Let us state a simple theorem.
Theorem: Suppose $f(x)$ and $g(x)$ are (sufficiently nice) function which are periodic, i.e. $f(x+2\pi) = f(x)$ and $g(x+2\pi) = g(x)$. If $u(x,t)$ is a function which solves the PDE, $u_{tt} = u_{xx}$ with the initial conditions that $u(x,0) = f(x)$ and $u_t(x,0) = g(x)$ then $u(x,t)$ would also be periodic function i.e. $u(x+2\pi,t) = u(x,t)$.
Proof: This follows immediately from d'Alembert's formula.
The key observation here is that the function $\psi(x)$ is already a periodic function, therefore, $u(x,t)$ will automatically be a periodic function. Furthermore, since the boundary condition is given by $u(0,t) = \phi(t)$, by the same function at the start of the problem, it means you do not need to do anything else to satisfy the boundary condition.
You would be correct that, in general, if the conditions $u(0,t)$ was entirely unrelated to $\psi(t)$ then this simple approach would not work and more effort is required to produce the solution.
Best Answer
Solve the two PDEs
$u_{tt}-4u_{xx}=xt$, $u(0,x)=0$ ,$u_t(0,x)=0$ $(1)$
and
$u_{tt}-4u_{xx}=0$, $u(0,x)=x^2$, $u_t(0,x)=x$ $(2)$
Solve (2) using D'Alembert's formula. For (1) use the transformation $\xi=x+2t,\eta=x-2t$ and reduce the problem to $u_{\xi\eta}=x(\xi,\eta)t(\xi,\eta)$. If $u_1$ is the solution of $(1)$ and $u_2$ is the solution of $(2)$ then $u_1+u_2$ will be the solution of your pde.