- #1

Another

- 104

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- 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##

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?