Let us go through all of the steps, once more. We'll choose our coordinate x to be the displacement from the equilibrium length of the spring, as shown on the diagram. The kinetic energy of the moving mass is then just

`T=\frac{m\dot{x}^2}{2}`

. The potential of a spring stretched (or compressed) x metres from its equilibrium length is given by `V=\frac{kx^2}{2}`

, where k is the spring constant (N m^-1).So, for our mass-spring system, the Lagrangian is

L = \frac{1}{2}\left(m\dot{x}^2 - kx^2\right)}

Applying the Euler-Lagrange equation, we obtain

\begin{multiline*}

-kx - \frac{d}{dt}\left(m\dot{x}\right) = 0 \\

\Rightarrow \ddot{x} - \frac{k}{m}x = 0

\end{multiline*}

Comparing the last line above with the general form of Simple Harmonic Motion (SHM),

`\ddot{q} - \omega^2 q = 0`

we can see that the equation of motion rendered by applying the Euler-Lagrange equation to the Lagrangian of a mass-spring system provides Simple Harmonic Motion with a frequency

\omega = \sqrt{\frac{k}{m}}

This is as expected for the case of a mass-spring system!

It is interesting to notice how different approaches imply different assumptions: when treating the mass-spring system in a Newtonian framework, you start by Hooke's Law of elasticity, while here you have to `make up' a potential energy term.

ReplyDeleteKeep up the good work!

(Oh and you forgot the right bracket when writing down the Lagrangian)

You don't make up the potential. You could treat it as also starting from Hooke's law: The force F = -kx is represented by the gradient of a potential, i.e. F = -dV/dx

ReplyDeleteTherefore the potential V = 1/2 kx^2 as used in the post, by direct integration of Hooke's law!

Bracket added, thanks for noticing.

From the point of view of the analysis, be it Newtonian or Lagrangian, Hooke's Law is `made up' too. Our knowledge of Hooke's Law is a result of forces being somehow more easily measurable than potentials.

ReplyDeleteSo imagine, for a moment, that you knew nothing about Newton's Second Law and forces, but had only ever used the Euler-Lagrange equation to solve problems in mechanics. Then, at least until you `discovered' how to derive Newton's Second Law from the Euler-Lagrange equation, you wouldn't be able to solve the simple spring-mass system unless you `made up' a potential energy term. But if you know nothing about forces, then clearly you don't know about the relationship between a force and a potential, so even if you knew that x'' = -kx for a spring (notice I am not mentioning forces) there would be no way to obtain a potential term, and you would have to look somewhere else!