Answer to Question #165824 in Physics for Cristian Constantinescu

Question #165824

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

Expert's answer

"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}."

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