Edits to “Statement”, “Solution”, “Answer”
en/3.2.37.md
+3 −3
| @@ -1,10 +1,10 @@ | |||
| ### Statement | |||
| − | $3.2.37.$ [Insert the problem statement] | ||
| + | $3.2.37.$ After the ship is loaded, its vertical oscillation period will increase from $7$ to $7.5$ s. What is the mass of the cargo? Waterline cross-section is $S=500$ m$^2$. Consider the nature of water involvement in motion unchanged during loading. | ||
| ### 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. Consequently, $\omega^2=\rho_wgS/m$, and | ||
| + | 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. | |||
| @@ -16,4 +16,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | Approximately $900$ tons. | ||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $3.2.37.$ [Insert the problem statement] | $3.2.37.$ After the ship is loaded, its vertical oscillation period will increase from $7$ to $7.5$ s. What is the mass of the cargo? Waterline cross-section is $S=500$ m$^2$. Consider the nature of water involvement in motion unchanged during loading. | ||
| ### Solution | ### 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. Consequently, $\omega^2=\rho_wgS/m$, and | 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 | ||
| \[m=\frac{\rho_wgS}{\omega^2}=\frac{\rho_wgST^2}{4\pi^2}.\] | \[m=\frac{\rho_wgS}{\omega^2}=\frac{\rho_wgST^2}{4\pi^2}.\] | ||
| Thus, | Thus, | ||
| \[\Delta m\approx\frac{\rho_wgS\cdot2T\Delta T}{4\pi^2}.\] | \[\Delta m\approx\frac{\rho_wgS\cdot2T\Delta T}{4\pi^2}.\] | ||
| Using $T=7$ s and $\Delta T=0.5$ s, we find that the mass of the additional cargo is approximately $900$ tons. | Using $T=7$ s and $\Delta T=0.5$ s, we find that the mass of the additional cargo is approximately $900$ tons. | ||
| @@ -16,4 +16,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | Approximately $900$ tons. | ||