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# 8.3: Applications of nonlinear systems

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Figure 8.7: Various possibilities for the motion of the pendulum.

The quantity

$\frac{1}{2} \omega^2 - \frac{g}{L} \cos \theta$

is conserved by any solution. This is the energy or the Hamiltonian of the system.

We have a conservative equation and so (exercise) the trajectories are given by

$\omega = \pm \sqrt{ \frac{2g}{L} \cos \theta + C} ,$

for various values of $$C$$. Let us look at the initial condition of $$(\theta_0,0)$$, that is, we take the pendulum to angle $$\theta_0$$, and just let it go (initial angular velocity 0). We plug the initial conditions into the above and solve for $$C$$ to obtain

$C = - \frac{2g}{L} \cos \theta_0 .$

Thus the expression for the trajectory is

$\omega = \pm \sqrt{ \frac{2g}{L}} \sqrt{ \cos \theta - \cos \theta_0 } .$

Let us figure out the period. That is, the time it takes for the pendulum to swing back and forth. We notice that the oscillation about the origin in the phase plane is symmetric about both the $$\theta$$ and the $$\omega$$ axis. That is, in terms of $$\theta$$, the time it takes from $$\theta_0$$ to $$-\theta_0$$ is the same as it takes from $$-\theta_0$$ back to $$\theta_0$$. Furthermore, the time it takes from $$-\theta_0$$ to $$0$$ is the same as to go from $$0$$ to $$\theta_0$$. Therefore, let us find how long it takes for the pendulum to go from angle 0 to angle $$\theta_0$$, which is a quarter of the full oscillation and then multiply by 4.

We figure out this time by finding $$\frac{dt}{d\theta}$$ and integrating from $$0$$ to $$\omega_0$$. The period is four times this integral. Let us stay in the region where $$\omega$$ is positive. Since $$\omega = \frac{d\theta}{dt}$$, inverting we get

$\frac{dt}{d\theta} = \sqrt{\frac{L}{2g}} \frac{1}{\sqrt{\cos \theta - \cos \theta_0 }} .$

Therefore the period $$T$$ is given by

$T = 4 \sqrt{\frac{L}{2g}} \int_0^{\theta_0} \frac{1}{\sqrt{\cos \theta - \cos \theta_0 }}\, d\theta .$

The integral is an improper integral, and we cannot in general evaluate it symbolically. We must resort to numerical approximation if we want to compute a particular $$T$$.

Recall from Ch. 2.4, the linearized equation $$\theta''+\frac{g}{L}\theta = 0$$ has period

$T_{\text{linear}} = 2\pi \sqrt{\frac{L}{g}} .$

We plot $$T$$, $$T_{\text{linear}}$$, and the relative error $$\frac{T-T_{\text{linear}}}{T}$$ in Figure 8.8. The relative error says how far is our approximation from the real period percentage-wise. Note that $$T_{\text{linear}}$$ is simply a constant, it does not change with the initial angle $$\theta_0$$. The actual period $$T$$ gets larger and larger as $$\theta_0$$ gets larger. Notice how the relative error is small when $$\theta_0$$ is small. It is still only $$15\%$$ when $$\theta_0 = \frac{\pi}{2}$$, that is, a 90 degree angle. The error is $$3.8\%$$ when starting at $$\frac{\pi}{4}$$, a 45 degree angle. At a 5 degree initial angle, the error is only $$0.048 \%$$.

Figure 8.8: The plot of $$T$$ and $$T_{\text{linear}}$$ with $$\frac{g}{L} = 1$$ (left), and the plot of the relative error $$\frac{T-T_{\text{linear}}}{T}$$ (right), for $$\theta_0$$ between 0 and $$\pi/2$$.

While it is not immediately obvious from the formula, it is true that

$\lim_{\theta_0 \uparrow \pi} T = \infty .$

That is, the period goes to infinity as the initial angle approaches the unstable equilibrium point. So if we put the pendulum almost upside down it may take a very long time before it gets down. This is consistent with the limiting behavior, where the exactly upside down pendulum never makes an oscillation, so we could think of that as infinite period.