Answer
$\mu_s=0.81$
Work Step by Step
The cup begins slipping when $A=5\times10^{-2}m$. The maximum acceleration at that point is $$a_{max}=A\omega^2=A(2\pi f)^2$$ We've got $f=2Hz$, so $a_{max}=7.9m/s^2$
When the tray moves back and forth, it produces a pushing force $P$ to move the cup, which can be calculated by $P=m_{cup}a$. The cup resists moving because static friction $f_s$ opposes motion.
Therefore, the cup only starts moving when $P=f_s^{max}$, which means $$ma_{max}=\mu_sF_N=\mu_smg$$ $$\mu_sg=a_{max}$$ $$\mu_s=\frac{a_{max}}{g}=0.81$$
