# Solution to Harmonic Oscillator

The damped oscillator is governed by the equation of motion: $\ddot{x} + 2\beta \dot{x}+\omega _{0}^2x=0$ where $\omega _{0}=\beta$ for an oscillator with critical damping. The solution is $x(t)= e^{-\beta t}[C_{1}e^{\sqrt{\beta ^{2}-\omega _{0}^{2}t}}+C_{2}e^{-\sqrt{\beta ^{2}-\omega _{0}^{2}t}}]$ . I see in the textbook I use an exercise to finding an alternative solution to the equation of motion by Considering $\beta =\omega _{0}+\epsilon$ and Expanding the solution in $\epsilon$ in order to demonstrate that for small values of $\epsilon/\omega _{0}$ , one can obtain the solution in the form $x(t)\approx e^{-\beta t}[(\tilde{C_{1}})+\tilde{C_{2}}t]$. I was wondering if anyone knew of alternative solutions to the equations, or how to represent this solution for small values of $\epsilon/\omega _{0}$.

• Is there a typo? $\omega_0 = \beta$? This isn't true in general. Are you asking about the critically-damped oscillator? – Bill N Oct 14 '15 at 2:51
• Are there factors of $i$ missing from the exponentials in these equations? – garyp Oct 14 '15 at 3:09
• I was asking about the critically damped oscillator. I don't think there are factors of i missing from the exponentials but correct me if I am wrong. – J Smith Oct 14 '15 at 4:21

• If $\beta > \omega_0$, then solution will be a linear combination of the basis $\{e^{-\beta t}e^{\omega t}, e^{-\beta t}e^{-\omega t}\}$.
• If $\beta = \omega_0$, then solution will be a linear combination of the basis $\{e^{\beta t}, te^{\beta t}\}$.
• If $\beta < \omega_0$, then solution will be a linear combination of $\{e^{-\beta t}\cos\omega t, e^{-\beta t}\sin\omega t\}$.