Diff Eq spring problem?

Question

A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 inch, and then set in motion with a downward velocity of 2 ft/sec, find the position of the mass at any time t. Assume that there is no damping and the only force acting upwards is the spring's force (F = -k(Delta(x))). Delta(x) = change of position from equilibrium. Determine the frequency (W), period (T), amplitude (A) and phase of motion (P). The answer by the way is x = (sqrt(2) * .25)*sin(8*sqrt(2)*t) - (1/12*cos(8*sqrt(2)*t)) ft. Where W = 8*sqrt(2), T = sqrt(2)*(pi/4), A = sqrt(11/288) and P = pi - arctan(3/sqrt(2)). I just don't know how to get this. Thanks.

Answer 1

The equation of motion is:

m d^2x/dt^2 +kx = mg = W where W is a constant and the weight of the mass.

Solve homogeneous equation:

md^2x/dt^2 +kx = 0 --> d^x/dt^2 +w^2x = 0 w = k/m

X = Asin(w*t)+Bcos(wt)

So particular solution is x + W/w^2 = Asin(wt) + Bcos(wt)+mW/w^2

Now applied initial conditions. These are given in the problem as:

x(0) =1/12 ft dx/dt|t=0 = -2 ft/s

x(0) =1/12 = B+W/w^2

dx/dt|t=0 = wA ---> A= 0

Then B = 1/12 -W/w^2

Now w^2 = k/m and k can be found from equillibrium condition

k*3/12 ft = W --> k =4W

So B=1/12 -W/(4W/mg) =1/12 - W/4 = 1/12 - 3/4 =-2/3

x = -2/3cos(wt) + 3/4


w = sqrt(k/m) =sqrt(4W/W/g)=2sqrt(g) = 4sqrt(2)

I'd check the provided solution. there should only be a cosine term since the non-zero initial velocity wipes out the sine term in x