[Solution by Claude - not from source] $\omega=\sqrt{k/m}$ increases as mass decreases. Leaking sand carries the box's instantaneous $x$ and $v$, so for the remaining mass $A^2=x^2+\frac{mv^2}{k}$ decreases as $m$ decreases. So $\omega$ rises and $A$ falls. Answer (2). (Judged from the graph shapes; verify against the official key.)