- #1
Knissp
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Homework Statement
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.
Homework Equations
[tex]\tau = I \frac{d^2 \theta}{dt^2}[/tex]
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.