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Finding K in a two spring system.

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    I did a lab experiment this past week and not sure how to determine k for a two spring system. We had an object on a frictionless surface and attached one spring on each side. I was able able to determine the spring constant for both springs individually but not sure how to determine the k for both springs acting together.



    2. Relevant equations



    3. The attempt at a solution

    Was thinking force= kx so k for both springs could I do force= K1x-K2x. K1 is spring constant for spring on the left. X is positive since the spring is elongating. K2 is the spring attached on the left side of the glider. X is negative because the spring is compression. Am I on the right path?
     
  2. jcsd
  3. Oct 16, 2011 #2
    To be more clear. I have the value for k for each spring and both springs working together. I curious in finding how the two individual spring constants relate to the spring constant of the two spring system.
     
  4. Oct 16, 2011 #3
    They should add. What values of k did you get for each situation?
     
  5. Oct 16, 2011 #4
    Makes sense the K are added. Think I over thought it . I can imagine the two springs as one spring once they are connected to the glider. The x wouldnt be negative for K2 since it one spring. X would just be the distance the glider moves from equilibrium. force=(k1+k2)*x

    for two springs system I got k= 7.9000
    left spring k= 3.655
    right spring k= 3.828

    So if I add it together it is roughly close to where I should be. The springs were pretty crappy so it was tough to get a constant k with them the more you used them. Please let me know if this thought pattern is correct.
     
  6. Oct 16, 2011 #5
    Yeah I mean if you pull the object one direction it is pushed by one spring in one direction and and pulled in that same direction by the other spring.
     
  7. Oct 16, 2011 #6
    That makes sense. Thanks for the clarification
     
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