Graphing Analysis LAB 1

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Northwest Vista College *

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1101

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Mechanical Engineering

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Apr 3, 2024

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Data Analysis and Graphing Lab Online Name: Course/Section: Instructor: 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 0.05 0.20 1.061 0.04 0.189 0.30 1.287 0.09 0.415 0.40 1.635 0.16 0.763 0.50 2.079 0.25 1.207 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) 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) 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) ∆ y / t 2 = 1 2 g The slope represents one half gravitational acceleration 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Displacement vs. Time Time (s) Displacement (m) 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 f(x) = 4.82 x − 0 Displacement vs. Time2 Time2 (s2) Displacement (m) 5. From the value of your slope determine your experimental value for g. (6 points) ∆ y / t 2 = 1 2 gt 2 4.8166 m/s = 1 2 g 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) 2

|(9.81-9.64)/9.81 | * 100% = 1.7% 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 0.05 20 100. 8.37 0.100 10 200. 4.02 0.200 5 400. 1.98 0.400 2.5 800. 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) 50. g * 1 kg/1000 g = 0.05 kg 1/0.05 = 20 kg - 1 2. Plot a vs m, where mass is in kilograms. Do NOT add a trendline to this graph. (10 points) 3. Plot a vs 1/m, where mass is in kilograms. Display the trendline on the graph. (10 points) 4. What is the value of the slope of the trendline, with its units? (4 points) (m/ s 2 )/(1/kg) m/s 2 *kg 0.7912 m/s 2 *kg 5. What physical quantity does the slope represent? What is the correct name for combination of units the slope possesses? (4 points) Newton, the quantity is FORCE 3

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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) 0 5 10 15 20 25 0 2 4 6 8 10 12 14 16 18 f(x) = 0.79 x + 0.07 Acceleration vs. Inverse Mass Inverse Mass (1/m) Acceleration (m/s2 4

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) 2. Fill in the T 2 column. Show some work to receive credit. (10 points) (0.910)(0.910) = 0.828 (1.26)(1.26) = 1.59 (1.58)(1.58) = 2.50 3. Make a graph of T 2 vs. l. Display the trendline on the graph. (10 points) 4. What is the value of the slope, with its units? (6 points) 4.21 s 2 /m 5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.5 1 1.5 2 2.5 Period vs. Length Length (m) Period (s) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 1 2 3 4 5 6 f(x) = 4.21 x − 0.04 Period2 vs. Length Length (m) Period2 (s2 ) 6

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