Absolute Uncertainty of 1/d^2 Homework

[Taniaz]

Homework Statement

Given values and uncertainties of d, find the absolute uncertainty in 1/d^2 (see table attached)

Homework Equations

I know that if it's a square function the absolute uncertainty will be 2 (delta d) / d but I'm not sure if it's correct for an inverse function.

The Attempt at a Solution

I read somewhere that it might be 2 (deIta d)/ d^3? If I use this equation, I get the range of answers they gave in the marking scheme which are 0.03 till 1. If I use only 2(delta d)/d I get uncertainties from 0.02 till 0.07.

Thank you

 

[BvU]

I read somewhere

Good place. The rule is simple if you know how to differentiate:$$\big(\Delta f(x) \big )^2 = (f'(x) \Delta x)^2$$

Another way to get the same:

percent uncertainty in 1/d = percent uncertainty in d
percent uncertainty in dⁿ = | n | x percent uncertainty in d​

 

[Taniaz]

So if it was d^2 it would just be 2(uncertainty in d)/d
And for 1/d^2= 2(uncertainty in d)/d^3
From differentiation

 

[haruspex]

So if it was d^2 it would just be 2(uncertainty in d)/d

No. Δf(x)≈f'(x)Δx.
If f(x)=x², what is f'(x)?

 

[Taniaz]

Is there a difference between absolute and percentage uncertainty? Because all the A'level Physics questions I've done so far have mentioned that if there's a power you just multiply the power by (change in uncertainty/actual value) so if it's d^3 for example, it will be 3 (delta d)/d. Is this just for percentage uncertainty?

 

[haruspex]

Is there a difference between absolute and percentage uncertainty? Because all the A'level Physics questions I've done so far have mentioned that if there's a power you just multiply the power by (change in uncertainty/actual value) so if it's d^3 for example, it will be 3 (delta d)/d. Is this just for percentage uncertainty?

Yes.
If f(x)=xⁿ then f'(x)=nx^n-1, Δf(x)=f'(x)Δx=nx^n-1Δx; Δf(x)/f(x)=nΔx/x.
But it would be better to call that fractional uncertainty. To make it percentage you would multiply that by 100%.

 

[mjc123]

I read somewhere that it might be 2 (deIta d)/ d^3? If I use this equation, I get the range of answers they gave in the marking scheme which are 0.03 till 1. If I use only 2(delta d)/d I get uncertainties from 0.02 till 0.07.

The absolute uncertainty of a quantity must have the same dimensions as the quantity. The relative uncertainty is dimensionless. Which of your expressions seems more appropriate for the absolute uncertainty?

 

[Taniaz]

The absolute uncertainty of a quantity must have the same dimensions as the quantity. The relative uncertainty is dimensionless. Which of your expressions seems more appropriate for the absolute uncertainty?

2(delta d)/d^3 has units of 1/m^2 and 2(delta d)/d has no units so the former is the absolute uncertainty and the latter is the fractional/relative/percentage uncertainty

 

[Taniaz]

So to get the absolute uncertainty you simply do
Delta f(x)=f'(x) (delta x)
But for fractional/relative/percentage uncertainty you divide the above by f(x)
Correct?

 

[haruspex]

So to get the absolute uncertainty you simply do
Delta f(x)=f'(x) (delta x)
But for fractional/relative/percentage uncertainty you divide the above by f(x)
Correct?

Yes.

 

[Taniaz]

The fractional/percentage uncertainy for 1/d^2 is just 2(Δd)/d right?

Thanks

 

[haruspex]

The fractional/percentage uncertainy for 1/d^2 is just 2(Δd)/d right?

Thanks

Yes.

 

[Taniaz]

Thank you and sorry just going back to the basics but table doesn't make sense, especially the second and third row. How are they even getting these uncertainties? This is much more basic than what I'm doing so that's why I'm confused.

 

[haruspex]

Thank you and sorry just going back to the basics but table doesn't make sense, especially the second and third row. How are they even getting these uncertainties? This is much more basic than what I'm doing so that's why I'm confused.

I'd need to see the text leading to this, but it looks to me as though the table is taking into account human visual accuity. E.g. in the first row, with 1mm markings, it is reasonable to judge a measurement to the nearest mm, but no better; in the third row, even with only 10cm markings, it should be possible to judge to within 10% of the distance between markings.

 

[Taniaz]

There's actually nothing in the text that relates to the table. They just say here's an example of some uncertainties unfortunately!

 

[haruspex]

There's actually nothing in the text that relates to the table. They just say here's an example of some uncertainties unfortunately!

In that case I would read it as possible entries in a lab report. The experimenter is recording the value measured, the gradations of the measuring instrument used, and her estimate of the uncertainty. There is no suggestion that the uncertainty is calculated by some algorithm from the other two.
As I described in my preceding post, the entries do make sense.

 

[Taniaz]

Ok thank you.

In the picture I attached, they asked us to measure the length of the wire using a ruler and then asked for the percentage uncertainty. In the mark scheme it says
"Percentage uncertainty in L based on absolute uncertainty of 2mm to 8mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty."

How did they get an absolute uncertainty of 2mm to 8mm?

Thank you

 

[BvU]

Am I right in assuming this has nothing to do with the thread thus far ?

Anyway:

How did they get an absolute uncertainty of 2mm to 8mm?

Where exactly does the clamp hold the wire ? And where exactly at the other end is the mass effectively acting ? A millimeter uncertainty at each end for those is a fair guess; add another for measuring accuracy at each end and you are in the range of 2 to 4 mm. 8 mm seems a it pessimistic to me.

 

[haruspex]

8 mm seems a it pessimistic to me.

Agreed, but maybe they are allowing for a range of students' physical abilities... Poor vision, coordination... What I don't like is that apparently they would reject an absolute error less than 2mm. At each end, a skilled operator could claim an absolute error of 0.5mm, then combine them to get a statistical total of 0.75mm.

 

[Taniaz]

Thank you for your replies.

So it's all a guess at the end? There's no definite way to find out the absolute uncertainty like we could when given an equation?

 

[haruspex]

Thank you for your replies.

So it's all a guess at the end? There's no definite way to find out the absolute uncertainty like we could when given an equation?

The uncertainty in this case is something for the experimenter to judge, based on a number of considerations. How fine are the gradations on the ruler? How well trusted are those gradations? How comfortable does the experimenter feel about judging the nearest gradation, or maybe estimating a position relative to neighbouring gradations? What about parallax? ...

How such judged uncertainties of different measurements are combined to produce an overall uncertainty in the answer is where mathematical theory comes in.

 

[Taniaz]

Alright, thank you!