Calculating Period of a Pendulum: 1.18 s

[Vandetah]

Homework Statement

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

Homework Equations

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

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 don't have square root?

 

[Simon Bridge]

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 :)