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Significant Figures

  1. Mar 17, 2006 #1
    The author of my physics book seems to be a little significant-figure-happy:
    m=150.0, k=225, A=.15, t=3.00, δ=π
    He does this : -A√k/m)sin(t√k/m)+δ)=-.15*1.22sin(3.00*1.22+π)=-.0907
    Whereas I do: -A√k/m)sin(t√k/m)+δ)=-A√k/m)sin(3.67+π)=-A√k/m)sin(6.81)=-.0924

    Isn't my way more accurate? These discrepancies are driving me nuts as they occur in almost every problem (and there is no method to his madness), so I just wanted to know.
     
  2. jcsd
  3. Mar 17, 2006 #2

    Chi Meson

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    How did (3.00*1.22) become 3.67? THis alone accounts for the discrepency.
     
  4. Mar 17, 2006 #3
    Here is a hint:

    3 * 1.22 is 3.66 not 3.67

    Does that help?
     
  5. Mar 17, 2006 #4
    t=3.00, k=225, m=150.0
    √k/m)=1.22
    and
    1.22t=3.66
    but
    t√k/m)=3.67

    And therein lies the problem. He's making a sig fig calculation before multiplying √k/m) by t. There's also a discrepancy in the ways we approach what's inside the sine function, thus exacerbating the already-present difference in our answers. To me, mine seems the more accurate way to do it, but I just wanted to make sure.
     
  6. Mar 17, 2006 #5

    Chi Meson

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    OK, I actually agree with you. When solving intermediate solutions it's a good idea to keep "sig plus one" figures to minimize rounding error. It doesn't matter too much here, since the final answer is limited to two sig figs due to the A = 0.15 factor. The discrepency is only .091 vs. .092 .
     
  7. Mar 17, 2006 #6
    When solving the intermediate, you give correct number of sig fig, but then continuing the path to the final answer without having rounded at all.
     
  8. Mar 17, 2006 #7

    Hootenanny

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    I think that it is always best so solve equations algerbraically first, then plug and chug the numbers :smile:
     
  9. Mar 17, 2006 #8
    'A' should actually be .150, so the difference between our answers is a bit bigger.

    Thanks guys.
     
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