Supplement to Significant Figures
How to decide Significant figures with the skill on page 2 and page 3 in your lab manual?
Adapted from Laboratory Manual of Freshmen Physics, National Tsinghua Univ., TAIWAN
Do it step by step!
Significant Figures of Multiplication or Division:
The following example ZERO is from Lab 5, Newton's 2nd Law of Motion, Part II, Phys2LA or Phys40A.
Now we have that 100.08 + 5.02 = 105.10 (gram) = 1.0510 ´ 10-1(kg). Refer to page 2 in your lab manual.
Now we have that 1.0510 ´ 10-1(kg) ´ 9.80(m/s2) = 1.02998 (kg-m/s2) = 1.03 ´ 100(Newton)
Now, we have a multiplication. Refer to page 3 in you lab manual.
|
| The answer is 9.80(m/s2). |
| The answer is 3. | |
| The final answer has 3 digits only. | |
| The answer is 3. | |
| The answer is 1.02998(kg-m/s2). | |
| The answer is 6. | |
| The answer is 3. | |
| They are 1, 0, 2. | |
| Those are 9, 9, 8. | |
| Turn 2 into 3 because of 9. | |
| Here, we do not have to. | |
| 1.03 ´ 100(Newton) |
Example 1:
We want to calculate the volume of a cube.
Here is the result measuring the length of the cube.
| Times of measurement |
Length, L, (mm) |
Deviation, d, (mm) |
Square of d, d2, (mm2) |
|
1 |
22.1 |
+0.17 |
0.029 |
|
2 |
22.0 |
+0.07 |
0.005 |
|
3 |
21.9 |
-0.03 |
0.001 |
|
4 |
21.8 |
-0.13 |
0.017 |
|
5 |
21.8 |
-0.13 |
0.017 |
|
6 |
21.7 |
-0.23 |
0.053 |
|
7 |
21.9 |
-0.03 |
0.001 |
|
8 |
22.0 |
+0.07 |
0.005 |
|
9 |
21.9 |
-0.03 |
0.001 |
|
10 |
22.3 |
+0.37 |
0.137 |
|
11 |
21.9 |
-0.03 |
0.001 |
|
12 |
22.1 |
+0.17 |
0.029 |
|
13 |
21.9 |
-0.03 |
0.001 |
|
14 |
21.8 |
-0.13 |
0.017 |
|
15 |
22.0 |
+0.07 |
0.005 |
|
16 |
21.8 |
-0.13 |
0.017 |
|
Total times, n=16 |
S L=350.9 |
S |d|=1.82 |
S |d|2=0.336 |
|
|
Average Length, Lave= 350.9/16= 21.93 |
Average Deviation, D= 1.82/16= 0.11 |
Standard Deviation, s = {0.336/(16-1)}(1/2)= 0.16 |
Mean value, Average Length, Lave =(S L)/n
Deviation, d = L - Lave
Average Deviation, D = (S |d|)/n
SD, Standard Deviation, s ={(S |d|2)/(n-1)}(1/2)
SDOM, Standard Deviation of Mean, s SDOM =s /(n)(1/2)=0.16/(16)(1/2)=0.04(mm)
The notation of Average Length, here, Lave=(21.93± 0.04)(mm)
Or, Lave=(21.93± 0.18%)(mm) where (0.04/21.93)´ 100%=0.18%
Remark: (0.04/21.93)´ 100%=0.18% but (0.04/21.93)´ 100¹ 0.18%! If you make this kind of mistakes, I will take your 0.1 points off, every time, consecutively!
From the notation, Lave=(21.93± 0.04)(mm), because of the calculated error (± 0.04)(mm), we know that the first three digits, "2", "1", and "9" are three justified figures. The last digit, "3" is an unjustified figure.
Now, the average volume Vave= (Lave)3 = (21.93)3 = 10546.68 (mm3). Please notice that this is not the correct result because of the incorrect significant figures.
Remark:
Here, s SDOM of Vave must be calculated from s SDOM of Lave, (± 0.04)(mm). The method to do so is in the "Advanced Supplement to Statistical Errors."
When quotients and products are taken, the number of significant figures of the result will be equal to that of the least precisely known factor. It is written in your lab manual on page 3.
Lave has 4 significant figures, 3 justified digits and 1unjustified digit. That is exactly the least precisely known factor. So, we keep the first 3 digits, "1", "0" and "5", of the average volume Vave as our 3 justified digits. And take one 1unjustified digit, "4" to complete the whole notation of 4 significant figures.
That is,
Vave= (Lave)3 = (21.93)3 = 10540 (mm3) or 10550 (mm3)
In other words, the precision of Vave= (Lave)3 = (21.93)3 = 10546.68 (mm3) is not allowed based on the data table of our measurement
Example 2:
We want to calculate the area of a desk, A. Length L here is 2.22(m). Width W here is 1.1(m).
L has 3 significant figures but W has 2 only. So, A = 2.4 (m2). The notation A= 2.44 (m2) is incorrect.
Example 3:
We want to calculate the volume of a box, V. Length L here is 2.22(m). Width W here is 1.1(m). Height H here is 0.1(m)
L has 3 significant figures. W has 2 but H has 1 only. So, V = 0.2 (m3). The notations like A= 0.2442 (m3), A= 0.24 (m3), or A= 0 (m3) are incorrect.
