Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

we are given a problem with spring with its own mass $m$. I am confused how to set up the PE term in the Lagrangian.

Assume the spring has length of $L_{0}$ when it is laying on a table horizontally. Now when we hold the spring upright and hang it vertically from a pivot in the ceiling to make a pendulum, it will stretch a little due to its own weight by $\frac{(m/3)g}{k}$ where $k$ is the spring constant. I use $m/3$ there since that is the common factor used to generate an effective mass used for a spring with its own weight.

Hence now the length of the spring is $L_{0}+\frac{(m/3)g}{k}$ before we even put the bob mass at the end of it. Should one use $L_{0}+\frac{(m/3)g}{k}$ as the relaxed length of the spring or use $L_{0}?$

Will the PE for the spring be $\frac{1}{2}k\left( y-L_{0}\right) ^{2}$ or $\frac{1}{2}k\left( y-(L_{0}+\frac{(m/3)g}{k})\right) ^{2}$?

Assuming one is measuring $y$ here from the ceiling where the spring is hanging from

                 \     |
                 /     |
                 \     |
                 /     | y(t)
                 \     |
                 /     |
                 M     V

Which term should be used for PE?

I know there will be additional PE term for the spring due to its own mass, which is $-mg\frac{y}{2}\cos\theta$

Hence total PE from the spring is

$PE=\frac{1}{2}k\left( y-L_{0}\right) ^{2}-mg\frac{y}{2}\cos\theta -Mgy\cos\theta$


$PE=\frac{1}{2}k\left( y-(L_{0}+\frac{(m/3)g}{k})\right) ^{2}-mg\frac{y} {2}\cos\theta-Mgy\cos\theta $

where $\theta$ is angle of rotation of the spring pendulum.

not sure which to use.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.