This does not fit as a comment.
A direct naive definition of a continuous time dynamical system on a discrete space may not exist. Here is one reason. Suppose the discrete space $S$ is finite. $\DeclareMathOperator{\Perm}{Perm}$ $\newcommand{\bR}{\mathbb{R}}$ Denote by $\Perm(S)$ the group of permutations of $S$. A dynamical system on $S$ would be a group morphism
$$\bR\ni t\mapsto \phi^t\in \Perm(S), $$
so $\phi^t$ is a permutation of $S$ for any $t$ and $\phi^t\circ \phi^s=\phi^{t+s}$ for any $s,t\in \bR$. For each $t\in \bR$ the permutation $\phi^t$ of $S$ is rather special: for any natural number $n$ it admits a $n$-th root,
$$\phi^t= \underbrace{ \phi^{t/n}\circ\cdots \circ\phi^{t/n}}_n. $$
Obviously the identity permutation has this property, but if $\phi^t$ is not trivial this thing is not possible. The forst of Joechen Glueck's comments explains why.
Here is a sketch of a proof.
Look at the cycles of $\phi^t$, or equivalently, at the orbits of the action of $\phi^t$. Observe first that $\phi^t$ cannot be cyclic, i.e., consists of a single cycle.
For any permutation $\phi\in \Perm(S)$ denote by $c(\phi)$ the size if the largest cycle of $\phi$. Note that if $\psi^n=\phi$ then $c(\psi)\geq c(\phi)$.
If $c(\phi^t)>1$ for some $t$, then $c(\phi^{t/n})\geq c(\phi^t)$, for any $n$ and we cannot have equality for all $n$'s since cyclic permutations do not have roots of any order. Thus $c(\phi^{t/n_1})>c(\phi^t) $ for some $n_1>0$. Set $t_1=t/n_1$. Run the above argument with $t$ replaced by $t_1$. Hat tip to Jochen Glueck for his useful comments.
If $S$ is infinite this may be possible, but I am not sure you are interested in this situation.
Now I come to my main point. I, as a mathematician, do not understand precisely the question.
For example, I do not understand what is that discrete state space: is it a finite set $X$ of oscillators or is it a set of pairs $(x,p)$, where $x\in X$ is an oscillator and $p$ is its period that belongs to a set $P$ of possible periods. In this case $S$ would by $X\times P$ and the set of periods could conceivably be infinite. Clarifying this point would help.
I also do not understand exactly what the transitions from one state to another are. In any case, if the set of periods is discrete, this situation reminds me of continuous time Markov chains where the state space is discrete, but the transitions are random and the waiting time $\delta t$ for the next transition to occur is also random. The book of J. R. Norris Markov chains has a nice description of such Markov chains.
On the other hand, the fact that you wrote that the period changes from $p +\delta p$ seems to suggest that you think of the set $P$ of periods a continuum, an interval. In this case you might have a continuous Markov process on a non-discrete state space.
These are pure speculations on my part, and it would help if you could provide a few more details about the problem you are studying.
