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 `16 pi^(2) A m`
B. The particle is acted upon by an impulse of magnitude `4 pi^(2) A m` at `t = 0 s` 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 force

Answer

a,b,d

Explanation

Here, `x = 0,` for `t lt 0s`
`x (t) = A sin 4 pi t` for `0 lt t lt (1)/(4)s`
`x = 0` for `t gt (1)/(4)s`
For `0lt t lt (1)/(4)s`
`upsilon (t) = (dx)/(dt) = 4piA cos 4pi t,a(t) = (d upsilon)/(dt) - 16 pi^(2) A sin 4 pi t`
At ` t= (1)/(8)s a (t) = - 16 pi^(2) A sin 4 pi xx (1)/(8) = - 16 pi^(2) A xx 1`
`F = ma (t) = - 16 pi^(2) Am`
Impulse = change in linear momentum) at `t = 0` is same as at `t = (1)/(4)sec`
Further as `F` depends on `A` which is not constant therefore the particle is not acted upon by a constant force .