PSC 151Laboratory Activity 3

Graphical Analysis IIA Nonlinear Graphs 1

andThe Acceleration Due to Gravity

Graphical Analysis ExerciseDetermining the Relationship between

Circumference and Diameter

Procedure:

1. Measure the Circumference and diameter of five circular objects.

2. Analyze data using graphical analysis.

diameter, cm Circumference, cmDATA

53.1

2.14.88.811.517

7.515.428.336.2

DIAMETER, cm

CIRCUMFERENCE vs. DIAMETER

2468

10

20

30

40

50

60

70

80

Plot a graph of Circumference versus diameter.

1. Is your graph a straight line?

CALCULATIONS AND OBSERVATIONS:

YES

2. Does the graph pass through the origin? YES…b = 0

3. Are circumference and diameter directly proportional? YES

4. Calculate the slope; Points Used:

(4.8cm,15.4cm) & (11.5cm,36.2cm)

slope =m=ΔYΔX =ΔC

Δd =36.2cm −15.4cm11.5cm−4.8cm =

20.8cm6.7cm =3.1

Slope has NO units

Y = m⋅X+b

What is the equation relating Circumference and diameter?

C d3.1⋅ 0+

C=3.1⋅dCompare slope = 3.1 to 314)

%error=experimental value-accepted value

accepted value×100%

% error =3.1 −3.14

3.14×100% = .04

3.14× 100% =1.3%

Non-Linear Graphs

What procedure do we follow if our graph is not a straight line?

Consider an experiment designed to investigate the motion of an object.

We want to determine the relationship between the object’s distance traveled and time.

We measure its distance each second for 10s.

Here is the resulting data.

time-t, s distance-d, m0 51 9.92 24.63 49.14 83.45 127.56 181.47 245.18 318.69 401.910 495

Data

We then plot a graph of distance versus time.

0

50

100

150

200

250

300

350

400

450

500

0 1 2 3 4 5 6 7 8 9 10 11

time, s

Distance versus Time

Not a straight line but is a uniform curve

Compare graph to graphs of other functions of the

independent variable

0

1

2

3

4

5

6

7

Y

0 1 2 3 4 5 6 7

X

y=x

Y =mX+b

Y ∝ X

0

10

20

30

40

Y

0 1 2 3 4 5 6 7

X

y=x2

Y ∝ X2

Y =mX2 +b

0

0.5

1

1.5

2

2.5

Y

0 2 4 6 8

X

y = x1/2

Y ∝ X

Y =m X +b

0

0.25

0.5

0.75

1

1.25

Y

0 1 2 3 4 5 6 7

X

y = 1/x

Y ∝ 1X

Y =m 1X +b

0

0.25

0.5

0.75

1

1.25

Y

0 1 2 3 4 5 6 7

X

y = 1/x 2

Y ∝ 1X2

Y =m 1X2 +b

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

Y

0 1 2 3 4 5 6 7

X

y = 1/x 1/2

Y ∝ 1X

Y =m 1X +b

0

1

2

3

4

5

6

7

Y

0 1 2 3 4 5 6 7

X

y=x

Y =mX+b

Y ∝ X

0

10

20

30

40

Y

0 1 2 3 4 5 6 7

X

y=x2

Y ∝ X2

Y =mX2 +b

0

0.5

1

1.5

2

2.5

Y

0 2 4 6 8

X

y = x1/2

Y ∝ X

Y =m X +b

0

0.25

0.5

0.75

1

1.25

Y

0 1 2 3 4 5 6 7

X

y = 1/x

Y ∝ 1X

Y =m 1X +b

0

0.25

0.5

0.75

1

1.25

Y

0 1 2 3 4 5 6 7

X

y = 1/x 2

Y ∝ 1X2

Y =m 1X2 +b

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

Y

0 1 2 3 4 5 6 7

X

y = 1/x 1/2

Y ∝ 1X

Y =m 1X +b

¸

0

50

100

150

200

250

300

350

400

450

500

0 1 2 3 4 5 6 7 8 9 10 11

time, s

Distance versus Time

Plot a new graph where time squared is the independent variable:

Distance, d versus Time Squared, t2

time-t,s distance-d,m0 51 9.92 24.63 49.14 83.45 127.56 181.47 245.18 318.69 401.910 495

Revised Data Table

0149

162536496481

100

time2-t2, s2

Y=m⋅X +b

Analysis of Graph

distance timesquared

d t2

d=m⋅t2 +b

Slope Calculation :m = ΔY

ΔX = ΔdΔt2

Points Chosen :

(4s2 , 24.6m) and (64s2 ,318.6m)

m = 318.6m −24.6m64s2 −4s2 =294m

60s2

m =4.9 ms2

With units of m/s2 the slope represents the acceleration of the object.

d=4.9 ms2 ⋅t2 +b

intercept,bIt will be difficult to

determine the intercept from the graph!

Two Other Methods for Determining the Intercept

1. The intercept is the value of the dependent variable where the graph intersects the vertical axis. At this point the value of the independent variable is zero.

Look at the data table to determine the value of d where t2 equals zero. t2 =0⇒ b=5m2. Start with the partial equation:

Solve for “b”:

d =4.9 ms2 ⋅t2 +b

b =d−4.9 ms2 ⋅t2

Choose any data pair and substitute the values of “d” and “t2” into the equation for “b”: (25s2, 127.5m)

b =127.5m−4.9 ms2 ⋅25s2 b =5m

Final Equation

d=4.9ms2 ⋅t2 +5m

Graphical Analysis of “Free-Fall” Motion

Determining the Acceleration Due to Gravity

Purpose:In this lab, you will determine the correct description of free-fall motion and to measure the value of the

acceleration due to gravity, g.

Introduction: The Greek natural philosopher Aristotle was one of the first to attempt a “natural” description of an object undergoing free-fall motion. Aristotle believed that objects moved according to their composition of four elements, earth, water, air, and fire. Each of these elements had a natural position with earth at the bottom, then water, then air, and fire at the top. If a rock, composed primarily of earth, was held in the air and then released its composition would cause it to return to the earth. Accordingly, Aristotle thought that objects fell with a constant speed which was proportional to the object's weight, that is, a heavier object would fall faster than a lighter one.

Motion at a constant speed can be described by the equation:

Comparing the equation above with the slope-intercept equation of a straight line, Y = mX + b,

where d is the distance fallen, v is the speed, t is the time the object has been falling, and d1 is the initial distance from the origin.

we see that a graph of distance fallen versus time should be a straight line with di as the y-intercept, and the slope of the line would give the speed, v, at which the object was falling.

d =v⋅t + di

d = v⋅ t + di

Y =m⋅X + b

where d is the distance fallen, a is the acceleration, t is the time the object has been falling, and di is the initial distance from the origin.

In the late 16th and early 17th centuries Galileo challenged much of the work of Aristotle. Working with objects rolling down inclined planes he demonstrated that objects fall with a constant acceleration that is independent of their weight. According to Galileo objects fell with a speed that changed uniformly and at the same rate for all objects.

Motion at a constant acceleration, starting from rest, can be described by the equation:

d =12 a⋅t2 + di

Comparing the equation above with the slope-intercept equation of a straight line, Y = mX + b,

we see that a graph of distance fallen versus time squared should be a straight line with d1 as the y-intercept and the slope of the line would equal one-half of the acceleration at which the object was falling.

Y m X +b

d=12 a⋅t2 + di

To find the true nature of Free-Fall:

Let a ball roll down an incline,

Measure the distance traveled after certain times,

Plot graphs of distance versus time and distance versus time-squared.

If distance versus time is a straight line then Free-Fall is at a constant velocity and the slope of the graph measures that velocity.

If distance versus time-squared is a straight line then Free-Fall is at a constant acceleration and the slope of the graph measures one-half of that acceleration.

v, velocity = m, slope

a, acceleration = 2m, 2 x slope

The Acceleration Due to Gravity

If distance versus time-squared is a straight line then Free-Fall is at a constant acceleration and the slope of the graph measures one-half of that acceleration.

The acceleration, a, found from the slope of the d vs t2 graph is related to but not equal to the acceleration due to gravity, g.

To find the actual value of g we must account for the effect of the incline.

g =a×Lh

L

h

CBR

CBL

Experimental Procedure

CBR / TI-83 Set-UPConnect the CBR to the TI-83

Press: APPS

Press “4” CBL/CBR

Press “Enter”

Press “2” Data Logger

Data Logger Set-Up

Probe

# SAMPLES

INTRVL (SEC)

UNITS

PLOT

DIRECTNS

GO...

Sonic

Enter: 20

Enter: .04

Select: m

Select: REAL TIME

Select: ON

Press "Enter"

Press: Enter

CBR/CBL Set-Upcontinued

Press “2” CBR

After CBR-CBL link has been tested:Press: “Enter”

After “Status OK”:Press: “Enter”

When you are ready to begin taking data”Press: “Enter”

After data collection is complete the TI-83 will plot a graph of distance versus time.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Time, s

Use this key to advance cursor

* * **

**

*

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Time, s

(0.4, 0.676)

(0.6, 0.847)

(0.8, 1.086)

(1, 1.393)

(1.2, 1.768)

Length, L = 162cm

Time-t, s Time Squared-t2, s2 Distance-d, m

Height, h = 27.7cm

Data Table 1

Plot graphs of:distance versus time

and distance versus time squared