Calculating Time to Reach 1/2x from Equilibrium for a Mass & Spring

[MIA6]

1. A mass attached to the end of a spring is stretched a distance x from equilibrium and released. At what distance from equilibrium will it have acceleration equal to half its maximum acceleration?

My answer was 1/2A, but the correct answer was 1/2x, I don't know if there is a difference between these two. But I mean x may not be the amplitude, but I think the answer should be the half of its Amplitude.

2. A mass of 2.62 kg stretches a vertical spring 0.315 m. If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?

Here, what does it mean by new equilibrium position? 0.315m is the equilibrium point?

Thanks for help.

 

[member 553083]

The correct answer to your first question is 1/2x, where x is the distance the spring is stretched from its equilibrium position. The new equilibrium position in the second question would be 0.445m (the original equilibrium position plus 0.130m). To find the time it takes for the mass to reach the new equilibrium position, you would need to know the spring constant for the spring.