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Small oscillations about equilibrium

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A rod of length L and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions. The springs have spring constant k, and at equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position.


    2. Relevant equations
    [tex]\tau = I \frac{d^2 \theta}{dt^2}[/tex]

    3. The attempt at a solution
    I define counterclockwise angular displacement to have positive angle.
    If the position of the rod is given a slight initial perturbation [tex]\theta[/tex], then the torques due to spring 1 (on top), spring 2 (at midpoint), and gravity are given by:

    [tex]\tau_g = mg\frac{L}{2}sin(\theta)[/tex]

    [tex]\tau_1 = -k(L sin(\theta)) L [/tex]

    [tex]\tau_2 = -k(\frac{L}{2} sin(\theta)) \frac{L}{2} [/tex]

    The moment of inertia I of a rod about one end is [tex] I = \frac{mL^2}{3} [/tex].

    Then [tex]\tau = mg\frac{L}{2}sin(\theta) - k(L sin(\theta)) L - k(\frac{L}{2} sin(\theta))\frac{L}{2} = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2} [/tex]

    Using small the angle approximation [tex]sin(\theta)=\theta[/tex],

    [tex]mg\frac{L}{2}\theta - kL^2\theta - k\frac{L^2}{4} \theta = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2} [/tex]

    [tex](\frac{3g}{2L} - \frac{3k}{mL} - \frac{3k}{4m}) \theta = \frac{d^2 \theta}{dt^2} [/tex]

    Is any of that right?

    Alternatively, I know that for forces, given a potential energy function U(x), the frequency
    of small oscillations is given by [tex]\omega = \sqrt{\frac{U''(x)}{m}}[/tex]. Perhaps a similar technique can be applied to angles?

    [tex]U(\theta) = \frac{1}{2}k x_1^2 + \frac{1}{2}k x_2^2 + mg\Delta h[/tex]

    [tex]x_1 = L sin(\theta)[/tex]

    [tex]x_2 = \frac{L}{2} sin(\theta)[/tex]

    [tex]\Delta h = h (1-cos(\theta))[/tex]

    Then using small angle approximations,

    [tex] U(\theta) = \frac{1}{2}k (L \theta)^2 + \frac{1}{2}k (\frac{L}{2}\theta)^2[/tex]

    [tex] U(\theta) = \frac{k}{2} L^2 \theta^2 + \frac{k}{2} \frac{L^2}{4}\theta^2[/tex]

    [tex] U(\theta) = \frac{k}{2} (L^2 \theta^2 + \frac{L^2}{4}\theta^2)[/tex]

    [tex] U(\theta) = \frac{k}{2} (\frac{5}{4}L^2 \theta^2)[/tex]

    [tex] U(\theta) = \frac{5k}{8}L^2 \theta^2[/tex]

    [tex] U''(\theta) = \frac{5k}{4}L^2[/tex]

    Then [tex]\omega = \sqrt{\frac{\frac{5k}{4}L^2}{m}}[/tex]

    [tex]=\frac{L}{2}\sqrt{\frac{5k}{m}}[/tex]

    So, I'm not sure which technique is valid or how to continue from here. Any help would be greatly appreciated. Thank you.
     

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  3. Nov 23, 2009 #2

    Redbelly98

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    Yes, though the middle term on the LHS should not have an L in the denominator. Perhaps just a typo?

    From here, you can get ω simply by inspection of the differential equation.
     
  4. Nov 23, 2009 #3
    Your period is contained in the term before your [tex]\theta[/tex].

    You should make the quick correction suggested, and see if you can get equivalent answers. If you can, then you're fine for either. If you cannot, try using Lagrange's equations to check a third method.
     
  5. Nov 23, 2009 #4

    Redbelly98

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    Just reread your post more carefully, now I realize that you are trying a different method here.

    Note that terms proportional to θ2 will contribute to U''(θ). For that reason we should approximate cos(θ) as (1-θ2/2), rather than simply 1, and include the gravitational term in U(θ).
     
  6. Nov 26, 2009 #5
    Er right that was a typo. Thanks for the help!
     
  7. Oct 8, 2011 #6
    I know this is 2 years old, but I did the same sum just now and got almost exactly what the OP has, except for one thing:
    [itex]\tau_g=\vec{r} \times \vec {mg} [/itex]
    [itex] =\frac{L}{2}\hat{r} \times mg (-\hat{j})[/itex]
    [itex] =\frac{mgL}{2}([cos\theta\hat{i} +sin\theta\hat{j}] \times -\hat{j})[/itex]
    [itex] =\frac{mgL}{2}cos\theta[/itex]

    and not [itex] \frac{mgL}{2}sin\theta[/itex]
    so after small angle approximation (sinθ≈θ, cosθ≈1)

    [itex]mg\frac{L}{2} - kL^2\theta - k\frac{L^2}{4} \theta = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}[/itex]

    As far as I can see, this isn't the simple harmonic form [itex]\frac{d^2 \theta}{dt^2}=k\theta[/itex]
    It is still solvable using what I know of differential equations, but would that be the correct frequency of oscillation or am I missing something ?

    Thank you.
     
  8. Oct 8, 2011 #7

    Redbelly98

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    The OP was solving for a vertical rod, while you appear to be solving for a horizontal rod.

    I suggest (1) combine the like terms (with θ in them) and simplify, and then (2) solve the equation for [itex]\frac{d^2 \theta}{dt^2}[/itex].

    Let's see what you get after doing that, then I'll take another look at it.
     
  9. Oct 9, 2011 #8
    Actually, the rod I'm talking about is also vertical. Here's an image:
    Untitled.png
    The rod is pivoted at O and torque calculations were done about this point.
    As far as I can see, the weight still contributes a cosθ component to torque.

    The equation simplifies to [itex]\ddot{\theta} + \frac{15k}{4m}\theta - \frac{3g}{2L} = 0[/itex]

    Whose solution is:
    [itex]\theta(t)= \frac{2mg}{5kL} + c_{1}sin(\sqrt{\frac{15k}{m}}t)+ c_{2}cos(\sqrt{\frac{15k}{m}}t) [/itex]

    Is this form still simple harmonic ?
    I can clearly see that this function does in fact have a period:
    [itex]ω = \sqrt{\frac{15k}{m}}[/itex]
    and that the only difference between this and the general SHM solution is the addition of the constant [itex]\frac{2mg}{5kL}[/itex] so the resulting periodic wave is shifted up the y axis a bit...
     
  10. Oct 12, 2011 #9
    Anyone ?
     
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