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 minium value of `omega` for which the mass just breaks off the plank and the moment it occurs first arter `t=0` are given by: `(y "is positive vertically upwards")`

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).

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).

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).

A particle moves in xy plane according to the law x = a sin omegat and y = a(1 – cos omega t) where a and omega are constants. The particle traces

Equations y = 2 A cos^(2)omega t and y = A (sin omegat + sqrt(3) cosomegat) represent the motion of two particles.

The displacement of two particle of same mass executing SHM are represented by the equations x_1=4 sin (10t+pi/6) and x_2=5 cos(omegat) The value of omega for which the energies of both the particle remain same is,

A point moves in the plane xy according to the law x=a sin omegat , y=a(1-cos omega t) , where a and omega are positive constants. Find: (a) the distance s traversed by the point during the time tau ,