# Nonlinear Pendulum

## Differential Equation of Oscillations

Pendulum is an ideal model in which the material point of mass $$m$$ is suspended on a weightless and inextensible string of length $$L.$$ In this system, there are periodic oscillations, which can be regarded as a rotation of the pendulum about the axis $$O$$ (Figure $$1$$).

Dynamics of rotational motion is described by the differential equation

$\varepsilon = \frac{{{d^2}\alpha }}{{d{t^2}}} = \frac{M}{I},$

where $$\varepsilon$$ is the angular acceleration, $$M$$ is the moment of the force that causes the rotation, $$I$$ is the moment of inertia about the axis of rotation.

In our case, the torque is determined by the projection of the force of gravity on the tangential direction, that is

$M = - mgL\sin \alpha .$

The minus sign indicates that at a positive angle of rotation $$\alpha$$ (counterclockwise), the torque of the forces causes rotation in the opposite direction.

The moment of inertia of the pendulum is given by

$I = m{L^2}.$

Then the dynamics equation takes the form:

$\frac{{{d^2}\alpha }}{{d{t^2}}} = \frac{{ - \cancel{m}g\cancel{L}\sin \alpha }}{{\cancel{m}{L^\cancel{2}}}} = - \frac{{g\sin \alpha }}{L},\;\; \Rightarrow \frac{{{d^2}\alpha }}{{d{t^2}}} + \frac{g}{L}\sin \alpha = 0.$

In the case of small oscillations, one can set $$\sin \alpha \approx \alpha.$$ As a result, we have a linear differential equation

$\frac{{{d^2}\alpha }}{{d{t^2}}} + \frac{g}{L}\alpha = 0\;\; \text{or}\;\;\frac{{{d^2}\alpha }}{{d{t^2}}} + {\omega ^2}\alpha = 0,$

where $$\omega = \sqrt {\frac{g}{L}}$$ is the angular frequency of oscillation.

The period of small oscillations is described by the well-known formula

$T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{L}{g}} .$

However, with increasing amplitude, the linear equation ceases to be valid. In this case, the correct description of the oscillating system implies solving the original nonlinear differential equation.

## Period of Oscillation of a Nonlinear Pendulum

Suppose that the pendulum is described by the nonlinear second order differential equation

$\frac{{{d^2}\alpha }}{{d{t^2}}} + \frac{g}{L}\sin \alpha = 0.$

We consider the oscillations under the following initial conditions

$\alpha \left( {t = 0} \right) = {\alpha _0},\;\;\; \frac{{d\alpha }}{{dt}}\left( {t = 0} \right) = 0.$

The angle $${\alpha _0}$$ is the amplitude of oscillation.

The order of the equation can be reduced, if we find a suitable integrating factor. Multiply this equation by the integrating factor $$\frac{{d\alpha }}{{dt}}.$$ This leads to the equation

$\frac{{{d^2}\alpha }}{{d{t^2}}}\frac{{d\alpha }}{{dt}} + \frac{g}{L}\sin \alpha \frac{{d\alpha }}{{dt}} = 0,\;\; \Rightarrow \frac{d}{{dt}}\left[ {\frac{1}{2}{{\left( {\frac{{d\alpha }}{{dt}}} \right)}^2} - \frac{g}{L}\cos\alpha } \right] = 0.$

After integration we obtain the first order differential equation:

${\left( {\frac{{d\alpha }}{{dt}}} \right)^2} - \frac{{2g}}{L}\cos\alpha = C.$

Given the initial conditions, we find the constant $$C:$$

$C = - \frac{{2g}}{L}\cos{\alpha _0}.$

Then the equation becomes:

${\left( {\frac{{d\alpha }}{{dt}}} \right)^2} = \frac{{2g}}{L}\left( {\cos\alpha - \cos{\alpha _0}} \right).$

Next, we apply the double angle identity

$\cos\alpha = 1 - 2\,{\sin ^2}\frac{\alpha }{2},$

which leads to the following differential equation:

${\left( {\frac{{d\alpha }}{{dt}}} \right)^2} = \frac{{4g}}{L} \left( {{{\sin }^2}\frac{{{\alpha _0}}}{2} - {{\sin }^2}\frac{\alpha }{2}} \right),\;\; \Rightarrow \frac{{d\alpha }}{{dt}} = 2\sqrt {\frac{g}{L}} \sqrt {{{\sin }^2}\frac{{{\alpha _0}}}{2} - {{\sin }^2}\frac{\alpha }{2}} .$

Integrating this equation, we obtain

$\int {\frac{{d\left( {\frac{\alpha }{2}} \right)}}{{\sqrt {{{\sin }^2}\frac{{{\alpha _0}}}{2} - {{\sin }^2}\frac{\alpha }{2}} }}} = \sqrt {\frac{g}{L}} \int {dt} .$

We denote $$\sin {\frac{{{\alpha _0}}}{2}} = k$$ and introduce the new variable $$\theta$$ instead of the angle $$\alpha:$$

$\sin \frac{\alpha }{2} = \sin \frac{{{\alpha _0}}}{2}\sin \theta = k\sin \theta .$

Then

$d\left( {\sin \frac{\alpha }{2}} \right) = \cos \frac{\alpha }{2}d\left( {\frac{\alpha }{2}} \right) = \sqrt {1 - {{\sin }^2}\frac{\alpha }{2}} d\left( {\frac{\alpha }{2}} \right) = \sqrt {1 - {k^2}\,{{\sin }^2}\theta } \,d\left( {\frac{\alpha }{2}} \right) = k\cos \theta d\theta .$

It follows that

$d\left( {\frac{\alpha }{2}} \right) = \frac{{k\cos \theta d\theta }}{{\sqrt {1 - {k^2}\,{{\sin }^2}\theta } }}.$

In the new notation, our equation can be written as

$\int {\frac{{\cancel{k\cos \theta} d\theta }}{{\sqrt {1 - {k^2}\,{{\sin }^2}\theta }\,\cancel{k\cos \theta} }}} = \sqrt {\frac{g}{L}} \int {dt} ,\;\; \Rightarrow \int {\frac{{d\theta }}{{\sqrt {1 - {k^2}\,{{\sin }^2}\theta } }}} = \sqrt {\frac{g}{L}} \int {dt} .$

Next, we discuss the limits of integration. The passage of the arc from the lowest point $$\alpha = 0$$ to the maximum deviation $$\alpha = {\alpha_0}$$ corresponds to a quarter of the oscillation period $$\frac{T}{4}.$$ It follows from the relationship between the angles $$\alpha$$ and $$\theta$$ that $$\sin \theta = 1$$ or $$\theta = {\frac{\pi}{2}}$$ at $$\alpha = {\alpha_0}.$$ Therefore, we obtain the following expression for the period of oscillation of the pendulum:

$\sqrt {\frac{g}{L}} \frac{T}{4} = \int\limits_0^{\frac{\pi }{2}} {\frac{{d\theta }}{{\sqrt {1 - {k^2}\,{{\sin }^2}\theta } }}} \;\; \text{or}\;\;T = 4\sqrt {\frac{L}{g}} \int\limits_0^{\frac{\pi }{2}} {\frac{{d\theta }}{{\sqrt {1 - {k^2}\,{{\sin }^2}\theta } }}} .$

The integral on the right cannot be expressed in terms of elementary functions. It is the so-called complete elliptic integral of the $$1$$st kind:

$K\left( k \right) = \int\limits_0^{\frac{\pi }{2}} {\frac{{d\theta }}{{\sqrt {1 - {k^2}\,{{\sin }^2}\theta } }}} .$

The function $$K\left( k \right)$$ is computed in most mathematical packages. Its graph is shown below in Figure $$2$$.

The function $$K\left( k \right)$$ can also be represented as a power series:

$K\left( k \right) = \frac{\pi }{2}\left\{ {1 + {{\left( {\frac{1}{2}} \right)}^2}{k^2} + {{\left( {\frac{{1 \cdot 3}}{{2 \cdot 4}}} \right)}^2}{k^4} + {{\left( {\frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}} \right)}^2}{k^6} + \ldots + {{\left[ {\frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}} \right]}^2}{k^{2n}} + \ldots } \right\},$

where the double factorials $${\left( {2n - 1} \right)!!}$$ and $${\left( {2n} \right)!!}$$ denote the product, respectively, of odd and even natural numbers.

Note that if we restrict ourselves to the zero term of the expansion, assuming that $$K\left( k \right) \approx {\frac{\pi }{2}},$$ we obtain the known formula for the period of small oscillations:

${T_0} = 4\sqrt {\frac{L}{g}} K\left( k \right) \approx 4\sqrt {\frac{L}{g}} \frac{\pi }{2} = 2\pi \sqrt {\frac{L}{g}} .$

Further terms of the series for $$n \ge 1$$ are just allow to consider the anharmonicity of the oscillations of the pendulum and the nonlinear dependence of the period $$T$$ on the oscillation amplitude $${\alpha_0}.$$

See solved problems on Page 2.