Find the equation of motion of a particle moving along theaxis if the potentialenergy is$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$. (This is a simple harmonic oscillator.)

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 the potential energy is$\mathrm{V}=\frac{1}{2}{\mathrm{kx}}^2$

Since in 1-dimensional system, the kinetic energy is trivial.

Then the kinetic energy will become:

$T=\frac{1}{2}m{\dot{x}}^2,$

And the potential energy will have the form:

$\mathrm{V}=\frac{1}{2}{\mathrm{kx}}^2$

According to the definition of the Lagrangian,

$$\begin{gathered} L=T-V \\ =\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2 \end{gathered}$$

So, the lagrangian is$L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2$

Since the lagrangian has only one degrees of freedom

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

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

Then the required derivative will be:

$$\begin{gathered} \frac{\partial \mathrm{L}}{\partial \dot{\mathrm{x}}}=\mathrm{m}\dot{\mathrm{x}} \\ \frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{x}}}=\mathrm{m}\ddot{\mathrm{x}} \\ \frac{\partial \mathrm{L}}{\partial \mathrm{x}}=-\mathrm{kx} \end{gathered}$$

Then the Euler equation will be:

$m\ddot{x}+kx=0$

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

$\ddot{x}+\frac{k}{m}x=0$

So, the equation of motion is$\ddot{x}+\frac{k}{m}x=0.$