What are the possible periods of the solutions to this ODE? 
Consider the non-autonomous equation $$\ddot x + x = \epsilon f(x,\dot x) + \epsilon \sin(3t)$$ Assume that this equation has (a) periodic solution(s). What are the possible periods?

I need to assume that this equation has (a) period solution(s), e.g., solutions $x = \phi(t)$ for which $\phi(t + T) = \phi(t)$ for some $T$ and I need to find for which $T$ this is possible. As a start I entered $\ddot x + x = \epsilon\sin(3t)$ ($f(x,\dot x) = 0$) in Wolfram for different $\epsilon$ and it seems as if $\epsilon$ doesn't have an impact on the period of the solution for this equation. I therefore think that the possible periods depend on the function $f(x,\dot x)$, but I'm not really sure which options for $f(x,\dot x)$ I should consider. Furthermore, I think that I should do something with $\sin(3t)$ since I think this term will definitely have an impact on the possible periods. 
My Question: I really have no clue how I should solve this problem, and I'm not necessarily looking for the direct solution. I would like to know where to start and which questions I should ask so that I can find the solution by myself.
Thanks in advance!

I'll follow the hint that I've left in the comments. Suppose that $x = \phi(t)$ is a periodic solution of an equation 
$$ \ddot{x} + x = \epsilon f(x, \dot{x}) + \epsilon \sin{3t}. $$
Let $\phi(t)$ be $T$-periodic, $\phi(t) \equiv \phi(t + T)$. Since $\phi(t)$ is a periodic function, its derivatives and compositions are also periodic; i.e., $\dot{\phi}(t)$, $\ddot{\phi}(t)$, $f(\phi(t), \dot{\phi}(t))$ are all periodic functions. So, the following 
$$  \ddot{\phi}(t) + \phi(t) \equiv \epsilon f(\phi(t), \dot{\phi}(t)) + \epsilon \sin{3t} $$
holds for all $t$, thus this
$$  \ddot{\phi}(t+T) + \phi(t+T) \equiv \epsilon f(\phi(t+T), \dot{\phi}(t+T)) + \epsilon \sin{3(t+T)} $$
also holds for any $t$.
Subtracting the first from the second and taking into account the periodicity of $\phi(t)$, we get
$$ \sin{3(t+T)} \equiv \sin{3t},$$
thus $T = \frac{2\pi}{3} k,\; k\in \mathbb{N}$. This essentially means that the only possible periods are multiples of the forcing period. 
Of course, the periods that can be really observed in such equation strongly depend on the $f(x, \dot{x})$. If $f(x, \dot{x}) = x$, you can have $T$-periodic solution for any $T = \frac{2\pi}{3} n$.
