# Differential Equations

## Second Order Equations # 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:

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

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

${\frac{{{d^2}\alpha }}{{d{t^2}}} + \frac{g}{L}\alpha = 0\;\;}\kern-0.3pt {\text{or}\;\;\frac{{{d^2}\alpha }}{{d{t^2}}} + {\omega ^2}\alpha = 0,}$

where $$\omega = \sqrt {\large\frac{g}{L}\normalsize}$$ 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

${\alpha \left( {t = 0} \right) = {\alpha _0},\;\;\;}\kern-0.3pt{\frac{{d\alpha }}{{dt}}\left( {t = 0} \right) = 0.}$

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 $$\large\frac{{d\alpha }}{{dt}}\normalsize.$$ This leads to the equation

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

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:

${{{\left( {\frac{{d\alpha }}{{dt}}} \right)^2} }={ \frac{{4g}}{L} \cdot}\kern0pt{ \left( {{{\sin }^2}\frac{{{\alpha _0}}}{2} – {{\sin }^2}\frac{\alpha }{2}} \right),\;\;}}\Rightarrow {{\frac{{d\alpha }}{{dt}} }={ 2\sqrt {\frac{g}{L}} \cdot}\kern0pt{ \sqrt {{{\sin }^2}\frac{{{\alpha _0}}}{2} – {{\sin }^2}\frac{\alpha }{2}} .}}$

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 $$\sin {\large\frac{{{\alpha _0}}}{2}\normalsize} = k$$ 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 $$\large\frac{T}{4}\normalsize.$$ It follows from the relationship between the angles $$\alpha$$ and $$\theta$$ that $$\sin \theta = 1$$ or $$\theta = {\large\frac{\pi}{2}\normalsize}$$ at $$\alpha = {\alpha_0}.$$ Therefore, we obtain the following expression for the period of oscillation of the pendulum:

${{\sqrt {\frac{g}{L}} \frac{T}{4} }={ \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\frac{{d\theta }}{{\sqrt {1 – {k^2}\,{{\sin }^2}\theta } }}} \;\;}}\kern-0.3pt {{\text{or}\;\;T = 4\sqrt {\frac{L}{g}} \cdot}\kern0pt{ \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\frac{{d\theta }}{{\sqrt {1 – {k^2}\,{{\sin }^2}\theta } }}} .}}$

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:

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

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:

${K\left( k \right) }={ \frac{\pi }{2}\left\{ {1 + {{\left( {\frac{1}{2}} \right)}^2}{k^2} }\right.}+{\left.{ {{\left( {\frac{{1 \cdot 3}}{{2 \cdot 4}}} \right)}^2}{k^4} }\right.}+{\left.{ {{\left( {\frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}} \right)}^2}{k^6} + \ldots } \right.}\kern0pt {\left. {+ {{\left[ {\frac{{\left( {2n – 1} \right)!!}}{{\left( {2n} \right)!!}}} \right]}^2}{k^{2n}} + \ldots } \right\},}$

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 $$K\left( k \right) \approx {\large\frac{\pi }{2}\normalsize},$$ 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 $${\alpha_0}.$$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Estimate the error in the calculation of the oscillation period of a simple pendulum at different amplitude $${\alpha_0}$$ when using the standard formula $${T_0} = 2\pi \sqrt {\large\frac{L}{g}\normalsize}.$$
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Concept
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Problem 1