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Simple Pendulum problem

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A pendulum with a swing amplitude of 150 cm is recorded to have a period of 2.1s. Calculate after 0.3s release from the furthest amplitude position.

    a) Angular velocity
    b) displacement
    c) Angular acceleration
    d) Length of the pendulum


    I've only had one lecture on circular motion and the motion of pendulums, so I'm a bit unsure of the correct steps for each calculation.


    2. Relevant equations
    [tex]\omega = \frac{2\pi}{T}[/tex]

    [tex]x = A cos (\omega t)[/tex]

    [tex]a = -\omega^{2}x[/tex]

    [tex]T = 2\pi\sqrt{\frac{l}{g}}[/tex]

    3. The attempt at a solution

    a)
    [tex]\omega = \frac{2\pi}{2.1} = 2.99 rad s^{-1}[/tex]

    b)
    [tex]x = 1.5 cos (2.99 x 0.3) = 1.5 m[/tex]

    c)
    [tex]a = -(2.99)^{2} x 1.5 = -13.41[/tex]

    d)
    [tex]l = \frac{T^{2}g}{2\pi^{2}} = 106.74 m[/tex]


    I'm not convinced that I'm getting b) right, as it seems odd for the displacement after 0.3 seconds to be the same as the amplitude. This makes me believe that I'm wrong in other places.

    See whats wrong? [I dont have the answers to check for myself]

    Any similar problems to practice on would be welcomed; I don't want to constantly feel that I'm missing something.
     
  2. jcsd
  3. Nov 18, 2009 #2
    one problem I see is that in part d) you should be dividing by 4, not 2
     
  4. Nov 18, 2009 #3
    also, I do not get the same answer as you do for part b). Make sure your calculator is in radiant, not degrees.

    (I am assuming the extra x in the cosine function was a typo)
     
  5. Nov 18, 2009 #4
    Oh yes, I think I made a typo when I typed the post; it's 4 PI in my notes. Regardless of that I made a calculation error on that part...

    I now have L = 1.1m for d).

    Thanks.
     
  6. Nov 18, 2009 #5
    also, check to see if your calculator is in radians. For part B)
     
  7. Nov 18, 2009 #6
    Ah... thank you.

    Neglected to think that I was working with radians.

    b) 0.9 m
    c) -8.05

    I assume that I've got the right set of answers for the question now.
     
  8. Nov 18, 2009 #7
    yes, your physics appears flawless.
     
  9. Nov 18, 2009 #8
    I think d) is 1.1m.

    I was getting 106.74 by forgetting to bracket the 4pi^2; it was dividing by 4 and then multiplying by pi^2. Don't know why, I thought that calculators multiplied before dividing.
     
  10. Nov 18, 2009 #9
    details my friend, details :smile:

    The physics is the important part.
     
  11. Nov 18, 2009 #10
    Thank you for your help; now I can move on and get stuck on something else. Midterm tests next week... :(
     
  12. Nov 18, 2009 #11
    Good luck my friend! may the [tex]ma[/tex] be with you :cool:
     
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