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Lab 1: Laboratory Fundamentals and Data Analysis Written by: Kayla Gallego Group A Partners: Cameron H, Daniel G, Tyja M

Introduction: No theory or physical model should be considered valid unless the prediction aligns with the results and reproducible measurements; scientific method, basis. It is necessary to have a standardized system of units, because it is the next necessary step to checking the reproducibility of any measurement. To interpret measurements, we must use statistical mathematics. The basic measurements can be converted by using, sample mean, sample deviation, sample variance, and standard deviation. Purpose: the purpose of this lab, is to know how become familiar with standards of measurement, and know how to report errors in a measurement.

Part 1: Hooke’s Law Data

Part 2: Statistical Analysis

Calculation of RMS Deviation Data point i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Value of Data xi 54 75 68 42 30 80 88 84 90 96 25 49 62 78 70

Deviation from mean -12.07 8.93 1.93 -24.07 -36.07 13.93 21.93 17.93 23.93 29.93 -41.07 -17.07 -4.07 11.93 10.93

Square deviation from Mean 145.7 79.74 3.72 499.36 1301 194 480.9 321.5 572.6 895.8 1687 291.4 16.56 142.3 119.4

Square root Deviations 12.07 8.93 1.93 24.07 36.07 13.93 21.93 17.93 23.93 29.93 41.07 17.07 4.07 11.93 10.93

Mean = 66.07

Mean Square Deviation = 445.07 (6675.98/15= 445.07)

Root Mean Square Deviation = 15.72

Standard Deviation = 3.96

Calculations: To calculate the Hooke’s Law the mass was multiplied by the acceleration due to gravity. I multiplied 0.050 kilograms by 9.8 meters per second squared and received 0.49 Newton’s.

In the calculation of RMS, The mean of the Value of Xi. I found the sum of all X- values 991 and divided by the total number of X-values 15 to get the mean 66.07 (991/15=66.07). Next, I found the deviation (di) from the mean by subtracting the mean of the X-Values from the X-values (54-66.07= -12.07). Then, I found the squared deviation by squaring the value of the deviation from the mean (-12.07^2= 145.7). To calculate the square root of the squared deviation I found the square root of the squared deviation ( √ 145.7 = 12.07 ). To calculate the mean squared deviation I added all of the squared deviations and divided it by the total number of squared deviation values (6675.98/15 = 445.07). To calculate the square root mean value I sum all of the square root values and divided the total by the number of square root values (235.86/15= 15.72). The standard deviation is obtained by finding the square root of the mean square root deviation ( √ 15.72= 3.96

Conclusion: We were able to utilize Hooke’s Law to determine the force needed to stretch a spring in a line with the length of the spring by first getting the force using F=-kx. For the second part of our experiment we calculated the RMS and concluded the result at the mean value with the mean value obtained ± one standard deviation.

Questions

1. What are the different types of errors encountered in an experiment? Give two examples of each. Absolute error: 1. 6.12 ± 0.01 m where 6.12m is the measurement and 0.01m is the uncertainty, also expressed in meters. 2. 7.12 ± 0.02 m, where 7.12m is the correct measurement and 0.02m is an uncertainty, because the measurement “m” was not initially stated. Relative error: 1. 6.12 m ±0.2%, where the same error is now shown as a percentage. 2. 7.12m ± 0.2%, the error is now shown as a percentage, because it was not stated. Statistical error: 1. the sample mean time is 480s and the sample variance is 25s2. One standard deviation is therefore 5s, and I report the result as 480 ± 5s. 2. I measure the length of a rod with a meter stick. The meter stick has markings no smaller than millimeters, and I estimate that Jam limited in the precision of my measure to one half of one millimeter. I report the length of the rod to be 0.325 ± 0.005 m. Systematic error: 1. I measure a series of weights with a spring scale that has not been properly calibrated. With no weight applied, the scale registers 2 lbs.

Therefore, everything I weigh on the scale will appear two pounds heavier than it really is. I must subtract two pounds from the result to correct it. 2. I get ready to weigh myself on a scale, I notice it is not on the “0”. I conclude it is 3 pounds off. So, when I get my weight, I subtract three pounds to get my “correct” weight.

2. How many significant figures are there in the numbers 6.429 x 103 and 3.18785 x 102 ? How many significant figures are there in the answer whenthese two numbers are multiplied together? Divided by one another? Added together? The number 6.429e3 contains 4 significant figures while the number 3.18785e2 contains 6 significant figures. The product of these two numbers would be 2.049e5, the quotient of these numbers were 20.17 (2.017e1), and the sum of these numbers would be 6748 (6.748e3). 3. Express the following numbers in scientific notation: 673402.2, 34.623, 0.00008730, and 1,232,000. The number 673402.2 becomes 6.734022e5 when expressed in scientific notation, the number 34.623 becomes 3.4623e1 in scientific notation, the number 0.000008730 becomes 8.730e-5 in scientific notation. The number 1232000 becomes 1.232e6 in scientific notation. 4. The accepted value for the acceleration of gravity at the surface of the earth is 9.8 m/s2 . In an experiment, I measure the acceleration of gravity, and get the result 9.6 ± 0.3 m/s2 . Determine the experimental discrepancy between my value and the accepted value, both a fraction and as a percentage. Is the experimental discrepancy consistent with the relative error of my measurement? Why or why not? .2/9.8= 2.04, Yes it is consistent with the relative error of my measurement. The reason for this is because 2.04 falls within +/- 3....