Significant Figures

  • Thread starter M_LeComte
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  • #1
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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.
 

Answers and Replies

  • #2
Chi Meson
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M_LeComte said:
The author of my physics book seems to be a little significant-figure-happy:
m=150.0, k=225, A=.15, t=3.00, ?=?
He does this : -A?k/m)sin(t?k/m)+?)=-.15*1.22sin(3.00*1.22+?)=-.0907
Whereas I do: -A?k/m)sin(t?k/m)+?)=-A?k/m)sin(3.67+?)=-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.
How did (3.00*1.22) become 3.67? THis alone accounts for the discrepency.
 
  • #3
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Here is a hint:

3 * 1.22 is 3.66 not 3.67

Does that help?
 
  • #4
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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.
 
  • #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 .
 
  • #6
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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.
 
  • #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:
 
  • #8
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'A' should actually be .150, so the difference between our answers is a bit bigger.

Thanks guys.
 

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