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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))/(6)(pi)/(sqrt(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))`

Text Solution

Verified by Experts

The correct Answer is:
A
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