Dynamics: Spring problem (Oscillation)

[Dean-o]

Homework Statement

The mass m = 4 kg and the spring constant k = 64 N/m. The spring is unstretched when x = 0.
At t = 0, x = 0 and the mass has a velocity of 2 m/s down the inclined surface. What is the value
of x at t = 0.8 s?
The angle of incline is 20 degrees, and with the mass moving down to the right with a spring attached on its left surface.

Homework Equations

ΣF_x = ma_x
x = Asin(wt)+Bcos(wt)

The Attempt at a Solution

d²x/dt² + (16 s^-2)x = 3.355 m/s²

I got to this point and I found that for the equation x = Asin(wt)+Bcos(wt) the solution gives x = Asin(wt)+Bcos(wt) + 0.210 m. I get that 0.210 = 3.355/16, but why is it included in this solution for the equation when all other examples of this problem for a vertically or horizontally held spring mass system use just the x = Asin(wt)+Bcos(wt) equation without this addition value? What is this value, and is it just disregarded as zero in the other forms of this problem?

 

[haruspex]

What is this value

It is the equilibrium position, i.e. the spring extension required to balance the downslope gravity.

 

[Dean-o]

It is the equilibrium position, i.e. the spring extension required to balance the downslope gravity.

Oh okay, that makes sense, thanks!