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.