Consider the second-order ordinary differential equation

$$\ddot{x}+2 k \dot{x}+\omega^{2} x=0,$$

where $x=x(t)$ and $k$ and $\omega$ are constants with $k>0$. Calculate the general solution in the cases (i) $k<\omega$, (ii) $k=\omega$, (iii) $k>\omega$.

Now consider the system

$$\ddot{x}+2 k \dot{x}+\omega^{2} x= \begin{cases}a & \text { when } \dot{x}>0 \\ 0 & \text { when } \dot{x} \leqslant 0\end{cases}$$

with $x(0)=x_{1}, \dot{x}(0)=0$, where $a$ and $x_{1}$ are positive constants. In the case $k<\omega$ find $x(t)$ in the ranges $0 \leqslant t \leqslant \pi / p$ and $\pi / p \leqslant t \leqslant 2 \pi / p$, where $p=\sqrt{\omega^{2}-k^{2}}$. Hence, determine the value of $x_{1}$ for which $x(t)$ is periodic. For $k>\omega$ can $x(t)$ ever be periodic? Justify your answer.