I get two different answers. Which one is correct?

  • Thread starter Thread starter carloz
  • Start date Start date
  • Tags Tags
    Uncertainty
AI Thread Summary
The discussion revolves around calculating the uncertainty in the expression M = (a-b)/2 + a, given specific values for a and b, along with their uncertainties. The initial attempt at a solution incorrectly applies the error propagation formula, questioning the independence of errors in a and b. A suggestion is made to simplify the expression to (3a-b)/2, allowing for a more straightforward application of error propagation rules without concern for the dependency of errors. Two methods for calculating the uncertainty in M yield different results, highlighting the complexity of error propagation in dependent variables. The conversation emphasizes the importance of correctly applying error propagation techniques in mathematical expressions.
carloz
Messages
4
Reaction score
0

Homework Statement



M = (a-b)/2 + a

a = 15
b=5

What is the uncertainty in M if the uncertainty in a and b is ±0.7?


Homework Equations



for c = a + b
Error in c =√[(error in a)^2 + (error in b)^2]

The Attempt at a Solution



Error in M = √[0.7^2 * 3] = 1.2124

The problem I am having is that we learn that the above formula can only work when the errors are independent of one another. the error in a is obviously not independent of the error in a. so i think I'm wrong.

What do you think?

Thank you.
 
Physics news on Phys.org
If I remember correctly, the errors work like this:

s= a+b ⇒ Δs=Δa + Δb

s= a-b ⇒ Δs= Δa + Δb

s=ab ⇒ Δs/s = Δa/a + Δb/b

So you can apply the first two as needed.
 
yes. but that only works when the uncertainties in a and b are independent. however in my equation for M, a appears twice. since the error of a is not independent of a, how do i go about finding the uncertainty?

thanks.
 
Why not just simplify your problem

(a-b)/2+a

into

(3a-b)/2

and then use the rules for the error of 3a+b. You don't need to worry about the independency of error a to error a.
This way I get

(3 Δa + Δb)/2 = 1.4

Or using the other rule [ √(Δa2 + Δb2) ]

( √(9*Δa2 + Δb2) ) / 2 ≈ 1.1068
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top