The motion of a particle of mass \(m\) is given by \(x=0\) for \(t<0 ~\text s\), \(x(t)=A~\text {sin}~4\pi t\) for \(0<t<(1/4) \text s ~(A>0)\), and \(x=0\) for \(t>(1/4) ~\text s\). Then:

(a) The force at \(t=(1/8) ~\text s \) on the particle is \(-16 \pi^2\text{Am}\)
(b) The particle is acted upon by an impulse of magnitude \(4 \pi^2 \text{Am}\) at \(t=0 ~\text s\) and \(t=(1/4) ~\text s\)
(c) The particle is not acted upon by any force
(d) The particle is not acted upon by a constant force
(e) There is no impulse acting on the particle

Which of the following statement/s is/are true?

1. (a, c, d, e) 2. (a, c)
3. (b, c, d) 4. (a, b, d)
(a, b, d) Hint: Using the equation of position of the particle, we can find the acceleration the particle.
Given,
Step 1: Find the force acting on the particle.
                                                             x=0 for t<0s                                                          x(t)=Asin4πt; for 0<t<14s                                                              x=0; for t>14s For, 0<t<14s                               v(t)=dxdt=4πAcos4πt
a(t)= acceleration =dv(t)dt=16π2Asin4πt At t=18s,a(t)=16π2Asin4π×18=16π2AF=ma(t)=16π2A×m=16π2mA= Change in linear momentum =F×t=(16π2Am)×14=4π2Am
Step 2: Find the impulse of the particle.
The impulse (Change in linear momentum)
                                   at t=0 is same as, t=14s
Clearly, toræ A which is not constant. Hence. force is also not constant.