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Consider a simple pendulum, having a bob attached to a string, that oscillated under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (L), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimension.

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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\(\sqrt{\frac{1}{mgL}}\)  [∵ = \(\frac{2\pi}{T}\)]

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