How Is the Limiting Displacement A0 Determined in a Spring System with Friction?

[azupol]

Homework Statement

A mass m is attached to a spring with spring constant k. There is a coefficient of static friction,
u_s
The coefficient of kinetic friction is u_k
Suppose you pull the mass to the right and release it from rest.
You find there is a limiting value of x = A₀ > 0 below which the
mass just sticks and does not move. For x > A₀ , it starts sliding
when you release it from rest. Find A₀ .

Homework Equations

$x''(t) + \omega x(t) - ( \mu mg)/k =0$

The Attempt at a Solution

I solved the ODE for Simple Harmonic Motion, and I get that $x(t)=B sin ( \omega t) + C ( \omega t) -umg/k$, but I'm not sure where to go from there. The derivative at x = A₀ must be zero, but how does that help me find A₀ itself?

 

[Dr.D]

The place to go is back to Newton's 2nd law. Your DE is wrong, and your thinking about static friction is in error. Please try again.

 

[azupol]

In light of your post, I figured this:
At x=A₀, the block is at rest, so the forces acting on it must be balanced. Thus, -kx=umg, and at A₀, -kA₀=umg, so solving for A₀ gives: A₀=-umg/k

Am I on the right track now?

 

[NATURE.M]

Wouldn't A₀ be positive since the question indicates A₀ > 0 ?