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

Answer

A

Explanation

`A_(R)=sqrt((sqrt3)^(2)+1^(2))=2` ` phi=tan^(-1)((sqrt3)/(1))=(pi)/(3)`
So equation of SHM is : `y=2sin(omegat+(pi)/(3))`
Maximum chance of break off is at extreme position. `mg-N=omega^(2)A`
For breakoff N=0 `Rightarrow omegasqrt((g)/(A))=sqrt((g)/(2))`
Also for y=A=2 `Rightarrow omegat+(pi)/(3)=(pi)/(2)Rightarrow t=(pi)/(6omega)=(pi)/(6)sqrt((2)/(g))`