To solve this problem, we need to use the formula for the acceleration of a simple harmonic oscillator.$\\$The acceleration $a$ of a simple harmonic oscillator is given by:$\\ a = -\omega^2 x \\$where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.$\\$Given:$\\ a = 20 \,$m/s$^2 \\ x = 5 \,$m$\\$Substitute the known values into the formula:$\\ 20 = \omega^2 \cdot 5 \\$Solve for $\omega^2: \\ \omega^2 = \frac{20}{5} \\ \omega^2 = 4 \\$Take the square root to find $\omega: \\ \omega = \sqrt{4} \\ \omega = 2 \,$rad/s$\\$The time period $T$ of a simple harmonic oscillator is given by:$\\ T = \frac{2\pi}{\omega} \\$Substitute the value of $\omega: \\ T = \frac{2\pi}{2} \\ T = \pi \,$s$\\$Therefore, the time period of oscillation is $\pi \,$s$. \\$This corresponds to Option 2.