1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measurement help

  1. Jun 22, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations

    v square = u square + 2as

    3. The attempt at a solution

    a) Let v square - u square be x
    delta x/x = 2(delta v/v) + 2(delta u/u)
    delta x/300 = 0.03
    delta x = 9

    b) i) a = (v square - u square)/2s
    delta a/a = 2(delta v/v) + 2(delta u/u) + deltas/s
    delta a/3 = 0.03 + 0.002
    delta a = 0.096 = 0.1 (round off to 1 sig fig)
    ans : 3.0 plus minus 0.1
    1. The problem statement, all variables and given/known data



    I think I did it correctly, but my answers seem to be wrong , because this is a question on my school website and I have to key in the answers and it'll tell me whether it is right or wrong. I thought maybe I did something wrong, so maybe somebody could help please?
     
  2. jcsd
  3. Jun 22, 2007 #2

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    a) well if we make a function:

    [tex] f(u,v) = v^{2} - u^{2} [/tex]

    then, from error propagation:

    [tex] \delta f = \sqrt{\left( \dfrac{\partial f}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial f}{\partial v} \delta v\right)^{2}}[/tex]

    So:


    [tex] \delta f = \sqrt{\left( -2*10 * 0.1 \right)^{2} + \left( 2*20 * 0.1 \right)^{2}} = 4.4721[/tex]

    the forumula you used is when f(u,v) is a product and quoitent only.
     
    Last edited: Jun 22, 2007
  4. Jun 22, 2007 #3

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    b) the expression for the acceleration is:

    [tex] a(v,y,s) = \dfrac{v^{2} - u^{2} }{2s}[/tex]

    then the uncertanty in a is:

    [tex] \delta a = \sqrt{\left( \dfrac{\partial a}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial a}{\partial v} \delta v\right)^{2} + \left( \dfrac{\partial a}{\partial s} \delta s\right)^{2}}[/tex]
     
    Last edited: Jun 22, 2007
  5. Jun 22, 2007 #4
    I don't get your equations, wouldnt the delta u and delta u or other symbols cancel each other out?
     
  6. Jun 22, 2007 #5

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Last edited: Jun 22, 2007
  7. Jun 22, 2007 #6
    I don't know the exact answers, I only know whether the answers would be right or wrong because we have to key in them inside a system after we found them.

    I've not learnt partial derivatives yet though, so I think there should be another easier way to do it using delta/fractional uncertainty?
     
  8. Jun 22, 2007 #7

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    partial derivatives are very easy.. you just do usual derivatives and treat the rest as constans. for example: if f(x,y) = 2xy. the f 'x =2y and f 'y = 2x

    According to my knowledges in science of errors, the functions we deal with here we must use error propagation as the formulas I have given.

    you can read alot of it on the web page I gave you.
     
  9. Jun 22, 2007 #8
    it's not easy for me XD

    for the 2nd part I tried using your method:
    so delta a = square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square + (0.1 x 50 x 2)square)] = root(150) which would give you a value of 12.2, much bigger than a=3 calculated. doesnt make sense, or did I do something wrongly?
     
  10. Jun 22, 2007 #9

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    well doing the partial derivative is just doing the usual treating the rest as constants.

    Well you did not to the derivatie for a vs s right.. try again. Remeber to trea't v and u just as constant (values)
     
  11. Jun 22, 2007 #10
    I have no idea how to do it..

    upon further thought, even without finding the derivative for a vs s, the answer already seems to be wrong from the 1st and 2nd part of the equation,

    square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square] would give you a value of around 4 which is still greater than a=3.
     
  12. Jun 22, 2007 #11

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    are you telling me that you can not differentiate functions like:

    f(x) = 1 / x ???

    if you wait a minute I can do the function for you and show how you do this..
     
    Last edited: Jun 22, 2007
  13. Jun 22, 2007 #12

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    [tex] \delta a = \sqrt{\left( \dfrac{\partial a}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial a}{\partial v} \delta v\right)^{2} + \left( \dfrac{\partial a}{\partial s} \delta s\right)^{2}}[/tex]

    [tex] a(v,y,s) = \dfrac{v^{2} - u^{2} }{2s}[/tex]

    [tex] \dfrac{\partial a}{\partial v} = \dfrac{2v-0}{2s} = \dfrac{v}{s} [/tex]

    [tex] \dfrac{\partial a}{\partial u} = \dfrac{0-2u}{2s} = \dfrac{-u}{s} [/tex]

    [tex] \dfrac{\partial a}{\partial s} = \dfrac{-(v^{2} - u^{2})}{2s^{2}} [/tex]

    so we get:
    [tex] \delta a = \sqrt{\left( \dfrac{-u}{s} \delta u\right)^{2} + \left( \dfrac{v}{s} \delta v\right)^{2} + \left( \dfrac{(u^{2} - v^{2})}{2s^{2}} \delta s\right)^{2}}[/tex]
     
    Last edited: Jun 22, 2007
  14. Jun 22, 2007 #13
    The answers are still wrong though, I used

    a) 4 (round off to one significant figure)
    b) 3.0 + - 0.2 (this was the value I calculated from the equation you gave me)
     
  15. Jun 22, 2007 #14

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Well I got (b) -> error in a = 0.045211

    are you sure you have the right formula for the acceleration?

    Sorry I can't help you more kid.

    And what kinda sucky school has an electronic answer machine lol

    it is also the way you get to the answer that is important, not the result itself. So you can basically try every number from 0 to 20 til it gets right :)
     
    Last edited: Jun 22, 2007
  16. Jun 22, 2007 #15
    hehe

    but the answers are still wrong :grumpy: I don't really get your method either, because I haven't learnt them yet. I think the school wouldn't have set something they hadn't taught yet, so I believe there's still another (easier( way ) of doing, maybe some other members might know?

    Thanks for your help and time anyway! at least I learnt something from you :cool:
     
  17. Jun 22, 2007 #16

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    then you show me the simpler way to do it. It is not the right one I will promise you.

    we have functions that are both multiplcation, addition and qouitent. And the errors are independent of each other, hence we must use the error propgation formula.
     
  18. Jun 22, 2007 #17
    If I knew the easier method I wouldnt be here asking :shy:

    Maybe I calculated something wrongly up there? Or maybe the school system just sucks :tongue2:
     
  19. Jun 22, 2007 #18

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper


    I bet your school sucks hehe :)

    Can´t you write donw all that we have discussed here, and maybe print out the relevant section from the www-page i showed you. Then go to the professor and ask "what is wrong?" :P
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Measurement help
  1. Quantum Measurement (Replies: 30)

Loading...