A pendulum bob on a 2 m string is displaced `60^(@)` from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path

Options

A. `sqrt(2) m//s`
B. `sqrt(9.8) m//s`
C. 4.43 m/s
D. `1//sqrt(2)` m/s

Answer

C

Explanation

`h = R - R cos theta`
`= R[ 1 - cos theta]`

From energy conervation
`mgh =1/2 mv^(2) rArr v^(2) =2gR[1- cos theta]`
`v = sqrt(2 xx 9.8 xx 2[1-cos 60^(@))) = sqrt(19.6) = 4.43 m//s`