Question

A horizontal plank (m = 2.0 kg, L = 1.0 m) is pivoted at one end. A spring. (k = 1.0 x 10. 3. N/m) is attached at the other end find the angular frequence for small oscillations?

Answer 1

There is a dynamic relationship between torque about the pivot and the rate of change of angular momentum, that can be written as a differential equation

If the linear deflection at the attachment point is X, the angular deflection A is

A = X/L

The angular momentum is

I dA/dt wherre I is the moment of inertia, (1/3) m L^2. The equation of motion is

Torque = -k X*L = -k L^2 A =
d/dt I dA/dt = (1/3) mL^2 d^2A/dt^2

d^2A/dt^2 + 3 k/m A = 0

The angular oscillation frequency that results from this differential equation is
w = sqrt (3k/m)

(The length cancels out). Note how this resembles the spring-mass natural oscillation frequency withough the hinge, which would be sqrt(k/m)

Answer 2

There is nothing wrong with this answer to what I know but here is how my comrade and I did it:
The angular frequency of a simple harmonic oscillator is as follows:
w=sqrt((k/I))
The system is called a torsional pendulum. and if one replaces I, the moment of inertia, with 1/3(m)(r^2) you will get the sqrt(k/(1/3(m)(r^2). and that leads to sqrt(3k/m(r^2)) and well since 1 =r^2 its sqrt(3k/m). Enjoy