A graph of potential energy V (x) verses x is shown in Fig. 6.12. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.
i.) Kinetic energy vs x
We know that,
At C, P.E =0,
∴ 
Till point D there is no change in P.E, ∴ the K.E remains constant
At F, 
∴ 
From B to C the potential decreases so the kinetic energy increases ∴ we get inverted curve of potential.
From C to D potential remains constant so the kinetic energy will also remain same and thus we get a straight line parallel to x- axis
From D to F the potential energy increases to Eo linearly so the kinetic energy must fall linearly to Eo.
∴ we get the following plot,
ii.) Velocity vs X
Now, 
-----(2)
From B to C, from above equation, velocity increases parabolically symmetric from x-axis.
From C to D there is no change in P.E energy so the velocity remains constant.
From D to F, the P.E increases linearly and thus the velocity decreases parabolically from equation (2)
Now since we get two values of velocities from equation (2), we will get a mirror image of the graph in the negative side of the fourth quadrant too.
∴ we obtain following the graph,
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
