Physics Lab - Graphing Analysis Worksheet
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Northwest Vista College *
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139
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Mathematics
Date
Feb 20, 2024
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1 Data Analysis and Graphing Lab Online Name Melissa Fernandez Ayala Course/Section PHY 1951 011 Instructor Christopher Dunn, TA: Amilcar Torres Quijano 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.
2 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
2
= 0.00 s
2 xf – xi (0.872 – 0.872) = 0.00 m 0.10 0.922 0.10
2
= 0.01 s
2 xf – xi (0.922 – 0.872) = 0.05 m 0.20 1.061 0.20
2 = 0.04 s
2 xf – xi (1.061 – 0.872) = 0.189 m 0.30 1.287 0.30
2 = 0.09 s
2 xf – xi (1.287 – 0.872) = 0.415 m 0.40 1.635 0.40
2 = 0.16 s
2 xf – xi (1.635 – 0.872) = 0.763 m 0.50 2.079 0.50
2 = 0.25 s
2 xf – xi (2.079 – 0.872) = 1.207 m 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 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) Slope according to graph = 4.8166 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) The slope on a displacement (m) vs. time
2
(s
2 ) graph indicates the object's acceleration due to gravity
5.
From the value of your slope determine your experimental value for g. (6 points) The experimental value of g can be determined from the equation Δy = gt
2
/2 We know g/2 = 4.8166 due to the slope found in our graph (it’s half the acceleration due to gravity) , and 2 x 4.8166 (we are multiplying it by two since the slope is half of the experimental value for g) allows us to determine: g = 9.6332 m/s
2
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4 6.
Find the percent error of the experimental value of g, using g = 9.81 m/s
2
as the accepted value. (2 points) Percent error = expected – experimental / expected x 100 = 9.81 – 9.6332 / 9.81 x 100 = 1.81% 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 kg 20 kg
-1
100. 8.37 0.100 kg 10 kg
-1
200. 4.02 0.200 kg 5 kg
-1
400. 1.98 0.400 kg 2.5 kg
-1
800. 1.03 0.800 kg 1.25 kg
-1
1,600 0.47 1.600 kg 0.625 kg
-1
1.
Complete the above data chart. Show some calculations to receive credit. (8 points) Calculation for g
kg: SI unit for kg = 10
3 50 g x 1 kg / 10
3
g = 0.050 kg 100 g x 1 kg / 10
3
g = 0.100 kg 200 g x 1 kg / 10
3
g = 0.200 kg 400 g x 1 kg / 10
3
g = 0.400 kg 800 g x 1 kg / 10
3
g = 0.800 kg 1,600 g x 1 kg / 10
3
g = 1.60 kg Calculation for kg
kg
-1
: 1/0.050 kg = 20 kg
-1
1/0.100 kg = 10 kg
-1
1/0.200 kg = 5 kg
-1
1/0.400 kg = 2.5 kg
-1
5 1/0.800 kg = 1.25 kg
-1
1/1.60 kg = 0.625 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)
6 4.
What is the value of the slope of the trendline, with its units? (4 points) The slope on the graph: 0.7912x kg
1
ms
-2
5.
What physical quantity does the slope represent? What is the correct name for combination of units the slope possesses? (4 points) The slope represents the force acting on the system/being exerted upon it (F = ma); the correct name for the combination of units is Newton and the units are kg x m/s
2
Exercise 3.
The period of a pendulum T
is given by the following equation. 𝑇 = 2𝜋ඨ
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 (0.910)
2 =
0.8281 s
2 0.400 1.26 (1.26)
2 = 1.5876 s
2 0.600 1.58 (1.58)
2 = 2.4964 s
2
0.800 1.80 (1.80)
2 = 3.24 s
2
1.00 2.08 (2.08)
2 = 4.3264 s
2
1.20 2.22 (2.22)
2 = 4.9284 s
2
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7 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) Work is written on the column. 3.
Make a graph of T
2
vs. l.
Display the trendline on the graph. (10 points)
8 4.
What is the value of the slope, with its units? (6 points) The slope shown on the plot is 4.2088 s
2
/m (rise over run, T/l) 𝑇 = 2𝜋ඨ
𝑙
𝑔
Considering this equation, to find g with the slope of the plot we just need to derive this equation: (considering that 4.2008 = l/T) 4.2088 =
ସగ
ଶ
𝑔 =
ସగ
ସ.ଶ଼଼
ଶ
g = 9.38ms
-2