Differential Equations

2nd Order Equations

Nonlinear Pendulum

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Problem 1

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 \(O\) (Fig. \(1\)).

Dynamics of rotational motion is described by the differential equation

\[\varepsilon = \frac{{{d^2}\alpha }}{{d{t^2}}} = \frac{M}{I},\]
Forces acting on a pendulum

Figure 1.

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, i.e.

\[M = – mgL\sin \alpha .\]

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\;\;}\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 \(\alpha:\)

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


{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.

Complete elliptic integral of the first kind

Figure 2.

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 on problem description to see 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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Problem 1