For a particle executing simple harmonic motion, the time period is...

The velocity of a particle executing simple harmonic motion is given by v = v_\max \cos(\omega t).

The time period of the oscillation is $T = \dfrac{2\pi}{\omega}$.

We need to find the velocity $\vec{v}$ for which the minimum time interval between two consecutive instances of attaining this specific velocity is exactly $T$.

If $v = 0$, the particle is at the extreme positions. It attains zero velocity twice in one complete cycle (at the positive and negative extremes). The time interval between these instances is $\dfrac{T}{2}$.

If 0 < |v| < v_\max, the particle attains this specific velocity vector at two different positions in one cycle. For example, if $\vec{v}$ is positive, it attains this velocity once while moving from the negative extreme to the mean position, and again while moving from the mean position to the positive extreme. The time intervals between these two instances are $\Delta t_1$ and $\Delta t_2$ such that $\Delta t_1 + \Delta t_2 = T$. Hence, the minimum time interval is strictly less than $T$.

If v = v_\max, the particle attains this maximum velocity only at the mean position while moving in the positive direction. This occurs exactly once per cycle. Therefore, the minimum time interval between attaining \vec{v} = v_\max is exactly $T$.