Harmonic oscillator

Lagrangian Mechanics viewpoint

It is a particle moving in a potential

u (x) = \frac{k}{2} x^{2}

In Lagrangian Mechanics approach we consider the Lagrangian

L = T - U = \frac{1}{2} m {\dot{x}}^{2} - \frac{1}{2} k x^{2}

Euler-Lagrange tell us

\frac{\partial L}{\partial x} = \frac{d}{d t} \frac{\partial L}{\partial \dot{x}}

\frac{\partial L}{\partial x} = - k x

\frac{d}{d t} \frac{\partial L}{\partial \dot{x}} = \frac{d}{d t} (m \dot{x}) = m \ddot{x}

and so

m \ddot{x} = - k x

Hamiltonian mechanics viewpoint

The Hamiltonian for a simple harmonic oscillator is given by:

H = \frac{p^{2}}{2 m} + \frac{1}{2} m ω^{2} q^{2}

where $p$ is the momentum, $m$ is the mass, $ω$ is the angular frequency, and $q$ is the position.
The corresponding Hamiltonian equations are:

\dot{p} = - \frac{\partial H}{\partial q} = - m ω^{2} q

and

\dot{q} = \frac{\partial H}{\partial p} = \frac{p}{m}

These equations describe the time evolution of the momentum and position of the harmonic oscillator. The first equation is essentially Newton's second law (force equals mass times acceleration), and the second equation is the definition of velocity in terms of momentum.

Energy

The total mechanical energy is:

E = \frac{1}{2} m {\dot{x}}^{2} + \frac{1}{2} k x^{2}

Differentiation with respect to time shows conservation:

\frac{d E}{d t} = \dot{x} (m \ddot{x} + k x)

Using the equation of motion ( $m \ddot{x} = - k x$ ), we see that $\frac{d E}{d t} = 0$ .
Since $E$ is constant, we can evaluate it at the turning points where $x = A$ and $\dot{x} = 0$ :

E = \frac{1}{2} k A^{2}