Answer
$a_m = 1727 \mathrm{~m/s^2}$
Work Step by Step
The acceleration of the equilibrium position of SHM is formed in this equation by
$$ a=-\omega^{2} x_{m} \cos (\omega t+\phi) $$
The term $\omega^{2} x_{m} $ represents the maximum acceleration $a_{max}$ at $\cos (\omega t+\phi) = 1 $.
$$a_m = \omega^{2} x_{m} $$
By using the expression of $\omega$ we get the maximum acceleration by
\begin{align*}
a_m =\omega^{2} x_{m}= (2 \pi /T )^{2} x_{m} = (2 \pi / 0.04 \,\text{s} )^{2} ( 0.07 \,\text{m} ) = \boxed{1727 \mathrm{~m/s^2}}
\end{align*}
