AIIMS 2019 Physics Simple Harmonic Motion MCQ Question

In simple harmonic motion (SHM), the acceleration $a$ is related to the displacement $x$ by the formula $a = -\omega^2 x$, where $\omega$ is the angular frequency. Given $a = 1.0 \, \text{m/s}^2$ and $x = 0.01 \, \text{m}$, we can rearrange the equation to find $\omega$: $\omega^2 = \frac{a}{-x} = \frac{1.0}{0.01} = 100$, leading to $\omega = 10 \, \text{rad/s}$. The period $T$ is then calculated using $T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \, \text{s}$. Other options do not satisfy the relationship between acceleration, displacement, and period in SHM, making them incorrect.