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Successive Approximation for Pendulum

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the pendulum equation

    [tex]\sin\frac{\theta}{2}=\sin\frac{\kappa}{2}\sin\phi=a\sin\phi[/tex]

    where

    [tex]\theta[/tex] oscillates between the limits [tex]\pm\kappa[/tex],


    [tex]\phi[/tex] is a new variable which runs from 0 to [tex]2\pi[/tex] for one cycle of [tex]\theta[/tex],

    and

    [tex]\phi+\frac{1}{8}a^2(2\phi-\sin2\phi)+... = {(\frac{g}{l})}^{\frac{1}{2}}t[/tex]

    Show, using successive approximations twice that

    [tex]\theta\doteq(\kappa+\frac{{\kappa}^3}{192})\sin\omega't+\frac{{\kappa}^3}{192}\sin3\omega't[/tex]

    where

    [tex]\omega'=(\frac{g}{l})^{\frac{1}{2}}(1-\frac{\kappa^2}{16}+...)[/tex]


    2. Relevant equations

    [tex]\sin{x} = x\mbox{ - }\frac{x^3}{3!}\mbox{ + }\frac{x^5}{5!}\mbox{ - ...}[/tex]

    [tex]\sin^3{\theta}=\frac{3\sin\theta-\sin3\theta}{4}[/tex]


    3. The attempt at a solution

    I found [tex]\omega'[/tex] by successive approximations of [tex]\phi[/tex]. For [tex]\theta[/tex], using the sine expansion to third order;

    [tex]\theta \mbox{ - } \frac{{\theta}^3}{24}=\mbox{2a}\sin\phi[/tex]

    and

    [tex]\theta=\mbox{2a}\sin\phi+\frac{{\theta}^3}{24}[/tex]

    The first order (in a) approximation is then

    [tex]\theta=\mbox{2a}\sin\phi[/tex]

    Substituting into the expanded expression for theta gives the first approximation:

    [tex]\theta=\mbox{2a}\sin\phi+\frac{{\mbox{(2a}\sin\phi})^3}{24}[/tex]

    This first appoximation is substituted into the expression for theta giving a second approximation. I have done so retaining sine terms up to the 3rd harmonic (the highest power of sine to reduce was to the nineth power) and sine expansion of kappa to third order, and I do not arrive at the desired answer. Is my approach correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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