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Lab Report Format

  • Thread starter catie1981
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Maybe someone can help explain this a little for me. In my lab reports, I am to show qualification of error and then quantify one of those values. Obviously, in some of hte labs, only certain causes of error can be reasonable shown. For example, on the angular velocity lab, I can say that the accuracy of the scale I used was only 0.1g, which could account for 2% of the error. However, I don't know how to show that it accounts for 2% of the error, nor what I am to do with that (how to re-calculate that information) once I have obtained it. If anyone can help, but maybe needs more explaination, I would most certainly be willing to help you help me!! Thanks in advance!!!
Catie
 

Answers and Replies

Andy Resnick
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Error analysis is a cornerstone of experimental science, and it's worthwhile understanding what you need to do.

There's a great little book by Taylor, "An Introduction to Error Analysis". It has a trainwreck on the cover.

A simple formula, derived in the book, describes how to calculate the total error in a measurement, if each of the different errors are independent from each other. Say you are calculating the kinetic energy: K = 1/2 m v^2. You independently measure the mass and velocity. You could take it a step further, if you measure the mass, distance, and elapsed time independently as well: K = 1/2 m (d/t)^2. In any case, the total error is given by:

[tex]\delta K = \sqrt{(\frac{\partial K}{\partial m} \delta m)^2+(\frac{\partial K}{\partial v} \delta v)^2}[/tex]

Does that help?
 
jtbell
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I can say that the accuracy of the scale I used was only 0.1g, which could account for 2% of the error. However, I don't know how to show that it accounts for 2% of the error,
The most direct way to do this is to re-calculate your result (the angular velocity) after changing your measured value for the mass by 0.1g, while leaving all the other measured numbers constant. If this causes your result to change by 2%, then you would say that the 0.1g uncertainty in the mass contributes 2% to the uncertainty in the result.

To continue with this, suppose you also have to measure the radius of something, and the accuracy of your ruler is 0.1 mm. To find out how much this contributes to the uncertainty of your result, you would proceed similarly: re-calculate the result after changing the radius by 0.1 mm, while leaving all the other measured numbers at their original values. (Use the original value of the mass, not the changed one you used in the previous paragraph.) Let's suppose you find that this changes the result by 1%. Then you would say that the uncertainty in the radius contributes 1% to the uncertainty in the result.

In order to find the total contribution to the uncertainty in the result, from these two sources, you would calculate

[tex]\sqrt{(2)^2 + (1)^2} = 2.2[/tex]

(all of these are percentages)
 
Last edited:
45
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Wow, thanks for the help, both of you....I think that will definitely get me going on my way to what the prof is looking for in our reports. Thanks!
 

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