To solve this problem, we need to use the properties of simple harmonic motion (SHM).$\\$In SHM, the velocity $v$ of a particle at a distance $x$ from the mean position is given by:$\\ v = \omega \sqrt{A^2 - x^2} \\$where $\omega$ is the angular frequency and $A$ is the amplitude of the motion.$\\$Given:$\\$• Velocities $V_1$ and $V_2$ at distances $x_1$ and $x_2$ from the mean position, respectively.$\\$We can write:$\\ V_1 = \omega \sqrt{A^2 - x_1^2} \\ V_2 = \omega \sqrt{A^2 - x_2^2} \\$Square both equations:$\\ V_1^2 = \omega^2 (A^2 - x_1^2) \\ V_2^2 = \omega^2 (A^2 - x_2^2) \\$Subtract the second equation from the first:$\\ V_1^2 - V_2^2 = \omega^2 (A^2 - x_1^2) - \omega^2 (A^2 - x_2^2) \\$Simplify:$\\ V_1^2 - V_2^2 = \omega^2 (x_2^2 - x_1^2) \\$Solve for $\omega^2: \\ \omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2} \\$The time period $T$ of SHM is given by:$\\ T = \frac{2\pi}{\omega} \\$Substitute $\omega^2$ into the expression for $T: \\ T = 2\pi \left( \frac{x_2^2 - x_1^2}{V_1^2 - V_2^2} \right)^{\frac{1}{2}} \\$Therefore, the correct option is:$\\$Option 1: $2\pi \left( \frac{x_2^2 - x_1^2}{V_1^2 - V_2^2} \right)^{\frac{1}{2}} \\$