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Spring (with mass) kinetic energy -- velocity assumption

  1. Sep 7, 2015 #1
    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. jcsd
  3. Sep 7, 2015 #2

    mfb

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    2016 Award

    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.
     
  4. Sep 8, 2015 #3
    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
     
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