= Solution
Assume $\omega>0$. The <characteristic roots> of the <damped harmonic oscillator> are $-k\pm\sqrt{k^2-\omega^2}$. Its general homogeneous solution is
$$
\boxed{\begin{array}{ll}
k<\omega:&x=e^{-kt}(C\cos pt+D\sin pt),\quad p=\sqrt{\omega^2-k^2},\\
k=\omega:&x=(C+Dt)e^{-kt},\\
k>\omega:&x=e^{-kt}(C\cosh qt+D\sinh qt),\quad q=\sqrt{k^2-\omega^2}.
\end{array}}
$$
These are respectively the underdamped, <critically damped> and overdamped regimes.
For the <velocity-switched driven oscillator>, the initial acceleration is $-\omega^2x_1<0$. Its velocity is therefore negative on the first half-cycle, and the forcing is zero:
$$
\boxed{x(t)=x_1e^{-kt}\left(\cos pt+\frac{k}{p}\sin pt\right),\qquad 0\leq t\leq\frac\pi p.}
$$
Indeed $\dot x=-(\omega^2x_1/p)e^{-kt}\sin pt<0$ in that open interval. Set $r=e^{-k\pi/p}$; at the first turning point, $x_2=-rx_1$ and $\dot x=0$.
Put $\tau=t-\pi/p$. On the second half-cycle the velocity is positive, so the particular solution is $a/\omega^2$. Matching position and velocity gives
$$
\boxed{x(t)=\frac a{\omega^2}
+\left(-rx_1-\frac a{\omega^2}\right)e^{-k\tau}
\left(\cos p\tau+\frac{k}{p}\sin p\tau\right),\quad
\frac\pi p\leq t\leq\frac{2\pi}p.}
$$
Its derivative is positive for $0<\tau<\pi/p$. At the next turning point,
$$
x_3=r^2x_1+(1+r)\frac a{\omega^2},\qquad \dot x_3=0.
$$
The <turning-point return map> therefore repeats the initial state exactly when
$$
\boxed{x_1=\frac{a}{\omega^2(1-r)},\qquad r=e^{-k\pi/p},\qquad T=\frac{2\pi}{p}.}
$$
This value exceeds $a/\omega^2$, so the subsequent acceleration initiates the required negative-velocity half-cycle again. The switched solution is continuous with continuous velocity, and is piecewise twice differentiable; acceleration has one-sided jumps where the forcing switches.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ia/paper-2-switched-oscillator.png]
{title=Periodic motion of the oscillator with forcing applied only during positive velocity}
For $k>\omega$, the specified positive initial value gives
$$
x=x_1e^{-kt}\left(\cosh qt+\frac{k}{q}\sinh qt\right),\qquad
\dot x=-\frac{\omega^2x_1}{q}e^{-kt}\sinh qt<0\quad(t>0).
$$
There is no next turning point: the forcing stays zero and $x$ approaches zero monotonically. \b[The motion from the specified initial state cannot be periodic.] More generally, a solution started at any positive maximum has this same obstruction; a nonconstant periodic orbit would require such a maximum. The zero equilibrium is a separate constant solution.
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