# Time required for a mass to move below its equilibrium position to a point above it

1. Jun 28, 2012

### Vandetah

1. The problem statement, all variables and given/known data

https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-prn1/540902_3288133970847_49118976_n.jpg [Broken]

2. Relevant equations
t = 2$\pi$ $\sqrt{}$$\frac{l}{g}$

3. The attempt at a solution
for t
t = 2$\pi$ $\sqrt{}$$\frac{l1}{g}$

t = 2$\pi$ $\sqrt{}$$\frac{0.15m}{9.8 m/s^{2}}$

t = 0.78 s

t = 2$\pi$ $\sqrt{}$$\frac{l2}{g}$

t = 2$\pi$ $\sqrt{}$$\frac{0.04m}{9.8 m/s^{2}}$

t = 0.4 s

total time t = 0.78 s + 0.4 s
total time t = 1.18 s

i changed the signs to positive because negatives dont have square root?

Last edited by a moderator: May 6, 2017
2. Jun 28, 2012

### Simon Bridge

Re: Time required for a mass to move below its equilibrium position to a point above

You've misapplied the equation:
$$T=2\pi \sqrt{ \frac{l}{g} }$$... is the equation to find the period T of a simple pendulum. The variable l is the distance from the pivot to the center of mass. So your working just figures out the natural period of pendulums with different length strings.
(I wrote the equation out like that so you could see how to write equations in PF :) - hit the quote button to see how I did that.)

It sounds like you need the equations of motion for a mass on a spring - something that looks like:$$mg-kx=m\ddot{x}$$... the solution will be like:$$x(t)=A \cos(\omega t) + B$$Since you are not provided with details for the system - you'll be looking at a ratio method or an approximation for very small distances. eg. is the speed close to the equilibrium point very nearly constant? Of course there may have been data that you have not given us :)

Last edited: Jun 29, 2012