Answer
An amplitude of $~~0.23~m~~$ would put the smaller block on the verge of slipping.
Work Step by Step
We can find $\omega$:
$\omega = \sqrt{\frac{k}{M+m}}$
$\omega = \sqrt{\frac{200~N/m}{10~kg+1.8~kg}}$
$\omega = 4.117~rad/s$
The magnitude of the maximum acceleration is $~~a = A~\omega^2$
We can find the maximum possible amplitude:
$F = ma$
$mg~\mu_s = m~A~\omega^2$
$g~\mu_s = A~\omega^2$
$A = \frac{g~\mu_s}{\omega^2}$
$A = \frac{(9.8~m/s^2)(0.40)}{(4.117~rad/s)^2}$
$A = 0.23~m$
An amplitude of $~~0.23~m~~$ would put the smaller block on the verge of slipping.
