Simple Pendulum question from MHT CET 2026 (14 May, Shift 1)

Answer: (b)

1. Path length $= 2A = 8\text{ cm} \Rightarrow A = 4\text{ cm}$.
2. $\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{\pi^2}{1}} = \pi\text{ rad/s}$
3. $v_{max} = A\omega = 4\pi\text{ cm/s}$
4. So the answer is (b) $4\pi$ cm/s.