Hey, guys. So for this video, I'll be giving you the solid equations that we'll need for mass spring systems and simple harmonic motions. Let's check it out. So remember that as the spring is moving back and forth along the spring, the acceleration is always changing. And because it's never constant, we can't use the kinematic equations from early on in physics. Now the equations 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 can actually depend on t on time for? So I'm not going to derive them, but they're right down here. So let's take a look at the subtle differences between them. They're all sinusoidal and the beginnings have \( a \omega \) and \( a \omega^2 \). Those are the front terms. Now, you have to be careful because the sign changes between these. We've got positive, negative, and negative, and we've also got changes between cosines and sines. So for those of you who've taken calculus, you'll be able to relate these functions using derivatives. There are 2 important things about these functions. The first is that these are 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 that into these equations to figure out what the position, velocity, and acceleration are. The second thing is that because we're using cosines and sines, just make sure that your calculator is in radians mode. Okay. So these were equations of time, whereas the equations that we've been dealing with so far, the old equations, are functions of position. So these are 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 relate this back to my simple harmonic motion diagram. I know that at the endpoints, 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, I can figure out their maximum forms. So I've got \( \pm k a \), and the acceleration is \( \pm \frac{k}{m} \times a \). So what about these bottom equations then? When are those things maximum? Remember that one's position and one's, for one's time. Well, if we take a look here, we've got cosines and sines, and cosines and sines 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, these are just going to be whatever's on the front terms here. So what that means is that \( x_{\text{max}} \) is just going to be \( \pm a \), which we actually know. The maximum displacement is the amplitude. The maximum velocity is \( \pm a \omega \) and the maximum acceleration is \( a \omega^2 \). Okay. So we've gotten all the maximum terms for these. One other thing that we can do is we've gotten two expressions for the maximum acceleration. One is a function of position and one of time. So we combine those two things together. They represent the same variable, and we can use that to figure out what this \( \omega \) term is. Now up until now, the \( \omega \), which is 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 \( \frac{k}{m} \). This is arguably the most important variable in this whole chapter, so make sure you remember that. \( \omega \) is always \( \sqrt{\frac{k}{m}} \). So to see how all of this stuff works, let's go ahead and do an example. So in this first example, I've got a 4 kilogram mass and it's at rest and it's attached to a spring. So I've got 4 kilograms. I'm given the k constant, and I'm told that it's pulled 2 meters. 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 \( \omega = 2\pi \times \text{frequency} \) and I'm not given any information about the frequency or the period. So what I'm going to use is this new \( \omega \) that gives me 7.07. And I know the units for that are radians per second. So now that I've got the angular frequency, now for this part b, I'm asked for when \( t = 0.5 \) 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 \( t \) as a function of t is that equation. So I've got \( -a \omega \) and then I've got \( \sin(\omega \times t) \). So I know what my \( \omega \) is, I just figured that out. I'm going to plug in that 0.5 seconds for \( t \). Now all I need to do is find the amplitude. So I'm told in this case that the mass has pulled 2 meters. So that means that I've got an amplitude that's equal to 2 meters. So it means that the velocity as a function of time is equal to \( -2 \times 7.07 \times \sin(7.07 \times 0.5) \). Now if you do this and you make sure that your calculator is in radians mode, we'll get a velocity when \( t = 0.5 \) seconds. That's equal to 5.42 meters per second. So all this means, notice how I didn't get a negative sign. This positive sign means the velocity at 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 to 0.5, and now we're supposed to find the acceleration. So now we're going to use the position formulas, the position functions. If I want to find out the acceleration as a function of position, then I'm just going to use \( -\frac{k}{m} \times x \). So the acceleration when the position is equal to 0.5 is going to be equal to \( -\frac{200}{4} \times 0.5 \). So I'm getting an acceleration of negative 25 meters per second squared. So that is the acceleration. And all that means is that at this particular instance, the now for this last one here, I'm supposed to figure out what the period of oscillation is. So now for this last one here, I'm supposed to figure out the period. 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 \( \omega \) contains that variable \( t \) in there. So I'm going to write that out. So \( \omega = 2\pi \times \text{frequency} = \frac{2\pi}{t} = \sqrt{\frac{k}{m}} \). So all of these things are equal to each other and I've got I've already got what 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, \( \omega = \frac{2\pi}{t} \), and now all I can do What I can do is just, just trade places. So the \( t \) will come up and then the \( \omega \) will take its place. So they'll just trade places there. So I've got \( t = \frac{2\pi}{\omega} \). So \( t = \frac{2\pi}{7.07} \). So what I get is a period of 0.89 seconds. Alright, guys. So that's it for this one. Let's keep going.
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17. Periodic Motion
Intro to Simple Harmonic Motion (Horizontal Springs)
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