When is the Particle at Equilibrium Position?

[Lengalicious]

Homework Statement

Position of particle is given by:
x(t) = 5 cos (3t +2)

At what time after t = 0 is the particle at the equilibrium position?

Homework Equations

The Attempt at a Solution

I understand the equilibrium position to be the point where x(t) = 0 since x(t) is thedisplacement from the equilibrium position is it not? But when i re-arrange for t, i get t = -0.14s, negative? Just doesn't seem right what am I doing wrong.

 

[Pengwuino]

Remember, the cosine function has an infinite number of 0s and your oscillator returns to the equilibrium position an infinite number of times (assuming it never slows down). Your function is at equilibrium whenever [itex]3t + 2 = {{\pi}\over{2}} + \pi n[/itex] where [itex] n = 0, \pm 1, \pm 2, \pm 3...[/itex]. Each of these values of [itex]n[/itex] gives you a different time where the particle will be at equilibrium. The meaning of a negative answer is simply that .14s in the past, it was at equilibrium.

When you get "realistic" about the problem in real life, however, you simply say that particle begins motion at t = 0 and you don't care about negative times. In fact, if you recall basic kinematics problems with cannons and all that good stuff, you would have had the same problem. Say you dropped a ball off a cliff and wanted to know at what time it would have reached a certain height. The math would tell you that there are 2 times where the ball would have reached that height, one being negative. You start the experiment at t = 0, though, so negative answers are unphysical or uninteresting to you at the least.

 

[Lengalicious]

Omg so obvious when you see it layed out infront of you, thanks a lot for the help.