A spring has a natural length l with one end fixed to the ceiling. The other end is fitted with a smooth ring which can slide on a horizontal rod fixed at distance I below the ceiling. Initially, the spring makes an angle of `60^@` with the vertical, when system is released from rest. Find the angle of the spring with the vertical, when the velocity of the ring reaches half of the maximum velocity, which the ring can attain during the motion.

Options

A. `30^@`
B. `cos^(-1) ( (2)/(2 + sqrt3) )`
C. `cos^(-1) ( (sqrt(3) -1) /(2) ) `
D. `45^@`

Explanation

From diagram

`cos 60^@ = l/h rArr h = (l)/(cos 60^@) rArr h = 2l`
Extension in spring = h - l = 2l - l = l
At mean position , if velocity is v and there is no friction, then
` 1/2 mv^2 = 1/2 kx^2 rArr v^2 = (k)/(m) l^2`
if angle with vertical is ` theta` when velocity is `v. = v/2`
Then , `1/2 mv.^2 = 1/2 kx.^2 1/2 m (v/2)^2 = 1/2 k ( (l)/(cos theta.) - 1)^2`
As, `cos theta. = (l)/(h.) rArr h. = (l)/(cos theta.)`
and extension, X. = h. - l =(l)/(cos theta.) - l`
Substituting value of `v^2` , we have
`1/2 m ( (hl^2)/(4m) ) = 1/2 k ( (l)/(cos theta.) -l)^2`
` rarr l/2 = (l)/(cos theta.) - l rArr (3l)/(2) = (l)/ (cos theta.)`
` cos theta. =2/3 " " theta . = cos^(-1) 2/3`
(No option is matching )