Question

A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 8 m/s. Find the equation of motion. x(t)

Answer 1

Answer:

x(t) = -8sin2t

Explanation:

See the attachment for solution

From my solving, we can deduce that w² = 4, and thus, w = 2

Therefore, the general solution is

x(t) = c1 cos2t + c2 sin2t

Considering the final variable, we can conclude that

x(0) = 0

x'(0) = -8 m/s

The final solution, thus

x(t) = -8sin2t