Answer to Question #141075 in Classical Mechanics for alwande hadebe

Question

A block of mass 10 kg is suspended from a fixed support by a spring of spring constant 2068 Nm−1. The block is subject to the vertical driving force 500cos((5rad/s)t) N. Let x be the downward displacement of the block (in metres), measured from the equilibrium position. What is the distance of the block from its equilibrium position at time t= 1.7 s? [Hint: Without damping you only need to consider the particular solution, xp(t).]

Answer 1

$$F=ma=-kx\\(2068)x=500\cos{5(1.7)}\\ x=0.24\ m$$