Vertical displacement of a plank with a body of mass m on it is varying according to the law `y=sinomegat+sqrt(3)cosomegat`. The minimum value of `omega` for which the mass just breaks off the plank and the moment it occurs first time after t=0, are given by (y is positive towards vertically upwards).

Options

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

Answer

A

Explanation

`y = sinomegat + sqrt(3) cos omegat = 2sin (omegat + (pi)/(3))`
To breaks off mg
`mg = momega_(min)^(2)A rArr g = 2omega_(min)^(2) rArr omega = sqrt((g)/(2))`
moment it occurs first after `t = 0`
`2 = 2sin(omegat_(1) + (pi)/(3)) rArr omegat_(1) = (pi)/(6) rArr t_(1) = (pi)/(6omega) = (pi)/(6)sqrt((2)/(g))`