# Dimensional Analysis: Period of a Pendulum to the length

## Homework Statement

Use unit analysis to show that the constant, 2π, is unitless.

T=2π√L/g[/B]

## The Attempt at a Solution

T= [T]
L= [L]
g= a= [L]/[T]^2= [L T^-2]

[T]= 2π√[L]/[L T^-2]
[/B]
Is this correct? I wasn't really sure how to do this. I'm using book examples to help me figure it out.
I already know that 2π is unitless by its own definition.

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haruspex
Homework Helper
Gold Member
Is this correct?
It would be correct, as far as you have gone, if you to use parentheses to show that the "/g" is inside the square root.
Next, simplify, treating the dimensionalities as normal algebraic variables.

It would be correct, as far as you have gone, if you to use parentheses to show that the "/g" is inside the square root.
Next, simplify, treating the dimensionalities as normal algebraic variables.
So.. cross out the Ls and leave T^2? It kinda confuses me with the lettering.

So.. cross out the Ls and leave T^2? It kinda confuses me with the lettering.
[T]^2= 2π⋅ [T]^2

Ray Vickson
Homework Helper
Dearly Missed
[T]^2= 2π⋅ [T]^2
$\sqrt{[T]^2} = [T].$

haruspex