The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement ` theta _ 0 ` . Its linear displacement at time t is given by :

Options

A. ` x = theta _ 0 sin 2pi sqrt ( ( L//g ) ) t `
B. ` x = L theta _ 0 cos 2pi sqrt ( ( g // L) ) t `
C. ` x = L theta _ 0 sin sqrt (( g//L )) t `
D. ` x = L theta _ 0 cos sqrt(( g//L )) t `

Answer

D

Explanation

As pendulum starts from EP
So equation of SHM is

` theta = theta _ 0 cos omega t" " ` where ` omega = ( 2pi ) /( T) `
` omega = ( 2pi ) / ( 2 pi sqrt((L)/(g))) = sqrt((g ) /(L)) `
` rArr theta = theta _ 0 cos sqrt ( (g)/(L)) t `
At time t
Linear displacement = ` x = L theta `
` x = L theta _ 0 cos sqrt ( (g ) /(L)) t `