Damped Driven Oscillator Dimensional Analysis

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

$$M\ddot{x}+\gamma \dot{x}+kx=F_0 \cos({\omega t})$$ 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:

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

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

$$t_c=\sqrt{\frac{M}{k-\frac{\gamma^2}{4M}}}$$ 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!

• Is $\Delta$ the discriminant which indicates that the damping is small, i.e. $\gamma^2/4-k/M$? Oct 12, 2015 at 22:46
• Also, shouldn't there in the formula for $t_c$ be $\gamma^2$ instead of $\gamma$? Oct 12, 2015 at 22:57
• Set $x_c=\frac{M}{F_0t_c^2}$ then see what you can do with $t_c$. Oct 12, 2015 at 23:16
• @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) Oct 13, 2015 at 8:19
• @David Are you sure I should do that? Why would you do that? Oct 13, 2015 at 8:20

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:
$$\frac{F_0 t_c^2}{Mx_c}=\lambda \Leftrightarrow x_c=\frac{F_0 t_c^2}{M \lambda}$$ 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: $$t_c=\sqrt{\frac{M}{k-\frac{\gamma^2}{4M}}}$$ or on terms of the frequency applied to the system by the drive force, therefore: $$t_c=\frac{1}{\omega}$$ One can use both, and I gave to myself a treat by using the second (and simpler) to reach the final result:
$$x_c=\frac{F_0}{\lambda M \omega^2}, \quad t_c=\frac{1}{\omega}$$