# Spring (with mass) kinetic energy -- velocity assumption

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1. Sep 7, 2015

### ELiT.Maxwell

why we assume that velocity decreases linearly in a spring (i.e. if one end is fixed, then velocity of a particle (of spring) at x from fixed =vx/l where v is the velocity of the free end) and why does it hold good too when the spring (linear mass density) is non uniform....

EDIT: spring has mass..
ref : https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system)

2. Sep 7, 2015

### Staff: Mentor

If every part of the spring compresses in the same way, that's the result.
Not all springs are uniform - if they are not, you might need a different relation between distance and speed.

3. Sep 8, 2015

### ELiT.Maxwell

by using spring constant inversely proportional to length and letting a element y from fixed end and Fnet=0 on it so,
K(L/y)x1=K(L/(L-y))x2
where L is the length of spring at any time t, now, y/x1=(L-y)/x2
x1+x2=x these are lim->0 inst changes and x1 = velocity of element and solving eqn , v=y/l(v of free end)

and, since we have not touched linear mass density (only used absolute lengths...) it works