Calculate the percentage difference

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The discussion revolves around calculating the percentage difference between two experimental values of a spring constant, k. The user initially struggles with a large discrepancy in their results, obtaining an unrealistic value when applying their equations. They question whether the units used for k could be affecting their calculations, as their error values seem small compared to the large constant. The user realizes that without a known correct value for comparison, they can use the average of their two experimental values to compute the percentage difference. Ultimately, they conclude that a more statistically sound approach would involve calculating the standard deviation and average from a larger dataset.
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Homework Statement


This is related to my last post. Again, I am getting a ridiculous answer hoping someone can help. I need to calculate discrepancy between two calculated values.


Homework Equations



EQN 1 difference between values = Value 2 - Value 1
EQN 2 error of values = square root of (error value 1 ^2 + error value 2 ^2)
discrepancy = EQN 1/EQN2

The Attempt at a Solution


When I use EQN 1, I'm subtracting one value of spring constant k from another. The difference between my values is in the 3000 range. For EQN 2, my error value is in 2 x 10^-3.

When I divide 3000/2x10^-3 I get 1500000 ! The answer should be relatively close to 2!

Now, my previous post was regarding units for k. Could this be the problem? The error values I calculated though were based on the data given in cm, g, seconds squared, etc. Does this matter? It seems strange to me that my calculate K is say 14000.00 +/- 0.003 (for example). Small error for such a large number!
 
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Are'nt you suppose to calculate the percentage difference, p, between the two values? If the "correct" value is k_o and the experimental value is k_e

p = 100 \times \frac{|k_o - k_e|}{k_o}
 
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I don't have a correct value to compare to. I have two experimental values I'm comparing.
 
Well the formula will give the percentage difference (with the average) if you take half the difference between the two and use the average for the "correct value". It is not really sensible to compare measurements with each other this way though. Statistically one normally calculate the standard deviation and the average of a larger set of measurements.
 
Last edited:
Thanks! Figured it out :smile:
 
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