= Solution
Let $Q(t)$ denote the <electric charge> on the initially positive <capacitor> plate and define the discharge <electric current> by $I=-\dot Q$. <Kirchhoff's voltage law> for the <series RLC circuit> gives the <damped harmonic oscillator> equation
$$
L\ddot Q+R\dot Q+\frac{Q}{C}=0.
$$
The initial conditions are $Q(0)=Q_0$ and $\dot Q(0)=0$: the open switch initially prevents <electric current>, and finite <inductance> prevents an instantaneous jump in it. Put $\gamma=R/(2L)$ and $\omega_0^2=1/(LC)$. The two <characteristic roots> are $s=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}$. An oscillatory <electric current> requires a nonzero imaginary part, hence
$$
\boxed{R^2<\frac{4L}{C},\quad\text{or}\quad R<2\sqrt{\frac LC},\qquad \omega_d=\sqrt{\frac1{LC}-\frac{R^2}{4L^2}}.}
$$
For positive $L,C$ and nonnegative $R$, this is the <underdamped> regime. The initial conditions give
$$
Q(t)=Q_0e^{-\gamma t}\left(\cos\omega_dt+\frac{\gamma}{\omega_d}\sin\omega_dt\right),\qquad I(t)=\frac{Q_0}{LC\omega_d}e^{-\gamma t}\sin\omega_dt.
$$
At equality there is <critical damping>, and above it there is an <overdamped> response, with no repeated changes of sign of the discharge <electric current>. A nonzero initial <electric charge> is understood; if $Q_0=0$, the stated initial conditions give no <electric current> at all.
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