A spring stretches 0.1 m when a 0.5 kg mass is hung from it. The spring is now oriented horizontally with the mass attached to one end. If it is stretched 0.2 m from the equilibrium length, determine the maximum velocity of the mass.

1.98 rad/s

Explanation:

Given that a spring stretches at a distance  0.1 m when a mass of 0.5 kg is attached to it.

The angular velocity

If it is stretched 0.2 m from the equilibrium length; we have:

The maximum velocity v = 0.2 × 9.9

v = 1.98 rad/s

Therefore; the maximum velocity of the mass =  1.98 rad/s

the area between the graph and the t axis

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..