Damped Harmonic Oscillator -

[Fjolvar]

Homework Statement

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.

Homework 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)

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.

 

[vela]

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.

 

[Fjolvar]

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.

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

 

[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?

 

[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.

 

[vela]

Your solution for the overdamped case is incorrect.

 

[Fjolvar]

Your solution for the overdamped case is incorrect.

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

 

[HallsofIvy]

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

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

 

[vela]

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.