Question about simple harmonic motion

[-=CN=-]

A square block, with a mass of 3.40 kg and edge lengths d = 6.00 cm, is mounted on an axle through its center. A spring of spring constant k = 1190 N/m connects the block's upper corner with a rigid wall. Initially the spring is at its rest length. If the block is rotated by 3° and then released, what is the period of the resulting SHM?

What type of problem should this be treated as?

 

[lightgrav]

It oscillates by rotation ... it's called a torsional oscillator.
You look at "restoring torque" which returns the object
(which responds slowly due to its rotational Inertia) to
the equilibrium orientation angle.

set torque = I alpha , get torque as function of theta.
Now it should operationally look like an oscillator eq'n.

Be careful to keep the omega_(orientation_change_rate)
distinct from the omega_(forward trig function argument)
omega_ocr has amplitude 3 degrees, while
omega_tfa is multiplied by time.

Enjoy it, this one is fun!

 

[-=CN=-]

there are two different omegas? I'm slightly confused. I know for a torsion oscillator, period is usually found using T = (2*pi)*(I/kappa)^(1/2)
Inertia can be calculated...but how should I go about getting kappa, setting the net torque = -k*theta?

 

[mukundpa]

Torque is force multiplied by perpendicular distance from axis of rotation.

Tau = -K(d/2)^2 @ sin@ =@ approx