Home
Class 12
PHYSICS
Vertical displacement of a plank with a ...

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
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SIMPLE HARONIC MOTION

    MOTION|Exercise EXERCISE -2 (Leve-I) ( SECTION - D ) ( Combination of springs )|5 Videos

  • SIMPLE HARONIC MOTION

    MOTION|Exercise EXERCISE -2 (Leve-I) ( SECTION - E,F ) ( Angu lar shm & s implependulum )|1 Videos

  • SIMPLE HARONIC MOTION

    MOTION|Exercise EXERCISE -2 (Leve-I) ( SECTION - B ) ( Time per iod and angu larfrequency in SHM )|7 Videos

  • SIMPLE HARMONIC MOTION

    MOTION|Exercise EXERCISE -3 Section - B Previous Year Problems | JEE MAIN|23 Videos

  • SOUND WAVES

    MOTION|Exercise Exercise - 3 (Section - B)|14 Videos

Similar Questions

Explore conceptually related problems

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.

Knowledge Check

  • 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")

    A
    `sqrt((g)/(2)),(sqrt2)/(6)(pi)/(sqrtg)`
    B
    `(g)/(sqrt2),(2)/(3)sqrt((pi)/(g))`
    C
    `sqrt((g)/(2)),(pi)/(3)sqrt((2)/(g))`
    D
    `sqrt(2g),sqrt((2pi)/(3g))`

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

  • Vertical displacement of a Planck with a body of mass m on it is varying according to law y = sin omegat + sqrt(3) cos omega t . The minimum value of w for which the mass just breaks off the Planck and the moment it occurs first after t = 0, are given by

    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)`

    • Similar Questions

    Explore conceptually related problems

    A point moves in the plane y according to the law x = A sin omegat, y = B cos omegat , where A,B & omega are positive constant. The velocity of particle is given by

    A point moves in the plane y according to the law x = A sin omegat, y = B cos omegat , where A,B & omega are positive constant. The acceleration of particle is given by [ :' barr is position vector]

    A point moves in the plane y according to the law x = A sin omegat, y = B cos omegat , where A,B & omega are positive constant. The equation for trajectroy for path taken by particle is

    The displacement of a particle moving in S.H.M. at any instant is given by y = a sin omegat . The accelreation after time t=T/4 os

    A point moves in the plane xy according to the law, x=a sin omegat,y=a(1-cosomegat) Answer the following question taking a and omega as positive constant The distance travelled by the point during the time T is

    Home
    Profile