I don't understand the solution to the following problem:(adsbygoogle = window.adsbygoogle || []).push({});

A horizontal plank of massmand lengthLis pivoted at one end. The plank's other end is supported by a spring of force constantk. The moment of inertia of the plank about the pivot is (1/3)m(L^2). The plank is displaced by a small anglethetafrom its horizontal equilibrium position. Show that it moves with simple harmonic motion with angular frequencyomega= 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 oftheta)

xis the vertical distance from the horizontal equilibrium position.

kxLsin(theta) = (1/3)m(L^2)*(second derivative oftheta)

kxL(theta) / (mL^2) = second derivative oftheta

becausethetais small, sin(theta) =theta

kx(theta)/ mL= second derivative oftheta

using this, I would getomegaas the square root of (3k/mL)

How do I get rid of thatLin my formula foromega? Where was I supposed to drop it?

Thanks.

**Physics Forums - The Fusion of Science and Community**

# Simple Harmonic Motion of a plank of mass

Have something to add?

- Similar discussions for: Simple Harmonic Motion of a plank of mass

Loading...

**Physics Forums - The Fusion of Science and Community**