Answer
$T=11.76\approx 12s$, $f=0.085Hz$
Work Step by Step
The distance from one side of the track, to the other, and back to the original position is twice the total distance, which is $2(5.0m)=10.m$. Remember that $$v=\frac{\Delta x}{\Delta t}$$ Therefore, solving for $\Delta t$ yields $$\Delta t=\frac{\Delta x}{v}$$ Substituting known values of $\Delta x=10.m$ and $v=0.85m/s$ yields a time, or period, of $$\Delta t=T=\frac{10.m}{0.85m/s}=11.76\approx 12s$$ Using the relation between frequency and period that $$f=\frac{1}{T}$$ and using the known value of $T=11.76s$ yields a frequency of $$f=\frac{1}{11.76s}=0.085Hz$$
