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

Options

A. `-4pi^(2) sin pi t`
B. `-4 pi^(2)cos pi t`
C. `4pi^(2)cos pi t`
D. None of these

Answer

B

Explanation

Let the line joining AB represents axis 'r' By the conditions given 'r' coordinates of the particle at time t is

`r = 2 sqrt(2) cos omega t`
`rArr omega = (2 pi)/(T) = (2 pi)/(2) = pi`
`:. R = 2 sqrt(2) cos pi t`
`x = r cos 45^(@) = (r)/(sqrt(2)) = 2 cos pi t`
`:. a_(x) = - omega_(2) x = - pi_(2) 2 cos pi t`
`:. F_(x) = ma_(x) = -4 pi_(2) cos pi t`.