Is the function $f(t)=\sin(\omega_0 t+\phi_0(t))$ periodic? Given the following function:
$$f(t)=\sin(\omega_0 t+\phi_0(t))$$
with:
$$\phi_0(t)=\sin(\omega_1t+\phi_1(t))$$
$$\phi_1(t)=\sin(\omega_2t+\phi_2(t))$$
$$\phi_k(t)=\sin(\omega_{k+1}t+\phi_{k+1}(t))$$
$$\phi_N(t)=\sin(\omega_{N+1}t+\phi_{N+1})$$
Under what conditions the $f(t)$ can be periodic?
Thanks.
 A: Some simple considerations. Leaving the question of convergence aside.
Start from the so-called 'trivial case' of $\omega_k=\omega_0$ for all $k$. It's obviously periodic, since at each recursive step the functions are periodic with the same period. The plot looks like this:

Now for the more general case:
$$\omega_{k+p}=\omega_k, \qquad p \in \mathbb{N}$$
In this case $\omega_p=\omega_0$ and the rest:
$$\omega_{j}=q_j \omega_0, \qquad j \in [1,p-1]$$

I conjecture that for $q_j \in \mathbb{Q}$, i.e. rational, and for any finite $p$ the function is periodic. 

It can be concluded from the usual trigonometric formulas for angle transformation, since they all involve rational multiples of the original angle/frequency.

Another conjecture is that if at least one $q_j$ is irrational then the resulting function is not periodic.

In principle $p$ can be infinite, but only if all $q_j$ are known and their denominators are bounded (see below).

Some examples:
$$p=2, \qquad \omega_1=2 \omega_0$$

$$p=3, \qquad \omega_1=3 \omega_0, \qquad \omega_2=2 \omega_0$$

$$p=2, \qquad \omega_1=\frac{3}{7} \omega_0$$

The period of the function becomes larger for $q_j$ not an integer because the original period related to $\omega_0$ has to be multiplied by the denominator.
