# Simple Harmonic Motion of a plank of mass

1. Apr 10, 2006

### 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.

2. Apr 10, 2006

### 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.

Last edited: Apr 10, 2006