Finding the Special Initial Velocity in a kx Force System

[treynolds147]

Homework Statement

A particle of mass m is subject to a force F(x) = kx, with k > 0. What is the most general form of x(t)? If the particle starts out at x₀, what is the one special value of the initial velocity for which the particle doesn't eventually get far away from the origin?

Homework Equations

m\ddot{x} = kx

The Attempt at a Solution

So the general solution is x(t) = Ae^{\omega t} + Be^{-\omega t}, with \omega = \sqrt{\frac{k}{m}} of course. My issue is with the second part of the question; as far as I understand it, it wants me to figure out for what value of v₀ makes it so that the first term of the general solution doesn't dominate (since that term would see x increase to infinity, whereas the second term would see x approaching the x-axis). If that's what the question is asking, I'm not quite sure how I'd go about that! Any hints and help would be much appreciated.

 

[HallsofIvy]

Yes, x will go indefinitely away from x_0 as long as A is not 0.

You have the initial values x(0)= x_0 and x'(0)= v_0. What value of v_0 gives A= 0?

 

[treynolds147]

Okay, so if A is going to be 0, I get
v_{0} = -B\omega.
Owing to the fact that x(0) = x_{0} = A + B, and seeing as how A = 0, this gives me a value of
v_{0} = -x_{0}\omega.
Does that sound right?