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 `-16pi^2A-m`
B. The particle is acted upon by on impulse of magnitude `4pi^2A-m` 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 force

Answer

A, B, D

Explanation

Given, `x=0` for `tlt0s`
`x(t)=Asin4pit`,`0lttlt(1)/(4)s`
`x=0`,for `tgt(1)/(4)s`
For, `0lttlt(1)/(4)sv(t)=(dx)/(dt)=4piAcos4pit`
`a(t)=`acceleration
`=(dv(t))/(dt)=-16pi^2Asin4pit`
At `t=(1)/(8)s`,`a(t)=-16pi^2Asin4pixx(1)/(8)=-o16pi^2A`
`F=ma(t)=-16pi^2Axxm=-16pi^2mA`
Impulse`=`Change in linear momentum
`F=xxt=(-16pi^2Am)xx(1)/(4)=-4pi^2Am`
The impulse (Change in linear momentum)
at `t=0` is same as,`t=(1)/(4)s`
Clearly, force depends upon A which in not constant. Hence, force is also not constant.