Question

Calculate the acceleration of the bob of the simple pendulum of time period 2s when the displacement from the equilibrium position is 42 mm 

Answer 1

$$a=A\omega^2=A(\dfrac{2\pi}{T})^2,$$ $$a=42\cdot10^{-3}\ m\cdot(\dfrac{2\pi}{2\ s})^2=0.414\ \dfrac{m}{s^2}.$$