Question

A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released from the equilibrium position with an upward velocity of 6 meters per second. Find an equation of the motion.

Answer 1

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t