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Simple Harmonic Motion

  1. Dec 6, 2009 #1
    A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

    what is its new period
    I= 1/12 mL^2

    equation: T= 2 pi * sqrt (I/k)
    where T= 5
    L= 1

    I solved for K in terms of m (mass)...where i got k=7.599 m

    then i replugged it in to the same equation but where
    T= unknown (trying to find)
    L=.76

    the mass cancels out and my answer gives me 3.8... which is wrong when i checked.

    The only thing i can see is that im using the wrong "I" for using the same equation 2nd time when i replugged it in. Should i be using the parallel axis theorem where

    I= 1/12 mL^2 + Md^2
    (since its not revolving around its center of mass)
    L=1
    d= .76

    when i tried that i got my answer to be 6.28...wrong again
     

    Attached Files:

  2. jcsd
  3. Dec 6, 2009 #2
    The eqn for I for a rectangular plate is m(l^2+h^2)/12 where l and h are length and height respectively. It is revolving about the center of mass in both cases so no need for parallel axis theorum. You may need to solve for height, if the factor 0.76^3 isn't giving the right answer (remember the mass is being reduced by 24% as well).

    BTW, neglecting the height, I got an answer of T=3.31s
     
    Last edited: Dec 6, 2009
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