Calculating Error in s for Free Fall Lab Experiment

[quicknote]

I am measuring the position of photogates in a free fall lab experiment.

The question is:
What is the error in s in terms of reading error? Now add in the error in the metal scale, a precision of one part in 4000, in quadrature to give an expression for the total error in s.

My reading error is +/- 0.02cm.
I'm confused about 'adding in the error in the metal scale, 1 part in 4000'.
Does this mean 1/4000 = 0.00025?

If so then would my total error would be $$\sqrt{0.00025^2 + 0.02^2}$$ =0.02 ... I don't think this makes much sense


Any help would be greatly appreciated.
Thanks!

 

[andrevdh]

One calculates an error in a value by taking into account the errors in the variables that the particular value depends on. So my question is do the time measurement depend on the measurement on the metal scale? If so what are the smallest divisions on the metal scale? It is most likely that the error in this variable is so small then that it has virtually no contribution to the error in the time measurement anyway. If the time value do not depend on the measurement on the metal scale it could not have any effect on the error in the time value and should not be included in the error calculation.

 

[quicknote]

I understand your confusion regarding the calculation of error in your free fall lab experiment. Let me explain the process in detail to help you better understand the concept.

Firstly, the reading error of +/- 0.02cm is the uncertainty in your measurement of position using the photogates. This means that the actual position could be anywhere within a range of +/- 0.02cm from the recorded value.

Now, for the metal scale, the precision of one part in 4000 means that the scale is divided into 4000 equal parts, and the smallest division represents 1/4000 of the total length. This translates to an uncertainty of 0.00025cm in each measurement.

To calculate the total error, we need to combine these two sources of error using the principle of quadrature. This means taking the square root of the sum of squares of the individual errors. In this case, the total error would be:

\sqrt{0.00025^2 + 0.02^2} = 0.0200000625cm

This may seem like a small value, but it is important to consider the relative magnitude of the error in relation to your measurements. In this case, the total error is still within the range of +/- 0.02cm, which is significant when compared to the scale of your measurements.

I hope this explanation helps you understand the concept of calculating error in your free fall lab experiment. It is important to always consider and account for sources of error in scientific experiments to ensure the accuracy and reliability of your results.