17. Periodic Motion

Simple Harmonic Motion of Pendulums

# Example

Patrick Ford

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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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