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LLLLLLLLLgc3 G = 9, C = 12 M2Write the stiffness matrix KWrite the matrix M-'KFind the eigenvalues and eigenveciors of MSmaller eigenvaluewith eigenvectorLarge...

Question

LLLLLLLLLgc3 G = 9, C = 12 M2Write the stiffness matrix KWrite the matrix M-'KFind the eigenvalues and eigenveciors of MSmaller eigenvaluewith eigenvectorLarger eigenvaluewith ergenvectorIf this spring system oscillates without any extemal forces present, then the position of each mass satisfies ihe following general formula;u(t) (at cOsC sqrt3t) +61 sin ( sqm342 COS} 5q1t146z siu( sqn14If the system begins oscillation with initial position u(0)and initial velocity u' (0)then the posit

LLLLLLLLL gc3 G = 9, C = 12 M2 Write the stiffness matrix K Write the matrix M-'K Find the eigenvalues and eigenveciors of M Smaller eigenvalue with eigenvector Larger eigenvalue with ergenvector If this spring system oscillates without any extemal forces present, then the position of each mass satisfies ihe following general formula; u(t) (at cOsC sqrt3 t) +61 sin ( sqm3 42 COS} 5q1t14 6z siu( sqn14 If the system begins oscillation with initial position u(0) and initial velocity u' (0) then the position of the masses at time IS given Dy Ui (t) "2(t) ~/11cos(sqmt3t) 1/11cos(sqmiat)



Answers

Consider the spring-mass system whose motion is governed by the differential equation $$\frac{d^{2} y}{d t^{2}}+2 \alpha \frac{d y}{d t}+y=0$$ Determine all values of the (positive) constant $\alpha$ for which the system is (i) underdamped, (ii) critically damped, and (iii) overdamped. In the case of overdamping, solve the system fully. If the initial velocity of the system is zero, determine if the mass passes through equilibrium.

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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.

So the information we have is the amplitude, is it? And the period is 0.5 seconds. So from there we can calculate be which would be to pie divided by P or that would be too high too. But as you're quite fine or foreplay and the next thing they're told us that AT T is equal to zero. Why is equal to negative eight? So what we're looking for is a function off the form negative course signed something so that function would be. Why is equal to negative it co sign four high tea.


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