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Hooke's law, again

  1. Jan 13, 2010 #1
    If I'm given a set of data such as...

    Mass(kg) Spring Length(cm)
    0.0 15.7
    1.0 16.5
    2.0 17.8
    3.0 19.3
    and so on...

    How do I determine whether or not the spring is obeying Hooke's law?
    I'm not sure what k is equal too.
    And is 15.7 the equilibrium?
     
  2. jcsd
  3. Jan 13, 2010 #2

    Born2bwire

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    Science Advisor
    Gold Member

    F = kx. If the force is zero, then the displacement should be zero, so you should take 15.7 cm as your equilibrium position. From there, it is simply seeing how well the force scales linearly with displacement. The force here being provided by the attached masses.
     
  4. Jan 13, 2010 #3
    You are hanging the mass from the spring vertically? If so, the equation is

    [tex]F=k(y_{0}-y)-mg[/tex]

    Where y_{0} is the point of equilibrium. g= 9.8 m/sec^2 (gravitational constant), m is the mass
    , and y is the measured length of the spring.

    If the force is zero, i.e., the mass is in equilibrium,
    [tex]k(y_{0}-y)-mg=0[/tex]

    Plug in the possible values for m and and y and if the equations are consistent(with y_{0}, and k being the unknowns), then hook's law is obeyed.
     
  5. Jan 14, 2010 #4
    From the data, draw a graph of mass of spring (horizontal axis) against extension (vertical).
    If it is a straight line then Hooke's Law is obeyed. The gradient is equal to the elastic constant, k.
     
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