Answer with Explanation:
We know from newton's second law that acceleration produced by a force 'F' in a body of mass 'm' is given by
![a=(Force)/(Mass)=\farc{F}{m}](https://img.qammunity.org/2020/formulas/physics/college/31363udi3g4abewh3k2l2kb8kzgwbwuexo.png)
In the given case the acceleration equals
![a=(F_(o)cos^(2)(\omega t))/(m)](https://img.qammunity.org/2020/formulas/physics/college/sjmuypti3m1zcmuwn2krj463ygpnya4bnv.png)
Now by definition of acceleration we have
![a=(dv)/(dt)\\\\\int dv=\int adt\\\\v=\int adt\\\\v=\int (F_(o)cos^(2)(\omega t))/(m)\cdot dt\\\\v=(F_(o))/(m)\int cos^(2)(\omega t)dt\\\\v=(F_(o))/(\omega m)\cdot ((\omega t)/(2)+(sin(2\omega t))/(4)+c)](https://img.qammunity.org/2020/formulas/physics/college/o0mnzz7066acy05tore2sp43hlassxzu4y.png)
Similarly by definition of position we have
![v=(dx)/(dt)\\\\\int dx=\int vdt\\\\x=\int vdt](https://img.qammunity.org/2020/formulas/physics/college/8jwh125ospr9998gz8oziwsbtmzcbt5gj1.png)
Upon further solving we get
![x=\int [(F_(o))/(\omega m)\cdot ((\omega t)/(2)+(sin(2\omega t))/(4)+c)]dt\\\\x=(F_(o))/(\omega m)\cdot ((\int (\omega t)/(2)+(sin(2\omega t))/(4)+c)dt)\\\\x(t)=(F_(o))/(\omega m)\cdot ((\omega t^(2))/(4)-(cos(2\omega t))/(8\omega )+ct+d)](https://img.qammunity.org/2020/formulas/physics/college/71uycd8lcdq9ie87c0gbsdwgg70xnr77qv.png)
The plots can be obtained depending upon the values of the constants.