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

Answer

A

Explanation

At "C", `T cos alpha = Mg`…(i)
`P.E_(B) = K.E_(A) rArr mgr = (1)/(2) mv_(A)^(2)`
hence at `"C", v = sqrt(v_(A)^(2) - 2gr(1 -cos alpha))`
`T-mg cos alpha = (Mv^(2))/(r) = (M.2 gr cos alpha)/(r)`.