Simple Harmonic Motion of a plank of mass

[Mishy]

I don't understand the solution to the following problem:

A horizontal plank of mass m and length L is pivoted at one end. The plank's other end is supported by a spring of force constant k. The moment of inertia of the plank about the pivot is (1/3)m(L^2). The plank is displaced by a small angle theta from its horizontal equilibrium position. Show that it moves with simple harmonic motion with angular frequency omega = square root (3k/m).

Here is my work:
net torque = I[I]*alpha [/I]
net torque = (1/3)m(L^2)*alpha
kxd = (1/3)m(L^2)*(second derivative of theta)
x is the vertical distance from the horizontal equilibrium position.
kxLsin(theta) = (1/3)m(L^2)*(second derivative of theta)
kxL(theta) / (mL^2) = second derivative of theta
because theta is small, sin(theta) = theta
kx(theta)/ mL = second derivative of theta
using this, I would get omega as the square root of (3k/mL)
How do I get rid of that L in my formula for omega? Where was I supposed to drop it?

Thanks.

 

[Physics Monkey]

Your formula for angular velocity can't be right because it doesn't have the right units to be an angular velocity. So what did you do wrong?

You have this equation $$\frac{d^2\theta}{dt^2} = \frac{3 k }{m L} x$$, but you can't just conclude that everything in front of the x is equal to $$\omega^2$$. To see why not, think about the units. What you need to do is relate x to $$\theta$$, then try to extract the angular frequency.

Hope this helps.