Consider a simple pendulum, having a bob attached to a string , that oscillates 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 is time period using method of dimension .

Explanation

(i) Let `theta` be the angle made by the string with vertical. When the bob is at the mean position `theta=0.`

(ii) There are two forces acting on the bob. The tension T along the vertical force due to gravity (=mg).
(iii) The force can be resolved into `mgcostheta` along a circle of length L and centre at this support point.
(iv) Its radial acceleration `(omega^(2)L)` and also tangential acceleration. So net radial force `= T - mg costheta` , While the tangential acceleration provided by `mg sintheta` the radial force gives zero torque.
(v) So torque provided by the tangential component.
`tau="Lmg"sintheta`
`tau=Ialpha` (by Newton's law of rotational motion)
`Ialpha=mgsinthetaL`
`alphaimplies("mgL")/(I)sintheta`
where I is the moment of inertia , `alpha` is angular acceleration,
(vi) The `theta` is to small,
`sintheta=theta`
`alpha=(mgL)/(I)theta,omega=sqrt((I)/(mgL))`
[for simple harmonic `a(t) =-omega^(2)txxtheta" is small ]"and T=2pisqrt((I)/(mgL))" "[:'omega=(2pi)/(T)]`