Question about Lagrangian density

[Another]

Homework Statement

Find
1. the Lagrangian density of massive spring system
2. the energy density of this system

Relevant Equations

divide massive spring to \rho = \frac{m}{L} = \frac{dm}{dy}
velocity of each small spring $dm$ in $v_i = \frac{V}{L} y_i$

this figure form ( https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system) )

massive spring ; m
K.E. of total spring equal to $K.E. = \frac{1}{2} \sum dm_i v_i^2 = \frac{1}{2} \sum \rho dy (Vy/L)^2$

V is the speed at the end of the spring and V are same speed of mass M at the end of spring, SO
$K.E. = \frac{1}{2} \rho \frac{V^2}{L^2} \int_{0}^{L} y^2 dy = \frac{1}{2} \frac{m}{3} V^2$
Potential of the spring when divided by $dm$
$P.E. = \sum dm_i g y_i$ at equilibrium $dm g = k dy_i$
$P.E. = \sum dm_i g y_i = \sum k y_i dy_i = \int_{0}^{L} k y dy$

P.E. of mass M at the end of spring is $P.E. = -MgL$ I think at any y $P.E.$ may be written $P.E. = -Mg dy$
K.E. of mass M $K.E. = \frac{1}{2} M V^2$

The lagrangian of this system is L = K.E. - P.E.
and The lagrangian density is L therefore ## L = \int L dy##

from above I can write K.E. and P.E. of massive and P.E. o mass M at the end of massive spring in term dy but i can't write K.E. of mass M in term dy How can i solve this problem?

 

[TSny]

Hello. For me, the problem statement does not clearly describe the system. Did you give the complete wording of the problem as it was given to you? Were you given a figure similar to the one that you posted from Wikipedia? Are you sure that the system in your problem is the same as the system in the Wikipedia article?

I ask this because a massive spring (without a lump of mass on the end) still makes an interesting dynamical system. Also, the problem statement does not mention that the system is hanging vertically in a gravitational field.