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SHM Spring System is Independent of Gravity?

  1. Apr 3, 2015 #1
    For a mass-spring system,
    Period, T = 2pi * root(m/k)

    So using hookes law,
    F = kx
    But if the spring is being stretched by a mass due to gravity,
    mg = kx
    So,
    k = mg/x

    But then this means,
    Period, T = 2pi * root(mx / mg)
    or,
    T = 2pi * root(x / g)

    Where have I gone wrong? I've been told countless times that a spring-mass system's period is independent of g, but it seems my proof states otherwise.

    Thanks
     
  2. jcsd
  3. Apr 3, 2015 #2

    TSny

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    Hlello, GameJammer. Welcome to PF!

    Just to make sure, what is the precise meaning of x in this equation? Is it an arbitrary value of x or some specific value of x?
     
  4. Apr 3, 2015 #3
    x is the extension of the spring, it's part of Hookes law
     
  5. Apr 3, 2015 #4

    TSny

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    Yes. But it's the extension of the spring under what condition?
     
  6. Apr 3, 2015 #5
    It's the extension of the string due to force F
     
  7. Apr 3, 2015 #6

    TSny

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    Does x in your equation k = mg/x represent how much the spring is stretched if you just hang the mass on the spring and let it sit in equilibrium? Or does it represent something else?
     
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