# Non-linear pendulum (Whittaker's treatise on analytical dynamics of particles)

Reading through Whittaker's "treatise on the analytical dynamics of particles and rigid bodies" I have a question regarding his analysis of the simple nonlinear pendulum at chapter IV. At some point he says that the equation of energy is:

$$a \dot \theta^{2} = 2g \cos\theta + \rm{constant} = - 4 g \sin^{2} \left( \frac{\theta}{2} \right) + \rm{constant}$$

Where $a$ is length of the pendulum or radius of the circle. Then he goes on and says "Suppose that when the particle is at the lowest point of the circle, the quantity $\frac{a^{2}\dot\theta^{2}}{2g}$ has the value $h$. Then this last equation can be written

$$a^{2} \dot \theta^{2} = 2gh - 4ga \sin^{2}\left(\frac{\theta}{2} \right)$$

I can't follow how he derived that last equation. It's important 'cause the analysis of the types of movements basically depends on $h$ being less, equal or greather than $2a$. Can someone clarify the issue?

He just computed the value of the constant based on a boundary condition.

\begin{align}a \dot \theta^{2} &= 2g cos \theta + \rm{C} \\ &= - 4 g \sin^{2} \left( \frac{\theta}{2} \right) + \rm{C}\end{align}\tag1

Multiply by $\frac{a}{2g}$:

$$\frac{a^2 \dot \theta^{2}}{2g} = - 2a \sin^{2} \left( \frac{\theta}{2} \right) + \frac{aC}{2g} = h$$

At the bottom, $\theta=0$ so the first term on the right is zero. This gives us an expression for $C$:

$$C = \frac{2gh}{a}$$

And substituting that into your earlier equation $(1)$ and multiplying by $a$ gives the expression you were looking for:

$$a^2\dot\theta^2 = 2gh - 2ga\sin^2\left(\frac{\theta}{2}\right)$$

• Thank you Floris! I managed to get the expression I was looking for. It was simpler than I thought. Just might add that your last equation should be: $a^{2} \dot\theta^{2} = 2gh - 4ga \sin^{2} \left( \frac{\theta}{2} \right)$ – Richard Costa Oct 20 '15 at 17:58