01b_PHY_205L_Error_Measurement_and_Graphing_S24

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Apr 3, 2024

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Physics 205L Measurement Error and Graphing© 40pts Introduction: This preliminary exercise ( Individualized, No Lab Partners) is designed to introduce error in measurements and why we graph data. Before beginning this important exercise, please read the "Graphical Analysis Handout", “The Lab Manual Handout”, StatHandoutLab and “The Vernier & Micrometer” handouts. THIS DOCUMENT MUST BE TYPED OUT DO NOT WRITE IN THE MARGINS. NEATNESS COUNTS! Part 1: Dimensions and Volume of a Block This exercise is designed to provide practice of data analysis which include uncertainties, error propagation, and interpretation of results. Enter the corresponding values below. WATCH UNITS!!! THIS DOCUMENT MUST BE TYPED OUT (20pt penalty if it is not). DO NOT WRITE IN THE MARGINS. NEATNESS COUNTS! Part 1A: Directions: Measure the dimensions of the block with a ruler, calculate their average dimensions L , W , H along with their associated uncertainty, and calculate the average volume of the block. Please note that the resolution (distance between two closest divisions) of the ruler is 0.1cm. Don’t forget to apply error propagation to determine the associated uncertainty in the average volume. Enter your results in Table 1. Show a sample of each type of calculation below the table. Note cells in Volume column are blacked out for a reason! Table 1. Dimensions of Block Number ____3 ____ Using a Ruler Trial Number Length (cm) Width (cm) Height (cm) Volume (cm 3 ) 1 1.90 1.89 1.90 2 1.91 1.90 1.89 3 1.90 1.90 1.91 4 1.92 1.91 1.91 5 1.90 1.90 1.90 AVERAGE 1.90±0.004 cm. 1.90±0.003 cm. 1.90± 0.004 cm. 6.85± 0.02 cm 3 σL = ( L L ) 2 ( N 1 ) L = 1 . 90 + 1 . 91 + 1 . 90 + 1 . 92 + 1 . 90 5 = 1.90cm . σ = ( 1.98 1.97 ) 2 +( 1.98 1.98 ) 2 +( 1.98 2.00 ) 2 +( 1.98 1.98 ) 2 +( 1.98 1.99 ) 2 4 = 0.01 cm δ = 0.01 5 = 0.004cm Experiment 1
Physics 205L Calabrese δV = ( H W ) 2 ( δL ) 2 + ( L H ) 2 ( δW ) 2 +( L W ) 2 ( δH ) 2 = ( 1.90 1.90 ) 2 ( 0.004 ) 2 + ( 1.90 1.90 ) 2 ( 0.003 ) 2 +( 1.90 1.90 ) 2 = 0.02 cm 3 Part 1B: Directions: Repeat part 1A with the same block, but this time use a vernier caliper instead of the ruler. Please note that the resolution of the vernier caliper is 0.002cm. Show a sample of each type of calculation below the table (2pts) Table 2. Dimensions of a Block Number ___6 ____ Using Vernier Calipers Trial Number Length (cm) Width (cm) Height (cm) Volume (cm 3 ) 1 2.000 1.980 1.990 2 2.010 1.990 1.992 3 2.000 1.980 1.990 4 2.000 1.980 1.990 5 2.010 1.990 1.988 AVERAGE 2.004± 0.002 cm 1.980± 0.01 cm 1.990±0.007cm 7.896± 0.05 cm 3 L = 2 . 000 + 2 . 010 + 2 . 000 + 2 . 000 + 2 . 010 5 = 2.004 cm . σL = ( L L ) 2 ( N 1 ) σL = ( 2.004 2.000 ) 2 +( 2.004 2.010 ) 2 +( 2.004 2.000 ) 2 +( 2.004 2.000 ) 2 +( 2.004 2.010 ) 2 4 = 0.005 cm δL = .005 5 = . 002cm δV = ( H W ) 2 ( δL ) 2 + ( L H ) 2 ( δW ) 2 +( L W ) 2 ( δH ) 2 = ( 1.990 1.980 ) 2 ( 0.002 ) 2 + ( 2.004 1.990 ) 2 ( 0.01 ) 2 +( 2.004 1 = 0.05 cm 3 Part 1C Directions: Repeat part 1A with the same block, but this time use a micrometer instead of the ruler. Please note that the resolution of the micrometer is 0.01mm. Show a sample of each type of calculation below the table (2pts). Table 3. Dimensions of a Block Number ___3 ___ Using Micrometer Trial Number Length (cm) Width (cm) Height (cm) Volume (cm 3 ) 1 1.990 1.994 1.986 2 1.988 1.998 1.988 3 1.988 1.998 1.986 4 1.990 1.992 1.990 5 1.990 1.990 1.986 2
Physics 205L Calabrese AVERAGE 1.989±0.0004 cm. 1.994±0.007 cm 1.987±0.003 cm 7.881± 0.04 cm 3 L = 1 . 990 + 1 . 988 + 1 . 988 + 1 . 990 + 1 . 990 5 = 1.989 cm . σL = ( L L ) 2 ( N 1 ) σL = ( 1.989 1.990 ) 2 +( 1.989 1.988 ) 2 +( 1.989 1.988 ) 2 +( 1.989 1.990 ) 2 +( 1.989 1.990 ) 2 4 = 0.001 cm δL = .001 5 = 0.0004 cm δV = ( H W ) 2 ( δL ) 2 + ( L H ) 2 ( δW ) 2 +( L W ) 2 ( δH ) 2 = ( 1.987 1.994 ) 2 ( 0.0004 ) 2 + ( 1.989 1.987 ) 2 ( 0.005 ) 2 +( 1.989 = 0.02 cm 3 Questions (answer in complete sentences) 1. Do the measured volumes agree with each other? If the values don’t agree identify the cause of the disagreement 2. Did the more precise tool give more consistent results for the block dimensions and volume? Explain your answer Part 2: Graphical Analysis of Data For many spring-mass systems, the amount a spring stretches when it suspends a mass is proportional to the amount of mass supported by the spring. This is called Hooke’s law where F=-kx , where k is the spring constant, which is a measure of the spring stiffness. We will see this here, except that we will not be measuring force, we will measure mass. The measured mass will be proportional to the force applied to the spring. Directions: Suspend a hanger on a spring and record the position of the bottom of the hanger. Add a known mass to the hanger and record the position of the hanger. WATCH SIG. FIGS.! Table 4- Mass vs. Position 3
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Physics 205L Calabrese Analysis 1. Calculate the average values and uncertainties of columns 3 and 4 in Table 4. Show your work for one of the uncertainty calculations below. σ Mass Position = ( Mass Position Mass Position ) 2 ( N 1 ) = ( 2.4136 0.549 ) 2 + ( 2.4136 1.068 ) 2 + ( 2.4136 1.518 ) 2 + ( 2.4136 1.943 ) 2 + ( 2.4136 2.343 ) 2 + ( 2.4136 2.712 ) 2 +( 2.4136 3.054 ) 2 +( 2.4136 3.374 ) 2 +( 2.4136 3.656 ) 2 +( 2.4136 3.919 ) 2 4 = 1.69 g cm δ Mass Position = 1.7 5 = 0.7 5 g cm 2. Plot mass of spring versus position of spring (y vs. x). Perform a fit using y=mx+b and correctly report the slope and y-intercept. Copy and paste the graph below. In your figure caption explain compare the slope of the graph to the average of the ratio column 3 of Table 4. In theory should they be the same (answer within your figure caption)? Explain any discrepancies (answer within your figure caption). 4 Mass (grams) Position (cm) Mass Position ( g cm ) Position Mass ( cm g ) 19.993 36.40 0.549 1.820 40.493 37.90 1.068 0.935 59.986 39.50 1.518 0.658 79.934 41.12 1.943 0.514 99.874 42.62 2.343 0.426 119.904 44.20 2.712 0.368 139.897 45.80 3.054 0.327 159.948 47.40 3.374 0.296 179.888 49.20 3.656 0.273 199.918 51.00 3.919 0.255 AVERAGE 2.4136 ± 0.7 g cm 0.5872 ± 0.3 g cm
Physics 205L Calabrese Slope= 12.396438 +/- 0.109836 g cm y-intercept= -429.435122 +/- 4.806357 g cm Figure 1: 3. Plot position of spring (y vs. x) versus mass. Perform a fit using y=mx+b and correctly report the slope and y-intercept. Copy and paste the graph below. In your figure caption explain compare the slope of the graph to the average of the ratio column 4 of Table 4. In theory should they be the same (answer within your figure caption)? Explain any discrepancies (answer within your figure caption). Slope=Slope= 0.080617 +/- 0.000714 cm g y-intercept= 34.647383 +/- 0.0886028 cm g 5
Physics 205L Calabrese Figure 2 : Questions (answer in complete sentences) 1. When you determine the average of columns 3 and 4 of Table 4 are you assuming that the two quantities are linearly related by the equation y=mx+b or y=mx? What is the meaning y-intercept in the second graph ? The two equations need to be y = mx + b because of the (b) in the equation. As shown in the graphs, each mass and measurement starts above the origin therefore needing (b) as a place of measurement for the spring position or it’s weight. 2. If your job is to determine the average of a physical constant that is linearly related to two physical measurements, is it better to average a bunch of measurements or determine the average from the slope of a graph? Which curve fit equation do you use, y=mx+b or y=mx? Using the average of a slope in a graph is more practical as it provides hard data in a specific range to find the average of measurement, especially if you can locate a constant rate on a graph. 6
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