1. Summary Part 1 Measured Value = True Value + Errors = True Value + Errors Errors = Random Errors + Systematic Errors How to minimize RE and SE: (a)RE – by taking MORE measurements! (b)SE – by SPOTTING! Precision  Accuracy

2. How to present the measured value ? In general, the result of any measurement of a quantity x is stated as following units (Best estimate  Uncertainty) units If possible  x is represented by the standard error!

3. How to present the measured value ? Rule for stating Uncertainties: Experimental uncertainties should almost always be rounded to one significant figure. Example:  x = 0.000153 = 0.0002 m

4. How to present the measured value ? Rule for stating Answers: The last significant figure in any stated answer should usually be of the same order of the same order of magnitude (in the same decimal position) as uncertainty. Example: x = 12.23452 m  x = 0.0002 m x ±  x = (12.2345 ± 0.0002) m

5. Statistical Formulas Mean Standard Deviation Standard Error

6. Error Propagation How to estimate the error of the quantity R from the errors associated with the primary measurement of x, y, z, …, respectively?

7. Summary Part 2 Presentation of final answer: (1.2  0.3) m Mean: average of the data Standard deviation: measures the spread of the data about the mean Standard error: standard deviation of the mean Use propagation of error formula to compute the combined error!

8. Example 1 In the determination of gravity, g by means of a simple pendulum (i.e. T = 2   (ℓ /g) ), the following data are obtained: Length of the string, ℓ: 99.40, 99.50, 99.30, 99.45, 99.35 cm. 99.45, 99.35 cm. Time for 20 oscillations, t: 40.0, 39.8, 40.2, 39.9, 40.1 sec. 39.9, 40.1 sec. Calculate g and its error.

9. Solution for Example 1 For ℓ, n = 5 Mean: Standard deviation of ℓ:  Standard error:

10. Solution for Example 1 For time of an oscillation T i = t i /20, n =5 Mean: Standard deviation: Standard error:

11. Solution for Example 1 The estimated uncertainty of g:

12. Solution for Example 1 From this function Take the partial derivative of g w.r.t. ℓ and T

13. Solution for Example 1

14. Which is the BEST line?

15. Least Squares Method (LSM) Finding the best straight line y best = m best x + c best to fit a set of measured points (x 1, y 1 ), (x 2, y 2 ), …, (x n, y n ). The following assumptions are made to find the best line: 1. 1.The uncertainty in our measurements of x is negligible but not in y 2. 2.The uncertainty in our measurements of y is the same (  i =  )

16. Least Squares Method (LSM) Let coordinates (x i, y i ) is the i-th data point that you have plotted in the graph and The best fit of a straight line takes the following form: y best (x i ) = m best x i + c best

17. Least Squares Method (LSM)

19. y best (x) = m best x + c best

20. Example 2 In the determination of g by means of a simple pendulum (i.e. T = 2   (ℓ /g)), the following data are obtained: ℓ 1 = 40.00 cm t 1 = 26.3, 25.5, 25.9 sec ℓ 2 = 60.00 cmt 2 = 31.8, 32.3, 32.7 sec ℓ 3 = 80.00 cmt 3 = 36.9, 36.5, 37.1 sec ℓ 4 = 100.00 cmt 4 = 42.6, 41.5, 41.7 sec ℓ 5 = 120.00 cmt 5 = 43.8, 44.8, 45.3 sec ℓ 6 = 140.00 cmt 6 = 48.4, 48.9, 49.1 sec ℓ 7 = 160.00 cmt 7 = 51.2, 51.5, 51.9 sec ℓ 8 = 180.00 cmt 8 = 54.5, 54.8, 54.3 sec where t i is the time for 20 oscillations. Determine g and its error.

21. Solution for Example 2

22. slope, m best = 0.040926 sec 2 /cm y-intercept, c best = 0.132583 sec 2  m_best = 0.000748 sec 2 /cm  c_best = 0.089178 sec 2 /cm m best  m_best = (0.0409  0.0008) sec 2 /cm c best  c_best = (0.13  0.09) sec 2 The best fit of a straight line, T 2 best = 0.0409 ℓ + 0.13

23. Solution for Example 2 4  2 / g = T best 2 / ℓ 4  2 / g = m best = 0.0409 sec 2 /cm  g best = g = 965.2425 cm/sec 2 How To Compute,  g ?

24. Solution for Example 2 How To Compute,  g : From 4  2 / g = m best  g = 4  2 / m best where   g = 18.88005 cm/sec 2   g = 6.67510 cm/sec 2   g   g = (9.65  0.07) m/sec 2