Vertical displacement of a plank with a body of mass `'m'` on it is varying according to law `y=sin omegat +sqrt(3) cos omegat`. The minimum value of `omega` for which the mass just breaks off the plank and the moment it occurs first after `t=0` are given by: `(y "is positive vertically upwards")`

Options

A. `sqrt((g)/(2),(pi)/(6)sqrt((2)/(g))`
B. `(g)/(sqrt(2)),(2)/(3)sqrt((pi)/(g))`
C. `sqrt((g)/(2)),(pi)/(3)sqrt((2)/(g))`
D. `sqrt(2g),sqrt((2pi)/(3g))`

Answer

A

Explanation

`A_(res) = 2 Units, g = A_(res) omega^(2)`
`y = 2sin (omegat +(pi)/(3))`
`t = (T)/(12)` to move from `(sqrt(3)A)/(2)` to `A`