- #1

Dean-o

- 2

- 0

## 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

d

^{2}x/dt

^{2}+ (16 s

^{-2})x = 3.355 m/s

^{2}

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?