A ball is suspended by a thread of length L at the point O on a wall which in indeed to the vertical by alpha. The thread with the ball is displaced by a small angle beta away from the vertical and also away from the wall. If the ball is released, the period of observation of the pendulum when beta gt alpha will be...........

A ball is suspended by a thread of length L at the point O on a wall w

1.	A ball is suspended by a thread of length L at the point O on a wall which in indeed to the vertical by alpha. The thread with the ball is displaced by a small angle beta away from the vertical and also away from the wall. If the ball is released, the period of observation of the pendulum when beta gt alpha will be...........
Answer» `SQRT((L)/(g)) [PI + 2 sin ^(-1) (ALPHA/beta)]` `sqrt((L)/(g)) [pi - 2 sin ^(-1) (alpha/beta)]` `sqrt((L)/(g)) [2 sin ^(-1) (alpha/beta)-pi]` `sqrt((L)/(g)) [2sin ^(-1) (alpha/beta)+pi]` Solution :Angular position `theta= theta_(0) sin omega t" and " omega = sqrt((g)/(L))` `therefore T= 2pi sqrt((L)/(g))` When `beta propto alpha`, time taken by pendulum from B to C and C to B `t_(1)= (T)/(2)= (1)/(2)XX 2pi sqrt((L)/(g))= pi sqrt((L)/(g))"""........"(1)` Time taken by pendulum from B to A and A to B, `t_(2)= 2t = (2)/(omega)sin^(-1) (alpha/beta)""(" USING " theta = theta_(0) sin omega t)` `therefore alpha= beta sin omega t" or "t= (1)/(omega) sin^(-1) (alpha/beta) = sqrt((L)/(g)) sin^(-1) (alpha)/(beta)` `t_(2)= 2t = 2sqrt((L)/(g)) sin^(-1) (alpha/beta)"""........"(2)` Time period of motion, `T= t_(1)+t_(2)` `= sqrt((L)/(g)) [pi + 2 sin^(-1) (alpha)/(beta)]`.