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Approximation to simple harmonic motion.

  1. Oct 26, 2007 #1
    [SOLVED] Approximation to simple harmonic motion.

    1. The problem statement, all variables and given/known data
    A small mass [tex]m[/tex], which carries a charge [tex]q[/tex], is constrained to move vertically inside a narrow, frictionless cylinder. At the bottom of the cylinder is a point mass of charge [tex]Q[/tex] having the same sign as [tex]q[/tex]. Show that if the mass [tex]m[/tex] is displaced by a small amount from its equilibrium position and released, it will exhibit simple harmonic motion with angular frequency
    [tex]\omega = \sqrt{\frac{2g}{y_0}}[/tex]
    where [tex]y_0[/tex] is the height, when the mass is in equilibrium.

    2. Relevant equations

    Simple harmonic motion:

    Coulomb's Law(one dimension, same sign of charges):
    [tex]F= \frac{kq_1q_2}{y^2}[/tex]

    Newton's Second Law:

    3. The attempt at a solution

    Force acting on the mass [tex]m[/tex] at a height [tex]y[/tex] is:
    [tex]F=\frac{kqQ}{y^2} - mg[/tex]

    I found the potential energy by:
    [tex] U=-\int F dy= \frac{kqQ}{y} + mgy [/tex]

    I found the equilibrium point:
    [tex] mg = \frac{kqQ}{y_0^2} [/tex]
    [tex] y_0 = \sqrt{\frac{kqQ}{mg}} [/tex]

    Near (stable) equilibrium I can approximate the potential energy by some parabola with equation:
    [tex]U_n = U(y_0) + B(y - y_0)^2 = 2\sqrt{kqQmg} + B(y - \sqrt{\frac{kQq}{mg}})^2[/tex]
    where [tex]B[/tex] is some (positive) constant.

    Than I can find the approximate force by taking the negative derivative of [tex]U_n[/tex]. That is
    [tex]F = -2B(y - y_0)[/tex]

    That is the same as simple harmonic motion and so we can say that
    [tex]\omega = \sqrt{\frac{2B}{m}}[/tex]

    Well my problem is that I have no idea how to find the [tex]B[/tex] or how to show that the [tex]B = \frac{mg}{y_0}[/tex]. The thing is I don't know how to mathematically describe that I want the parabola of the approximate potential energy function to be the most similar to the original function near equilibrium.

    I will really appreciate any help from you.
    Last edited: Oct 27, 2007
  2. jcsd
  3. Oct 26, 2007 #2


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    You said
    [tex] U(y) = \frac{kqQ}{y} + mgy [/tex]
    (which I didn't check, but I assume it is correct).

    I would consider making a Taylor expansion around the equilibrium point, i.e.
    [tex]U(y) = U(y_0) + U'(y_0)(y - y_0) + \frac12 U''(y_0) (y - y_0)^2 + \mathcal{O}((y - y_0)^3). [/tex]
    Since it's an equilibrium point, the linear term will vanish. Then approximate the potential by throwing away the (y - y0)^3 terms and you will have a parabola, from which you can easily read off the coefficient.

    I quickly tried this and it will give you the B you stated, so it should work. But there might be another way.
  4. Oct 27, 2007 #3
    Hi...I had never heard about Taylor's polynomials, but I looked it up, when I read your reply and it's a great stuff. Exactly what I needed and how you said - it works. So thanks a lot. I really appreciate it.
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