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 theexpression for is time period using method of dimension .

Consider a simple pendulum, having a bob attached to a string , that o

1.	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 theexpression for is time period using method of dimension .
Answer» Solution :(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 providedby 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)]`