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A uniform rod of length 1.00 m is suspen...

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.

Text Solution

Verified by Experts

For small amplitude the angular motion is nearly simle harmonic and the time period is given by
`T=2pisqrt(I/(mgl))=2pi(sqrt((mL^2/3))/(mglL/2))`
`=2pi(sqrt(2L)/(3g)=2pi(sqrt(xx1.00m)/(3xx9.80ms^-2))=1.64s`.
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