The force of static friction, $F_s$, provides the necessary horizontal acceleration, $a$, for the block. From Newton's second law, $F_s = ma$.

The maximum possible static friction force is $F_{s,max} = \mu_s N$. On a horizontal surface, the normal force $N$ equals the gravitational force $mg$, so $F_{s,max} = \mu_s mg$.

For the block not to slip, the required force cannot exceed the maximum available static friction:

$$ma \le \mu_s mg \implies a \le \mu_s g$$

The acceleration in SHM is $a(t) = -A\omega^2\cos(\omega t)$, with a maximum magnitude of $a_{max} = A\omega^2$. The angular frequency is $\omega = 2\pi f$.

The no-slip condition requires that the maximum acceleration does not exceed $\mu_s g$:

$$a_{max} \le \mu_s g$$ $$A(2\pi f)^2 \le \mu_s g$$

The maximum amplitude $A_{max}$ is found when the equality holds:

$$A_{max} = \frac{\mu_s g}{(2\pi f)^2}$$