1. ©Gioko ®2007 Uncertainties in calculated results

2. ©Gioko ®2007 Uncertainties. Absolute uncertainty is the value of random uncertainty reported in a measurement (the larger of the reading error or the standard deviation of the measurements). This value carries the same units as the measurement. Fractional uncertainty equals the absolute error divided by the mean value of the measurement (it has no units). Percentage uncertainty is the fractional uncertainty multiplied by 100 to make it a percentage (it has no units).

3. ©Gioko ®2007 Determine the uncertainties in results. A simple approximate method rather than root mean squared calculations is sufficient to determine maximum uncertainties For functions such as addition and subtraction absolute uncertainties may be added When adding or subtracting measurements, the uncertainty in the sum is the sum of the absolute uncertainty in each measurement taken An example: suppose the length of a rectangle is measured as 26 ± 3 cm and the width is 10 ± 2 cm. The perimeter of the rectangle is calculated as L + L + W + W = (26 + 26 + 10 + 10) ± (3 + 3 + 2 + 2) = 72 ± 10 cm

4. ©Gioko ®2007 Determine the uncertainties in results. Cont.. A simple approximate method rather than root mean squared calculations is sufficient to determine maximum uncertainties For multiplication, division and powers percentage uncertainties may be added When multiplying and dividing, add the fractional or percentage uncertainties of the measurements. The absolute uncertainty is then the fraction or percentage of the most probable answer. An example: find the area of the same rectangle. Area = L x W. The most probable answer is 26 x 10 = 260 cm 2. To find the absolute uncertainty in this answer, we must work with fractional uncertainties. The fractional uncertainty in the length is 3/26. The fractional uncertainty in the width is 2/10. The fractional uncertainty in the area is the sum of these two = 3/26 + 2/10 = 41/130. The absolute uncertainty in the area is found by multiply this fraction times the most probable answer = (41/130) x 260 cm2 = 82 cm 2. Finally, the area is 260 ± 82 cm 2.

5. ©Gioko ®2007 Determine the uncertainties in results. Cont.. A simple approximate method rather than root mean squared calculations is sufficient to determine maximum uncertainties For other functions (for example, trigonometric functions) the mean, highest and lowest possible answers may be calculated to obtain the uncertainty range. For functions of the form x n, the fractional uncertainty in the result is equal to n x fractional uncertainty in x. An example: find the volume of a cube with a side of length 5 ±.25 cm. Volume = L 3. So, the most probable value is 5 3 = 125 cm 3. The fractional uncertainty in the length is.25/5, so the fractional uncertainty in the volume is 3 x.25/5 =.75/5. Then, the absolute uncertainty in the volume is (.75/5) x 125 = 18.75 cm 3. Finally, the volume is reported as 125 ± 19 cm 3. If one uncertainty is much larger than others, the approximate uncertainty in the calculated result may be taken as due to that quantity alone.

6. ©Gioko ®2007 Uncertainties in graphs Based on the ideas above about drawing trend lines it is possible to draw several different trend lines through the data. To find the range of values for the slope & intercept you should draw the trend line with the largest value for the slope that still crosses all of the error bars and the trend line with the smallest value for the slope. Then simply measure the slope and intercept for each of these new lines. These represent the range of values for the slope and intercept. Take the average of these two values as the absolute uncertainty for the value.

7. ©Gioko ®2007 Transform equations into generic straight line form (y = mx + b) and plot the corresponding graph To use the slope, intercept, or area under a graph to determine the value of physical quantities, it is easiest to work with a straight- line graph. Most graphs can be transformed from a curved trend to a linear trend by calculating new variables. If the relationship between two variables is known in theory, then the known relationship can be used to transform a graph into a straight- line graph.

8. ©Gioko ®2007 An example: Suppose that the speed of a falling ball is measured at various distances as the ball falls from a height of 10 m toward the ground on the moon. From energy conservation, we can predict that the speed of the ball is related to the height above the ground as v 2 = 2gh. If we were to plot v vs. h we would get a sideways facing parabola. How can we make this a linear equation? If we calculate a new quantity (call it w) so that w = v 2, then our relationship becomes w = 2gh. If we plot w vs. h, we will get a straight line. (see the graphs below). This straight line can be used to find g on the moon (the slope of the graph = 2g).

9. ©Gioko ®2007 Analyze a straight-line graph to determine the equation relating the variables Once the straight-line graph has been achieved, using the intercept and slope the equation of the trend line can be written (y = mx + b). Then substitute into the linear equation for the quantity that was used as the y and/or x variable to produce the straight-line graph. In our example, the equation of the trend line is y = 3.2x. This means that our relationship is v 2 = 3.2h. And, the slope of the trend line (3.2) should be twice the acceleration of gravity on the moon, so 2g = 3.2, or g on the moon = 1.2 ms -2.

10. ©Gioko ®2007 Transform equations involving power laws and exponentials -into the generic straight line form y = mx + b plot the corresponding -log-log -semi-log graphs from the data

11. ©Gioko ®2007 The relationship between two variables may not be readily seen and if the relationship between the variables is not linear, then it might be difficult to find the correct relationship. One method to find the relationship is guess and check. If y vs. x is not a straight line, then maybe y vs. x 2 or y vs. x 3 might work. Luckily, mathematics provides us a more reliable method. Analyze log-log and semi-log graphs to determine the equation relating two variables

12. ©Gioko ®2007 Analyze log-log and semi-log graphs to determine the equation relating two variables If the relationship between two quantities, x and y, is of the form y = kx n, where k is a constant and n represents some unknown exponent, then we can use logs (either base 10 logs or ln may be used) to transform this equation into a linear relationship as follows. take the log of both sides of the equation: log y = log (kx n ) use the properties of logs to simplify: log y = log k + n log x substitute log y = Y and log x = X: Y = log k + n X

13. ©Gioko ®2007 Graph of Y vs. X, we would get a straight line. AND, the slope of that line is equal to n, the exponent in our original relationship. The vertical intercept (b) = log k. The value of the vertical intercept, b, can be used to find the constant k: b = log k, so 10b = 10log k = k. Simply substitute the values of n and k determined from the transformed log-log graph into the original relationship to get the equation relating the two variables, y and x.

14. ©Gioko ®2007 y = ke nx Again, this will produce a curved graph and we can use logs (this time it has to be ln) to transform the graph into a straight line. Let's see what logs will do for us this time. take the ln of both sides of the equation: ln y = ln (ke nx ) used the properties of ln to simplify: ln y = ln k + (nx)ln e = ln k + nx substitue ln y = Y Y = ln k + nx. if we were to make a graph of Y vs x, we would get a straight line. AND, the slope of that line is equal to n (the constant in the exponent of our relationship). The vertical intercept (b) = ln k. The value of the vertical intercept, b, can be used to find the constant k: b = ln k, so eb = eln k = k.

15. ©Gioko ®2007 This type of graph is called a semi-log graph because we only had to use the ln function to transform one of the variables. Semi-log graphs will transform exponential relationships into straight-line graphs. Simply substitute the values of n and k determined from the transformed semi-log graph into the original relationship to get the equation relating the two variables, y and x.

16. ©Gioko ®2007 NEXT SESSION MECHANICS REVISE ON MEASUREMENT NOTES UPLOADED IN THE CLASS BLOG. PRESENTATIONS CLICK ON THE LINKS IN THE BLOGS Good Day