# 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?

## Answers and Replies

Homework Helper
Gold Member
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.