Fall22 Data Analysis and Graphing Online_8.16.22

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University of Texas *

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1951

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Mechanical Engineering

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Jan 9, 2024

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Data Analysis and Graphing Lab Online Name_____________________________________ Course/Section______PHY 1951_________________________ Instructor_________________________________ Introduction The purpose of this exercise is to learn some basic techniques of data analysis: conversion of units, plotting data, finding the slope of a graph, determining the units of the slope, using the units of the slope to determine the physical quantity the slope represents, and calculating the percent error of your results. Instructions for Graphing 1. You are to use a graphing program such as Excel, or something similar to make your graphs. a. Turn in all your graphs with this lab worksheet. 2. The graph needs to be titled as such, (First physical quantity vs Second physical quantity) 3. All graphs are to be plotted as y vs x. a. The first physical quantity goes on the y – axis (Vertical). b. The second physical quantity goes on the x – axis (Horizontal). 4. Each axis needs to be labeled by its physical quantity and with its units. a. As an example, an axis representing displacement where the units of measurement are meters needs to be labeled as such: Displacement (m). 5. Add a trendline (also called a Best-Fit Line) on the graph itself. a. Unless you are specifically told which trendline (best-Fit line) to use, use the one that best fits your data. b. Do not choose the option that “connects the dots”. This is not a Best-Fit Line. A Best-Fit Line is a line the “best” fits all of the data points. 1

Exercise 1. Table 1 show the data collected by a motion sensor for a ball, initially at rest, then allowed to freely fall straight downward. Table 1 Time, t(s) Distance from the sensor (m) t 2 (s 2 ) Displacement, Δy(m) 0 0.872 0 0.872 0.10 0.922 0.01 0.922 0.20 1.061 0.04 1.061 0.30 1.287 0.09 1.287 0.40 1.635 0.16 1.635 0.50 2.079 0.25 2.079 1. Fill in the t 2 , and the displacement columns. Remember that displacement is direct line length directed from the initial position to the current position. (8 points) 2. Plot displacement vs time (Δy vs. t). This means that Δy is the ordinate (vertical axis) and t is the abscissa (horizontal axis). Do NOT add a trendline to this graph. (6 points) 0 0.1 0.2 0.3 0.4 0.5 0.6 0 0.5 1 1.5 2 2.5 Displacement vs. Time Time (s) Displacement (m) 2

3. Plot Δy vs. t 2 . Apply a Best-Fit Line through the data points. Determine the value of the slope of this line, and its units. ( Do not calculate the slope !) (10 points) 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.5 1 1.5 2 2.5 f(x) = 4.82 x + 0.87 R² = 1 Displacement vs. Time Squared Time Squared (s2 ) Displacement (m) The slope of the best-fit line is 4.82 m/s 2 . 4. What physical quantity (velocity, acceleration, etc.) does the slope of this graph represent? (Note: you are NOT being asked to describe the relationship between displacement and the square of the time shown by the graph) Here is a hint: The magnitude of the displacement of a freely falling mass with the initial velocity of zero is given by ∆ y = 1 2 gt 2 . (6 points) By the given equation, ∆ y t 2 = 1 2 g. Since ∆ y t 2 , displacement divided by time squared, denotes the slope of the graph from question 3, that slope represents the physical quantity of acceleration. 5. From the value of your slope determine your experimental value for g. (6 points) By the given equation, ∆ y t 2 = 1 2 g and ∆ y t 2 = ¿ 4.82 m/s 2 ; thereby, the experimental value for g is 9.64 m/s 2 . 6. Find the percent error of the experimental value of g, using g = 9.81 m/s 2 as the accepted value. (2 points) 3

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percent error = | 9.64 m s 2 − 9.81 m s 2 | 9.81 m s 2 x 100 = ¿ 1.73% Exercise 2. Table 2 shows the acceleration of different masses on a level surface, with the same constant force being applied to each mass separately. Table 2 Mass, m (g) Acceleration, a (m/s 2 ) Mass, m (kg) 1/m (kg -1 ) 50. 15.72 0.050 20. 100. 8.37 0.100 10.0 200. 4.02 0.200 5.00 400. 1.98 0.400 2.50 800. 1.03 0.800 1.25 1,600. 0.47 1.600 0.6250 1. Complete the above data chart. Show some calculations to receive credit. (8 points) Grams to Kilograms Conversion: Grams x .001 = Kilograms 1/m Calculation: 1 Mass ∈ Kilograms = 1 m ∈ kg − 1 2. Plot a vs m, where mass is in kilograms. Do NOT add a trendline to this graph. (10 points) 4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 2 4 6 8 10 12 14 16 18 Acceleration vs. Mass Mass (kg) Acceleration (m/s2) 3. Plot a vs 1/m, where mass is in kilograms. Display the trendline on the graph. (10 points) 0 5 10 15 20 0 2 4 6 8 10 12 14 16 18 f(x) = 0.79 x + 0.07 Acceleration vs. 1/Mass 1/Mass (kg-1) Acceleration (m/s2) 4. What is the value of the slope of the trendline, with its units? (4 points) The slope of the trendline is 0.79 m ∗ kg s 2 . 5. What physical quantity does the slope represent? What is the correct name for combination of units the slope possesses? (4 points) 5

The slope represents the physical quantity of force. The correct name for this combination of units is a newton (N). Exercise 3. The period of a pendulum T is given by the following equation. T = 2 π √ l g Where l is the length of the pendulum, and g is the gravitational acceleration 9.81 m/s 2 . Table 3 shows the data of the period of a pendulum as a function of length. Table 3 l(m) T(s) T 2 (s 2 ) 0.200 0.910 0.828 0.400 1.26 1.59 0.600 1.58 2.50 0.800 1.80 3.24 1.00 2.08 4.33 1.20 2.22 4.93 1. Plot T vs. l. Do NOT add a trendline to this graph. (10 points) 6

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0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.5 1 1.5 2 2.5 Time vs. Length Length (m) Time (s) 2. Fill in the T 2 column. Show some work to receive credit. (10 points) T 2 =( T ∈ seconds ) 2 Ex. ( 0.910 ) 2 = 0.8281 s 2 ≈ 0.828 s 2 3. Make a graph of T 2 vs. l. Display the trendline on the graph. (10 points) 0 0.2 0.4 0.6 0.8 1 1.2 0 1 2 3 4 5 6 f(x) = 4.21 x − 0.04 R² = 1 Time Squared vs. Length Length (m) Time Squared (s2) 4. What is the value of the slope, with its units? (6 points) 7