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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")`

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)`

Text Solution

Verified by Experts

The correct Answer is:
A
`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))`
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