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## Homework Statement

A thin uniform rod is pivoted about its mid-point, with the lower end attached to a spring. See the figure. At equilibrium the rod is vertical. Show that on a (small) displacement from equilibrium the rod will oscillate with simple harmonic motion at a frequency:

[tex]

f = \frac{1}{2 \pi} \sqrt{\frac{3k}{M}}

[/tex]

Where [itex]k[/itex] is the spring constant and [itex]M[/itex] is the mass of the rod.

## Homework Equations

[tex]

F = ma\\

\tau = I \alpha\\

I = \frac{1}{12} ML^{2}\\

f = \frac{\omega}{2 \pi}\\

\sin(\theta) \approx \theta

[/tex]

I believe this is the moment of inertia for a uniform rod rotating about its mid-point

## The Attempt at a Solution

I tried to relate the displacement [itex]x[/itex] to the displacement [itex]\theta[/itex] using geometry and then write a well known differential equation with a well known solution to get [itex]\omega[/itex] and then do some massaging to get [itex]f[/itex]. However my answer doesn't match.

[tex]

\tau = I \alpha \\

-kx = \frac{1}{12}ML^{2} \alpha \\

\sin(\theta) = \frac{2x}{L}\\

x = \frac{L\theta}{2}\\

-k \frac{L\theta}{2} = \frac{1}{12}ML^{2} \ddot{\theta}\\

0 = \frac{1}{6}ML\ddot{\theta} + k\theta\\

\omega = \left(\frac{6k}{ML}\right)^{\frac{1}{2}}\\

f = \frac{1}{2\pi}\left(\frac{6k}{ML}\right)^{\frac{1}{2}}\\

[/tex]