- #1

KevinFan

- 47

- 0

## Homework Statement

A very light, rigid rod with a length of 0.620m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation.

(a)Determine the period of oscillation.

(b)By what percentage does this differ from a one meter long simple pendulum?

## Homework Equations

omega=sqrt((mgh)/i)

omega=sqrt((g/l))

omega=(2Pi)/T

## The Attempt at a Solution

I got the correct period for part a, and it's 2.19s

For the second part, I used omega=sqrt((g/l)), and omega is 3.13, and T is 2.007s.

I used (2.19-2.007)/2.19 to find the percentage which is 0.0835. However, it's wrong...

This seems like a simple question but I'm keep getting the second part wrong. Any suggestions?