Thread: Harmonic oscillators (Laplace transform)

Problem states: Find the frequency, periodic time and solution for each of the following harmonic oscillators:
(a). f''(t) +f(t)=0 given that f(0)=0 and f'=1
(b). 6f''(t)+2f'(t)+9f(t)=0 given that f(0)=0 and f'(0)=3

Okay. So I know to do the substitutions (L{f''(t)} = s^2.F(s) - sf(0) - f'(0)) but I have trouble with doing the inverse Laplace (i.e. the solution XD). I didn't get to try this because I got stuck on a similar problem and couldn't solve that one either! But this one I have to present in class I don't want to let them down!!!!

D:

I guess frequency would be equal to sqrt(b/a) radians of equation af''(t)+bf(t)=0? And would periodic time be equal to 2*pi/sqrt(b/a) or would that be the term in the unit step? Thanks for any help you can give me.

Alrighty. For the first I got the solution into the form F(s)= 1/(s^2 + 1) which fits the identity for sin(x). I did this by Taking the Laplace which I gave me:
[s^2*F(s) - f(0) -f'(0)] + F(s) = 0

For the second again I took the Laplace and I got: 6[S^2*F(s)-f(0)-f'(o)] +2[sF(s) - f(0)] +9F(s) = 0
which I simplified into: (6s^2 + 2s +9)F(s) -18 = 0
and further: (6s-2)(s+4)F(s) +17 = 18
and then I had: F(s) = 1/(6s-2)(s+4)
so I use partial fractions: 1= a/(6s-2) + b/(s+4)
I got a=3/13 and b=-1/26
so: F(s) = 3/13(6s-2) + 1/-26(s+4)
Which I think gives: F(t) = 1/26(e^x/3 - e^-4x)
does that look right?

Which apparently is totally wrong for the second. The back of the book says
F(t) = [18e^(-t/6)/sqrt(53)]sin(sqrt(53)/6)*t

totally clueless

HOLD THE PHONE! I see what I did wrong!
F(s)= 18/(6s^2 +2s+9) it figures this problem would be a lot harder than it appears. don't know how to simplify