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Homework Help: Uncertainty with Equations Question

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    You would calculate the average velocity of the ball bearing in the fluid as

    v = h/t

    How would you calculate the absolute uncertainty on v?
    a) h(Δh/t) + (Δt/t2)

    b) (Δh/t) + h(Δt/t2)

    c) (Δh/h) + (Δt/t)

    d) h(Δh/t) - (Δt/t2)

    e) t(Δh/h) + h(Δt/t)

    2. Relevant equations

    I don't think you need to use any equations on this as it's an uncertainty question. Just math.

    3. The attempt at a solution

    I have no idea where to start. I can't find any questions like this anywhere else and I am unsure as to how to relate the uncertainty rules to an equation.
  2. jcsd
  3. Jan 15, 2012 #2


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    hmm. The experiment involves measuring total time and total distance, so each of these is a possible source of error. From there, you need to use propagation of errors to find the error on the calculated velocity. I'm guessing your teacher has taught you guys about propagation of errors?
  4. Jan 15, 2012 #3
    Use calculus, Δv = ... something involving partial derivatives and Δt and Δh,, also remember that adding uncertainties you add them both as positive.
  5. Jan 15, 2012 #4
    No she hasn't. She gave us an appendix on uncertainties, and propagation of error is one of the topics inside of it, but it's a little confusing.

    If I summarize it here, it basically says this:

    When adding or subtracting, add the relative uncertainties to get the absolute uncertainty.

    Δ(A+B) = ΔA + ΔB (Where delta A and B are the absolute uncertainties)
    Δ(A-B) = ΔA + ΔB

    When multiplying or dividing, add the relative uncertainties to get the relative uncertainty.

    (Δ(AB)/AB) = (ΔA/A) + (ΔB/B)
    (Δ(A/B)/(A/B)) = A/B((ΔA/A) + (ΔB/B))

    Then to get the absolute uncertainty in the quantity, multiply the relative uncertainty by the quantity.

    (blah blah blah)

    If the function is more complicated, use the derivative method (which I have no idea how to do, and I don't think it's necessary for this question or my course, so I'm not going worry about it unless it makes the question much easier).

    Is it really as easy as just punching the numbers into that equation? My problem with this particular question is that there aren't any numbers given in the actual question, which is where I get a little bit confused; it's like you're adding variables together instead of numbers.
  6. Jan 15, 2012 #5
    Ok thanks I'll check that out.
  7. Jan 15, 2012 #6

    Simon Bridge

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    Rule to divide measurements is that the relative uncertainties are added.

    Thus: [tex]\frac{\Delta v}{v}=\frac{\Delta h}{h}+\frac{\Delta t}{t}[/tex]... solve for [itex]\Delta v[/itex] and simplify.

    technically if the measurements are independent, you sum the squares... what you are doing is adding two normally distributed variables.

    if [itex]y=f(x)[/itex] then [itex]\Delta y = f^\prime (x)\Delta x[/itex]
    easy to prove by representing f(x) as a power series and applying the basic rules to each term.
    Last edited: Jan 15, 2012
  8. Jan 15, 2012 #7
    Ok hold on I think I figured it out... can't believe I didn't see this before, I should have just done the question lol.

    = (Δh/h) + (Δt/t)
    = h/t ((Δh/h) + (Δt/t))
    = (Δh/t) + h(Δt/t^2)

    Okay so that one was pretty easy. I have another harder one that I really don't understand that I'll post in a new thread.
  9. Jan 15, 2012 #8

    Simon Bridge

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    Your second line does not follow from the first one ... but you are just using lazy notation right?

    Very often just proceeding naively will net the right answer ... you just have to put pen to paper and start doing the math without knowing that it is the right way to go and see where it leads you. It can take a bit of courage to do this ... don't be scared of being wrong, you learn more that way.

    Note: this method will over-estimate the uncertainties.
  10. Jan 15, 2012 #9
    Not using lazy notation. In my appendix, it says to multiply the relative uncertainty by the quantity to get the absolute uncertainty. Unless that is what you define as lazy notation, haha. Also, according to my problem set that is the correct answer.

    What do you mean by overestimating the uncertainties? Does that mean it's wrong?
  11. Jan 15, 2012 #10

    Simon Bridge

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    But you wrote:
    which says (literally): (Δh/h) + (Δt/t) = h/t ((Δh/h) + (Δt/t))
    which is false.

    It should have been:
    (Δv/v) = (Δh/h) + (Δt/t)
    => Δv= h/t ((Δh/h) + (Δt/t))

    ... which is correct.
    Leaving off the bit before the equals sign is the lazy part of the notation... you seemed to be jotting noted to keep track of your reasoning rather than doing math.

    Just saying: in general, it is not a good idea to do that even if it does lead you to the right answer. It won't be too long before you have to deal with problems where nobody knows the right answer and the disciplines you learn now will determine how good you get.

    This was just multi-choice, so you are not marked on your working.
    But you are also communicating your understanding in a multi-national forum.

    Yes - basically. I think you are being given a simple model of uncertainties while you are still coming to grips with the basics.

    The theory of uncertainties is the theory of statistics and hypothesis testing.
    We model a measurement as representing the mean of a sampled normal distribution with a standard deviation equal to the "absolute error".

    When we say that some length was 5cm ±0.2cm we are saying that we don't know what the actual length was, but we are 95% confident that it will be somewhere between 4.6cm and 5.4cm ... see what I mean?

    When you add two distributions, the result is a new one with mean equal to the sum of the individual means and a variance equal to the sum of the individual variances. This is provided the measurements don't depend on each other... which I gather is the case here.

    So if I measure two things and one is 5cm and the other 5.7cm, I can ask if these two things are actually the same length and the difference is just a result of having an inaccurate measuring method.

    That "these two things are the same length" is the hypothesis H we are testing.

    The difference between them is 0.7m ± 0.3m, so we can be 95% confident that these two lengths are actually different (rejecting H because 0m lies outside 2sd of the mean). However, if we use the method you've been given the difference becomes 0.7m ± 0.4m, putting the 0m mark inside the confidence interval, causing us to accept H, perhaps, too easily.

    I suspect you'll be covering this stuff later.
  12. Jan 15, 2012 #11


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    Yeah, I don't get this. I always thought that you should sum the squares, so I would have done:
    [tex](\frac{\Delta v}{v})^2 = (\frac{\Delta h}{h})^2 + (\frac{\Delta t}{t})^2 [/tex]
    But this doesn't agree with any of the possible answers. Unless there is some strange approximation that I don't know about...
  13. Jan 15, 2012 #12

    Simon Bridge

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    That's right Bruce. Summing the squares is valid when the measurements don't affect each other - when they do, you just add them.

    An example would be [itex]y=x^2[/itex] - in this case, x depends on itself (well doh!) so [tex]\frac{\Delta y}{y} = 2\frac{\Delta x}{x}\Rightarrow \Delta y = 2x\Delta x[/tex]... which you'll see is consistent with the result [itex]\Delta y = f^\prime(x)\Delta x[/itex] for [itex]y=f(x)[/itex].

    Similarly, you may measure a long something by making multiple placements of your ruler - each time you shift your ruler adds an uncertainty to the end result. (Get 10 people to measure the length of a sports-ground with a meter ruler and you'll get 10 different results. More different than if they used a tape measure.) Or you may measure the period of a pendulum by timing many swings and dividing by the number of swings... this method will divide the error making it a more accurate measurement. (Uncertainties can be used to work out how much effort you need to put into being accurate.)

    I prefer the notation: [itex]\sigma_y[/itex] instead of [itex]\Delta y[/itex] since the latter implies [itex]y_f-y_i[/itex] which is not what is going on. Anyway, using sigma reminds us that these are supposed to be standard deviations.
  14. Jan 15, 2012 #13
    I think you sum the squares when you're doing a standard deviation and you sum when you only have Δy (you're measurement is off by at most some distance). is that correct?
  15. Jan 15, 2012 #14

    Simon Bridge

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    This would mean you are assuming a flat distribution ... when you sum two flat distributions, you get a trapezoidal one...
    A demo scroll down to the sum of uniform probability distributions.

    You should realize this because a sum of a large number of these measurements should give you a normal distribution (central limit theorem).

    But you are onto something - the simple addition will act as a rule of thumb to find new absolute limits.

    so [itex](x\pm\Deltax) + (y\pm\Delta y)[/itex] would have absolute limits [itex]\Delta x + \Delta y[/itex] ... we'd refer to this as "margin for error". Perhaps this is the context of the problem?

    In which case it is not so bad - thought you are still better off with the statistical description where you usually take half the 95% confidence interval or the interquartile range as your margin for error.
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