Use Lagrange’s equations to find the equation of motion of a simple pendulum. (SeeChapter $\mathbf{7}$, Problem$\mathbf{2}\mathbf{.}\mathbf{13}$.)

An ordinary first-order differential equation that is linear in the independent variable and unknown function but not solved for the derivative is termed as Lagrange's equation.

A function that equals the difference between potential and kinetic energy and characterises the state of a dynamic system in terms of position coordinates and their time derivatives is termed as Lagrangian.

Given a simple pendulum

The equilibrium position is denoted by polar coordinates with $\varphi =0$.

Now it is aconstraint that the length is fixed. This means that $r=l$where$l$is the length of the pendulum string and thus localid="1664362325356" $\dot{r}=0$

Then the kinetic energy will become:

And the potential energy will have the form:

$$\begin{gathered} V=mgh \\ =mgl\left(1-\cos \varphi \right) \end{gathered}$$

According to the definition of the Lagrangian,

$$\begin{gathered} L=T-V \\ =\frac{1}{2}m{l}^2{\dot{\varphi }}^2-mgl\left(1-\cos \varphi \right) \end{gathered}$$

So, the lagrangian is$L=\frac{1}{2}m{l}^2{\dot{\varphi }}^2-mgl\left(1-\cos \varphi \right)$

Since the lagrangian has only one degrees of freedom

The $\mathrm{\varphi }$degree of freedom, the Euler equation will be given by:

$\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}-\frac{\partial L}{\partial \varphi }=0$

Then the required derivative will be:

Then the Euler equation will be:

Divide the above Euler equation by $\mathrm{l}$, the result will be

$\ddot{\varphi }+\frac{g}{l}\sin \varphi =0$

So, the Lagrange’s equation is$\ddot{\varphi }+\frac{g}{l}\sin \varphi =0.$