Answer
$12g$
Work Step by Step
We can find the maximum acceleration as
$\alpha_{max}=A(2\pi f)^2$
$\alpha=(0.25mm)(2\pi \times110)^2=119\frac{m}{s^2}$
Now we can write this acceleration as a multiple of g as
$\alpha_{max}=119(\frac{g}{9.81})=12g$
