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

[itex]m\ddot{x} = kx[/itex]

The Attempt at a Solution

So the general solution is [itex]x(t) = Ae^{\omega t} + Be^{-\omega t}[/itex], with [itex]\omega = \sqrt{\frac{k}{m}}[/itex] 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 [itex]x_0[/itex] as long as A is not 0.

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

 

[treynolds147]

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