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Simple Harmonic Motion (Pendulum)

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Two pendula of length 1.00m are set in motion at the same time. One pendula has a bob of mass 0.050kg and the other has a mass of 0.100kg.

    1. What is the ratio of the periods of oscillation?

    2. What is the period of oscillation if the initial angular displacement is small?

    3. What is the period of oscillation if the initial angular displacement is 60.0°? Calculate the series out to three terms.


    2. Relevant equations

    1. T=2π√(L/g)

    2. T=2π√(L/g)[1+1^2/2^2 〖sin〗^2 (θ/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (θ/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (θ/2)]

    3. The attempt at a solution

    1. 1:2 or 2:1 depending on which you consider mass 1 or mass 2.

    2. T=2π√(L/g)
    T=2π√(1.00m/〖9.808m/s〗^2 )
    T=2.01s

    3. T=2π√(L/g)[1+1^2/2^2 〖sin〗^2 (θ/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (θ/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (θ/2)]

    T=2π√(1.00m/〖9.808m/s〗^2 )[1+1^2/2^2 〖sin〗^2 (60/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (60/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (60/2)]

    T=2.006[1+0.5000+0.0088+0.00153]

    T=2.006[1.51033]

    T=3.03s
     
  2. jcsd
  3. Oct 24, 2013 #2
    Are my solutions correct?
     
  4. Oct 25, 2013 #3

    rude man

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    Is m included in the formula for T?

    How is this formula derived? Where did you get it?
     
  5. Oct 25, 2013 #4
    You should revise your concepts
     
  6. Oct 25, 2013 #5
    The formulas used were those provided by my lab manual.
     
  7. Oct 25, 2013 #6

    rude man

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    OK, let's start with part 1. Your formula is correct. So where do you get the idea that T is a function of mass?

    In part 2, I realize you were given that formula. I probably should not have commented at all, but what I had in mind is that you're being given a formula without its prior derivation, and no way could you have been shown its derivation. So, bottom line, never mind on part 2.
     
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