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Mathematical Modelling

  1. Dec 28, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m moves along the positive x-axis with a potential energy given by


    where C is a postive constant.

    Calulate the equilibrium position X0 of the particle.

    2. Relevant equations


    3. The attempt at a solution

    I realise that at the equilibruim position x=0. However the C part has thrwon me, am i supposed to integrate or differentiate this??? Thanks.
  2. jcsd
  3. Dec 28, 2007 #2

    Andrew Mason

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    How is potential energy related to force? How is force related to equilibrium position?

    [Hint: what is the force in terms of the rate of change of potential energy with respect to distance?]

  4. Dec 28, 2007 #3
    Potential energy is the rate of change of potential energy. Therefore I have to differentiate the potential energy, to my mind this comes to: F=10/4x.
    However if I set this equal to 0 (which I though was the condition for equilibrium) x is equal to zero. Is this correct?
  5. Dec 28, 2007 #4


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    That's a pretty strange looking derivative. How did you get that??
  6. Dec 28, 2007 #5

    D H

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    I think you might have typed a bit too fast here. That doesn't make any sense.

    Rate of change with respect to what? Usually when physicists say "rate of change" they are talking about differentiating with respect to time. That is not the correct relationship between potential energy and force. A very important tool that physicists use is "dimensional analysis". This can be of some help here.

    The kinetic energy of some object is [itex]T=1/2mv^2[/itex]. Velocity has units length/time, so in terms of basic units, kinetic energy has units mass*length2/time2, or [itex]ML^2/T^2[/itex] for short. (We throw away unitless constants like 1/2 when doing dimensional analysis). Potential energy is just another form of energy; it has the same units as kinetic energy: [itex]E \approx ML^2/T^2[/itex]. By Newton's second law, [itex]F=ma[/itex]. In terms of dimensions then, force has units [itex]F \approx ML/T^2[/itex]. The only difference between energy and force in terms of dimensions is that energy has extra factor of length compared to force. This is a time for one of those ah ah! moments: "Ah ah! Now I remember: force is the spacial derivative of potential energy."

    Dimensional analysis is an extremely powerful tool. You can use it as a sanity check on your answers (units should match up). If you are very smart, you can use it to estimate the yield of an atomic bomb just from a couple of photos of an A bomb explosion.

    That's just wrong. Please show your work here and we'll help you get to the right answer.
  7. Dec 28, 2007 #6

    Andrew Mason

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    Is your function:

    [tex]V(x) = C + \frac{4}{x^2} + x^2[/tex] ?

    If so, this is equivalent to:

    [tex]V(x) = C + 4x^{-2} + x^2[/tex]

    Differentiate V(x) with respect to x.

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