 Solution

Let $q(t)=\mathbf{r}(t)\cdot \mathbf{B}$. Since the field is constant, dotting the equation of motion with $\mathbf{B}$ eliminates the magnetic term:This is a harmonic oscillator with frequency $\omega =\sqrt{k/m}$. Initially $q(0)={\mathbf{r}}_0\cdot \mathbf{B}$ and $\dot{q}(0)=0$, because the particle starts at rest. ThereforeThe magnetic force modifies the transverse motion but not this projection. The formula also holds trivially when $\mathbf{B}=0$.