(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

ΣF_{x}= ma_{x}

x = Asin(wt)+Bcos(wt)

3. 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?

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# Homework Help: Dynamics: Spring problem (Oscillation)

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