Let θ be the angle made by the string with vertical when the bob is at exteme position.
There are two forces acting on the bob; the tension T along the vertical force due to gravity (=mg). The force can be resolved into mgcos θ along a circle of length L and centre at this support point. Its radial acceleration (ω2L) and also tangential acceleration. So net radial force = T – mg cos θ, while the tangential acceleration provided by mg sin θ. Since the radial force gives zero torque. So torque provided by the tangential component.
τ = L mgsin θ ...(i)
τ = Iα (by Newton’s law of rotational motion) …(ii)
∴ Iα = mgsin θL
α = \(\frac{mgL}{I}sin \,\theta\)
where, I is the moment of inertia, α is angular acceleration.
If θ is too small
∴ sin θ = θ
α = -\(\frac{mgL}{I}\theta\) [for simple harmonic α = -ω2θ] [θ is expressed in radian].
ω = \(\sqrt{\frac{mgL}{I}}\)
and T = 2π\(\sqrt{\frac{1}{mgL}}\) [∵ = \(\frac{2\pi}{T}\)]