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# Evaluating a Torsional Pendulum experiment

Extracts from this document...

Introduction

Evaluation:

I will firstly work out the overall experimental error and how far it was from the true value, using the same formula used in the preliminary.

=2π  = 10.36

Therefore the total error from what the true value should be is [(11.368-10.36)/11.368] x 100= 8.89%

This shows that my experimental results had an overall 8.89% error, where as in my preliminary I had an error of 15.89%, therefore I believe my improvements have improved the accuracy of my results.

From the 2 graphs above I can see that the result for 0.1 meter length seems to be the furthest away from the line of best fit, and may be considered as an anomalous result, however I don’t think it’s necessary to remove this result. The reason for this error could be any of the ones stated below, or possibly as it was the first reading I took, there could have been an initial fault in my experiment set up.

Even though I have improved the accuracy of my experiment there are still many errors which will have decreased the accuracy of my results. I will now state each one and estimate percentage errors for the reading error and also experimental error if possible.

• The meter ruler is accurate to ±0.5mm, therefore error on the smallest length would be (0.5/100)x100=0.5% and largest length (0.5/500)x100=0.1% . Therefore the error here can be no greater than 0.5%, so this is not a very significant error. However there is also a large span for experimental error, the length of string may not have been fully straight due to not being stretched fully, and also every time I change the length of the wire there will be a new random error generated. These can’t be avoided but overall these experimental errors may have been about ±0.3cm , meaning the maximum error would be (3/100)x100=3% error, which is therefore very significant.
• The micrometer is accurate to ±0.005mm, therefore the error on my diameter of 0.49mm was (0.005/0.49)x100=1.02%, this shows a reduced error that of the preliminary, however a 1% error on the diameter can still be a major factor. This is due to the fact that the diameter is raised to the power of 4 in the equation. Therefore a very small change in the diameter may cause a larger than expected change in time period. Therefore I think the error of the diameter may have been the most significant error. If the diameter had been 0.48mm then the percentage error calculated above would have been only 5%, this shows how significant it was. The experimental error is also a factor due to the fact I had to twist two wires together to make a larger diameter. After taking 5 readings of the diameter, which were 0.49, 0.49, 0.48,0.49,0.47mm. I decided to use 0.49 as my value being the mode, however the fact that the diameter varied slightly meant there was an error. The range was 0.02mm, this could therefore have caused an error (0.02/0.49)x100=4.08%, therefore also very significant. Also the fact that I twisted two wires together, after some use, parts of the wire may have untwisted meaning the diameter would change again, this again contributes to the error above.

Middle

• The value for shear modulus I used was 44.7x109GPa, however when doing research for this value, there were more than one of the same value, so there is no guarantee that the value I used was the value of my copper wire. The following website gave me a range of 40-47GPa.

http://www.efunda.com/materials/common_matl/common_matl.cfm?MatlPhase=Solid&MatlProp=Mechanical However, as the shear modulus is so large the error will  be so small. It’s difficult to work out the percentage error, therefore my error is just a range of 40-47GPa

• The scale is accurate to ±0.05 grams so error on my bar was (0.05/196.3)=0.0254% error, there is no real experimental error in this reading.

The percentage errors above show that the overall error should have decreased, where time period is now a very small error, reduced from about 8% in the preliminary.

From my log log graph I got 0.4532 as my gradient. However theoretically it should have been 0.5. I also found that if I exclude the 0.1meter length and time period from the log log graph then my gradient would change from 0.4532 to 0.4963, which is very close to 0.5. This again shows that the 0.1m length may be considered as an anomalous result.

Conclusion

Improvements to final method

If I was to perform this experiment again I would try to further decrease the reading and experimental errors in the following ways.

• As I found diameter to be the largest error I would ensure that the wire I am using has a constant diameter, by using only one wire and ensuring it has not been stretched in any way before using it. I would then also measure the diameter of the wire at least 5 times for each length, as when the length is decreased the weight will be pulling down on a wire of shorter length, and may stretch the wire more. Therefore I will record the diameter for each length I do and if it changes take these new diameters into consideration.
• The shear modulus of copper ranged from 40-47GPa, therefore I was unable to even come up with an actual error for this. Therefore to reduce the error to almost zero I would measure the actual shear modulus of the copper wire I am using to do the experiment. This can be done using the following formulae  G = E / [2(1+ν)]
where G is the shear modulus, E is the tensile modulus, and ν is the Poisson's ratio of the material.
http://www.ides.com/property_descriptions/ISO537.asp

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