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 $\epsilon/\omega _{0}$.

  • $\begingroup$ Is there a typo? $\omega_0 = \beta$? This isn't true in general. Are you asking about the critically-damped oscillator? $\endgroup$
    – Bill N
    Commented Oct 14, 2015 at 2:51
  • $\begingroup$ Are there factors of $i$ missing from the exponentials in these equations? $\endgroup$
    – garyp
    Commented Oct 14, 2015 at 3:09
  • $\begingroup$ 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. $\endgroup$
    – J Smith
    Commented Oct 14, 2015 at 4:21

1 Answer 1


The solution set of a second order differential equation must be the combination of two linear independent solutions. When solving it, one arrives at 3 possible outcomes:

  • 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 $\{e^{-\beta t}\cos\omega t, e^{-\beta t}\sin\omega t\}$.

So, basically, there are only those solutions. Its easy to mathematically formally prove this statement.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.