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A graph of potential energy V(x) verses ...

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.

Text Solution

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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.
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Knowledge Check

  • The potential energy function for a particle executing linear SHM is given by V(x)= 1/2kx^2 where k is the force constant of the oscillator. For k = 0.5 Nm^(-1) , the graph of V(x) versus x is shown in the figure A particle of total energy E turns back when it reaches x=pmx_m .if V and K indicate the potential energy and kinetic energy respectively of the particle at x= +x_m ,then which of the following is correct?

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    B
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    momentum
    D
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