Answer to Question #280503 in Physics for Faith

Question #280503

A 0.38-kg glider is attached to a horizontal spring with stiffness of 130 N/m. If the equilibrium position is at x = 0, what is the acceleration of the glider at x = 0.12 m?

Expert's answer

"a=\\omega^2A=\\dfrac{kA}{m}=\\dfrac{130\\ \\dfrac{N}{m}\\times0.12\\ m}{0.38\\ kg}=41\\ \\dfrac{m}{s^2}."

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