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I Simple Harmonic Motion/Period of a Physical Pendulum

  1. May 13, 2017 #1
    I'm studying the motion of a physical pendulum, could someone help me make the final step in figuring out how to find the period so I can make predictions before carrying out a practical? Basically I have a meter rule with holes drilled along the length and will be pivoting it at various points.

    I've done lots of reading and watched lots of videos, I know simple calculus, I understand moment of inertia and torque, small angle approximations and parallel axis theory. I've done dimensional analysis which resulted in:

    [tex]T=\sqrt{\frac{I}{mgb}}[/tex]

    where T = Period, I = Moment of inertia, m = Mass and b = distance from center of mass to pivot.

    I know dimensional analysis doesn't give any constants, and I've seen in books that the actual equation is

    [tex]T=2\pi\sqrt{\frac{I}{mgb}}[/tex]

    but I'd like to be able to prove this. I know it has to do with angular acceleration and angular frequency, but not sure how. Can anyone point me in the right direction? Most the videos I watch just make the jump and say "we've previously shown why... [tex]\omega^2=\sqrt{\frac{mgb}{I}}[/tex] but I can't find any previous videos that show! :D
    I think it's the [tex]\omega^2[/tex] that I can't see where it comes from.

    Any help much appreciated!

    K


    (p.s. If the Latex is messed up I'm probably working on fixing it, I've not used it much)
     
  2. jcsd
  3. May 13, 2017 #2
    You made a typo it should be ##\omega## not ##\omega^2##.
     
  4. May 13, 2017 #3
    The easiest derivations of this result requires you to be able to solve a second order differential equation.
    Another method using the conservation of energy requires you to solve a first - order differential equation, or equivalently solving a trigonometric integral like integral dx over square root of (a squared - x squared) with a constant. This is usually attacked about second semester calculus.
     
  5. May 13, 2017 #4
    I've seen it written [tex]\omega^2[/tex] in several places though, what makes you say that? Here's a screen clipping of a video from (credit to Dan Fullerton on Youtube)




    upload_2017-5-14_0-11-36.png


    I don't know where [tex]\theta=A cos(\omega t)[/tex] comes from either, but heard there are two uses for [tex]\omega[/tex] in rotional motion?
     
  6. May 13, 2017 #5
    Ah, I'm not at university. I know integration and differentiation, but don't know how to solve equations, I'll turn my attention there, thank you!
     
  7. May 13, 2017 #6
    In general torque = moment of inertia times the angular acceleration = I alpha = I d / dt angular velocity = I second derivative of theta with respect to time.
    This torque is also equal to Force ( - mg ) times moment arm (b sin theta). Now for small angles sin theta is approximated by theta, so the torque is -mgb theta.
    equating the torques gives a second order differential equation for theta. I times second derivative (d 2 theta / d t squared) = - m g b times theta

    Now most text "guess" the solution theta = A cos w t. (w is the frequency)

    Then they show the second derivative of this solution using calculus is : d 2 theta / d t squared = - A w squared time theta.
    we also have derivative (d 2 theta / d t squared) = - (m g b / I ) times theta
    This allows us to make the identification of w squared with mgb / I
     
  8. May 14, 2017 #7
    Thanks for all your input, I think I need to learn how to solve differential equations, or at least how to prove something is the solution of one. Probably the same thing :)

    Just to clear up:
    [tex]\tau=I\alpha[/tex]
    [tex]\tau=I\frac{d}{dt}\mbox{angular velocity (sometimes }\omega\mbox{)}[/tex]
    [tex]\tau=I\frac{d^2\theta}{dt^2}[/tex]
    Because of small angle approximations we can get rid of 'sin' in the calculation of the torque cased by mg:
    [tex]\tau=-mgb\theta[/tex]
    Equating the torques gives
    [tex]-mgb\theta=I\frac{d^2\theta}{dt^2}[/tex]
    rearranged to
    [tex]\frac{d^2\theta}{dt^2}+\frac{mgb}{I}\theta =0[/tex]

    So what I need to do is learn how to prove that solving the above gives
    [tex]\theta =A\cos \ \omega \ t\mbox{ (this time }\omega\mbox{ is frequency)}[/tex]
    and then learn how to show the second derivative of that is
    [tex]\frac{d^2\theta}{dt^2}=-\frac{mgb}{I}\theta[/tex]
    which I can do from reading textbooks/videos?

    (again, if the Latex is screwy I'm working on it :D )
     
  9. May 14, 2017 #8
    Is ##\omega^2=\sqrt{\dfrac{mgb}{I}}## correct ?

    Sorry I don't know much.
     
  10. May 15, 2017 #9
    it is ##\omega=\sqrt{\dfrac{mgb}{I}}##, that's because a general ##\alpha(t)=-\omega^2*x(t)##
     
  11. May 16, 2017 #10

    kith

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    If you are specifically interested in where the factor of [itex]2\pi[/itex] comes into play, have a look at a simple harmonic motion where the displacement [itex]s(t)[/itex] is described by a sine function.

    [itex]T[/itex] is the period of the motion, so we need to have [itex]s(T) = s(0)[/itex]. Since the period of the sine is [itex]2\pi[/itex], this number needs to be involved somehow. Now check that the following ansatz
    [tex]s (t) = A \sin\left(\frac{t}{T} 2\pi\right)[/tex]
    has the desired properties: as [itex]t[/itex] goes from [itex]0[/itex] to [itex]T[/itex], the argument for the sine goes from [itex]0[/itex] to [itex]2\pi[/itex] (sketching the function may help).

    From this, you see that [itex]\omega = \frac{2\pi}{T}[/itex].
     
  12. May 26, 2017 #11
    Thanks to everyone that commented, it's starting to make much more sense now I've been reading up and done some kitchen practicals! Think I'll have a nice grasp soon. Buffu, I see what you mean with my typo now in post #1, I'd mixed the two equations together, even the screenshot I posted in post #3 showed I was wrong!! Kith and mpresic you've been really helpful.

    We had a lot of fun doing the practicals and we're now trying to get a practical as close as possible to the prediction. Materials and fabrication have helped a lot :) , but could someone help me with error calculations? I watched a Walter Lewin video where he talks about a simple pendulum, and the uncertainty in the length being 1%, and once the root is taken it becomes 0.5%. My equations is much more complicated, I have simplified [tex] T=\sqrt{\frac{I}{mgb}}[/tex] to cancel the mass, and include the parallel axis theorem, giving [tex] T=\sqrt{\frac{L^2+12b^2}{12gb}}[/tex]

    Is there a simple explanation as to what to do with my percentage errors in measurement? I have the uncertainty of L to be 0.1% and b varies so will need to be calculated at each pivot point, but I'm sure will follow 1 rule for any position?
     
  13. May 26, 2017 #12

    kith

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    Science Advisor

    There's the general formula and many special cases. A superb text about this is "An introduction to error analysis" by Taylor. Or search for things like "gaussian error propagation" or "error analysis introductory lab".
     
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