Successive Approximation for Pendulum

[chrisk]

Homework Statement

Given the pendulum equation

$$\sin\frac{\theta}{2}=\sin\frac{\kappa}{2}\sin\phi=a\sin\phi$$

where

$$\theta$$ oscillates between the limits $$\pm\kappa$$,


$$\phi$$ is a new variable which runs from 0 to $$2\pi$$ for one cycle of $$\theta$$,

and

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

Show, using successive approximations twice that

$$\theta\doteq(\kappa+\frac{{\kappa}^3}{192})\sin\omega't+\frac{{\kappa}^3}{192}\sin3\omega't$$

where

$$\omega'=(\frac{g}{l})^{\frac{1}{2}}(1-\frac{\kappa^2}{16}+...)$$

Homework Equations

$$\sin{x} = x\mbox{ - }\frac{x^3}{3!}\mbox{ + }\frac{x^5}{5!}\mbox{ - ...}$$

$$\sin^3{\theta}=\frac{3\sin\theta-\sin3\theta}{4}$$

The Attempt at a Solution

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

$$\theta \mbox{ - } \frac{{\theta}^3}{24}=\mbox{2a}\sin\phi$$

and

$$\theta=\mbox{2a}\sin\phi+\frac{{\theta}^3}{24}$$

The first order (in a) approximation is then

$$\theta=\mbox{2a}\sin\phi$$

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

$$\theta=\mbox{2a}\sin\phi+\frac{{\mbox{(2a}\sin\phi})^3}{24}$$

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?

 

[Jhero]

Your approach seems to be on the right track, but there are a few issues with your calculations. Firstly, in your first approximation for theta, you have used the sine expansion to the third order, but then in your second approximation, you have only retained terms up to the 3rd harmonic. This could lead to errors in your final result.

Another issue is that you have not taken into account the limits of \theta in your calculations. Remember that \theta oscillates between \pm\kappa, so you will need to consider the sine expansion for both positive and negative values of \theta.

Finally, it seems like you have made a mistake in your substitution for the first approximation of theta. When substituting \theta=\mbox{2a}\sin\phi into the expanded expression for theta, you should get:

\theta=\mbox{2a}\sin\phi+\frac{{(\mbox{2a}\sin\phi)}^3}{24}+\frac{{(\mbox{2a}\sin\phi)}^5}{24}+...

Hopefully, these tips will help you arrive at the correct answer. Keep in mind that successive approximations can be a tricky concept, so don't get discouraged if it takes a few attempts to get the correct result. Good luck!