SHM of simple pendulums

0:00- [Instructor] This pendulum helps us detect Earth's rotation. But what's important for us is that these kinds of pendulums are called simple pendulums. What makes them simple is that the mass of the string or the rod is negligible compared to the mass hanging from it.

0:16So we can model it as a point mass attached to a massless string or a rod. That's what makes it a simple pendulum. In contrast, if you had say, a ruler, and you nailed it down over here and you made it oscillate, well, now you cannot model this to be a point mass attached over here because the mass is uniformly distributed.

0:38This is what we call a physical pendulum. So let's analyze simple pendulums. If the pendulum is placed here at rest, it remains at rest, right? So we'll call this position the equilibrium position. This is the position where the angle that the pendulum makes with the vertical is zero.

0:55And while we're at it, we can also choose a coordinate system. Let's choose counterclockwise as positive direction and clockwise as negative direction. Now, if we were to pull this bulb and let it go, it swings about the equilibrium position. And our goal here is to analyze this motion.

1:13Well, we can start by drawing a free body diagram. So what are the different forces acting on this bulb? Well, there's a force of gravity and there's a force of tension. And since the two forces are not in the same direction, we can resolve the force of gravity along the tangential part and the radial part.

1:29So we get a radial component, we'll call it as Fg.radial, and we have a tangential component. Can we calculate these two components? Yeah, this angle should be equal to this angle, theta. They are corresponding angles. Can you see that? You can see that these two are vertical lines that are intersected by one single line over here.

1:49And we can do a sense check. If this angle becomes smaller, look, this angle also becomes smaller. If the angle becomes larger, this angle becomes larger. And we know that this force of gravity is just mg, then this becomes mg-cos-theta, and then this becomes mg-sin-theta.

2:05Okay, so let's start by analyzing the radial components. Does the tension force balance mg-cos-theta? No, because remember, this bulb is going in a circular path, which means there must be an acceleration towards the center, the centripetal acceleration. So we expect the tension to be higher than mg-cos-theta.

2:24And that makes things a little bit complicated. I mean, you have an acceleration in the radial direction. The bulb also is accelerating in the tangential direction. And these accelerations keep changing with time. So it's a little complicated. So what do we do?

2:40Well, instead of thinking that the bulb is undergoing a translational motion, we can think that the bulb is actually rotating about this point. And instead of thinking in terms of forces, we can start to think in terms of torques. But I'm sure we might be wondering, how does that help?

2:57Well, let's see. Let's quickly recap torques. If there's a force acting on a bar at some distance r from the pivot point, then the magnitude of the torque produced by this force equals r times the perpendicular component of that force. What's the perpendicular component?

3:11Well, it's the component that's perpendicular to r. And that makes sense, right? That's the component that is trying to make the whole thing rotate. There's also the paddle component that doesn't do anything that is just pushing it and that does not make it rotate it, so it's the perpendicular component that matters.

3:29This means if this angle between the force and r, if this angle is large then the perpendicular component would be very high and the torque would be very high. If that angle is small, then the perpendicular component will be very low and the torque would also be pretty low.

3:43And if that angle is zero, then the perpendicular component goes to zero. Again, that makes sense if you push it this way, that's not gonna make things rotate. Similarly, if that force was aligned this way, that would also not make it rotate, so the torque would be zero.

3:58So if the force is directly pointing away from the pivot or pointing directly towards the pivot, the perpendicular component is zero and it does not produce any torque. Okay, so now let's come back over here. There are three forces that we have to think about.

4:12Let's think about the torques that they are producing about this point. If you look at the force of tension, it's directed directly along the pivot point, which means it does not produce any torque and I can stop worrying about it. What about this radial component of gravity?

4:28Hey, that's directly pointed away from the pivot point. That too does not produce any torque, which means there's only one force producing torque. That is this one. So notice thinking in terms of torque actually greatly simplifies this. Okay, so what is the torque produced by this component of the force on this bulb about this point?

4:49Well, that torque is going to be r. Well, in this particular case, r is basically the length of the pendulum. Let's just call that as l. And multiplied by the perpendicular component of this force, well, mg-sin-theta itself is the perpendicular component, right?

5:04This is the tangential component. Tangents are perpendicular to the radius. So that is the perpendicular component. So it's multiplied by mg-sin-theta. And this is the only torque that is acting, which means this is the net torque acting on this bulb. But this is just the magnitude of the torque.

5:20What is the direction? Well, notice that when theta is positive, the torque is negative. And if theta becomes negative, well, the torque would be positive, which means the torque is in the opposite direction of theta itself. And for that, we can put a negative sign over here.

5:37So notice the torque always points towards the equilibrium position. And the farther you go, the magnitude of the torque increases because as theta increases, sin-theta increases and this value will increase. So this is a restoring torque, which means the torque tries to restore the bulb to its equilibrium position.

6:00And that's why the bulb oscillates. But now the main question is how exactly does the bulb oscillate? Or in other words, how does theta change with time? Well, let's compare this to something that we have probably already seen before, springs. We've already seen that if a stretched spring will experience a net force, which is also restoring.

6:23It's also directed towards the equilibrium position, and it's proportional to the displacement from the equilibrium position. When that happens, the mass oscillates sinusoidally, right? And this oscillation is what we call a simple harmonic motion. Now let's compare this with our pendulum case.

6:44In the case of pendulum, we have something very similar, but instead of the force, we have a torque. And instead of thinking in terms of linear displacement, we're thinking in terms of angular displacement. So the question is, will this bulb also oscillate as a simple harmonic motion sinusoidally?

7:02Pause and think about this. Well, let's see. The main criteria for the simple harmonic motion is that the restoring force must be proportional to the displacement from the equilibrium position. But in our case, look, the restoring torque is proportional to the sine of the angular displacement.

7:23It's not directly proportional to angular displacement. It's proportional to the sine of the angular displacement. So although this is a periodic motion, it's an oscillatory motion, it is not an SHM. Which means the oscillations are not exactly sinusoidal. However, what will happen if we restrict theta to small values?

7:47Think about this for a second. If we call the length of this pendulum to be 1, okay, then this perpendicular distance becomes sin-theta, because sin-theta is opposite side divided by the hypotenuse. Hypotenuse is 1, so sin-theta directly becomes the opposite side, right?

8:05And this arc length, by definition, is theta in radiance. And so, in general, you can see theta is bigger than sin-theta. Well, because this is an arc over here, right? In fact, for very large angles, you can clearly see that this arc length is much bigger than sin-theta.

8:26But what if we restricted theta to small values? Let's say between negative 30 to positive 30 degrees. Then notice the sin-theta and theta value come very close to each other. Theta is still bigger than sin-theta, but we can approximate it to be very close to each other, right, we can say theta in radians is very close to sin-theta.

8:46And just to get some feeling for numbers, these are the values of theta in degrees. These are the values of theta in radians and these are the values of sin-theta. Notice that as long as you're below 30 degrees the values are equal to the nearest 10th, but about 30 degrees notice that's no longer true.

9:05So as long as we're okay with this level of precision, we can say sin-theta equals theta for theta less than 30 degrees. Of course, if you wanted higher precision then we would have to restrict it to even smaller angles. Anyways, this means if we restrict our angles to less than 30 degrees sin-theta can be approximated to theta to the nearest 10th, and look what happens to our restoring torque now, that becomes proportional to theta, the angular displacement from the equilibrium position.

9:32Very similar to that of the springs, which means now, we will get a simple harmonic motion. So for small angle approximations, the pendulum executes a simple harmonic motion. All right, so let's clean this all up. And since we know theta executes an SHM, we can now draw a graph for it.

9:52So if we graph theta as a function of time, what do we expect? Well, if we pull this bulb to some angle theta-naught, then that becomes the amplitude. It'll now swing between theta-naught and minus theta-naught. We're assuming there are no drag forces or any other resistive forces so that it'll keep swinging forever.

10:12And so now, the oscillations would be sinusoidal. So right now, we are over here at the positive amplitude. As time goes by, it goes through the equilibrium position, negative amplitude, back to the equilibrium position, positive amplitude, and so on and so forth.

10:29And the period of this oscillation is given by this particular expression. The last question over here is, what about energies? What are the different kinds of energies involved? Well, since the bulb is moving, there is kinetic energy associated with it. And because of gravity, we can also, the Bulb Earth system also has gravitational potential energy.

10:52And because there are no drag forces, there is no external work being done on our Bulb Earth system. I mean, there is a force of tension, for example, but the tension force doesn't do any work because it's always going to be perpendicular to the direction of the motion.

11:04And since there is no external work being done, the total mechanical energy stays conserved. And so the potential energy plus kinetic energy will always be the same. So great idea to pause the video and think about where the, you know, at what positions would we expect maximum gravitational potential energies?

11:24And at what positions would we expect maximum kinetic energies? So pause and think about this for a while. Okay, so we know that the bulb comes to a rest at the amplitude positions. So that's where kinetic energy is zero. That means all the energy at those positions must be potential, gravitational potential.

11:44So at these amplitude positions, we have maximum gravitational potential energy. And then when it comes to the equilibrium position, that's when the bulb has the maximum speed, because until then it speeds up and then it slows down. And so at that point, we would expect it to have maximum speed, maximum kinetic energy.

12:05So you get maximum kinetic energy at the equilibrium position.