2. (25 points) A typical mass-spring system is governed by the second order diff

Question

2. (25 points) A typical mass-spring system is governed by the second order differential equation:

mx?? + cx? + kx = F(t). (1)

To simplify, let F (t) = 0.

(a) Compute the critical damping value ccr in terms of m and k.

(b) Write out the characteristics equation for (1). What the roots would be if c = ccr?

(c) Find the general solution to (1) using the roots you found in part (b).

(d) Let u = (x,y)T, where y = x?, find the matrix A such that

u? = Au.

(e) For the matrix A, verify that det(A ? r1I) = 0 and det(A ? r2I) = 0 where r1 and

r2 are the two roots you found in part (b).

Explanation / Answer

M*(d2x/dt2) + C*(dx/dt) + K*x = Mg

Mx''+Cx'+Kx=Mg

This is a second order linear non homogeneous differential equation.
Mx''+Cx'+Kx=g(t)
where g(t) is a constant.

Now my question is, how do I solve it for x?

I tried using the undetermined coefficients method, but I do not get an equation that implies that x is oscillating. Shouldn't I get a sin or cos somewhere in the answer?

Here is what I tried:

corresponding Homogeneous Equation:

Mx''+Cx'+Kx=0

characteristic equation = M*(r^2) + C*r + K

General solution to the homogeneous equation:

C1(e^pt)+C2(e^qt)
where
p=(-C+SQRT((C^2)-4*M*K))/2M
q=(-C-SQRT((C^2)-4*M*K))/2M //from the quadratic equation

Now to calculate the particular solution:

Mx''+Cx'+Kx=Mg

The right hand side is a polynomial of degree 0. try a polynomial as the solution

x=Ax^2 + Bx + C
x'=2Ax +B
x''=2A

Sub these in to the differential equation and you end up with:

x=Mg/K

So the general solution to the differential equation is:

x(t) = C1(e^pt)+C2(e^qt) + Mg/K

This can't be correct, because x should oscillate over time.
C1, C2, p, q, ang Mg/K are all constants so this equation does not describe an oscillation

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