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.