Homework Help: Damped Harmonic Oscillator - Help Please

1. Oct 11, 2011

Fjolvar

1. The problem statement, all variables and given/known data

A damped harmonic oscillator is displaced a distance xo from equilibrium and released with zero initial velocity. Find the motion in the underdamped, critically damped, and overdamped case.

2. Relevant equations

d2x/dt2 + 2K dx/dt + ω2x = 0

Underdamped: x = C*e-Ktcos(ωt-$\gamma$)

Overdamped: x = A*e-K-t+B*e-K+t

Critically Damped: a*e-Kt*(1+bKt)

3. The attempt at a solution

I haven't attempted the solution because I'm not sure how to incorporate xo into the equations. I understand that at time t=0, x=xo, but how do I use this fact? Any help would be greatly appreciated. Thank you in advance.

2. Oct 11, 2011

vela

Staff Emeritus
Each of the solutions has two arbitrary constants. You need to determine those using the initial conditions. In the critically damped case, for example, when t=0, you have x(0)=a; therefore, you know that a=x0. By using the fact that the oscillator starts at rest, you can similarly solve for b.

3. Oct 12, 2011

Fjolvar

Hmm, is the equation I have for under damped correct then, since there is only one constant?

4. Oct 12, 2011

Fjolvar

I solved the equations of motion for the Under-Damped and Critical-Damped conditions, but I'm having trouble finding the Over-Damped constants.. Any advice?

Last edited: Oct 12, 2011
5. Oct 12, 2011

Fjolvar

If I write the equation for over-damping using the initial conditions for t=0, I get: Xo=Ae^(-K) + Be^(-K)... which just gives me e^(-K)*(A+B). I'm not sure how to solve for A and B given this.. Any help before class would be greatly appreciated.. again thanks in advance.

6. Oct 12, 2011

vela

Staff Emeritus
Your solution for the overdamped case is incorrect.

7. Oct 12, 2011

Fjolvar

I believe that is just the solution for the real part? I'll have to look for the general solution..

8. Oct 13, 2011

HallsofIvy

Your "underdamped" solution does have two constants, C and $\gamma$

Your "overdamped" solution, $x= Ae^{-K-t}+Be^{-K+t}$ should be $x= Ae^{Kt}+ Be^{-Kt}$.

9. Oct 13, 2011

vela

Staff Emeritus
HallsofIvy has uncharacteristically made a mistake regarding the overdamped solution. In the overdamped case, the characteristic equation, $r^2+2Kr+\omega^2=0$, has two real roots
\begin{align*}
r_+ &= {-K}+\sqrt{K^2-\omega^2} \\
r_- &= {-K}-\sqrt{K^2-\omega^2}
\end{align*}so the solution is
\begin{align*}
x(t) &= Ae^{r_+ t}+Be^{r_- t} \\
&= Ae^{(-K+\sqrt{K^2-\omega^2})t}+Be^{(-K-\sqrt{K^2-\omega^2})t}
\end{align*}So when t=0, you get x(0) = A+B.