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 12 m/s. Find the equation of motion.

Answer 1

-6sin(2t)

Step-by-step explanation:

fayemioluwatomisin is correct until the end.

x(0)=0

`x(t)=C_1cos(2t)+C_2sin(2t)\\0=C_1cos(2(0))+C_2sin(2(0))\\C_1=0`

x'(0)=-12

`x'(t)=2C_2cos(2t)\\-12=2C_2cos(2(0))\\-12=2C_2\\C_2=-6`

`x(t)=-6sin(2t)`

Answer 2

Step-by-step explanation:

I will try to use newton’s second law and its’ concept of spring. The detailed solution is shown it the documents and I also used some mathematical concept which is highlighted.