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Help deriving expression for the error in velocity.

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Derive the error expression [tex]\delta v_{}x [/tex] from the equation [tex]v _{x}[/tex]=[tex]\frac{s}{sqrt(\frac{2h}{g})}[/tex]

    3. The attempt at a solution

    I've derived error expressions before, but I have a history of getting the calculations right and the error equations wrong. So, if possible I'd like to have someone tell me what I am doing wrong (if anything) when I derive this.

    For any two numbers divided by one another (2h/g), the error z is:

    [tex]\delta[/tex]z=|z|([tex]\frac{\delta x}{|x|} + \frac{\delta y}{|y|}[/tex])
    So:
    [tex]\delta[/tex]z=|z|([tex]\frac{\delta h}{|h|} + \frac{\delta g}{|g|}[/tex])

    At this point intuition tells me that I should just multiply that by two to get the error in 2(h/g), but I think I've done that before and got it wrong.

    I have no idea what the error of a square root is. I'm going to guess it's the same as the error in squaring something, so the error in the initial equation would be:

    [tex]\delta[/tex]v=|v|([tex]\frac{2 \delta h}{|h|} + \frac{2 \delta g}{|g|} + \frac{\delta s}{|s|}[/tex])

    So, am I right? If not, what am I doing wrong? It's fairly crucial to understand these rules, I think, so any help would be wonderful.

    Thanks.
     
    Last edited: Oct 16, 2008
  2. jcsd
  3. Oct 17, 2008 #2

    LowlyPion

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    One thing I would note is that g is a gravitational constant that unless you are measuring it and using it generally doesn't carry an error or if it does carries a relative error of 0.
     
  4. Oct 17, 2008 #3
    Well, the equation is a lab equation and they always give gravity along with its uncertainty. But is the rest of it right?
     
  5. Oct 17, 2008 #4

    LowlyPion

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    Some treatments of uncertainty for independently measured variables take the Root Sum of the Squares of the relative uncertainties for multiplication division and the absolute uncertainties for addition subtraction. You will have to be the judge of what your lab may be wanting from you.

    Your answer would be the more conservative approach as it would yield a greater resultant uncertainty.
     
  6. Oct 17, 2008 #5

    LowlyPion

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    As for exponents I think you take the exponent and multiply it times the relative uncertainty of the quantity being taken to the exponent. So for sqrt I think you take 1/2 the relative error quantity.
     
  7. Oct 17, 2008 #6
    Oh, right. Square root is the same as taking it to the 1/2 exponent. I keep forgetting that, thanks.
     
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