Finding Time for Spring to Reach Equilibrium Length

[HunterDX77M]

Homework Statement

A 20 kg mass is attached to a 120 N/m spring and is free to slide across a horizontal
frictionless surface. If the spring is originally at its equilibrium length when the mass is given an initial velocity to the right, after how much time is the spring once again at its equilibrium length?

Homework Equations

F_s = -kx
ω² = k/m

The Attempt at a Solution

Well I thought that I'd just get ω and solve the simple equation. But I'm not getting the right answer.

\omega = \sqrt{120/20} = \sqrt{6}
x(t) = Acos(\omega t) = 0
0/A = cos(\omega t) = 0
arccos(0) = \omega t
\pi \div 2 = \omega t
t = \frac{\pi}{2 \sqrt{6}} = 0.641 s

But the correct answer is twice that. Why?

 

[PeterO]

Homework Statement

A 20 kg mass is attached to a 120 N/m spring and is free to slide across a horizontal
frictionless surface. If the spring is originally at its equilibrium length when the mass is given an initial velocity to the right, after how much time is the spring once again at its equilibrium length?

Homework Equations

F_s = -kx
ω² = k/m

The Attempt at a Solution

Well I thought that I'd just get ω and solve the simple equation. But I'm not getting the right answer.

\omega = \sqrt{120/20} = \sqrt{6}
x(t) = Acos(\omega t) = 0
0/A = cos(\omega t) = 0
arccos(0) = \omega t
\pi \div 2 = \omega t
t = \frac{\pi}{2 \sqrt{6}} = 0.641 s

But the correct answer is twice that. Why?

The mass will be put into SHM, with a period able to be calculated with the usual formulas.
During each cycle of SHM, the mass will pass through the equilibrium position twice - so when does the first of them occur?

 

[HunterDX77M]

Ah, I think I get you.

I've already solved for ω. And the period T = 2π/ω. Therefore, T = 2.57 sec. And since it returns to equilibrium in half a period, it returns at 1.28 seconds. Thanks!