Answer with Explanation:
From newton's second law the acceleration produced by a force on a mass 'm' is given by
[tex]Acceleration=\frac{Force}{Mass}[/tex]
Applying the given values in the above equation we get
[tex]Acceleration=\frac{kvx}{m}[/tex]
Also we know that acceelration of a particle can ve mathem,atically written as
[tex]a=\frac{v\cdot dv}{dx}[/tex]
Applying the given values in the above equation we get
[tex]\frac{kvx}{m}=\frac{v\cdot dv}{dx}\\\\\Rightarrow {kx}\cdot dx=m\cdot dv\\\\\int kxdx=\int mdv\\\\\frac{kx^2}{2}=mv-c[/tex]
'c' is the constant of integration
whose value is found that at x =0 v= [tex]v_o[/tex]
Thus
[tex]c={mv_o}[/tex]
Thus the velocity as a function of position is
[tex]v=\frac{1}{m}(\frac{kx^2}{2}+c)[/tex]
Now by definition of velocity we have
[tex]v=\frac{dx}{dt}[/tex]
Using the function of velocity in the above relation we get
[tex]\frac{dx}{kx^{2}+\sqrt{2c}}=\frac{dt}{2m}\\\\\int \frac{dx}{(\sqrt{k})^2x^{2}+(\sqrt{2c})^2}=\int \frac{dt}{2m}\\\\\frac{1}{\sqrt{2kc}}\cdot tan^{-1}(\frac{(\sqrt{k})x}{\sqrt{2c}})=\frac{t}{2m}+\phi \\\\[/tex]
where
[tex]\phi [/tex] is constant of integration
Now it is given that at t = 0 ,x = 0
thus from the above equation of position and time we get [tex]\phi =0[/tex]
Thus the position as a function of time is
[tex]x(t)=\sqrt{\frac{2c}{k}}\cdot tan(\frac{kct}{\sqrt{2}m})[/tex]
where c=[tex]mv_o[/tex]
