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
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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
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  • #6
haruspex
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[T]^2= 2π⋅ [T]^2
One more simplification to make.
You can write a dimensionless term as [1].
 

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