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Spring Constant Coursework

  1. Feb 10, 2015 #1
    • Warnining given Re: use of template in homework sections.
    Find the spring constant by timing simple harmonic motion. I have done a Hooke's law experiment already and found out about initial tension. What I learned was:
    force ≠ spring constant x displacement
    force = initial tension + (spring constant x displacement)

    So using F=kx would not give an accurate value for k, it would over-estimate it by including initial tension. My question is would the equation T=2π√(m/k) give the 'true' value of k or would it give the over-estimated value? The two are so close I can't tell which my results are showing.
     
  2. jcsd
  3. Feb 10, 2015 #2

    gneill

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    Draw a free-body diagram for the mass at the equilibrium position. Oscillation occur around the equilibrium position. Now displace the mass by a small amount Δx. What's the net restoring force that the mass feels?
     
  4. Feb 10, 2015 #3
    I'm not too sure what you mean, would it be equal to kx?
     
  5. Feb 10, 2015 #4

    gneill

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    Simple Harmonic Motion occurs when a displacement from equilibrium results in a force that want to push the object back to the equilibrium position.

    Draw a free body diagram for the case that you are having trouble with, say a mass hanging at the end of a spring so that when at rest, the spring is stretched to some new equilibrium location.

    Draw the free body diagram for the forces when the mass is at rest. What's the net force on the mass?

    Draw the free body diagram for the forces when the block is displaced by some small amount Δx. What's the net force on the mass? How does it compare to that of a spring/mass system that is not "pre-stretched"?
     
  6. Feb 10, 2015 #5
    When it's at equilibrium, there's no net force. Gravity and tension are equal. If you stretch it, then you provide a force in the direction of gravity and tension increases equally so that when you let go of the spring the initial force is equal to change in tension and proportionate to displacement from equilibrium. Whereas in a non-stretched system, you first have to overcome initial tension. So would T=2π√(m/k) give the lower value for spring constant where initial tension is not included? Am I along the right lines at least?
     
  7. Feb 10, 2015 #6

    gneill

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    The spring constant remains the same always. It's a constant!

    The point is that for small displacements around an equilibrium position, the resulting restoring force is the same in both cases. Thus the differential equation describing the motion is the same, and the resulting expression for the period is the same.
     
  8. Feb 10, 2015 #7
    I get that it's a constant its just that if you don't know about initial tension and use F=kx to work it out, you will get a slightly high value, not the true value. I just can't get my head round whether simple harmonic motion would give you the slightly high value or the real one. Think of a graph of extension against load, you'd have a flat bit until you overcome initial tension, then a straight line. The gradient of that line is the spring constant yeah? But if during your oscillation the spring is moving up and compressing and extension dips into that flat region of the graph, surely then initial tension starts messing with your results for spring constant?
     
  9. Feb 10, 2015 #8

    gneill

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    When you write the equation of motion for the mass you don't use the spring extension x (the total length of the spring at any time). You use the displacement from equilibrium. The equations all hold together just fine when you do that and no corrections for initial loading or pre-stretching are required.
     
  10. Feb 10, 2015 #9
    That makes sense for a Hooke's law but I'm still not seeing it for harmonic motion. I guess I'll have to accept it anyway though, thanks for your help
     
  11. Feb 10, 2015 #10

    gneill

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    Well, SHM results from writing the equation of motion for a mass/spring system that obeys Hooke's law.
    Sorry I can't think of a better way to show that the initial spring tension does not affect the resulting SHM frequency.
     
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