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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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