So, I am studying the damped driven oscillator with the following drive force:

\begin{equation} M\ddot{x}+\gamma \dot{x}+kx=F_0 \cos({\omega t}) \end{equation} where $M$ is the mass, $\gamma$ is the damping force, $k$ is the spring constant and $\omega$ is constant.

What I am trying to do is find a dimensionless expression of the equation above which can be done by applying the transformation:

\begin{align} & x \to \xi=\frac{x}{x_c} \\ & t \to \tau=\frac{t}{t_c} \end{align} where $x_c, t_c$ are some "x, t characteristic". Now, I found that the expression of the oscillator in those variables would be:

\begin{equation} \ddot{\xi}+\left( \frac{\gamma t_c}{M} \right) \dot{\xi}+\left( \frac{k t_c^2}{M} \right) \xi=\left( \frac{F_0}{M} \right)\frac{t_c^2}{x_c}\cos({\omega t}) \end{equation}

and by solving the initial equation I can acquire the analytical solution from which I could deduct that (for the case of $\Delta <0$):

\begin{equation} t_c=\sqrt{\frac{M}{k-\frac{\gamma^2}{4M}}} \end{equation} which proves that everything is in the right place since it does match the period of the damped oscillator.

My question is the following

What about $x_c$? I am not able to find an expression for it, and I can see that it is directly involved in the ODE.

Thank you all!

  • $\begingroup$ Is $\Delta$ the discriminant which indicates that the damping is small, i.e. $\gamma^2/4-k/M$? $\endgroup$ Oct 12, 2015 at 22:46
  • $\begingroup$ Also, shouldn't there in the formula for $t_c$ be $\gamma^2$ instead of $\gamma$? $\endgroup$ Oct 12, 2015 at 22:57
  • $\begingroup$ Set $x_c=\frac{M}{F_0t_c^2}$ then see what you can do with $t_c$. $\endgroup$
    – David
    Oct 12, 2015 at 23:16
  • $\begingroup$ @Sobanoodles yeap its $\gamma ^2$ :) And $\Delta$ is the discriminant of the characteristic polynomial for the initial equation. (I also studied the overdamped and the critical case but only the underdamped case has appears to have oscillations obviously) $\endgroup$
    – Bazinga
    Oct 13, 2015 at 8:19
  • $\begingroup$ @David Are you sure I should do that? Why would you do that? $\endgroup$
    – Bazinga
    Oct 13, 2015 at 8:20

1 Answer 1


So, I think I figured it out.

For the dimensionless expression, the constant coefficient $\frac{F_0 t_c^2}{Mx_c}$ of the drive force $\cos({\omega t})$, stands for the oscillation length. Taking that into account, along with the fact that we do demand $x_c$ to be dimensionless, one could think:

\begin{equation} \frac{F_0 t_c^2}{Mx_c}=\lambda \Leftrightarrow x_c=\frac{F_0 t_c^2}{M \lambda} \end{equation} where $\lambda$ of course is a dimensionless constant.

Next, something that did not occure to me while posting this question, is that $t_c$ could be expressed either on terms of the system's own period like above: \begin{equation} t_c=\sqrt{\frac{M}{k-\frac{\gamma^2}{4M}}} \end{equation} or on terms of the frequency applied to the system by the drive force, therefore: \begin{equation} t_c=\frac{1}{\omega} \end{equation} One can use both, and I gave to myself a treat by using the second (and simpler) to reach the final result:

\begin{equation} x_c=\frac{F_0}{\lambda M \omega^2}, \quad t_c=\frac{1}{\omega} \end{equation}


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