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.