Calculate the percentage difference

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SUMMARY

The discussion focuses on calculating the percentage difference between two experimental values of the spring constant (k). The user initially applied the equations for discrepancy and error but encountered an unrealistic result of 1,500,000. The correct approach involves using the formula for percentage difference, defined as p = 100 × |k_o - k_e| / k_o, where k_o is the average of the two experimental values. The user concluded that statistical methods, such as calculating the standard deviation and average, are essential for accurate comparisons.

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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}
 
Last edited:
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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