Time period of a block hanging from a pulley

[randomgamernerd]

Homework Statement

:[/B] The figure shows a pulley block system in equilibrium. If the block is displaced down slightly from its position and released, find the time period of the oscillation. Assume friction is sufficient.
I: moment of inertia of pulley
k: spring constant
R: Radius of pulley
m: mass of hanging block

Homework Equations

: [/B]F=kx(for spring force)
T=2π√(x/a)

The Attempt at a Solution

:[/B]
I have done the sum using concervation of energy and the answer matched. But when I tried it by balancing torque and force, I'm unable to get the answer.
My Attempt:

Initially, when the system was at equilibrium,
T=mg
⇒kx = mg
⇒x=mg/k
(x is the stretch produced in the spring when the system is at equilibrium)
After we disturb the equilibrium,
Let us suppose at any instant(when the spring is restoring its position) the additional stretch is x' (I used additional as already the spring was stretched by an amount x)
So by equating forces on the block, we get:
ma = F_s + T - mg
(a= acceleration of block at the instant)
From equating torque on the pulley we get
F_sR - TR = Iα
α is the angular acceleration of the puley.
and
α= a/R.
I just want to know if my equations are correct.
I will solve the rest on my own.

I have also attached the FBD sketched by me.

 

[TSny]

Your free-body diagram for the block is incorrect. Reconsider how many forces act on the block.

 

[randomgamernerd]

Your free-body diagram for the block is incorrect. Reconsider how many forces act on the block.

Okay, So the spring force is not acting..right?

 

[TSny]

Okay, So the spring force is not acting..right?

That's right, the spring force does not act on the block.

 

[randomgamernerd]

That's right, the spring force does not act on the block.

okay, one more question, initially I equated T= mg and then T= kx.
I could do that because the system was in equilibrium and the string is massless, right?

 

[TSny]

okay, one more question, initially I equated T= mg and then T= kx.
I could do that because the system was in equilibrium and the string is massless, right?

Yes

 

[randomgamernerd]

Yes

ok, thanks for helping me again