Answer
$a_m = 0.2 g$
Work Step by Step
The maximum acceleration is at $\cos(\omega t +\phi)=1$ and is given by
$$a_{m}=\omega^{2} x_{m}=(2 \pi f)^{2} x_{m}$$
Substitute with the values of $f$ and $x_m$
$$a_{m}=(2 \pi f)^{2} x_{m}=(2 \pi \times 2.2 \mathrm{Hz})^{2}(0.01 \mathrm{m})=1.91 \mathrm{m} / \mathrm{s}^{2}$$
And in the tern of $g$, it will be 0.2 g as next
$$a_{m}=\left(1.91 \mathrm{m} / \mathrm{s}^{2}\right)\left(\frac{1 g}{9.8 \mathrm{m} / \mathrm{s}^{2}}\right)=0.2 g$$
