Answer
a) To reduce the maximum acceleration, we should increase the period.
b) 11 s
Work Step by Step
(a) We know that $\alpha_{max}=A(\frac{2\pi}{T})^2$. This equation shows that acceleration is inversely proportional to the time period. Thus, to reduce the maximum acceleration, we should increase the period.
(b) We know that $\alpha_{max}=A(\frac{2\pi}{T})^2$
This can be rearranged as:
$T=2\pi \sqrt{\frac{A}{\alpha_{max}}}$
We plug in the known values to obtain:
$T=2\pi \sqrt{\frac{30.0}{9.81}}$
$T=11s$
