A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation.

Explanation

For small amplitude the angular motion is nearly simple harmonic and the time period is given by
`T=2pisqrt((1)/(mg(l//2)))=2pisqrt(((ml^(2)//3))/(mg(l//2)))=2pisqrt((2l)/(3g))=pisqrt((2xx1.00m)/(3xx10m//s^(2)))=(2pi)/(sqrt(15))s`.