A mass of 1 slug, when attatched to a spring stretches it 16 feet (so spring con

Question

A mass of 1 slug, when attatched to a spring stretches it 16 feet (so spring constant k=2) and then comes to rest in the equilibrium position. Starting at t=0, an external force equal to f(t)= -13sin(4t) is applied to the system. Find the equation of motion if the surrounding medium offers a damping force equal to 2 times the instantaneous velocity. Analyze the motion at t approaches infinity. I get as far as the auxiliary equation, but then I get lost because of the imaginary numbers in the roots!

Explanation / Answer

my'' + by' + ky = f(t)

m =1 1 slug
So, m = 1

b = damping force = 2

F = kx
1*32 = 16k
So, 32 = 16k
k = 2

So, the equation becomes :

y'' + 2y' + 2y = -13sin(4t)

m^2 + 2m + 2 = 0

m = (-2 +/- sqrt(4 - 8)) / 2

m = (-2 +/- 2i) / 2

m = -1 +/- i

So, yc = Ae^(-t)cos(t) + Be^(-t)sin(t)

Now, on the right, we have -13sin(4t)

So, yp(t) = Asin(4t) + Bcos(4t)

yp' = 4Acos(4t) - 4Bsin(4t)

yp'' = -16Asin(4t) - 16Bcos(4t)

yp'' + 2y' + 2y = -13sin(4t)

-16Asin(4t) - 16Bcos(4t) + 8Acos(4t) - 8Bsin(4t) + 2Asin(4t) + 2Bcos(4t) = -13sin(4t)

Comparing coefficients :

-14A - 8B = -13
8A - 14B = 0

Solving, we get :

A = 7/10 and B = 4/10

So, with this, we get :

yp(t) = (7/10)sin(4t) + (2/5)cos(4t)

So, now, y = yc + yp

y = Ae^(-t)cos(t) + Be^(-t)sin(t) + (7/10)sin(4t) + (2/5)cos(4t)

y(0) = 0 :

0 = A + 2/5

A = -2/5

So, y = (-2/5)e^(-t)cos(t) + Be^(-t)sin(t) + (7/10)sin(4t) + (2/5)cos(4t)

Now, y'(0) = 0 :

y' can be found and using initial condition B = -16/5

So, final solution is :

y = (-2/5)e^(-t)cos(t) - (16/5)e^(-t)sin(t) + (7/10)sin(4t) + (2/5)cos(4t) ---> ANSWER

As t approaches infinity, e^(-infinity) would equal 0

So, y(infinity) = 0 - 0 + (7/10)sin(4t) + (2/5)cos(4t)

y(infinity) = (7/10)sin(4t) + (2/5)cos(4t)

We know sin(4t) and cos(4t) must lie within [-1 , 1]

So, when t ---> infinity, the displacement would be :

y = (7/10)sin(4t) + (2/5)cos(4t) ----> Analysis when t --> infinity

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