# Simple harmonic oscillation: uniform rod

• vetgirl1990

## Homework Statement

A uniform rod of mass m and length L is freely pivoted at one end. What is the period of its oscillations? Icm for a uniform rod rotating about its centre of mass is 1/12mL2

(a) √3g/2L
(b) 2π √3L/2g
(c) 2π √2L/3g
(d) 2π √L/g
(e) none of the above

ω2 = mgL/I

Icm = 1/12mL2
I = Icm +1/12mL2

Period: T = 2π/ω

## The Attempt at a Solution

I'm fairly certain that the answer is "none of the above", but I'd just like to make sure I'm not neglecting anything in what seems to be a simple plug-and-chug question.

I = Icm +1/12mL2 = 1/12mL2 + mL2 = 13/12 mL2

ω = √mgL/I= √(mgL)/(13/12)mL2 = √12g/13L

T = 2π/ω = 2π √13L/12g

Last edited:

What is the distance between the Centre of mass of rod and the end? Answer this and you'll realize what you did wrong.

vetgirl1990
What is the distance between the Centre of mass of rod and the end? Answer this and you'll realize what you did wrong.
I just realized what I did wrong. The distance is L/2, so the moment of inertia equation turns into: I = 1/3mL2

So then ω2 = mg(L/2) / (1/3)mL2

Therefore T = 2π √2L/3g

Thanks!

Thanks!
Welcome :)