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Calculus/Differential Equation- displacment

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Expert: - 10/6/2009


Question
Hi Scotto,

I have the displament of a damped spring system and it is described by the Differential Equation

m d^2y/dt^2 + c dy/dt + ky =0

y =displacment, m =mass, c =constant, k =spring constant
If c^2 = 4mk critical damping occurs and the suspension behaves well.
Find the general solution y(t) for the case of critial damping

Thanks for your help

Matt

Answer
To solve this, the equation that arises is mx² + cx + k = 0.
Use the quadratic equation to find the zeroes.

Suppose it was y" + 3y' + 2y = 0.

The equation would be x² + 3x + 2 = 0.
This factors into (x+1)(x+2) = 0.
Since the factors of this are -1 and -2, the solution would be of the form
y = Ae^(-x) + Be^(-2x).

Note that y' = -Ae^(-x) - 2Be^(-2x) and y" = Ae^(-x) + 4Be^(-2x).
y" + 3y' + 2y would then be
[Ae^(-x) + 4Be^(-2x)] + [-3Ae^(-x) - 6Be^(-2x)] + [2Ae^(-x) + 2Be^(-2x)].
For the Ae^(-x), we have 1 -3 + 2 = 0.
For the Be^(-2x), we have 4 - 6 + 2 = 0.

Using this, we have seen that the results are correct.

The A, B in this equation can be found by the quadratic equation x = (-b±√(b²-4ac))/(2a)
where a=m, b=c, and c=k.  Note that c on the left is for the quadratic and c on the right is from your equation.

Note that you could also replace your c with 2√(mk).

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