A force-time (F-t) graph for a linear motionis shown in following...

From graph, the force is varying sinusoidally with time by equation, $F=-2 \cdot \cos \omega t$ ...(i) where, $\omega$ be the angular frequency of variation of force. Since, the graph is completing one cycle in $8 \mathrm{~s}$, hence time period, $T=8 \mathrm{~s}$ Using expression of change of momentum, $\Delta p=\int_0^8 F d t=\int_0^8-2 \cos \omega t d t$[from Eq. (i)] $$\begin{aligned} & =-\frac{2}{\omega}[\sin \omega t]_0^8 \\ & =\frac{-2}{\omega}[\sin \omega 8-\sin 0]\end{aligned}$$ $=\frac{-2}{\omega}\left[\sin \frac{2 \pi}{8} 8-0\right]$ \left(\because \omega=\frac{2 \pi}{T}\right)$=\frac{-2}{\omega}[\sin 2 \pi]=0$ $[\because \sin 0=\sin 2 \pi=0]$ Hence, the gain or change of linear momentum is zero.