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.
You are on the right track
$$ \frac{{\rm d}}{{\rm d}t} = \frac{{\rm d}\tau}{{\rm d}t}\frac{{\rm d}}{{\rm d}\tau} = \frac{1}{T}\frac{{\rm d}}{{\rm d}\tau} $$
And
$$ \frac{{\rm d}^2}{{\rm d}t^2} = \frac{{\rm d}}{{\rm d}t}\left(\frac{{\rm d}}{{\rm d}t}\right) = \frac{1}{T}\frac{{\rm d}}{{\rm d}\tau} \left(\frac{1}{T}\frac{{\rm d}}{{\rm d}\tau} \right) = \frac{1}{T^2}\frac{{\rm d}}{{\rm d}\tau}\left(\frac{{\rm d}}{{\rm d}\tau} \right) = \frac{1}{T^2}\frac{{\rm d}^2}{{\rm d}\tau^2} $$