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A particle of mass m = 2kg executes SHM ...

A particle of mass `m = 2kg` executes `SHM` in `xy`- plane between point A and B under action of force `vecF = F_(x)hati+F_(y)hatj`. Minimum time taken by particle to move from A to B is 1 sec. At `t = 0` the particle is at `x = 2` and `y = 2`. Then `F_(x)` as function of time t is

A

`-4pi^(2) sin pit`

B

`-4pi^(2) cos pit`

C

`4pi^(2) cos pit`

D

None of these

Text Solution

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The correct Answer is:
B
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