Solution

When the ship is displaced by the distance $x$ deeper into the water, the extra buoyancy force $\rho_wgSx$ serves as the restoring force, where $\rho_w$ is the density of water. Consequently, $\omega^2=\rho_wgS/m$, and

$$m=\frac{\rho_wgS}{\omega^2}=\frac{\rho_wgST^2}{4\pi^2}.$$

Thus,

$$\Delta m\approx\frac{\rho_wgS\cdot2T\Delta T}{4\pi^2}.$$

Using $T=7$ s and $\Delta T=0.5$ s, we find that the mass of the additional cargo is approximately $900$ tons.