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

(N/A) The total mechanical energy $E$ of the particle is conserved and is given by $E = K + V(x)$,where $K$ is the kinetic energy and $V(x)$ is the potential energy. Given $E = E_0$,we have $K = E_0 - V(x)$.
$1$. Kinetic Energy $(K)$ versus $x$ graph:
- At points $A$ and $F$,$V(x) = E_0$,so $K = E_0 - E_0 = 0$.
- Between $C$ and $D$,$V(x) = 0$,so $K = E_0$ (maximum kinetic energy).
- At point $B$,$V(x) > 0$,so $K = E_0 - V(x) < E_0$.
- The graph of $K$ versus $x$ will be the mirror image of the potential energy graph shifted upwards by $E_0$.
$2$. Velocity $(v)$ versus $x$ graph:
- Since $K = \frac{1}{2} m v^2$,we have $v = \pm \sqrt{\frac{2K}{m}}$.
- At $A$ and $F$,$K = 0$,so $v = 0$.
- At $C$ and $D$,$K$ is maximum,so $v$ is maximum $(v = \pm \sqrt{\frac{2E_0}{m}})$.
- The velocity graph will show both positive and negative values for $v$ corresponding to the particle moving in opposite directions.