How Does a Rod Attached to a Spring Undergo Simple Harmonic Motion?

[K.QMUL]

Homework Statement

The figure shows a thin uniform rod of mass M and length 2L that is pivoted without friction about an axis through its midpoint.A horizontal light spring ofspring constant k is attached to the lower end of the rod.The spring is at its equilibrium length when the angleθ with respect to the vertical is zero.Show that for oscillations of small amplitude, the rod will undergo SHM with a period of 2π√M/3k. The moment of inertia of the rod about its midpoint is ML2/3.(Assume the small angle approximations)

The Attempt at a Solution

My book gives the following answer but is not indepth, could someone help me out by going through step by step how they came to this answer...

Couple acting on rod = −kLsinθ×Lcosθ=−kL2θ for smallθ.
Hence, I(d^2θ/dt^2)=−kL^2θ

 

[BvU]

Sure. My telepathic powers are limited, however, so I can't see the figure clearly. Is the spring dangling loose or is it attached to something fixed ?

As I said, limited powers, so I can't read what you intended to fill in under 2., either. Could you oblige ?

Then: the english word for couple is torque.

 

[K.QMUL]

I think I've solved it, never mind, thanks for your offer

 

[dauto]

You should also learn to use superscripts. L2 and L² are not the same thing.