Spring launcher equation

  • #26
How did you calculate x0 anyway? I did some more tests with hanging weights, I now I have more detailed information about the spring.
Deformation (cm) | Force (N)
======================
2.3 | 5.39
3.9 | 8.33
4.7 | 10.29
7.2 | 14.7
8.0 | 15.19
11.0 | 20.58
13.9 | 25.48
16.5 | 30.87
Surprisingly, the spring obeys hooke's law quite well! I must've done a miscalculation in the last post, because one of the data points was quite off from the line of best fit.
I calculated the new k value, for the same spring, to be around 175 N/m. But how did you calculate x0? I tried to calculate x0 by dividing the x-intercept of the line by the k value, from which I got approximately 0.91 cm.
 
  • #27
collinsmark
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How did you calculate x0 anyway? I did some more tests with hanging weights, I now I have more detailed information about the spring.
Deformation (cm) | Force (N)
======================
2.3 | 5.39
3.9 | 8.33
4.7 | 10.29
7.2 | 14.7
8.0 | 15.19
11.0 | 20.58
13.9 | 25.48
16.5 | 30.87
Surprisingly, the spring obeys hooke's law quite well! I must've done a miscalculation in the last post, because one of the data points was quite off from the line of best fit.
I calculated the new k value, for the same spring, to be around 175 N/m. But how did you calculate x0? I tried to calculate x0 by dividing the x-intercept of the line by the k value, from which I got approximately 0.91 cm.
The assumption is that x = 0 is not quite at the equilibrium point of the spring. The equilibrium point of the spring is somewhere behind x = 0. Somewhere not accessible. The spring collapses at some point distance a little greater than the equilibrium point.

Which gives you
F = k(x + x0) = kx + kx0.
Now think about the equation for a line,
y = mx + b
You can solve for m and b as long as you have two points for (x,y). So go ahead and plug in two points and you can solve for k and x0.
 

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