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Question
A graph of potential energy V(x) verses x is shown in figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.
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Solution
KE versus x graph
We know that Total ME = KE + PE
⇒ E0 = KE + V(x)
⇒ KE = E0 – V(x)
At A1x = 0, V(x) = E0
⇒ KE = E0 – E0 = 0
At B1 V(x) < E0
⇒ KE > 0
At C and D1V(x) = 0
⇒ KE is maximum at F1V(x) = E0
Hence, KE = 0
The variation is shown in the adjacent diagram.
Velocity versus x graph
As KE = `1/2` mv2
∴ At A and F, where KE = 0, v = 0
At C and D, KE is maximum.
Therefore, v is ± max.
At B, KE is positive but not maximum.
Therefore, v is ± some value .....(< max)
The variation is shown in the diagram.
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