Answer
$0.259$ m
Work Step by Step
We know that:
$v_{max}=A \omega$
First, we need to find $\omega$:
$\omega = 2 \pi f = 2\pi(3.2 Hz)=20.11 s^{-1}$
Since the block's speed is measured at equilibrium, it is at maximum speed. We find $A$ as:
$A=\frac{v_{max}}{\omega}=\frac{5.2}{20.11}=0.259m$
