# 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

Staff Emeritus
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.