A simple pendulum is oscillating with an angular amplitude of `90^(@)` as shown in the figure. The value of `theta` for which the resulting acceleration of the bob is directed (i) vertically downward, (ii) vertically upward and (iii) horizontally is`

Options

A. `90^(@),0^(@),sin^(-1)(1//sqrt(3))`
B. `0^(@),90^(@),sin^(-1)(1/sqrt3)`
C. `90^(@),0^(@),cos^(-1)(1//sqrt3)`
D. `0^(@),90^(@),cos^(-1)(1//sqrt3)`

Answer

C

Explanation

`v^(2)=2gLcostheta`
`a_(c)=(v^(2))/(L)=2g cos theta`
`a_(t)g sin theta`
since the acceleration is in the horizontal direction,
`a_(c) cos theta=a_(t)sin theta`
`2g cos^(2)theta=g sin^(2)theta`
`tan^(2)theta=2impliestan theta=sqrt(2)`
`cos theta=(1)/sqrt(3)impliestheta=cos^(-1)((1)/sqrt(3))`
when `theta=0`, we have
total acceleration is vertically upward.
when `theta=90^(@)`, we have
total acceleration is vertically downward.