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Investigating pendulums for physics

  1. Jul 18, 2009 #1
    Hi I'm investigating pendulums for physics and I was wondering if someone could explain how this equation works and why: T=2π√(L/g) Where T is period, π is pi, L is length of string and g is gravitational force:9.8 m/s. I need to understand the physics for this formula because i was hoping to create an E.E.I. where I investigate the effects of changing length of the string to the period. Thank you for any possible help =], Oh I'm also in year 11 so I'm not very advanced in physics.
     
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  3. Jul 18, 2009 #2
  4. Aug 22, 2009 #3
    Re: Pendulums

    ok, could someone explain the physics concept for why length effects the period of a pendulum? i've got this equation but i know it will not be enough to support my hypothesis:T=2π√(L/g) Where T is period, π is pi, L is length of string and g is gravitational force:9.8 m/s. My hypothesis is that increasing the length of a pendulum will result in the period to increase.
     
  5. Aug 22, 2009 #4
  6. Aug 22, 2009 #5
    Re: Pendulums

    lol what happened to the period of simple harmonic motion T=2π√(m/k) Where T is period, π is pi, m is mass and k is spring constant. Cause in my textbook that is where the equation for period of a pendulum derived from. Because in a pendulum k=mg/L. Thus subbing in mg/L T=2π√(mL/mg) . And those two m's cancel each other out. thus giving the equation for period of a pendulum. I have a second question. In an experiment investigation, in the discussion and conclusion is using this equation: T=2π√(L/g) acceptable when justifying why longer lengths equal greater periods? Because i don't see the connection between length and period. I've look at conservation of energy and i know that increasing the length will make the system have more energy overall.
     
    Last edited: Aug 22, 2009
  7. Aug 22, 2009 #6

    ideasrule

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    Re: Pendulums

    T=2π√(m/k) is also derived by applying Newton's second law to an oscillatory system and solving the resulting differential equation. Same as the pendulum formula. There's no intuitive reason that k would equal mg/L, so that's not a way to get the pendulum formula.
     
  8. Aug 22, 2009 #7
    Re: Pendulums

    That's right, and if you go to your parents (if they're not physicists :P) and tell them which pendulum is faster, the one with 5 kg or 10 kg, they'll intuitively say: "Of course the one with 10 kg" which is heavier. This is just like predicting that free fall is mass-dependent, the mystery that Newton and Galileo have solved ;)
     
  9. Aug 22, 2009 #8

    ideasrule

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    Re: Pendulums

    Unfortunately for physics, the 10 kg mass will be faster because its inertia means it's less affected by air resistance and friction. Aristotle thought freefall was mass-dependent for the same reason; control of variables was not part of his reasoning, even though it ought to have been obvious that air exerts resistance and that heavier objects have more inertia than lighter ones.
     
  10. Aug 22, 2009 #9
    Re: Pendulums

    @ ideasrule what do you mean by "There's no intuitive reason that k would equal mg/L, so that's not a way to get the pendulum formula." In my text book it uses hooks law F=-kx (F=restoring force, k=spring constant and x=displacement) . It states in the case of a pendulum the restoring force is the component of the weight tangent to the arc. So F=-mg sin[tex]\Theta[/tex] . It then goes on to explain that if [tex]\Theta[/tex] is small then sin [tex]\Theta[/tex] is approximately equal to [tex]\Theta[/tex]. So F=-mg sin[tex]\Theta[/tex][tex]\approx[/tex]-mg[tex]\Theta[/tex]. And then the book goes on to say that displacement x=L[tex]\Theta[/tex]: rearrange this x/L=[tex]\Theta[/tex], so that you get F[tex]\approx[/tex]-mgx/L. Then it says that F=-kx[tex]\approx[/tex]-mgx/L. If you cancel out the -x you get k[tex]\approx[/tex]mg/L.

    sorry if that's too much equation talk
     
  11. Aug 22, 2009 #10
    Re: Pendulums

    tbh, i am doing an experimental investigation on why increasing the length will increase the period of a pendulum. I'm just having difficulty finding a physics concept that will explain why this occurs, and i was wondering if using that equation was sufficient in justifying this relationship. I also have graphs and other results to show that this is true. Am i over-complicating this?
     
  12. Aug 22, 2009 #11
    Re: Pendulums

    I'm talking about simple pendulum, period is 2pi sqrt(L/g) under small angle approximation
     
  13. Aug 22, 2009 #12
    Re: Pendulums

    yes, me too. okay let me rephrase my question T_T, thx for helping. What is the explaination for why increasing the length will result in an increase in period.
     
  14. Aug 22, 2009 #13
    Re: Pendulums

    OK, let me try to answer in some intuitive sense. You can say because the moment of inertia will increase for the system (I= m l^2), and therefore the gravity will need more time to exert more force on the system. Is that good enough? hope so :)
     
  15. Aug 22, 2009 #14

    kuruman

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    Re: Pendulums

    Here is a simple explanation without equations. Consider two simple pendula, one longer than the other. Start them swinging at the same initial angle. Which one will return to the initial angle first? Clearly the short pendulum. Why? Note that at any given angle the tangential acceleration of each pendulum will be the same, g sinθ. Now the shorter pendulum has a shorter arc to cover while its tangential acceleration as a function of angle matches that of the longer pendulum. So the shorter pendulum takes less time for a full swing, therefore its period is shorter.
     
  16. Aug 22, 2009 #15
    Re: Pendulums

    okies thx for the help, i got a fair idea now. So the equation for arc length is: x=L[tex]\Theta[/tex], sorry this is probably dumb but how come x is not equal to: [tex]\Theta[/tex]/360*2[tex]\prod[/tex]L. Just to clarify, tangential acceleration is the restoring force? And one final question for tangential acceleration how did you get the equation g sin[tex]\Theta[/tex], my textbook says the restoring force is mgsin[tex]\Theta[/tex].
     
  17. Aug 22, 2009 #16
    Re: Pendulums

    We are using small angle approximation.

    g sin(t) is tangential acceleration mg sin(t) is tangential force
     
  18. Aug 22, 2009 #17
    Re: Pendulums

    ohhh okay sorry , lol F=ma so the a is g sin(t) >.<. Thank you for your help;D, it really helped to make sense of what i was doing.And so is this explanation legitimate for proving my hypothesis
     
    Last edited: Aug 22, 2009
  19. Aug 22, 2009 #18
    Re: Pendulums

    Welcome :)

    And remember, there is no dumb question, but there is dumb answer ;)
     
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