Speed and Angular Momentum for Harmonic Central Potential

To determine the speed $v$ and angular momentum $L$ of the particle, we analyze the central force acting on it.

The potential energy of the particle as a function of the radial distance $r$ from the fixed origin $O$ is given by: $V(r) = \frac{1}{2} k r^2$

The conservative central force acting on the particle is directed radially inward and is given by: $\vec{F} = -\frac{dV}{dr}\hat{r} = -k r \hat{r}$

Thus, the magnitude of the attractive central force at a distance $r = R$ is: $F = k R$

For the particle to execute a uniform circular motion of radius $R$ with speed $v$, this attractive central force must provide the necessary centripetal force: $\frac{m v^2}{R} = k R$

Solving for the speed $v$: $v^2 = \frac{k}{m} R^2 \implies v = \sqrt{\frac{k}{m}} R$

The magnitude of the angular momentum $L$ of the particle about the origin $O$ is: $L = |\vec{r} \times \vec{p}| = m v R \sin(90^\circ) = m v R$

Substituting the expression for $v$ into the formula for $L$: $L = m \left(\sqrt{\frac{k}{m}} R\right) R = \sqrt{mk} R^2$

Therefore, the correct statements are given by options B and C.