Hey, guys. So for this video, we're gonna take a look at a different kind of simple harmonic motion, the simple pendulum. And what we're gonna see is that just like mass spring systems, pendulums also display simple harmonic motion. So it means we're gonna see the same kind of equations. It's just the letters are gonna be slightly different. So let's take a look.

When we had a mass spring system, we would take this mass and pull it out and then let it go. And so given some spring constant here, and given the fact that we pulled it out to some amplitude, it would just go back and forth between those amplitudes. And we say that those amplitudes were basically equal to the maximum displacements that we pulled them away.

So for pendulums, it's slightly similar. So except we've got a pendulum of length \( l \), and now we're gonna pull this mass out to some theta, some angle \( \theta \), and we're gonna let it go. So now what happens it's gonna oscillate between these two points, and these are the amplitudes. So the positive \( a \) and negative \( a \). And this is basically just equal to the maximum angle that you pull it out for.

In mass spring systems, as soon as you let it go, the restoring force was back towards the center and that restoring force was equal to \( -kx \), which means that the acceleration was \( -\frac{k}{m} \times x \). Well, what is that restoring force for pendulums? Well, for pendulums, we've got the gravity of the object that's hanging, and then we've also got a tension force. So what happens is that we've got a component of gravity that wants to pull it back towards the center. That is that restoring force. It's the \( x \) component of gravity. And so that force is equal to \( -mg \times \sin(\theta) \), which means that our acceleration is equal to \( -g \times \sin(\theta) \).

Now there's a couple of things that we need to remember because we're dealing with thetas, and so we have to just make sure our theta and our calculator are in radians. So just go ahead and make sure that you're in radians mode. So the other variable with mass spring systems, the other really important variable was \( \omega \). And we related \( \omega \) using linear frequency and period, and we said that for mass spring systems, that \( \omega \) was \( \sqrt{\frac{k}{m}} \). Well, for pendulums, the same relationships still hold except now we have a different formula. \( \sqrt{\frac{g}{l}} \). So we've got \( \frac{k}{m} \) and \( \frac{g}{l} \). The way to remember that is that \( k \) comes before \( m \) in the alphabet and then \( g \) comes before \( l \) in the alphabet as well. So that's one way you can remember that.

So these are the equations for the omegas of both of these pendulum and mass spring systems. Make sure you memorize these because these are really important.

Okay. So we've got an issue here. So for simple harmonic motion, we need that the restoring force has to be proportional to the deformation. So for mass spring systems, what happens is we have force. Net force was proportional to the distance, and then \( k \) was just some constant. Over pendulums, we've got \( mg \). Those are just constants here. But now we've got \( f \) that is directly proportional to the sine of theta, not theta itself. So this \( \theta \) is like wrapped up inside this sine function here.

Well fortunately what happens is for small angles, assuming that we have radians, the \( \sin(\theta) \) is approximately equal to \( \theta \) itself. So we can make a simplification. The restoring force is not \( mg \sin(\theta) \), it's just \( -mg\theta \). So that takes care of that issue.

So that means that we can sort of simplify this as \( -mg\theta \) is the restoring force, and the acceleration is \( -g \times \theta \). But honestly, most of these questions are actually gonna come from these relationships here.

So let's take a look at an example. So in this example, we're given the length of a pendulum. That pendulum is equal to 0.25. We've got the mass which is equal to 4 kilograms. Four kilograms. I'm just gonna write a 4 there. And we're pulling this thing out by 3.5 degrees and we're supposed to find a couple of things like the restoring force. So for part a, I'm just gonna take a look at my equations for restoring force.

Well, for a pendulum, that restoring force is equal to \( mg\theta \). So I'm gonna say that the magnitude of the restoring force is equal to \( mg\theta \). I've got \( m \). I've got \( g \). And now I just need \( \theta \). But is \( \theta \) the 3.5 degrees? Well, no, because we have to remember that \( \theta \) must be in radians. So the first thing we have to do is we have to convert the 3.5 degrees into radians. And so what we do is we wanna get rid of the degrees and we want radians to, go on the top. So you got something radians divided by something degrees.

So that relationship is \( \pi \) divided by 180. So remember that there are \( \pi \) radians inside of a 180 degrees. So if we wanted this in radians, this is just gonna be 0.061 rads. And this is a very very small number. So that's what we're gonna use for our equation. So we've got the magnitude of the force is just gonna be the mass times gravity, which is 9.8, and then 0.061. So what we get is a restoring force that's equal to 2.39 newtons. That's our restoring force.

So what is the period of oscillation look like? So now for part b, we're looking for the period of oscillation, which is that \( t \) variable. So let's take a look at our equations. So if I'm looking for \( t \), I'm just gonna look at my angular frequency equation. So I've got a \( t \) in here and I've got this whole entire, expression. So let me go ahead and write it out.

So if I'm looking for \( t \), then I've got \( \omega \) equals \( 2\pi \) frequency equals \( \frac{2\pi}{t} \) equals square roots of \( \frac{g}{l} \). Now I'm not told anything about the frequency or angular frequency. So if I wanted to figure out the \( t \), I have the length of the pendulum, and so I can use basically these relationships here. Those those last two terms. So here's what I'm gonna do. If I want \( t \) on top, I could basically take these two things and flip them. I can flip the fractions. And if I do that, I'm gonna get \( \frac{t}{2\pi} \) because I flipped that one, but then I've got to flip the other side. So I'm gonna get \( \sqrt{\frac{l}{g}} \). So even though we're starting off with \( \sqrt{\frac{g}{l}} \), sometimes depending on what we're solving for, we can end up with \( \frac{l}{g} \). But just remember that \( \omega \) is always \( \frac{g}{l} \).

So what we get is that the period is equal to \( 2\pi \) times the square root of \( \frac{l}{g} \). So now we can just go ahead and solve for the period. So \( t \) is equal to \( 2\pi \), and I've got the square root of 0.25 divided by 9.8. And what you'll end up with is a period of 1.00 seconds. So that's the period of oscillation of this pendulum here.

Okay. So now for this final thing, for this final, question, we're asked for the time that it takes for the mass to reach its maximum speed. So let's see what's going on here. So as this pendulum is swinging, it's making some some oscillations and some cycles. So for mass spring systems, this the period was all the way to the other side and then all the way back. Well, for a pendulum, it's gonna be the exact same thing. So for our pendulum, the half period is to get all the way to the other side. That's \( \frac{t}{2} \), and then it's gotta go all the way back, and that's gonna be another \( \frac{t}{2} \) for the full period. So that means that the point from its amplitude all the way down to its lowest position, that is actually going to be a quarter period. And at this lowest position, and with is when this thing's gonna have its maximum speed. So what we're really looking for is we're really looking for the quarter period. And so, if we were looking for the quarter period, then we're just gonna take a quarter of one second, and so that quarter period is just going to be 0.25 seconds. That's how long it takes to swing down to reach its maximum speed.

Alright, guys. So that's it for this one. Let's take a look at a different