Oscillating spring experiment uncertainty

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  • #1
maxim07
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Homework Statement:
I am trying to calculate the uncertainty in the spring constant calculated by timing the period of an oscillating spring. The period is T. Plotting a graph if T^2 against m means that the uncertainty in the spring constant is the uncertainty in the gradient. To plot the error bars I need to calculate the uncertainty in T^2.

In the experiment I recorded 10T, repeat 10T, mean 10T, T, T^2

Uncertainty in mean 10T = 1/2 range / mean

In order to find the uncertainty in T do I just divide the absolute uncertainty by 10, or do I need to calculate the % uncertainty in mean 10T then divide this by 10?

Then do I double the % uncertainty in T and use it to calculate the absolute uncertainty in T^2. Or do I calculate the absolute uncertainty in T then double it for the uncertainty in T^2?
Relevant Equations:
Uncertainty = 1/2 range /mean
I know how to do the individual uncertainties but just not sure of the order.
 

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  • #2
BvU
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Hi,

In order to find the uncertainty in T do I just divide the absolute uncertainty by 10, or do I need to calculate the % uncertainty in mean 10T then divide this by 10?
Is there a difference ? In the first case you get the absolute uncertainty in T, in the second case you get the relative uncertainty in 10 T, which is the same as the relative uncertainty in T (since the relative uncertainty in 10 is zero).
 
  • #3
maxim07
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So if both of them are the same, do any make calculating uncertainty in T^2 easier. Do you need to double the relative uncertainty in T to get the relative uncertainty in T^2 then convert this to the absolute uncertainty. Or is it the same thing to just double the absolute uncertainty in T to get the absolute uncertainty in T^2.

basic is the relative uncertainty necessary at any point, as I would prefer not to have to calculate this and just use absolute uncertainty in each step as there would be no need for converting between relative and absolute.
 
  • #4
hutchphd
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You should use the relative uncertainties. Using the absolute uncertainties is possible but the way you describe it is not correct (think about the units) . In my experience the relative uncertainties are much easier and more useful.
 
  • #5
maxim07
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So you think i should calculate the absolute uncertainty in the mean 10T, then convert this to a % uncertainty, then divide it by 10, for the % uncertainty in T, then multiply that by 2 for the % uncertainty in T^2 then convert this into an absolute uncertainty for error bars on a graph
 
  • #6
BvU
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Or is it the same thing to just double the absolute uncertainty in T to get the absolute uncertainty in T^2.
No it's not !
I don't know where you are in your curriculum, but if you know about differentiation things are fairly straightforward: the rule is that $$\Biggl ( \Delta f (x)\Biggr)^2 = \left (df\over dx\right )^2 (\Delta x) ^2 $$
so ## \Delta T^2 = 2T\Delta T ## or -- in terms of relative errors -- ## \quad \displaystyle {{\Delta T^2 \over T^2}= 2{\Delta T\over T} }##

And if you don't know about differentiation: to first order $$(T+\Delta T)^2 = T^2 + 2\Delta T + {\mathcal O} (\Delta T)^2 $$

##\ ##
 
  • #7
maxim07
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I’m at alevel, I just need to know what the correct method is as it doesn’t tell me in my textbook.
 
  • #8
BvU
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I'm sure you are at a level, but for me as a foreigener that doesn't tell me much :smile: .
There's a table of error propagation formulae here

From that you should e.g. be able to reduce a measurement of 10 T = 2.3 s ##\pm## 0.2 s to T = 0.23 s ##\pm## 0.02 s and then T2 = 0.053 ##\pm## 0.005 s2.

[edit] Oops o:) ! Someone should have corrected me ! ##\Delta T/T = 0.09## so ##\Delta T^2/T^2 = 0.17## and​
then T2 = 0.053 ##\pm## 0.009 s2.​

If you have a series of observations of T2 versus m, you make a plot with error bars.
And if the errors in T2 are not too wildly different you might assume they are the same and try to draw a best fit straight line. If they are not : In principle the weights of the individual points are ##1/\sigma_i^2## and there are expressions for slope and error in slope. (from this thread).

##\ ##
 
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  • #9
hutchphd
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I’m at alevel, I just need to know what the correct method is as it doesn’t tell me in my textbook.

So you think i should calculate the absolute uncertainty in the mean 10T, then convert this to a % uncertainty, then divide it by 10, for the % uncertainty in T, then multiply that by 2 for the % uncertainty in T^2 then convert this into an absolute uncertainty for error bars on a graph
If you ask a question, and a definitive answer is given, please do not ask me if I really really mean the answer.
It is different if there is a misunderstanding of some kind but yes I really really mean every answer.
Yikes.
 
  • #10
JoeyBob
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Im confused by the statement of an "absolute certainty" here. Dont all uncertainties have a certain confidence to them?

Ex. A 99% confidence interval will provide a much wider uncertainty than a 95% confidence interval? Isnt the uncertainty also dependent on the confidence interval you pick?

Absolute certainty seems to imply, at least to me, a 100% confidence interval. I don't understand how that's possible unless the interval has infinite range.
 
  • #11
hutchphd
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There is no absolute certainty. Engineering is the art of doing the best we can.
 
  • #12
JoeyBob
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There is no absolute certainty. Engineering is the art of doing the best we can.
Then why is the term used?
 
  • #13
haruspex
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Uncertainty in mean 10T = 1/2 range / mean
Please explain this more.

If you timed one set of 10 oscillations for each value of m, what does the range refer to?

If you took repeated measures of ten oscillations for each at the same mass, no, the uncertainty in the mean of the time for ten oscillations is not half range over mean.
If you did N sets of ten oscillations, with times t1 to tN, the uncertainty in the mean of those is (sample) standard deviation divided by sqrt N.
Note that using range/mean will give a value that tends to increase with sample size instead of decreasing.
 
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  • #14
maxim07
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If you ask a question, and a definitive answer is given, please do not ask me if I really really mean the answer.
I was just trying to ensure I had understood what you meant.

I took 2 sets of readings for 10T At the same mass. At A level I am only supposed to know the most basic ways of calculating uncertainty, so I am not supposed to use standard deviation, only 1/2 range /mean.
 
  • #15
haruspex
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I was just trying to ensure I had understood what you meant.

I took 2 sets of readings for 10T At the same mass. At A level I am only supposed to know the most basic ways of calculating uncertainty, so I am not supposed to use standard deviation, only 1/2 range /mean.
In the special case of only two measurements, it turns out that the standard method and the half range approximation give the same answer.
Since you are dividing the half range by the mean you already have the fractional uncertainty, not the absolute uncertainty. The fractional uncertainty in T is the same as the fractional uncertainty in 10T,
 
  • #16
BvU
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Then why is the term used?

The term is used to distinguish uncertainty in a variable (a quantity with the same dimension as the variable itself) from relative uncertainty in a variable (a dimensionless quantity). So not in the meaning 'exact' as you seem to have concluded.

Even exact sciences have to communicate in some language that can easily be misunderstood.

##\ ##
 

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