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Abhisar

  • one year ago

A particle moves in a potential region given by \(\sf U=8x^2-4x+400\) J. Its state of equilibrium will be

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  1. Abhisar
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    we have to request that the subsequent condition holds: \[\Large \frac{{\partial U}}{{\partial x}} = 0\]

  3. Abhisar
    • one year ago
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    Oh, so you mean we have to solve the equation for u=0?

  4. Michele_Laino
    • one year ago
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    not for U=0, its first derivative with respect to x has to be equal to zero, since, in a field of force coming from a potential, the relationship between force and potential energy is: \[\Large {\mathbf{F}} = - \nabla U\]

  5. Abhisar
    • one year ago
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    Oh, one min....

  6. Abhisar
    • one year ago
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    Ok, yes. Thanks a bunch c:

  7. Michele_Laino
    • one year ago
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    thus we get the subsequent condition: \[\Large {x_0} = \frac{1}{4}\] as equilibrium position

  8. Michele_Laino
    • one year ago
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    :)

  9. Abhisar
    • one year ago
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    Yes... c:

  10. Michele_Laino
    • one year ago
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    :)

  11. UnkleRhaukus
    • one year ago
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    now, the sign of the second derivative at this point, will determine whether this equilibrium point is stable or unstable

  12. Abhisar
    • one year ago
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    Oh I see, thanks for the info Felix c:

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