# Laplace Transforms with an Undamped Spring with Unit Impulse

1. Nov 5, 2013

### Rapier

Ok, so I start out with the basics and find k.

F=ma=kx
(1.5 kg)(9.8 m/s^2) = k (4.9 m)
k = 3 N/m

I also know from the problem that c=0 (no damping) and x(0) = 2m and x'(0) = 0.

ƩF = ma
ma = -cx' - kx, a = x''
mx'' + cx' + kx = 0 and since c=0
mx'' + kx = 0

Nothing out of the ordinary or unusual...but where the heck/how the heck do I include the unit impulse? From reading the textbook I have the impression that

mx'' + kx = ∂(t - 2∏)

and laplace [∂(t - 2∏)] = e^(-2∏s)

If my assumption is correct, I can do the problem.

1.5x'' + 3x = ∂(t - 2∏)
laplace (1.5x'') + laplace (3x) = laplace (∂(t - 2∏))
X(s^2) -s x(0) - x'(0) + 3X = e^(-2∏s)
X (s^2 + 3) - 2s - 0 = e^(-2∏s)
X (s^2 + 3) = 2s + e^(-2∏s)
X = [2s + e^(-2∏s)]/(s^2 + 3)
inverse laplace (X) = inverse laplace [(2s + e^(-2∏s)) / (s^2 + 3)]
x(t) = bleh.

Wolfram alpha (which instructor says is a legit method to solve this) says that my laplace is invalid, so I think somewhere I don't understand the unit impulse thing.

Pls help. kk thnx.