A uniform rod of length l is suspended by end and is made to undego small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

Answer

(2l)/3

Explanation

Let Ararr suspension point Brarr Centre of Gravity, `l=1/2h=1/2` moment of inertia about A is
`I_(CG)+mh^2=(ml^2)/12+m l^2/4`
`=(ml^2)/12+(ml^2)/4=(ml^2)/3`
`:.T=2pisqrt(((I)/(mgl)))`
`=2pisqrt((2ml^2)/(3mgl))=2pi sqrt((2l)/(3g))`
let the time period T is equal to the time
period of simple pendulum of lenght `x`
`:.T=2pisqrt(x/g)`
So, `(2l)/(3g)=x/y`
`rarr x=(2l)/3`
:. lenght of the simple pendulum
=(2l)/3`