Damped HO and Laplace transform method

[h0dgey84bc]

Hi,

I am trying to solve the damped harmonic oscillator:

$\frac{d^2y}{dt^2}+\frac{b}{m} \frac{dy}{dt}+\frac{k}{m}y=0$

and I thought using the Laplace transform might do the trick. Anyway so I did the LT (and inserted the initial conditions that at t=0 y=A, and dy/dt=0) and obtained:

$Y(p)=\frac{A(p+\frac{b}{m})}{p^2+\frac{b}{m}p+\frac{k}{m} }$

I can't seem to get this into a simple form so I can use the lookup tables to do the inverse transform. So I guess I have to somehow use the Bromwhich integral. Obviously the two poles are $p=-\frac{b}{2m} \pm sqrt{ \frac{b^2}{4m^2}-\frac{k}{m}$ and I recognise this from the actual solution so I think I'm on the right track.

I just don't know how to get the residues and complete the inverse

 

[HallsofIvy]

Why go through all that?

This is a linear, homogeneous d.e. with constant coefficients. It's characteristic equation is r²+ (b/m)r+ k/m= 0. Solve for r using the quadratic formula.

If the roots are a$\pm$ bi, then the general solution is y(t)= e^at(C cos(bt)+ D sin(bt))

 

[h0dgey84bc]

I know it can be done by simpler means, it's just that I'm currently revising Laplace transforms and thought this would be a good exercise for me...until I got stuck :(

 

[Ben Niehoff]

Using the two roots for p, you can factor the denominator and then decompose into partial fractions.

It's a lot of algebra, but it should work.

(Aside: Your tables don't have examples with a quadratic in the denominator?)