Hooke's law, again

  • Thread starter audreylynn
  • Start date
  • #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?
 

Answers and Replies

  • #2
Born2bwire
Science Advisor
Gold Member
1,779
18
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.
 
  • #3
259
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.
 
  • #4
648
2
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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