Consider a system of differential equations $$ \begin{cases} \dot{p}(t)= \frac{1}{R} p(t) \times q(t)\\ \dot{q}(t)= - p(t) \times q(t) \end{cases} \qquad (1) $$ for $p(t),q(t)\in \mathbb{R}^3$.
Of course one may notice that $\dot{q}(t)=-R\dot{p}(t)$ giving $q(t)=-Rp(t)+d$ for some $d\in\mathbb{R}^3$ constant. Then the above system reduces to the single equation $$ \dot{p}(t) = \frac{1}{R} p(t) \times q(t) = \frac{1}{R} p(t) \times (-Rp(t)+d) = -\frac{1}{R}d \times p(t) = c \times p(t) \qquad (2) $$ for $c=-\frac{1}{R}d$.
This answer provides the solution to (2): $$ p(t)=c\frac{c\cdot p(0)}{c\cdot c}+\left(p(0)-c\frac{c\cdot p(0)}{c\cdot c}\right)\cos\left(|c|\,t\right)+\frac{c}{|c|}\times p(0)\sin\left(|c|\,t\right).\qquad (3) $$
The solution (2) is of period $\frac{2\pi}{\vert c\vert} = \frac{2\pi R}{\vert p(0)+Rq(0)\vert}$.
How can we modify (1) only minimally (i.e. keeping a similar form) in order to get solutions $p(t)$ (and also $q(t)$) which are of fix period $T\in\mathbb{R}$ independently from the initial conditions $p(0)$ and $q(0)$?
Edit: Corrected small errors in the formulation of (2).
