SHM

use calculus

The time period of a compound pendulum -

T = 2 π ( Im g L ) ^{1 / 2}

Here , " I " is the Moment of Inertia of the rigid body about the point of suspension .

So , by " Parallel Axis " theorem ,

I = I ' + m L ²

But , I ' = Moment of Inertia of the rigid body about its " Center of Gravity " = m K ²

Where , " K " - Radius of Gyration

So , T = 2 Ï€ ( m K ² + m L ²m g L ) ^{1 / 2}

= 2 Ï€ ( K ² + L ²g L ) ^{1 / 2}

For " T " to be minimum , " F = K ² + L ²L " must also be minimum .

For , let us put : - dFdL = 0

Or , K ²L ² = 1

Or , K = L

This is the required condition which we wanted to prove .

thnx dude..