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.
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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`.