Damped & Negatively damped harmonic oscillator [not physics question] $$\frac{d^2Q(t)}{dt^2} + \frac{\omega(t)}{2\pi} \frac{dQ(t)}{dt} + \omega^{2}(t)Q(t) = 0, $$
where $ \omega(t)= \frac{2\pi}{v(t)} \frac{dv}{dt}  $ and $v>0$ is a known parameter which is given from a data feed.
I have already built some visualization by using mathematica.
Can I transform this equation by using the variable $T(t)$ instead of $t$?:
$$\frac{1}{T(t)}= \frac{1}{v(t)} \frac{dv}{dt} .$$
 A: This transformation is possible iff $T(t)$ is a bijection. You can for example guarantee this by demanding that
$$0< \frac{d}{dt} T(t)^{-1} = \frac{d^2}{dt^2} \ln \nu(t)\,.$$
Assuming  that the above holds, we can invert the relation and obtain $t(T)$. You are asking what differential equation $f(T)=Q(t(T))$ obeys. The relevant chain rule reads
$$\frac{dQ}{dt}= \frac{df(T(t))}{dt} =  \frac{df}{dT} \frac{dT}{dt}
 = -\frac{df}{dT} \frac{2\pi \omega'}{\omega^2}$$
where we have used that
$$\frac{dT}{dt} = \frac{d}{dt} \left(\frac{2\pi}{\omega} \right)= -\frac{2\pi \omega'}{\omega^2} \,.$$
Similarly, we obtain
$$ \frac{d^2 Q}{dt^2} = \frac{d}{dt}\left( -\frac{df}{dT} \frac{2\pi \omega'}{\omega^2}\right) = \frac{d^2 f}{dT^2}\left(\frac{2\pi \omega'}{\omega^2}\right)^2 -\frac{df}{dT} \underbrace{\frac{d}{dt} \left(\frac{2\pi \omega'}{\omega^2}\right)}_{=2\pi(\omega \omega''-2 \omega')/\omega^3}\,. $$
As a result, your differential equation transforms into
$$\left(\frac{2\pi \omega'}{\omega^2}\right)^2f''-\frac{\omega^2 \omega'+2 \pi  (\omega \omega
   ''-2  \omega'^2)}{\omega^3} f' + \omega^2 f =0\,.$$
