Solution

For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So, $$x(t) = A~\sin{~\omega t}$$ where $A$ is the amplitude such that $A = 1~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then, $$x_0 = A~\sin{~\omega t_1} \quad(1)$$ $$A = A~\sin{~\omega t_2} \Leftrightarrow \boxed{\sin{~\omega t_2} = 1}$$ Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$.

As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$, $$t_1 = \frac{T}{4} - \Delta t \quad(2)$$
Putting $(2)$ into $(1)$ and separating $T$, it is obtained $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin {\frac{x_0}{A}}}$$