Question

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 versus x for one complete cycle AFA.

Answer 1

We know that Total ME=KE+PE
`Rightarrow E_(0)=KE+V(x)` ltbRgt `Rightarrow KE=E_(0)-V(x)`
at `A_(1) x=0,V(x)=E_(0)`
`Rightarrow KE=E_(0)-E_(0)=0`
at `B_(1) V(x) gt E_(0)`
`Rightarrow KE gt0`
at C and `D_(1) V(x)=0`
`Rightarrow KE is maximum at `F_(1)V(x)=E_(0)`
Hence, KE=0
Hence, KE=0 The variation is shown in adjacent diagram.
Velocity versus x graph
As `KE=(1)/(2)mv^(2)`

`therefore` At A and F, where KE=0, v=0
At C and D, KE is maximum. Therefore, v is `pm` max.
At B, KE is positive but not maximum
Therefore, `v is pm "some value"`
The variation is shown in the diagram.