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Simple Harmonic Motion of Pendulums

Patrick Ford
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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 it 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. So in mass spring systems, as soon as you let it go, the restoring force, the restoring force was back towards the center and that restoring force was equal to negative K. X. Which means that the acceleration was negative K over 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 negative MG times the sine of theta. Which means that acceleration is equal to negative G times sine of data. Now there's a couple of things that we need to remember because we're dealing with datas and so we have to make sure our theta and our calculator are in radiance. So just go ahead and make sure that you're in radiance mode. So the other 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 the square root of K over M. Well, for pendulums, the same relationships to hold except now we have a different formula, the square root of G over L. So we got K over M and G over L. The way to remember that is that K comes before M in the alphabet and then G comes before L. And 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 had a force net force was proportional to the distance and then K was just some constant. Well, for pendulums we've got M. G. Those are just constants here, but now we've got F that is directly proportional to the sine of theta, not data itself. So this this data is like wrapped up inside the sine function here. So we've got an issue there. Well, fortunately what happens is for small angles, assuming that we have radiance, the sine of theta is approximately equal to theta itself. So we can make a simplification. The restoring force is not MG sine theta. It's just negative MG data. So that takes care of that issue. So that means that we can sort of simplify this as negative MG theta as the restoring force and the acceleration is negative G. Times data. But honestly, most of these questions are actually going to 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 four kg 4 kg. I'm just gonna write a four there and we're pulling this thing out by 3. 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 em I've got G. And now I just need Theta. But its data, the 3.5 degrees will know because we have to remember that data must be in radiance. So the first thing we have to do, We have to convert the 3.5° into radiance. And so what we do is we want to get rid of the degrees and we want radiance to to go on the top. So you got something radiance divided by something degrees. So that relationship is Pi divided by 180. So remember that there are pi radiance inside of 180°. So if we wanted this in radiance, this is just going to be 0.061 rads. 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 we get is a restoring force. That's equal to 2.39 Newtons, that's a 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 tea then I've got omega equals two pi frequency equals two pi over T equals square roots of G. Over 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 tea on top, I can basically take these two things and flip them. I can flip the fractions. And if I do that I'm gonna get t divided by two pi because I flip that one. But then I gotta flip the other side. So I'm gonna get square root of L. Over G. So even though we're starting off with squared of G over L. Sometimes depending what we're solving for, we can end up with L. Over G. But just remember that omega is always G. Over L. So what we get is that the period is equal to two pi times the square root of L. Over G. So now we can just go ahead and solve the period. So t. Is equal to two pi and I've got the square root of 0. divided by 9.8. And what you'll end up with is a period of 1.0 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 the period was all the way to the other side and then all the way back. Well for a pendulum it's going to be the exact same thing. So for our pendulum, the half period is to get all the way to the other side. That's T. Over two and then it's got to go all the way back and that's gonna be another T. Over two 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 is when this thing is going to have its maximum speed. So what we're really looking for is we're really looking for the quarter period. And so if we're looking for the quarter period then we're just going to 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, reach its maximum speed. Alright guys, so that's it for this one. Let's take a look at a different exam.