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Applying Dimensional Analysis to correct the equation

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Check if the equation is dimensionally correct
    The frequency f of vibration of a pendulum of length L is given by the equation:

    f = L/2π √(g/L)

    I'm assuming 'g' is the gravitation field strength (?), which is [L]1[T]-2

    3. The attempt at a solution

    I did not know what the physical quantity was for f, so i looked up and it was [T]-1
    so i did:

    [T]-1 = [L] x √([L]1[T]-2/[L])
    simply
    [T]-1 = [L] x √([T]-2)


    From here i'm stuck on what to do with that square-root.
     
    Last edited: Jan 30, 2016
  2. jcsd
  3. Jan 30, 2016 #2

    Samy_A

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    I assume the equation to check is ##f=\frac{L}{2\pi}\sqrt{\frac{g}{L}}##.

    EDIT: please write the complete question in part 1 of the template next time. For the rest, see haruspex post.
     
    Last edited: Jan 30, 2016
  4. Jan 30, 2016 #3

    haruspex

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    You have not stated what you are tasked with. Is the idea to check the given equation is dimensionally correct? Was the equation provided as part of the problem statement? You might want to check you have copied it out correctly.
    With regard to how to proceed from your last equation above, be careful about the scope of the square root sign. The way you have written it it only appears to apply to the leading [L].
    There is some cancellation you can do inside the square root.
     
  5. Jan 30, 2016 #4
    Thanks, Ive edited the question in. It was to dimensionally see if the equation was correct
    I've followed your advice and ended up at
    [T]-1 = [L] x √([T]-2)
    Could you please help me from here?

    Thanks!
     
  6. Jan 30, 2016 #5

    haruspex

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    You might find it obvious if you write T-2 as 1/T2
     
  7. Jan 30, 2016 #6
    Thanks for the reply!

    So i've gone ahead where you left me from,
    [T]-1 = [L] x √([T]-2)
    [T]-1 = [L] x √(1/[T]2)
    [T]-1 = [L] x 1/√T2
    Rationalized it
    [T]-1 = [L] x [T]2

    Does that mean the equation at the start is not dimensionally correct?
     
  8. Jan 30, 2016 #7

    haruspex

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    you were doing well until that last step. What happened to the square root?
     
  9. Jan 30, 2016 #8
    Oh oops!!

    is it
    [T]-1 = [L] x √(1/[T]2)
    [T]-1 = [L]/√T2

    Therefore it not dimensionally correct?
     
    Last edited: Jan 30, 2016
  10. Jan 30, 2016 #9

    haruspex

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    Correct, but you should simply the √T2.
     
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