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concept

## Simple Harmonic Motion of Vertical Springs

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But we're gonna see that vertical systems are very very similar to horizontal and the only difference has to do with the equilibrium position. But rather than tell you, I just want to show you using this example. So let's take a look. We've got a 0.5 m spring and it's hanging from the ceiling. So I've got that distance 0.5 m. And that's gonna be the original length of the spring and then it stretches down by some distance after I attach a mass to it. Now, why does it stretch down and stop? Well, we've got an extra force to consider because now we have the objects weight that pulls it down. Whereas originally we didn't have that before in horizontal systems. So as the thing is pulling the spring down, the restoring force gets higher and higher upwards. And so what ends up happening is that the restoring force ends up canceling out with the gravity. And so it reaches a new equilibrium position. So in horizontal mass spring systems we said that X equals zero is where like no forces are acting on it. But in vertical mass spring systems, the equilibrium is where these forces will cancel out. That's the important part. And so we call that distance that hanging distance until it reaches equilibrium DELTA L. So that in this case is equal to 0.2. So again that Delta L. Really represents the the spring systems hanging deformation or like hanging distance. It's the amount that you need to stretch the spring so that you reach equilibrium. And equilibrium condition is where these two forces balance out. Well the upward force is gonna be K times delta L. And the downward force is going to be M. G. So that means at equilibrium we've got K. Delta L equals MG. So let's look at the second part of this problem. So after all of this stuff happens, we're gonna pull this spring system an additional 0.3 m downwards. So now just like in horizontal spring systems, your additional push or pull was the amplitude. So that means that once you pull this thing downwards 0.3, now this thing is just going to go up and down between these two amplitudes. So this is gonna be the positive and that's gonna be the negative amplitude. So it's just gonna oscillate up and down like that. And so it's important to remember that this amplitude is the additional push or pull and it's not the delta L. So it's not the natural hanging distance or defamation, it's the additional pusher pool. So in problems, what you'll see is that like you're gonna attach a mass and it's gonna stretch by some distance. What that represents is delta L. And then you're gonna pull it down an additional something and that is going to be the amplitude. So don't confuse those two. Alright, so now basically we have everything we need to solve the problem. This first part is asking us for the force constant. So it's asking us for K. So we're just gonna use this new equilibrium equation that we have. So we've got K times delta L. Is equal to MG. Now I know what MG and delta L. R. I just have to figure out what this K constant is. So I've got K once I just move this to the other side, I've got K is equal to five, which is the mass. And I've got, I want to use 10 for gravity. And then the delta L. Is what will the delta L. We said that natural stretching distance was 0.2 m. So that means I gotta K constant of 250 newtons per meter. And that's it. So what about the second part here? The second part is now asking us at its maximum height. So it's oscillating at maximum height. How far is that ceiling from the block? Great. So let's check it out. So we've got this this thing here, uh and so I'm going to sort of represent this whole entire line here as this motion. The original distance, the original sort of length of the spring, was that black dot. And then you hang it down and reaches some equilibrium position and then you're gonna pull it down an additional 0.3 m. So that's the amplitude. And it's just going to oscillate between these two points. Okay, so we're being asked for basically this distance right here, what is the distance between the ceiling and its maximum heights? That's where it reaches that maximum amplitude. And so that distance here is going to be the letter D. Well, we're told that this thing has an original length of 0.5 m. And the equilibrium position is when it stretches an additional two. So that means that the Length of equilibrium I'll call this Q is equal to 0.7 m. So now what happens is it goes 0.3 down and then 0.3 up. So at this bottom part here, this length here is 1.0 and at the top point right here it's 0.7 -0.3. And so we say that at maximum height, so at the max height, The distance d away from the ceiling is equal to 0.4 m. And that's it. Alright guys, that's it for vertical oscillations.

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Problem

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

A

.17 kg; .56 m/s

B

.63 kg; .29 m/s

C

.17 kg; 1.12 m/s

D

.63 kg; .15 m/s

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example

## Example

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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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Problem

A chair of mass 30 kg on top of a spring oscillates with a period of 2s. (a) Find the spring’s force constant. You place an object on top of the chair, and it now oscillates with a period of 3s. (b) Find the object’s mass.

A

296 N/m; 1286.9 kg

B

188 N/m; 42.9 kg

C

188 N/m; 817.4 kg

D

296 N/m; 38.1 kg