is equal to 0.5 seconds is gonna be negative six times the sign of three pi times 0.5. Make sure that you're in radiance mode and should get a velocity that's equal to 6 m per second, and that's it. Let me know if you guys have any questions, and if not, let's keep going.