How to construct and oscillation with exponentially growing period times? I'm searching for the (maybe even smooth) "oscillating" function
$$f(t)=A\sin{\left(g(t)\right)},$$
such that there are zeroes at times $t_n=T^n$ for some fixed number $T$. So this will not really be periodic, it will be a motion which makes one full turn at exponentially growing gaps, like for example 
$$t_1=2\ \ \text{sec},\ \ t_2=4\ \text{sec},\ \ t_3=8\ \text{sec},\ \ t_4=16\ \text{sec},\ ...$$
Which function does that? Is there a corresponding Newtonian equation of motion?
 A: Consider the equation of motion for a simple harmonic oscillator,
$$y''(x)+y(x) = 0, \hspace{5ex} y(0) = 0,
\hspace{5ex} y'(0) = 1$$
with solution 
$y(x) = \sin x.$
Change coordinates, let $x(t) = \log t$ and $f(t) = y( x(t)) = \sin(\log t)$.
The zeros of $f(t)$ occur when $\log t_n = n \pi$, that is, when $t_n = (e^\pi)^n$.
Notice that 
$$\frac{d}{d x} = \frac{d t}{d x} \frac{d}{dt} = t \frac{d}{dt}
\hspace{5ex} \textrm{and so}\hspace{5ex} 
\frac{d^2}{dx^2} = \left(t \frac{d}{dt}\right)\left(t \frac{d}{dt}\right)
= t^2 \frac{d^2}{dt^2} + t \frac{d}{dt}.$$
Therefore, the function $f(t)$ satisfies the differential equation
$$t^2 f''(t) + t f'(t) + f(t) = 0,
\hspace{5ex} f(1) = 0,\hspace{5ex} f'(1) = 1.$$
As indicated in the comments, if instead we choose $x(t) = \pi \log_T t$, the zeros will be at $t_n = T^n$.
The differential equation satisfied by $f(t) = \sin(\pi \log_T t)$ is then
$$ t^2 f''(t) + t f'(t) + \left(\frac{\pi}{\log T}\right)^2 f(t) = 0,$$
We can think of this roughly as a harmonic oscillator with time-dependent restoring force.
Addendum: 
Find below a plot of $f(t) = \sin(\pi \log_2 t)$. 

