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= A "sin" 2pisqrt(l//g) xx t`
B. `x = A "cos" 2pi sqrt(l//g)xx t`
C. `x = A "cos" sqrt(g//l) xx t`
D. `x = A "sin" sqrt(g//l) xx t`

Answer

C

Explanation

The bob starts from the extreme position.
As the angular displacement is very small, the bob performs a linear S.H.M. and its equation is
`x=A "cos" omega t =A "cos" ((2pi t)/(T))`
But `T=2pi sqrt((l)/(g))`
`x = A "cos" [(2pit )/(2pisqrt((l)/(g)))]=A "cos" ((sqrt((g)/(l)))t`