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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:


    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


    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!


    (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)


    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\frac{d}{dt}\mbox{angular velocity (sometimes }\omega\mbox{)}[/tex]
    Because of small angle approximations we can get rid of 'sin' in the calculation of the torque cased by mg:
    Equating the torques gives
    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
    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


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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


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    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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