Laplace Transforms with an Undamped Spring with Unit Impulse

[Rapier]

A mass of 1.5 kg is attached to an undamped spring which causes to stretch 4.9m. This is set into motion with initial displacement of 2m and no initial velocity. The mass is subjected to unit impulse at time t = 2∏ . Find the equation of motion. Use g=9.8 m/s^2

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.

 

[jvicens]

Hello! It looks like you are on the right track with your equations and calculations. The unit impulse, or delta function, represents an instantaneous force applied at a specific point in time. In this case, the unit impulse is applied at t = 2∏, so it would be represented as δ(t - 2∏).

To incorporate this into your equation, you would have:

mx'' + kx = δ(t - 2∏)

Then, when you take the Laplace transform, you would use the property:

L(δ(t - a)) = e^(-as)

So your final equation would be:

X (s^2 + 3) = 2s + e^(-2∏s)

And you can continue with your inverse Laplace transform to solve for x(t). I hope this helps clarify the use of the unit impulse in your equation. Keep up the good work!