1

concept

## Simple Harmonic Motion of Pendulums

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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.

2

example

## Example

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Hey, astronauts, let's do an example. So this is actually really common type of problem with pendulums your landing on an unfamiliar planet. You're constructing a simple pendulum. So let's go and check it out. We've got this pendulum that's a length of 3 m and a mass of 4 kg. So I've got l equals three and Mass is equal to four. Now, we're gonna release this pendulum. We're gonna pull it back to 10 degrees from the vertical right there, and they were gonna let it go. So it's just gonna oscillate back and forth. So that's our initial push or pull. That's Data Max, and that's equal to 10 degrees. Uh, just remember that because this is in degrees, anytime we're gonna be working with data we're gonna have in radiance. So just make just be aware that you might have to do some conversions, okay? And the last thing we're told is that it's one full cycle at two seconds. What does that mean? It takes two seconds for this thing to go back and then all the way back to the other side work started from So that is the full period, right? That is t equals two 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 E Just remember that in Earth G equals 9.8. You probably know that by now. But that any other planet, we cannot take that for granted. So and any other planets G does not equal 98 Right? So just be aware of that. Sometimes you just look at Gene, you're like, Oh, that's 9.8, but it's not in these situations. Okay, So which equation or we're gonna use because we actually have a lot of them involved. 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, we've got mass here, but I have the theta. Okay, But its data Max. Okay, so it might be able to use it here, but I don't know how the restoring force, so I wouldn't be able to solve Fergie. 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, Theo. Nly other equation that I could use is this big Omega equation involving squared of D 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 you'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 got to pie divided by the period. The periods equal to two seconds equals the square root of the gravity on that planet, divided by the length of the pendulum, which is what we're given that it's three. So let's go and do some simplification. We've got the twos. Cancel You got pi equals square root of G over three. So now, in order to get rid of the square root, it's gotta square. Both sides got pi squared, equals G planted over three. And so then I just gotta move the three over so the G planet is equal to three pi squared. So that's equal to 29.6 m 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.

3

concept

## Pendulum Equations

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Hey, guys. So we talked about some of the differences between mass spring systems and pendulums. We talked about the differences between F A and Omega. So now you might be wondered what the other equations look like. I'm gonna give them to you in this video. So remember that we have a pendulum here. It's some length l and you're gonna take this mass and you're gonna pull it out to some angle, which is data you're gonna let it go, and it's gonna 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 right. It's the same exact thing for pendulums. So I'm gonna 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 trick to involve here. I've got a triangle, so I've got the high pot news, I've got the angle and I've got the opposite side. So from so Kyoto, uh, I can relate this X at any point during the pendulum swing by L Times, they've sign of data. But remember that when we're talking about pendulums sign of data and data are just about the same thing. So I could make this sort of simple simple by saying that X is equal to L. A Times data. And remember that we're gonna be working with radiance when we're doing this right? So X equals health data. So he 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 amplitude. So now it happens. This is the same thing for pendulums. So you're gonna swing. You're gonna take this thing all the way back to some theta. And once data 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 waas. So it's the same thing. But because X is equal to L Times theta thing, I mean the amplitude X max is going to be at l Times theta max. So that means that the for pendulums, the maximum X is the amplitude, and that is equal to L a Times data, Max. Okay, so we also had a couple of other things about mass spring systems. We have the maximum values for X V N A. So he said here at the very middle that the velocity is at its maximum and at the end points, the acceleration is at its maximum, and it points towards the center. Right? So that's a max 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 is now something slightly different. Remember that now A is equal to l Time State a max. So what I'm gonna do here, I'm gonna just replace these ays with El Fattah Max So V Max is a Omega. Then we can also say that it's equal to L. A Times data Max Omega Just replacing the A with Al Qaeda Max. And so a Omega squared is a is L X max or theta Max Omega squared. OK, so the other thing is that the Omegas are slightly different for pendulums as well. Remember that this is G over L Square rooted, whereas for mass springs, it's k over m 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 ever asked for what data is and you're given what T is, you're gonna be using this equation. Tha tha t it looks just like the VT and 80 equations are, um 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 a tanking from the spring, so we've got this pendulum system here so I'm gonna draw 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 gonna be, like over here somewhere. Okay, so we're told a couple of things were given that the mass is equal to g, so that's equal to 0.5 kg. Just remember, everything has to be a Nessie. The length of the string, which is the length of the pendulum, is equal to 0.4 m again were given centimeters and they were told that the object as a speed of 0.25 m 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 they're really telling us about this. 0.25 m per second is that V Max is equal to 0.2 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 data does this reach in degrees. So just remember that we're working with all of these equations were gonna get radiance. And so I'm just gonna 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 gonna take a look at all of my equations that have theta max in them, so I don't wanna go all the way up there. I'm just gonna paste all of these equations right here. So I've got all my theta max. Is Aaron these equations? I got theta Max here. Betamax, Betamax and data 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 that the length is I've got this length right here. Let's take a look. I do know what the maximum velocity is. Uh, I'm gonna go ahead and circle that I've got the max velocity. I don't know what the amplitude is, so we're just gonna cross it all the A's. And I also don't know what the Amax is. The last thing is that I could Onley use this bottom equation if I'm given something about time. But I'm not given any time. And because I'm not giving at a time and I'm supposed to find what data Max is, I have two 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 gonna 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 you've got V. Max is equal to I've got l times theta max, then times omega. So if I wanted to figure out what this Theta Max is, let me just go ahead, rearrange that equation. So I've got Theta Max is equal to v Max, divided by L. Omega. I'm just gonna move these guys to the other side, right? So I'm just gonna 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 last thing I just have to figure out is what Omega is. So I'm gonna go over here and do that. Well, Omega is equal to once Let me write out that big omega equation that I have. I've got two pi frequency to pie divided by T and then I've got that is equal to square root of G over L. Now, I'm not giving anything about frequency or the period, so I'm just gonna have to use this last one squared of geo over l. So I've got omega equals square roots. 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 that's gonna 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 rats per second. Great. So I'm just gonna plug that back into here. Okay, so I've got Theta Max now is equal to 0.25 then I've got divided by the length of the pendulum 0.4 and then multiply that by 4.95. So what I'm gonna get is Theta Max. Remember, Betamax is gonna be in radiance, and that's gonna equal 0126 That's new rads. So now the last step is just to convert it back to degrees. So how do I do that? So I've got data. Max is equal 2.126 and now in a multiplied by something that's gonna get rid of the rads. So I've got 180 degrees divided by Pi Radiance. So that's gonna cancel out, and then we're gonna get 7.2 degrees. That's the maximum angle that we make for the pendulum. So let me know if you guys have any questions, let's keep going on for now.

4

example

## Example

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What's up, guys? Let's take a look at another pendulum problem. So we've got a 100 grand mass. It's on a 1 m long string and it's pulled seven degrees to one side and then it's let go. It's gonna pull this thing back out to 77 degrees, let it go. It's gonna swing back and forth, and we're supposed to figure out how long it takes to reach four degrees in the opposite side. So what does that mean? So I'm just gonna draw a sketch of my pendulum real quick. So I 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 a seven degrees, I know it's not to scale. Now this thing is just gonna go back forth and back and forth until it reaches the opposite side. Okay, so we're supposed to figure out is if it starts at seven degrees here? How long does it take for to swing through this angle right here and then reach on the opposite side where this angle is going to be four degrees? But because this is the equilibrium position at the very at the very middle point here. This actually is gonna pick up a negative sign. It's important. It's kind of like if you go back and forth to the mass spring system equilibrium zero. 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. So I know that the mass is equal to 0.1. I've got the length of the pendulum is one. And then I've got Theta Max, which is my initial push out or pull out whatever. My theta max is equal to seven degrees and then it's supposed to be four degrees in the opposite side. So I've got, like, this fate. A final is supposed to be negative for degrees. Okay, so let's take a look at our equations. I've got data all over the place, but I only have teas in two places, so let's take a look. Either can use the end cycles equation, which has this t variable here, or I can use this data T equation Now, in order to use the end 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 with data. Max is and theta at some later time is kind of what I'm looking for. And this has got some omegas and tease. So let's go ahead and use that equation. Let me go ahead, write it out. So I've got data at some later time is gonna be equal to theta Max times the co sign of Omega T. Because we're working this coast, I just make sure that you're in radiance mode. Okay, So this data of tea here is really like the final theta, right? It's state at some later time, and that's equal to negative four degrees and then state a max we said was equal to seven degrees. Now we just have to convert these into radiance mode. So I'm gonna go ahead and do that. Eso Let's go ahead and actually move that over here. So if I want to convert this four degrees into radiance, I've gotta multiplied by pi over 1 80. And so what I get is sorry. Negative four degrees. What I get is negative 0.7 if I do the same exact thing for the seven. So this is gonna be negative 0.7 If I do the same thing for the seven degrees, I am going to get 0 12 and that's gonna be positive. OK, so now let's go ahead and just fill out those just plugging those numbers. So I've got negative. 0.7 equals 0.12. Then I got co sign of Omega Times T and this tea 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. Um, So I'm gonna go ahead and move this 0.12 over to the other side because I want to start getting this cosine omega t by itself. I wanna get this t by itself, but it's like all wrapped up in this coastline function. So let's go ahead and move everything else to one side. If you go and do that, you divide those two decimals over and you're gonna get negative 0.58. And that's equal to cosign times Omega T Now again, I want this t But it's all, like, wrapped up inside this cosine function. So how do I get it out? Well, if you have a co signing to get rid of it, you have to multiply by the inverse cosine. So what I'm gonna 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 radiance, and what you're gonna get out of this is you're gonna get to 0.19 and that's gonna be in rads. And now you're gonna 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 figure and find that. So I've got Omega's all over the place in my equations chart. So I've got all these like the Max is a maxes and X Max is that all have Omega's, but I'm not giving any information about V. Max is or a Max is What do I know instead? Well, let's see. I've got the length of the pendulum. So I'm gonna use this omega that relates the frequency period and 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 thes square roots of G is 9.8 over. And then what's the length of the pendulum? It's just one. So that's gonna be 3.13 radiance per second. So now I just take that number. So I've got to 0.19 equals 3.13 t now just divided to the other side, and we get the 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. All right, that's it for this one

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PRACTICE PROBLEMS AND ACTIVITIES (6)

- A simple pendulum 2.00 m long swings through a maximum angle of 30.0° with the vertical. Calculate its period ...
- A simple pendulum 2.00 m long swings through a maximum angle of 30.0° with the vertical. Calculate its period ...
- A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, wher...
- A building in San Francisco has light fixtures consisting of small 2.35-kg bulbs with shades hanging from the ...
- You pull a simple pendulum 0.240 m long to the side through an angle of 3.50° and release it. (b) How much tim...
- You pull a simple pendulum 0.240 m long to the side through an angle of 3.50° and release it. (a) How much tim...