Answer
$T = 3.1~ms$
Work Step by Step
Note that the relation defining maximum acceleration is $$a_m=A\omega ^2$$ Solving for omega yields $$\omega = \sqrt{\frac{a_m}{A}}$$ Substituting known values of $a_m=8.0\times 10^3 m/s$ (maximum acceleration) and $A=0.0020m$ (amplitude) to get an omega value of $$\omega=\sqrt{\frac{8.0\times 10^3m/s}{0.0020m}}=2.0\times 10^3~rad/s$$ Omega and the period can be related using the equation $$T=\frac{2\pi}{\omega}$$ Substituting the known value of $\omega=2.0\times 10^3rad/s$ yields a period of $$T=\frac{2\pi}{2.0\times 10^3rad/s}=3.1\times 10^{-3}s=3.1 ~ms$$
