Question

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? 

Answer 1

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