Modeling spring-mass oscillators

0:00- [Instructor] If you stretch a spring and let it go, it'll start oscillating sinusoidally about the equilibrium position. We call this a simple harmonic motion. And the period of the oscillation is given by this particular expression. But this is an empirical result, meaning the way we found it out was experimentally.

0:20But how do we predict these results mathematically? I mean, how do we predict exactly how the position changes as a function of time and how do we derive the expression for period? Let's find out. So once again, let's stretch the spring and let it go. Now our goal is to figure out how the position of this mass changes as a function of time.

0:46If we call that position the X coordinate, for example, then all we need to do is figure out how the X coordinate changes with respect to time. How do we do that? How do we figure this out? Well, the best place to start is Newton's Second Law, which says that net force acting on any object equals mass times acceleration, okay?

1:05So what are the forces that are acting on this mass? There's a force of gravity acting downwards, but it's completely balanced by the normal force. So the forces in the vertical are completely balanced, so let's ignore them. What about the forces in the horizontal?

1:19Well, let's ignore friction and air resistance. We're keeping this simple, okay? Then there's only one force acting in the horizontal, the spring force. Because the spring is stretched, it'll pull back on this mass. And so that's the spring force. And the spring force depends on how much the spring got stretched from the equilibrium position.

1:37So if that stretch is delta x, for example, then the spring force would be in the opposite direction, and it's given by this expression. Where k is the stiffness constant or the spring constant, it tells you how stiff the spring is. And this force is often called the restoring force because the force is always directed towards the equilibrium position, the force tends to restore the mass back to the equilibrium position.

2:01Okay, now since there are no other horizontal forces acting on this mass, well, then this itself is the net force. And so for net force, I can plug in negative k times delta x, that equals ma. And so I have x in the picture. That's what I need, I want to find out how x changes as a function of time, right?

2:23Okay, but I have delta x. How do I get x from delta x? Well, we can just choose a coordinate system. We can call the equilibrium position as our x equals 0. And then the displacement from the equilibrium position can be just called as x. And so we can just call this as x now because of this coordinate system.

2:43And since we are, you know, since this motion is in one dimension only, we can forget about these vector rotations and we can use signs to represent direction. We can choose one direction as positive, another direction as negative. So the signs will take care of the direction, so I can just get rid of the vectors over here.

3:00I can say negative kx equals ma because this is in one dimension. Okay, so I have my equation with x, but I also have another unknown, a, which is the acceleration. And both are changing with time. Both are functions of time. So what do I do? Well, we know acceleration is, you know, derivative of velocity.

3:19It's, you know, you differentiate velocity with respect to time, you get acceleration. But velocity itself is a time derivative of position, and therefore acceleration becomes a second derivative, second time derivative of position. And so I can plug that in over here, so that our equation becomes negative kx equals mass times secondary derivative of position.

3:39And if we rearrange this, we now have an equation with just one variable, the position, which we want, and now we can try and solve this. But notice, this is not a regular algebraic equation. This is a differential equation. And differential equations are not that easy to solve.

3:56They require a lot of mathematical tricks and that comes from experience in solving lots and lots of them. So what we are going to do over here is we're going to try and guess a solution. That's also one of the tricks, okay? So what I mean by guessing over here is that we need a function which when, you know, when you take a second derivative, should be proportional to negative times itself.

4:26That's what we want. That's the function that we are looking for, right? And so we need to guess which kind, you know, which type of function gives me that result. So let's start with some guesswork. Let's say x equals t cubed. I'm guessing this because I need x to be some function of time.

4:45So let's just start with a polynomial t cubed. Let's see if this works. So the way to check whether it works is let's just differentiate this with respect to time. dx over dt would be just 3 t squared. And since I want a second derivative, let's differentiate one more time.

4:59You get 6t. Which means the second derivative is not proportional to negative times itself. So this function does not work. And hopefully you can see any polynomial function won't work because when you, you know, take derivatives of it, the power reduces and it just can't be proportional to itself.

5:20So this tells us that this can't be a polynomial function. Okay. What about an exponential function? That's another usual suspect. Well, let's see. If you take e to the power negative t, I could have chosen e to the power t as well, but I'm choosing negative t because I want a negative sign, so I'm guessing, I'm wondering whether this works.

5:39And remember that derivatives of exponential functions are itself, and so from the chain rule, again, the negative pops out. And so you get a negative e to the power negative t. And if you differentiate that one more time, another negative pops out making this whole thing positive.

5:54So this time we do get the second derivative proportional to, in fact, equal to itself, which is awesome, except that we don't get the negative sign. Oh, we were so close. We wanted the negative sign. Even if you choose, if you take e to the power plus t, you would get the same result.

6:13So this is close. This is a much better guess than before, but this, again, won't work. What other functions do we have? Well, another usual suspect is a trigonometric function. So let's try x equals cos t. You could've to have tried sine t as well. It would be a great idea to pause the video and see if this works yourself.

6:33All right? So what's the derivative of cos? It's a negative sine. And what's the derivative of sine? It's cos. And so the minus sign remains. And look, we do get that to be equal to negative times x, which means this works. This works. And so if you're taking sine function, that would have also worked.

6:55And so this tells us that our solution has to be either a sine or a cos function, meaning it is a sinusoid. And so we have mathematically convinced ourselves that it's a sinusoid. It pops out of this differential equation. Okay, but this is not a general solution because remember in our general solution, the second derivative should be equal to negative times some constant k divided by m times x.

7:28So clearly there must be some constants over here, some k's and m's and maybe some other constants over here. So this is the backbone, but the exact solution would have some constants. And so now the question is what would those constants be? Okay, so let's look at this sinusoid in a little bit more detail.

7:48So we have our x and we have our t, and the general equation for a sinusoid is given this way. We have introduced three constants. So let's look at each one of them. The A represents the amplitude, the maximum value. In our case, it literally represents the maximum distance of that mass from the equilibrium position.

8:12If the amplitude is lower, the height of this graph becomes lower. If the amplitude is higher, the height of the graph becomes higher. So that's the amplitude. What is this? This is the Greek symbol omega. And this controls the spacing, or in this particular case, the period.

8:28So changing that will change the spacing of our sinusoid. And we will see exactly how it changes in a while, but as of now, that's all that matters to us. That changes the spacing. Okay. What about this one? This is phi, and this controls the initial phase. Changing phi value basically shifts the graph left and right.

8:53All right, so how do we find the values of A, omega, and phi? Well, A and phi, their values purely depend on the initial conditions. Initially, when you pull that mass, you have two choices, what the initial position would be and what the initial velocity would be, right?

9:10Those two choices will directly tell us, will directly result in the amplitude and the phi value. But now the main question is, what about omega? Because the omega that controls the spacing, it controls the period. So how do we figure what omega is? Well, let's just substitute it in this particular equation, or we can just differentiate it twice and compare it with this equation and see if we can find the expression for omega.

9:36Great idea to pause the video and see if you can try this yourself. Okay, so if you take the first derivative, dx over dt, A is a constant, derivative of cos is a negative sign, and from the chain rule, the omega pops out. So I get negative omega A sine, omega plus phi.

9:52And if I differentiate this one more time, derivative of sine is cos. So the negative sign stays, another omega pops out, so I get negative omega squared, A times cos omega plus phi and A cos omega plus phi is x. So I can just substitute that over here. And if I compare this with this, what do we notice?

10:12Well, we see that omega squared equals k over m. And so from there, we have found omega. That's the square root of k over m. Okay, but how do we figure out the period from omega? Well, let's say our initial condition is such that phi equals 0, okay? That just simplifies the whole thing.

10:38Then x will be A cos omega t. And so this graph will be shifted to get a cos function, right? And we know this is the period, the time it takes to finish one full cycle. But we also know when cos finishes one full cycle, the argument becomes 2 pi, right? At this point, the argument is 0, and when it finishes one full cycle, the argument reaches 2 pi.

11:04which means when t equals capital T, this entire thing inside should be 2 pi. In other words, omega times capital T should equals 2 pi. From here, T should equals 2 pi divide by omega. And since I know what omega is, I can just plug that in over here and rearrange, and that gives me the period.

11:27This is exactly the expression that we found empirically. But that's not all. We can now understand what omega even is. See from here, omega is 2 pi over T. 1 over T is f, the frequency. So omega is 2 pi times f. What does that even mean? Well, frequency represents number of oscillations per second, number of cycles per second.

11:55Well, omega is just 2 pi radians times that. It gives us radians per second. So frequency is number of cycles per second. Omega is radians per second. So omega is also often called the angular frequency. I used to always wonder why we use radians over here when there are no angles or circles.

12:15Well, that's because trigonometric functions take radians as the argument, right? So that's why we are forced to work with the radians. That's where it comes from. But what's important is we have mathematically solved for the position. We've shown that it's a sinusoid, and we have derived the expression for the period.

12:31But we can go one step further. We can now write down expressions for velocity and acceleration. Velocity is just the derivative of this. So if you differentiate x with respect to time, you know, derivative of cos is, hit a sine one more time. Again, omega pops out.

12:49So we get negative omega A, sine omega t. And acceleration will be derivative of this, derivative of velocity. And that will be, again, well, you can differentiate this and you get negative omega squared A times cos omega t. And from this, we even figure out what the maximum values for velocity and acceleration are.

13:09The maximum value for velocity is omega A. You can call that as the velocity amplitude, which is basically the maximum speed of that mass. And over here, omega squared A represents the acceleration amplitude. That is the maximum acceleration of our mass.