A graph of potential energy $V(x)$ verses $x$ is shown in figure. A particle of energy $E_0$ is executing motion in it. Draw graph of velocity and kinetic energy verses $x$ for one complete cycle $AFA$.

$KE$ versus $x$ graph :

We know that

Total $\mathrm{E}=\mathrm{K}+\mathrm{U}$

$\begin{array}{ll}\mathrm{E}=\mathrm{K}+\mathrm{V}(x) & (\because \mathrm{U}=\mathrm{V}(x)) \\ \mathrm{K}=\mathrm{E}_{0}-\mathrm{V}(x) & \left(\because \mathrm{E}=\mathrm{E}_{0}\right)\end{array}$

at $\mathrm{A}, x=0, \mathrm{~V}(x)=\mathrm{E}_{0}$

$\therefore \quad \mathrm{K}=\mathrm{E}_{0}-\mathrm{E}_{0}=0$

at $\mathrm{B}, \mathrm{V}(x)<\mathrm{E}_{0}$

$\begin{array}{ll}\therefore & \mathrm{K}>0 \\ \text { At } & \mathrm{C} \text { and } \mathrm{D}, \mathrm{V}(x)=0\end{array}$ (Positive)

(Maximum)

$\mathrm{K}=0$

At $F$

The variation is as below,

Velocity versus $x$ graph :

As $\mathrm{KE}=\frac{1}{2} m v^{2}$

$\therefore$ At $\mathrm{A}$ and $\mathrm{F}$, where $\mathrm{KE}=0, \nu=0$. At $\mathrm{C}$ and $\mathrm{D}, \mathrm{KE}$ is maximum. Therefore, $v$ is $\pm \max$. At B, KE is positive but not maximum Therefore, $v$ is $\pm$ some value $(<\max .)$