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How did scientists compose the equation for period?

  1. Sep 15, 2009 #1
    T=2pi√(l/g)

    Is it in radian*seconds?
     
  2. jcsd
  3. Sep 15, 2009 #2
    Hi Kashiark-

    What are the units of sqrt(L/g)? of sqrt(g/L)?

    Bob S
     
  4. Sep 16, 2009 #3
    √(l/g) would be √m/(√m/s) or √m*(s/√m) = s, and I was just guessing that the 2pi was radians; isn't it?
     
  5. Sep 16, 2009 #4
    Err...no. Sqrt(L/g) is in units of seconds per radian, snd sqrt(g/L) is in units of radians per second. 2 pi has units of radians per revolution.
     
  6. Sep 16, 2009 #5
    Why is √(l/g) in units of seconds per radian? Length is in meters, and g is in m/s². Where did the radians come from?
     
  7. Sep 16, 2009 #6
    Pi have no physical units. The radian is defined as a ratio (length of the arc/radius).
    It's just a matter of convention to say that angular speed is in radians/s. The actual physical unit for angular speed is just 1/s.
    Look at v=omega*r
    v is not in m*rad/s, is it?
     
  8. Sep 16, 2009 #7
    Ok, that makes sense, but why is there a 2pi in the equation? What was the argument made when someone said, "Hey, I think period equals 2 pi times the square root of length over the acceleration due to gravity!"
     
  9. Sep 16, 2009 #8

    rock.freak667

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    you should read about the http://en.wikipedia.org/wiki/Pendulum_(mathematics)" [Broken] then.
     
    Last edited by a moderator: May 4, 2017
  10. Sep 16, 2009 #9
    Wow... I'm kind of overwhelmed. I can follow it from s=lΘ to d²Θ/dt²+(g/l)sinΘ=0 ..... From there I'm too ignorant about differential equations to follow it, but why does F=-mgsinΘ ?
     
  11. Sep 16, 2009 #10

    ideasrule

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    There's a vector diagram to the right of that equation which explains it pretty clearly. Gravity exerts -mg on the pendulum; the component parallel to the pendulum's motion is -mgsinΘ.
     
  12. Sep 16, 2009 #11

    rock.freak667

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    The force mg acts at an angle θ, so the components are one tangentially mgsinθ and the other mgcosθ. mgsinθ points in the opposite direction of motion so it is -mgsinθ

    that is how F= -mgsinθ and we know F=ma

    so ma = -mgsinθ => a= -gsinθ for small θ, sinθ≈θ so a=-gθ. θ=x/l (arc length/radius)

    so a = -(g/l)x

    which is of the form a= -ω2x, so it exhibits SHM, for ω2=g/l. And for SHM T=2π/ ω which works out as:

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


    this is the simplest way to show it IMO
     
  13. Sep 16, 2009 #12
    What does tangentially mean? I'm familiar with the function itself, but I'm not sure what you mean by it. After there, I can follow you until a=-ω²x; where did the ω come from? I have another problem after there, but we'll get to that later I suppose if you guys still want to help me :)
     
  14. Sep 16, 2009 #13
    Forget the first part, I just got that from thinking about it while brushing my teeth, but I still don't understand why it should be a negative mgsinθ
    Edit: Ok, I figured that out as well, but I'm still stuck with the ω. It wasn't in any equation one second, and then it was. What does it represent anyway?
     
  15. Sep 16, 2009 #14

    rock.freak667

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    angular frequency
     
  16. Sep 18, 2009 #15
    What's the x stand for?
     
  17. Sep 18, 2009 #16

    rock.freak667

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    at the angle θ, the string marks out an arc length x
     
  18. Sep 18, 2009 #17
    I get it! Thanks!
     
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