Time a spring takes to slow a mass

[LANS]

Homework Statement

A mass M is attached to a spring with spring constant K. At the equilibrium point of the spring, the mass has a velocity of V.
M = 8.07 kg
K = 113 N/m
V_o = 0.638 m/s

How far does the mass travel until it stops? How long (in seconds) does it take for the mass to travel from the equilibrium point until it stops?

Homework Equations

\frac{1}{2}MV_o^2 = \frac{1}{2}Kx^2 - equation 1

F(x) = kx - spring force

V(x) = \sqrt{V_o^2 - \frac{Kx^2}{m}} - from energy.

The Attempt at a Solution

Using equation 1, I can solve part 1 easily. I plug in M,K,V to equation 1and solve for x, which gives me x = 0.1705m

I have no idea how to solve part 2. I've tried using power, but that doesn't go anywhere meaningful.
P(x) = F(x)*V(x)

Integrating for total power gives me
\frac{MV_o^2}{2t} = \int F(x)*V(x)

Simplifying the integral:
\frac{MV_o^2}{2t} = \int \sqrt{K^2 x^2 V_o^2 - \frac{K^3 x^4}{m}}

I've tried solving that for t, and it doesn't give me the right answer. I haven't thought of it yet, but I feel like there should be an easier solution to this problem. Any help is appreciated.

 

[denverdoc]

It might be easier to cast this as a SHO formulation where T can be found directly from the other variables. in other words can you convert this to the form

x(t)=A sin (wt) and solve for T. hint: w=sqrt(k/m) and the time to stop is 0.5T.

 

[ideasrule]

Have you learned the formula for the period of a mass-spring system? The time it takes to go from equilibrium to rest is actually 1/4T (not 0.5T).

 

[denverdoc]

Have you learned the formula for the period of a mass-spring system? The time it takes to go from equilibrium to rest is actually 1/4T (not 0.5T).

my bad. I forgot it was a quarter cycle and not 1/2, thanks for the correction.