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Simple Harmonic Motion

  1. Dec 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A very light, rigid rod with a length of 0.620m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation.
    (a)Determine the period of oscillation.
    (b)By what percentage does this differ from a one meter long simple pendulum?


    2. Relevant equations
    omega=sqrt((mgh)/i)
    omega=sqrt((g/l))
    omega=(2Pi)/T
    3. The attempt at a solution
    I got the correct period for part a, and it's 2.19s
    For the second part, I used omega=sqrt((g/l)), and omega is 3.13, and T is 2.007s.
    I used (2.19-2.007)/2.19 to find the percentage which is 0.0835. However, it's wrong...
    This seems like a simple question but I'm keep getting the second part wrong. Any suggestions???
     
  2. jcsd
  3. Dec 3, 2016 #2

    rock.freak667

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    I believe the denominator should be the period of the 1 m pendulum .
     
  4. Dec 3, 2016 #3
    The system said it's incorrect
     
  5. Dec 3, 2016 #4

    haruspex

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    Do you mean that you tried that as well, and the system said both answers are incorrect, or are you under the mistaken impression that rock.freak's suggestion is what you originally posted?
     
  6. Dec 3, 2016 #5
    I tried both answers and they are all incorrect
     
  7. Dec 3, 2016 #6
    Never mind, rock.freak667's method is correct. I forgot to multiply my answer by 100...
    Thanks for all the help
     
  8. Dec 4, 2016 #7

    rock.freak667

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    Good that you found your mistake. Those errors are simple to make and easily overlooked at times!
     
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