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A point moves in the plane y according t...

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

A

`x^(2) +y^(2) =A^(2)`

B

`(x^(2))/(A^(2)) +(y^(2))/(B^(2)) =1`

C

`y = BA`

D

`y =Ax +Bx^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`sin^(2) theta +cos^(2) theta = 1`
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Knowledge Check

  • 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
    `v = sqrt(A^(2) +B^(2)) omega`
    B
    `v = sqrt(A^(2)cos^(2) omegat +B^(2) sin^(2) omegat)omega`
    C
    `v = omega sqrt((A^(2)+B^(2))-(x^(2)+y^(2)))`
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    `v = omega sqrt((A+B)^(2)-(x+y)^(2))`

  • 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
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    `bar(a) = omega bar(r )`
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