17. Periodic Motion

Simple Harmonic Motion of Vertical Springs

# Example

Patrick Ford

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Yeah. Hey, guys, let's check out this problem. So we've got an elastic cord and we're told that it's 65 centimeters long when when some weight hangs from it and then we attach mawr weight from it and it stretches down even farther. So I'm just gonna go ahead and draw a little picture for that. So in this case, uh, for this first box, let's go ahead and draw out the forces. We know that there's a weight acting on this box, and that's m one g. But M one G is the weight of this object, and we're told that that's equal to 75 now. When that happens, it's an equilibrium with the elastic cord, which you kind of just think about is the spring and the equilibrium condition is that we have K times Delta L one. Right. So we have the restoring force that acts upwards now, for the second situation is the exact same thing. So in this case, we've got m two G, and we know that that's equal to 180. And the restoring force acting in this direction is now K Times Delta L two. Because this spring has now gotten longer. Now that Delta Els even farther, it's sagging even farther. So we've got to like equilibrium conditions to consider here. We always gonna gonna start with our equilibrium condition that K Times Delta l is equal to m G. So for the first box K times Delta L one is equal to M one g. And now for the second situation, we've got K times Delta l two equals M two g. Now let's look at all of our variables. What are we looking for? We're looking for the spring Constant k. So really what we're gonna be solving this problem is Kaye in the case is the same for both of the elastic cords because it's the same thing. Okay, so I've got m one g is equal to 75 I've got M two g is equal to 180 now. What about this l one and l two? So what about these distances that I'm told is the 65 centimeters the L one and the 85 centimeters the l two of the Delta l two? Well, no, we can't. We can't say that because the Delta Els represent the change in distance. So remember that Delta l So I want to write this out. That delta l is actually just the length final minus length initial. That's what Delta l represents. So really, what they're actually giving us is the final length of this spring. This is the number that they're giving us, and that is equal to 75 for the second one. We have k times. Uh, l f to minus l. I is equal to 180. Okay, so we've got now we've got two variables to consider. We've we don't know what The Delta We know what? We don't know what the delta l is because we don't know what the initial length of the spring is. So these K's and these allies are unknowns. What I am given, though, is that LF one is equal to 0.65 65 centimeters and LF two is equal to 0 85. So, really, what this becomes is an equation like this. So okay. Times 0.65 minus l I equals 75 then K times 750.85 minus l I is equal to 180. Okay, so here we've got into a situation where I have two unknowns, right? I've got this k that's unknown. And the ally that's unknown. K and l I on the rest of just numbers. So what have we done in the past? To sort of deal with the situation? We always either do substitution or we add the two equations together. But first, what I'm gonna do is sort of expand this and I'm just gonna multiply this K straight throughout. So I get K Times 0. minus k times L I equal to 75 in this formula, and then I've got 0.85 k minus k l. I is equal to 180. That's just distributing the case into both of these terms. Okay, so now what I'm gonna dio is if I'm gonna do substitution or I'm just gonna stack these equations on top of each other. Let's see what I guess I'm gonna bring these two together, so I've got 0.85 k minus k l. I equals 1 80 then we got 0.65 k minus k l. I equals 75. So what do I do with these two equations? To get rid of those two unknowns, I just have to subtract those two equations. So if you subtract these two equations, what's gonna happen is that these two things a minus on a minus, they're gonna cancel out and then just have thio subtract straight down. So 20.85 minus 0.65 point 20 k and then 1 minus 75 is 15 So now K is just equal to 525 Newtons per meter. So that is our force constant. So let me know if you guys have any questions for this one. I know this is kind of a weird problem.

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