Simple Harmonic Motion of Pendulums - Online Tutor, Practice Problems & Exam Prep

Simple harmonic motion is exemplified by pendulums, which oscillate between maximum angles (amplitudes) when displaced. The restoring force is derived from gravity, expressed as F = -mgsin(θ), simplifying to F = -mgθ for small angles. The period of oscillation is given by T = 2πlg. Understanding these relationships is crucial for analyzing pendulum dynamics.

Simple Harmonic Motion of Pendulums

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

Example

Hey, astronauts. Let's do an example. So this is actually a really common type of problem with pendulums. You're landing on an unfamiliar planet. You're constructing a simple pendulum. So let's go ahead and check it out. We've got this pendulum that's a length of 3 meters and a mass of 4 kilograms. So I've got l=3 and mass is equal to 4. Now we're gonna release this pendulum. We're gonna pull it back to 10 degrees from the vertical right there, and then we're gonna let it go. So it's just gonna oscillate back and forth. So that's our initial push or pull. That's theta max, and that's equal to 10 degrees. Just remember that because this is in degrees, anytime we're gonna be working with theta, we're gonna have it in radians. So just be aware that you might have to do some conversions. Okay. And the last thing we were told is that it's one full cycle at 2 seconds. So what does that mean? It takes 2 seconds for this thing to go back and then all the way back to the other side where it started from. So that is the full period. Right? That is t=2 seconds. And now what we're supposed to do is calculate the acceleration due to gravity at the surface of this planet. What does that mean? Well, I just want you to be aware of this. But just remember that on Earth, g=9.8. You probably know that by now. But that at any other planet, we cannot take that for granted. So at any other planets, g does not equal 9.8. Right? So just be aware of that. Sometimes you just look at g and you're like, oh, that's 9.8, but it's not in these situations. Okay. So which equation are we gonna use? Because we actually have a lot of them involve g. So I've got one here, I've got one here, and I've got one here. Well, let's take a look at all my variables. I know the length of the pendulum. I'm told the mass. I'm told the period and I'm told the theta max. Okay. Well, I've got mass here, but I have the theta. Okay. But it's theta max. Okay. So I might be able to use it here, but I don't have the restoring force, so I wouldn't be able to solve for g. And in this case, I also don't have the acceleration. So it means I can't use the force of the acceleration because I'm not given that information. The only other equation that I can use is this big omega equation involving square root of g over l. So if I'm trying to figure out what g of the planet is, I'm gonna go ahead and use that big omega equation. So I'm gonna write it out. So I've got square root of g over l. Okay. And we're looking for what the gravity on this particular planet is. So again, what do we have? Well, I have the length of the pendulum and I also have the period. So I can actually just go ahead and use these last two equations because I only have one unknown. So let me go ahead and write that out. So I've got 2 pi divided by the period. The period is equal to 2 seconds equals the square root of the gravity on that planet divided by the length of which is what? We're given that it's 3. So let's go ahead and do some simplification. We've got the twos cancel and you got pi equals square root of g over 3. So now, in order to get through to the square root, just got to square both sides. Okay. π=2 equals g planet over 3, And so then I just gotta move the 3 over. So the g planet is equal to 3π=2. So that's equal to 29.6 meters per second squared, which is much higher than our 9.8. Okay. So that's it for this one. Let me know if you guys have any issues, and let's keep going.

Pendulum Equations

Hey, guys. So we talked about some of the differences between mass spring systems and pendulums. We discussed the differences between f, a, and omega. So now you might be wondering what the other equations look like. I'm going to give them to you in this video. So remember that we have a pendulum here at some length l and you're going to take this mass and you're going to pull it out to some angle, which is theta. You're going to let it go and it's going to keep going back and forth. Right? And so in a mass spring system, whenever you pulled something back away from the equilibrium position, that distance was x. It's the same exact thing for pendulums. So I'm going to draw this line here, and this x distance represents the distance away, the deformation from the equilibrium. But now in order to solve that, I've got some trig to involve here. I've got a triangle. I've got the hypotenuse. I've got the angle and I've got the opposite side. So from SOHCAHTOA, I can relate this x at any point during the pendulum swing by l⁢sin⁢θ. But remember that when we're talking about pendulums, sine of theta and theta are, like, just about the same thing. So I can simplify this by saying that x=l⁢θ. And remember that we're going to be working with radians when we're doing this. Right? So x=l⁢θ. So we said that in a mass-spring system, you're pulling this thing all the way back, and its maximum displacement or its maximum deformation was equal to the amplitude. Right? Swings back and forth between the two amplitudes. So now what happens, this it's the same thing for pendulums. So you're going to swing you're going to take this thing all the way back to some theta. And once theta is at its maximum value, x is at its maximum value away from the equilibrium position. And that's what we said that the amplitude was. So it's the same thing. But because x=l⁢θ, then that mean the amplitude x max is going to be at l times theta max. So that means that for pendulums, the maximum x is the amplitude and that is equal to l times theta max. Okay. So we also had a couple of other things about mass spring systems. We have the maximum values for x, v and a. So we said here at the very middle that the velocity is at its maximum, and at the end points, the on either on either side. It's the same thing for pendulums, except for now, now that what's happening is at the bottom of its swing, we've got a velocity in this direction, and that is going to be the maximum. And then we've got a max over here that wants to pull it back towards the center. So it's just the same exact thing for pendulums. Except now, so our equation for mass spring systems was that v max was a omega and a max was a omega squared. It's the same thing for pendulums. We've got a omega and a omega squared. The only difference is that a=l⁢θmmax. So what I'm going to do here is I'm going to just replace these a's with lθmmax. So v max is a omega, then we can also say that it's equal to l⁢θmmax⁢ω, just replacing the a with l theta max. And so a omega squared is a=lxmmaxorθmmax⁢ω2. Okay. So the other thing is that the omegas are slightly different for pendulums as well. Remember that ω is g/l, whereas for mass springs, it's k/m. So just remember those two. Okay. So it's very rare for you actually to be asked what the horizontal displacement is away from the equilibrium. You most likely won't be asked that. What's more common is you'll be asked what the angle is at a certain time. And so if you're ever asked for what theta is and you're given what t is, you're going to be using this equation, theta t. It looks just like the XT, VT, and a at equations are. That's it. So that's basically it. So let's go ahead and take a look at an example. So we've got this mass and it's hanging from the spring, so we've got this pendulum system here. So I'm going to go ahead and draw that out. I've got the equilibrium positions down here, and this thing is just going to swing back and forth until it reaches the other side. So that's going to be like over here somewhere. Okay. So we're told a couple of things. We're given that the mass is equal to 500 grams, so that's equal to 0.5 kilograms. Just remember everything has to be in SI. The length of the string, which is the length of the pendulum, is equal to 0.4 meters. Again, we're given centimeters. And then we're told that the object has a speed of 0.25 meters per second as it passes through the lowest point. What does that mean? Remember, we just said at the lowest point, this thing has its maximum speed. So what they're really telling us about this 0.25 meters per second is that v max is equal to 0.25. And we're supposed to figure out what the maximum angle is in degrees from the equilibrium position. So we're supposed to figure out basically how far in terms of theta does this reach in degrees. So just remember that when we're working with all of these equations, we're going to get radians. And so I'm just going to have to do a conversion to get it back into degrees at the very end. So we're looking for theta max. Right? So I'm going to take a look at all of my equations that have theta max in them. So I don't want to go all the way up there. I'm just going to paste all of these equations right here. So I've got all my theta maxes are in these equations. I got theta max here, theta max, theta max, and theta max. So let's take a look at all of these equations. So I don't have anything about X max or the amplitude, but I do know what the length of the pendulum is. So I know what the length is. I've got this length right here. Let's take a look. I do know what the maximum velocity is. So I'm going to go ahead and circle that. I've got the max velocity. I don't know what the amplitude is, so I'm just going to cross out all the As. And I also don't know what the a max is. The last thing is that I can only use this bottom equation if I'm given something about time. But I'm not given any time. And because I'm not given at a time and I'm supposed to find what theta max is, I have 2 unknowns, and I can't use that equation either. So let's look at the equation that I know the most about. I know what v max is, and I know what l is. So in order to find theta max, I'm just going to need to find out what this omega is. So that's basically it. Let me write out that equation for v max. So let's just try that. So we've got vmax=l⁢θmax⁢ω. So if I wanted to figure out what this theta max is, let me just go ahead and rearrange that equation. So I've got θmax=vmax/l⁢ω. I'm just going to move these guys to the other side. Right? So I'm just going to divide them over to the other side. Okay. So I've got everything. I know what this v max is, and I've got l. The the last thing I just have to figure out is what omega is. So I'm going to go over here and do that. Well, omega is equal to what? Let me write out that big omega equation that I have. I've got 2 pi frequency, 2 pi divided by t, and then I've got that is equal to square root of g over l. Now, I'm not given anything about frequency or the period, so I'm just going to have to use this last one, square root of g over l. So I've got ω=gl. Well, g is equal to 9.8 and then I've got l is equal to 0.4. So what I get is that omega is equal to 4.39, and that's going to be rads per second. I think that's what I get. So I get 4.39 rads per second. So, sorry, not 4.39. I get, 4.95. So I got 4.95 rads per second. Great. So now I'm just going to plug that back into here. Okay. So I've got θmax=vmax/l⁢ω. Remember, theta max is going to be in radians, and that's going to equal, 0.126. That's in rads. So now the last step is just to convert it back to degrees. So how do I do that? So I've got θmax=0.126⁢180degrees/piradians. So that's going to cancel out, and then we're going to get 7.2 degrees. That's the maximum angle that we make for the pendulum. So let me know if you if you guys had any questions. Let's keep going for now.

Example

What's up, guys? Let's take a look at another pendulum problem. So we've got a 100 gram mass. It's on a 1 meter long string, and it's pulled 7 degrees to one side, and then it's let go. So we pull this thing back out to 7 degrees, let it go, and it's going to swing back and forth. And we're supposed to figure out how long it takes to reach 4 degrees on the opposite side. So what does that mean? So I'm just going to draw a sketch of my pendulum real quick. So I've got this little pendulum right here. Got the equilibrium position that's over here. And what happens is when I pull this thing out to 7 degrees, I know it's not to scale. Now this thing is just going to go back and forth until it reaches the opposite side. Okay. So what we're supposed to figure out is if it starts at 7 degrees here, how long does it take for it to swing through this angle right here and then reach on the opposite side where this angle is going to be 4 degrees. But because this is the equilibrium position at the very middle point here, this actually is going to pick up a negative sign. It's important. It's kind of like if you go back and forth to the mass-spring system, the equilibrium is 0, so it goes from negative to positive. It's the same thing. So we're basically trying to figure out how long does it take for it to get from there to there. So let's take a look at our formulas and our variables. Mass is equal to 0.1. I've got the length of the pendulum as 1 and then I've got theta max, which is my initial push out or pull out whatever. My theta max is equal to 7 degrees. And then it's supposed to be 4 degrees on the opposite side. So I've got like this theta final is supposed to be negative 4 degrees. Okay. So let's take a look at our equations. I've got thetas all over the place, but I only have t's in 2 places. So let's take a look. I either can use the n cycles equation, which has this t variable here, or I can use this theta t equation. Now in order to use the n cycles equation, I would have to know some information about cycles, the period, or the frequency. But I don't know any of those things. Whereas if I look at this equation here, I have what theta max is, and theta at some later time is kind of what I'm looking for. And this has got some omegas and t's. So let's go ahead and use that equation. Let me go ahead and write it out. So I've got theta at some later time is going to be equal to theta max times the cosine of omega t. Because we're working at this cosine just to make sure that you're in radians mode. Okay. So this theta of t here is really like the final theta, right? It's theta at some later time and that's equal to negative 4 degrees. And then theta max, we said was equal to 7 degrees. Now we just have to convert these into radians mode. So I'm going to go ahead and do that. So let's go ahead and actually move that over here. So if I want to convert this 4 degrees into radians, I've got to multiply it by pi over 180. And so what I get is, sorry, negative 4 degrees. What I get is negative 0.07. If I do the same exact thing for the 7, so this is going to be negative 0.07. If I do the same thing for the 7 degrees, I am going to get 0.12, and that's going to be positive. Okay. So now let's go ahead and just fill out those, just plug in those numbers. So I've got negative 0.07 equals 0.12, and I've got a cosine of omega times t. And this t is really what I'm looking for. That's my target variable. So now, the only thing I need, is to go ahead and start solving some of this stuff. So I'm going to go ahead and move this point 12 over to the other side because I want to start getting this cosine omega t by itself. I want to get this t by itself, but it's like all wrapped up in this cosine function. So let's go ahead and move everything else to one side. If you go ahead and do that, you divide those 2 decimals over and you're going to get negative 0.58. And that's equal to cosine times omega t. Now again, I want this t, but it's all wrapped up inside this cosine function. So how do I get it out? Well, if you have a cosine, you need to get rid of it. You have to multiply by the inverse cosine. So what I'm going to do is multiply by the inverse cosine of negative 0.58. And if I do the same thing to this side, the cosine will go away. So it means I just get omega t on that side. So just make sure that you're in radians and what you're going to get out of this is you're going to get 2.19 and that's going to be in rads. And now you're going to have omega times t. So I'm really close. All I have to do is just figure out what this omega variable is. So how do I figure out what omega is? Let's go over here and find that. So I've got omegas all over the place in my equations chart. So I've got all these like v maxes, a maxes, and x maxes that all have omegas. But I'm not given any information about v maxes or amaxes. What do I know instead? Well, let's see. I've got the length of the pendulum. So I'm going to use this omega that relates the frequency period and the length of the pendulum together. So I don't have the frequency and the period, so I can't use those, but I can use the square root of g over l. So let's go ahead and plug that in. So the square root of g is 9.8 over, and then what's the length of the pendulum? It's just 1. So that's going to be 3.13 radians per second. So now, I just take that number. So I've got 2.19 equals 3.13 t. Now it just divides it over to the other side, and we get a t is equal to 0.7 seconds. And that is the answer to the problem. Let me know if you guys have any questions. I'll be happy to help you out. Alright. That's it for this one.

Here’s what students ask on this topic:

What is the formula for the period of a simple pendulum?

The period of a simple pendulum, which is the time it takes for one complete oscillation, is given by the formula:

$T=2\pi \sqrt{\frac{l}{g}}$

Here, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity (approximately 9.8 m/s² on Earth). This formula assumes that the amplitude of the oscillation is small, so the approximation $\sin \theta \approx \theta$ holds true.

How does the length of a pendulum affect its period?

The period of a pendulum is directly related to the square root of its length. According to the formula:

$T=2\pi \sqrt{\frac{l}{g}}$

As the length $l$ increases, the period $T$ also increases. Specifically, if you double the length of the pendulum, the period increases by a factor of $\sqrt{2}$. This relationship shows that longer pendulums swing more slowly than shorter ones.

What is the restoring force in a simple pendulum?

The restoring force in a simple pendulum is the component of the gravitational force that acts to bring the pendulum back to its equilibrium position. For small angles, this force can be approximated as:

$F=-\mathrm{mg}\theta$

Here, $m$ is the mass of the pendulum bob, $g$ is the acceleration due to gravity, and $\theta$ is the angular displacement in radians. This approximation holds true for small angles where $\sin \theta \approx \theta$.

How do you convert the angle of displacement from degrees to radians in a pendulum problem?

To convert the angle of displacement from degrees to radians, you can use the conversion factor:

$\theta =\frac{\pi \theta °}{180}$

Here, $\theta$ is the angle in radians and $\theta °$ is the angle in degrees. For example, to convert 3.5 degrees to radians:

$\theta =\frac{3.5\pi }{180}\approx 0.061$

What is the relationship between angular frequency and the length of a pendulum?

The angular frequency $\omega$ of a simple pendulum is related to the length of the pendulum by the formula:

$\omega =\sqrt{\frac{g}{l}}$

Here, $g$ is the acceleration due to gravity and $l$ is the length of the pendulum. This shows that the angular frequency is inversely proportional to the square root of the length. As the length increases, the angular frequency decreases, meaning the pendulum swings more slowly.