• Support PF! Buy your school textbooks, materials and every day products Here!

Dimensional Analysis: Period of a Pendulum to the length

  • #1

Homework Statement




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

Homework Equations



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.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,532
4,974
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.
 
  • #3
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.
 
  • #4
So.. cross out the Ls and leave T^2? It kinda confuses me with the lettering.
[T]^2= 2π⋅ [T]^2
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
[T]^2= 2π⋅ [T]^2
##\sqrt{[T]^2} = [T].##
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,532
4,974
[T]^2= 2π⋅ [T]^2
One more simplification to make.
You can write a dimensionless term as [1].
 

Related Threads for: Dimensional Analysis: Period of a Pendulum to the length

  • Last Post
Replies
6
Views
674
Replies
3
Views
15K
Replies
3
Views
4K
Replies
2
Views
8K
Replies
1
Views
6K
Replies
3
Views
2K
Top