Answer to Question #165824 in Physics for Cristian Constantinescu

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

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