# Nonlinear Pendulum

## Solved Problems

### 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 ${T_0} = 2\pi \sqrt {\frac{L}{g}}.$

Solution.

We use the solution of the nonlinear equation of the pendulum, in which the expression for the period $$T$$ is represented as a series. Taking into account the term $$n = 1,$$ the formula $$T\left( {{\alpha _0}} \right)$$ looks like

${T_1}\left( {{\alpha _0}} \right) = 4\sqrt {\frac{L}{g}} K\left( k \right) = 4\sqrt {\frac{L}{g}} K\left( {\sin {\alpha _0}} \right) = 4\sqrt {\frac{L}{g}} \left[ {\frac{\pi }{2}\left( {1 + \frac{1}{4}{{\sin }^2}\frac{{{\alpha _0}}}{2}} \right)} \right] = {T_0}\left( {1 + \frac{1}{4}{{\sin }^2}\frac{{{\alpha _0}}}{2}} \right),$

where $${T_0}$$ is the period of oscillation, calculated by the standard formula

${T_0} = 2\pi \sqrt {\frac{L}{g}} .$

Thus, the term $${{\frac{1}{4}} {{\sin }^2}{\frac{{{\alpha _0}}}{2}}}$$ immediately shows the deviation from the standard formula (expressed as a decimal), depending on the angle $${\alpha_0}.$$

Similarly, we take into account the contribution of the following terms of the series for $$n = 2$$ and $$n = 3.$$ The corresponding formulas have the form:

${T_2}\left( {{\alpha _0}} \right) = {T_0}\left( {1 + \frac{1}{4}{{\sin }^2}\frac{{{\alpha _0}}}{2} + \frac{9}{{64}}{{\sin }^4}\frac{{{\alpha _0}}}{2}} \right),$
${T_3}\left( {{\alpha _0}} \right) = {T_0}\left( {1 + \frac{1}{4}{{\sin }^2}\frac{{{\alpha _0}}}{2} + \frac{9}{{64}}{{\sin }^4}\frac{{{\alpha _0}}}{2} + \frac{{225}}{{2304}}{{\sin }^6}\frac{{{\alpha _0}}}{2}} \right).$

The graphs presented in Figure $$3$$ show the value of the expression in the square brackets for the functions $${T_1}$$ and $${T_2}$$ (in percentage). In fact, they give the error in determining the period of oscillation when using the standard formula $${T_0}$$ compared with more accurate approximations.

It is seen that the power series converges well, and in the range of angles up to $${\alpha_0} = 20^{\circ}$$ it is possible to restrict the series expansion by the first term $$n = 1$$ to ensure the accuracy rate of about $$1\%.$$