# Second derivative of $\phi$ in nondimensionalization problem.

I am reading a section in my differential equations textbook and am trying to fill in the left out steps in their explananation of how to nondimensionalize the second order system $$mr \ddot{\phi}=-b\dot{\phi}-mg\sin \phi + m r \omega^{2}+mr\omega^{2}\sin \omega \cos \phi$$

Where $\dot{\phi}=\frac{d\phi }{dt }$ and $\ddot{\phi}=\frac{d^{2}\phi }{dt^{2}}$ (what the other things mean is not relevant to my question and so I am leaving those details out).

To find the nondimensionalized form of just the $\dot{\phi}$ part, we let $t=\tau T$. Then, $dt=T d\tau$ implies that $\frac{d\tau}{dt}=\frac{1}{T}$, so that $$\dot{\phi}=\frac{d\phi}{dt}=\frac{d\phi}{d\tau}\frac{d\tau}{dt}=\frac{d\phi}{d\tau}\cdot \frac{1}{T}$$

This I understand how to do.

However, I would like to show similarly, by means of the chain rule, that $$\ddot{\phi}=\frac{1}{T^{2}}\frac{d^{2}\phi}{d\tau^{2}}.$$

Those details are NOT included in my text, and I'm not going to be able to understand anything about dimensional analysis and scaling until I can figure out how they got this and reproduce it myself. Can someone please show me how they did this? Using the chain rule, please.

$$\frac{{\rm d}}{{\rm d}t} = \frac{{\rm d}\tau}{{\rm d}t}\frac{{\rm d}}{{\rm d}\tau} = \frac{1}{T}\frac{{\rm d}}{{\rm d}\tau}$$
$$\frac{{\rm d}^2}{{\rm d}t^2} = \frac{{\rm d}}{{\rm d}t}\left(\frac{{\rm d}}{{\rm d}t}\right) = \frac{1}{T}\frac{{\rm d}}{{\rm d}\tau} \left(\frac{1}{T}\frac{{\rm d}}{{\rm d}\tau} \right) = \frac{1}{T^2}\frac{{\rm d}}{{\rm d}\tau}\left(\frac{{\rm d}}{{\rm d}\tau} \right) = \frac{1}{T^2}\frac{{\rm d}^2}{{\rm d}\tau^2}$$