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Quettionobject of the mass m-1 [#g} is attoched t0 . the Kull by @ spring of the stiffness nfij:sketch free body diagram the spring and tfie object: Derue- the equa...

Question

Quettionobject of the mass m-1 [#g} is attoched t0 . the Kull by @ spring of the stiffness nfij:sketch free body diagram the spring and tfie object: Derue- the equation ot motion ol the object: Solve the equation of motion for the Initial position of I[m] and velocity of Hrs] Write the solution the forrn x(t)-Asinlwttf}, where w Is the angull velocity and f is the phase.

Quettion object of the mass m-1 [#g} is attoched t0 . the Kull by @ spring of the stiffness nfij: sketch free body diagram the spring and tfie object: Derue- the equation ot motion ol the object: Solve the equation of motion for the Initial position of I[m] and velocity of Hrs] Write the solution the forrn x(t)-Asinlwttf}, where w Is the angull velocity and f is the phase.



Answers

(II) A block of mass $m$ is attached to the end of a spring (spring stiffness constant $k ),$ Fig. $6-40$ . The block is given an initial displacement $x_{0}$ , after which it oscillates back and forth. Write a formula for the total mechanical energy (ignore friction and the mass of the spring) in terms of $x_{0}$ , position $x$ , and speed $v$ .

Okay, so in this problem to start finding the displacement we're going to see Oh, first find the auxiliary equation. That's p of our is gonna be go to R squared plus three R plus two. We said that equals zero. If you notice we can factor this. So the factors of two R two and one which does add up to three So are factors are gonna be our is to r plus one. Is he with zero So that our roots are going to be negative one or negative too a negative one. So our general solution why a t is gonna look like C one e to the negative to t plus C to ease of the negative t Now we're going to use our initial conditions to solve for C one and C two. So to solve for C one C two. Now we're going to first find why Primacy socially negative. To see one eats a negative two teeth and then minus C to eat on the negative. T Then, um, we're gonna plug in zero. So why of zero is able to C one plus C two, which is evil to one, then my prime of teeth are zero it's gonna be. And then again, uh, so I eat of the zero is one we get negative to see one minus si two is equal to negative three. So if you notice, we can add these two and the CT's will cancel out. So we'll get negative C one. It's gonna be able to negative two. So see, one is equal to two. So since C one is equal to two, so we have two plus C two equals one. Move that over and then we get C two is equal to negative one. Okay, see, two is equal to negative one. So our equation emotion, why of tea It's gonna be equal to C one is too. So to e to the negative to t and then minus e to the negative t like So okay, now for part B, we need to determine the time when the mass passes of the equilibrium position. So that's when we set why of tea equal to zero. So that's to eat on the negative to t minus e to the negative. T said that able to zero so we can move this over to the other side. We have to e to the negative to t it's gonna be able to eat to the t Sorry eats of the negative. T Then we can multiply both sides by each of the two t here. Um, so then we just get that two is equal to e to the t. So then we get when we take, the longer both sides that t it's going to be equal to log of to. So this is when our equilibrium or when our mass passes through the equilibrium point here. Now we need to make a sketch of the general emotion for part C. So we start at one and we go to or we start going down, and then we do pass to the equilibrium. So we start at one, we do pass through the equilibrium. So that's gonna be, um, again, we start going downward first, and then we come back up towards equilibrium. Uh, so this passes through the equilibrium at the time log to So this is the sketch of our system here

Okay, so in the beginning of this problem part A we are looking for the equation for the potential energy of a particle that is being stretched. It's being pulled the distance X. The original length of these springs is capital L. And I've just written here the new length of the springs with lower case. Al so by using trigonometry here, where you can say that lower case L the new length is equal to the square root, L squared plus X squared or X squared plus health squared. And what we're really interested in is how far this spring has stretched. So that distance is right here. The difference between the new length and the original length that's going to help us get to that potential energy equation. So the original potential energy equation is just one half K X squared for a spring, the elastic potential energy. And so for one spring we have one half K. And this is the stretch length L minus capital L quantity squared. Can we have two springs? So we just add those two together. We double this. That takes care of the haves. And then if we actually multiply out this squared here we get lower case L squared minus two times both lengths plus the original length square. And then we can plug in this relationship right here for this lower case. L I'm going to get this right here because this is all squared. It's just capital all squared plus X squared. And then this lower case L. We just put that in right there, we can pull out this K X squared on its own. And then that will leave us with this and then we have to L squared minus two L. Times this quantity. So we can pull a two L out of this of each of these terms. And we'll have this relationship right here, which is what we were trying to prove just with these in the opposite order. So that is parting part B. We need to plot this and so I did that using just some just a spreadsheet put in the values of 0 40 newton meters for K and 1.2 for L. and I plotted this. And so you see as we get to 1.2, a negative 1.2 that maxes out the potential energy and it's going to have a minimum value potential energy just at X equals zero. So that is the only equilibrium point. Now, for part C, we need to find, oh, the final speed of the particle. And we can't actually do that numerically, but we'll have it in in terms of its mass. So we can use the conservation of energy here or the work energy theorem. And we have one half Mv squared and then the right side of this equation, it's just gonna be our potential energy relation that we found before. And then we can plug in the values that we are given here. So we know the Distance. It's stretched his .5 m. We know the K value is 40 Newton m. We know the length of the spring and stretched as 1.2 m. We plugged that into this side of the equation. We're going to get .4 jewels. And then if we sell this for velocity, We're gonna get two times at her .8 divided by the mass. And So our final velocity for the is equal to the square root 0.8 over him. And whatever this mass is, that both give us what our velocity is.

Okay, so in this problem, let's first determine if it's under dime, critically damned or overdone. So that requires looking at the roots of this problem here. Or a few, Um, the auxiliary equation. So Axler equation is r squared, plus two are plus one is equal to zero. So if we take a look at, um, well, the factors of one here for one and one this actually factors into our plus one squared is equal to zero. So we get that are is going to be equal to negative one with a multiplicity, too. Since we have a repeated route. Repeated route means that we are critically damped here. So this is critically damped. Okay, Now that we've determined that this is critically damped, we can sketch the solution from here. So we, uh, start with negative one here, um, and then we're going up, actually. Ah, sorry. Actually. Gonna look something like it's gonna go up once and then down, actually, let's let's find the general solution first, actually, or yes. Actually, we can. We can graph this first, um, so we start at negative one, and then we're going upward, so we're gonna go up and then down like that. So this is critically. Don't. Okay, Um, so that's negative one. Here. Let's find our general solution. Our general solution is going to be of the form. So why of tea? It was gonna be able to see one e to the negative t and then plus C two t e to the negative t because of our, um, multiple city, too. So to solve for C one and C two, we're going to need to find d y d t and D y or the squared Y d t swear. So we'll get the first derivative here. There's gonna be negative. C one e to the negative t. Actually, first of all, to simplify this, we can go ahead and plug in our first initial condition. That's why of zero. So why of zero is equal to let me write this in a different color. Let me do this in green here. Okay. Y of zero. It's going to be equal to well, this become zero. So we have C one either Negative t. So we just have C one is equal to negative one, so Well, uh, we can go ahead and plug in that here. So again, why? Prime t is so That's negative one here. So this is going to become e to the negative t each of the negative teeth and then, ah, plus, then we're gonna need to use product will here. So we do first, um, so C two e to the negative t and then, well, we have minus C two t e to the negative t again. We're gonna need to plug in. Why? Prime a zero. So why prime a zero? Here, this becomes one. Here we have one. And then plus this C two. Ah, sorry. That also becomes one. So have one plus C two. This goes to zero, and then this is equal to and then we have to hear so one plus C two is equal to two. Therefore, see, too is equal to um So I see two is equal to one. So are, um the motion is governed by y of tea. It's gonna be equal to negative e to the negative. T plus t e to the negative t. So this is our solution here

Okay, So to solve this initial value problem first, let's take a look at the auxiliary creation. That's p of our listen, Beagle two R squared plus two are plus one is equal to zero. If you notice we can factor this into our plus one squared is equal to zero. So there are some people who negative one with a multiplicity, the multiplicity of to So therefore, that means we can. Or that means we have a critically damped system system critically damped. OK, so that's our classifications now, um, Teoh Now we need to find the, uh, general solution. So that's gonna be Wyatt T. It's going to be able to see one e to the negative t and then plus C two t e to the negative t to use the initial condition we need to find B, um, derivatives. That's why Primer t b negative c one e to the negative t and then plus, and then we're going to use product will. So first time's derivative second is gonna be minus C two ah t e to the negative t and then plus C two e to the negative t It's so now we need to find y zero y zero, it's gonna be equal to. So this goes to one. This goes to zero because of this here. So we just get that C one is equal to negative one. So now we can plug in negative one here, so this becomes positive. One s. We have positive one. And then this goes to zero, and this is also one. So we get that C two, uh, plus one. Okay, so why prime Miss Euro. First, I'm gonna get one plus c two. It's gonna be equal to to. Okay, So that means see, too, it's going to be able to one. So our general solution is going to take the forum. Why of t is equal to and then see one executive one so negative e to the negative t and then plus t e to the negative t If we notice here, the, uh, coefficients are of different signs. So when we graph the solution, Jay, we start at negative one because of this initial condition here, and we're going up. Thank you. So the derivative is positive. So we're going up, and then the two different signs tells us that we're going to pass through the equilibrium once we're gonna go up and then back down and then go towards the, um, go towards the equilibrium position at zero. Okay, so this is these are our final answers here.


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