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Simple harmonic motion

  1. Mar 16, 2010 #1
    A uniform rod of length , pivoted at one end, is set into small oscillation in a vertical plane. Show that the period of oscillation is [tex]T=2\pi\sqrt{\frac{3L}{2g}}[/tex]
    Shouldn't the period be [tex]
    T = 2\pi\sqrt{\frac{L}{g}}
    [/tex]regardless of the mass and dimension of the object?
     
    Last edited: Mar 16, 2010
  2. jcsd
  3. Mar 16, 2010 #2

    kuruman

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    Homework Helper
    Gold Member

    Look up "physical pendulum".
     
  4. Mar 16, 2010 #3

    Jmf

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    Some quick working out gives me the time period as:

    [tex]T = 2\pi\sqrt{\frac{2L}{3g}}[/tex]

    rather than what you quoted. Or was it just formatted a little oddly? It looked like 2 to the power pi to me.

    How much maths and mechanics do you know? To 'show' that I set up a differential equation and assumed a solution which was a phasor - I'm not sure if there's an easier way.

    EDIT: An easier way would be to use a general equation for the time period, if you have one.
     
  5. Mar 16, 2010 #4
    I got it using the general equation of time period but how did you do it using phasor?
     
  6. Mar 16, 2010 #5

    Jmf

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    Since you've got to the answer yourself, then I suppose there's no harm in me posting my solution. So here goes - from first principles:

    Our rod is hanging downwards. If we assume it is a small angle [tex]\theta[/tex] away from vertical, then from Newton's second law:

    [tex]T = I\frac{d^2 \theta}{dt^2}[/tex]

    T is torque (moment) and I the moment of inertia of the rod, both taken about the same point.

    I'd suggest drawing a diagram of the rod at this point.

    Since we can take the motion of the rod to be rotation about it's pivot (the pivot is always the instantaneous centre of rotation), if we take our torque and I about this pivot then:

    [tex]I = \frac{1}{3}ML^2[/tex]

    [tex]T = \frac{gML sin\theta}{2}[/tex]

    'I' there is simply the standard result for the moment of inertia of a uniform rod about it's end, and T is found by drawing a free-body diagram and taking moments about the pivot.

    For small angles, [tex]\theta \approx sin \theta[/tex] so substituting this into the above, and then substituting into [tex]T = I\frac{d^2 \theta}{dt^2}[/tex] yields the differential equation:

    [tex]\frac{d^2 \theta}{dt^2} + \frac{3g}{2L}\theta = 0[/tex]

    If we assume a solution in the form [tex]\theta = e^{j\omega t}[/tex] (a phasor) then we can substitute this into the ODE, noting that the second derivative is [tex]-\omega^2\theta[/tex] and we get:

    [tex]-\omega^2 e^{j\omega t}+\frac{3g}{2L} e^{j\omega t} = 0[/tex]

    thus after cancelling the exponentials:

    [tex]-\omega^2+\frac{3g}{2L} = 0[/tex]

    Solving for omega gives:

    [tex]\omega = \sqrt{\frac{3g}{2L}}[/tex]

    and then noting that [tex]T = \frac{1}{f}[/tex] and [tex]2\pi f = \omega[/tex] gives us:

    [tex]T = \frac{1}{2\pi} \sqrt{\frac{2L}{3g}}[/tex]

    as needed.

    If this seems overly complicated, then don't worry, just use the expression for the time period. :)
     
  7. Mar 17, 2010 #6
    yeah its the same procedure as to how the general equation is derived however, we have to assume the rod is making a small angle w.r.t the vertical axis. Can we do the question without this assumption or how should we do it if the question says something like the rod is hanging vertically/orthogonal w.r.t the horizontal axis?
     
  8. Mar 17, 2010 #7

    Jmf

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    We can't really do it (analytically, anyway) if we didn't assume the small angle somewhere, since our differential equation would then be:

    [tex]
    \frac{d^2 \theta}{dt^2} + \frac{3g}{2L}sin(\theta) = 0
    [/tex]

    - A nonlinear ODE - which, as far as I know, doesn't have a general solution that can be expressed in terms of elementary functions - so we really have to resort to numerical methods (e.g. getting a computer simulation to work out the motion step-by-step) if we wanted to know what happens for large angles.

    Interestingly though, if the rod were horizontal then we can say that if [tex]\delta\theta[/tex] is small, [tex]sin(90^\circ + \delta\theta) \approx 1[/tex] which means that if our rod is approximately horizontal we can write:

    [tex]
    \frac{d^2 \theta}{dt^2} + \frac{3g}{2L} = 0
    [/tex]

    For our differential equation, which gives:

    [tex]
    \frac{d^2 \theta}{dt^2} = - \frac{3g}{2L}
    [/tex]

    This is a constant acceleration downwards, giving a parabolic trajectory (as in free-fall). When you think about it, this is what you would expect if the rod were horizontal.
     
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