The motion of a particle of mass m is given by `x=0` for `tlt0s`,`x(t)=Asin4pit` for `0lttlt((1)/(4))s(Agt0)`,and `x=0` for `tgt((1)/(4))s`.

Options

A. The force at `t = (1//8)s` on the particle is `m16pi^(2)A`.
B. The particle is acted upon by an impulse of magnitude `m4 pi^(2) A` at `t =0s` and `t = (1//4)s`.
C. The particle is not acted upon by any force.
D. The particle is not acted upon by a constant

Answer

A, B, D

Explanation

Given `x =0` for `t lt 0s`
`x (t) =A sin 4pit, for 0 lt t lt (1)/(4)s`
`x =0,` for `t gt (1)/(4)s`
For `0 lt t lt (1)/(4) s`
`v (t) = (dx)/(dt) =4pi A cos 4pi t`
`a (t) = (dv(t))/(dt) = - 16pi^(2) a sin 4pit`
`At = (1)/(8) s,a (t) = - 16pi^(2) A sin 4pi xx (1)/(8) = - 16pi^(2) A`
`F =ma (t) = - 16pi^(2) A xx m = - 16 pi^(2) mA`
impulse =Change in linear mo-mentum
`I =F xx t = (-16 pi^(2) Am) xx (1)/(4)`
` =- 4pi^(2) Am`
the impulse (Change in linear momentum)
`at t =0` is same
`as, t = (1)/(4)s`
Clearly force depends upon A which is not constant Hence forces is also not constant .