Answer
$0.72$
Work Step by Step
We can determine the required coefficient of static friction as follows:
We know that
$F_x=f_{s, max}=m_{a_{x,max}}$
$\implies \mu_s n=m\omega^2 A$
$\implies \mu_s (mg)=m(\frac{2\pi}{T})^2A$
This simplifies to:
$\mu_s=\frac{(\frac{2\pi}{T})^2A}{g}$
We plug in the known values to obtain:
$\mu_s=\frac{(\frac{2\pi}{1.5s})^2(0.4m)}{9.8m/s^2}$
$\implies \mu_s=0.72$
