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Math Help - maximum error percentage

  1. #1
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    maximum error percentage

    period T of a simple pendulum with small oscillations is calculated:

     T=2\pi \sqrt{\frac {L}{g}}

    L = length of pendulum, with 0.5% max error value
    g = acceleration due to gravity, with 0.1% max error value

    Use differentiation to approximate the max % error in the calculated value of T.

    any help gratefully received!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Dr Zoidburg View Post
    period T of a simple pendulum with small oscillations is calculated:

     T=2\pi \sqrt{\frac {L}{g}}

    L = length of pendulum, with 0.5% max error value
    g = acceleration due to gravity, with 0.1% max error value

    Use differentiation to approximate the max % error in the calculated value of T.

    any help gratefully received!
    We have  T=2\pi \sqrt{\tfrac {L}{g}} so

    \ln T=\ln\left( 2\pi\right) + \ln \sqrt{L}-\ln \sqrt{g}=\ln\left( 2\pi\right) + \frac{\ln L}{2}-\frac{\ln g}{2}.

    Differentiate :

    \frac{\mathrm{d}T}{T} = 0 + \frac12\cdot\frac{\mathrm{d}L}{L} - \frac12\cdot\frac{\mathrm{d}g}{g}=\frac12\cdot\fra  c{\mathrm{d}L}{L} - \frac12\cdot\frac{\mathrm{d}g}{g}

    Can you take it from here ?
    Last edited by flyingsquirrel; September 5th 2008 at 03:57 AM. Reason: dS/S --> dT/T
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  3. #3
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    I think so.
    Is
    \frac{dL}{L} = \frac{1}{2L}
    and
    \frac{dg}{g} = -\frac{1}{2g}
    ?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Dr Zoidburg View Post
    I think so.
    Is
    \frac{dL}{L} = \frac{1}{2L}
    and
    \frac{dg}{g} = -\frac{1}{2g}
    ?
    No, you're given relative errors, not absolute ones. From

    \frac{\mathrm{d}T}{T}=\frac12\cdot\frac{\mathrm{d}  L}{L} - \frac12\cdot\frac{\mathrm{d}g}{g}

    we derive

    \frac{\Delta T}{T}=\frac12\cdot\frac{\Delta L}{L} {\color{blue}+} \frac12\cdot\frac{\Delta g}{g}

    where \Delta T, \Delta L, \Delta g are the absolute error in T, L and g, respectively. By definition, the ratios \tfrac{\Delta T}{T}, \tfrac{\Delta L}{L} and \tfrac{\Delta g}{g} are relative errors ; as we are given \tfrac{\Delta L}{L}=0.5\% and \tfrac{\Delta g}{g}=0.1\%, we can compute the value of \tfrac{\Delta T}{T}.
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