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

no simpler subs form complex ones. synthesise = make

i think the answer is a and b

i wish you luck