To solve this problem, we need to determine the spring constant and then use it to find the time period of oscillation.$\\$Given:$\\$• The spring is stretched by $5$ cm (or $0.05$ m) by a force of $10$ N.$\\$• A mass of $2$ kg is suspended from the spring.$\\$First, we find the spring constant $k$ using Hooke's Law:$\\ F = k \cdot x \\$Where:$\\ F = 10$ N (force applied)$\\ x = 0.05$ m (displacement)$\\$Rearrange the formula to solve for $k: \\ k = \frac{F}{x} \\$Substitute the known values:$\\ k = \frac{10}{0.05} \\ k = 200$ N/m$\\$Now, we use the formula for the time period $T$ of a mass-spring system:$\\ T = 2\pi \sqrt{\frac{m}{k}} \\$Where:$\\ m = 2$ kg (mass)$\\ k = 200$ N/m (spring constant)$\\$Substitute the known values into the formula:$\\ T = 2\pi \sqrt{\frac{2}{200}} \\ T = 2\pi \sqrt{0.01} \\ T = 2\pi \times 0.1 \\ T = 0.2\pi \\$Calculate the numerical value:$\\ T \approx 0.2 \times 3.14 \\ T \approx 0.628$ s$\\$Therefore, the time period of the oscillations is $0.628$ s. This corresponds to Option 4.