P 251
Laboratory Activity III
Graphical Analysis II
Nonlinear Graphs
Graphical Analysis Exercise
Determining 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
2.1
4.8
8.8
11.5
17
DATA
Circumference, cm
7.5
15.4
28.3
36.2
53.1
Plot a graph of Circumference versus diameter.
CALCULATIONS AND OBSERVATIONS:
1. Is your graph a straight line?
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  Y  C  36.2cm  15.4cm 
X d 11.5cm  4.8cm
20.8cm  3.1
6.7cm
Slope has NO units
What is the equation relating Circumference and diameter?
Y = mX+b
C 3.1 d + 0
C 3.1d
Compare slope = 3.1 to p (3.14)
% error 
experimental value - accepted value
accepted value
3.1  3.14
.04  100%

% error 
 100% 3.14
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.
Data
time-t, s
0
1
2
3
4
5
6
7
8
9
10
distance-d, m
5
9.9
24.6
49.1
83.4
127.5
181.4
245.1
318.6
401.9
495
We then plot a graph of distance versus time.
Distance versus Time
500
Not a straight line
but is a uniform curve
450
400
distance, m
350
300
250
200
150
100
50
0
0
1
2
3
4
5
6
time, s
7
8
9
10
11
Compare graph to
graphs of other
functions of the
independent variable
7
y =x
Y X
6
Y  mX  b
5
4
Y
3
2
1
0
0
1
2
3
4
X
5
6
7
40
y =x2
YX
Y  mX  b
2
30
Y
2
20
10
0
0
1
2
3
4
X
5
6
7
y = x1/2
2.5
2
1.5
Y X
Y
Y  m X b
1
0.5
0
0
2
4
X
6
8
1.25
y = 1/x
1
Y
X
1
Y  m 1 b
X
0.75
Y
0.5
0.25
0
0
1
2
3
4
X
5
6
7
1.25
y = 1/x2
1
Y 2
X
Y  m 12  b
X
1
0.75
Y
0.5
0.25
0
0
1
2
3
4
X
5
6
7
1.1
y = 1/x1/2
1
1
Y
X
0.9
1
Ym
b
X
0.8
Y
0.7
0.6
0.5
0.4
0
1
2
3
4
X
5
6
7
7
40
Y X
y =x
Y  X2
6
Y  mX  b
5
y = x1/2
2.5
y =x2
2
Y  mX  b
30
2
Y X
1.5
4
Y

20
Y
3
2
10
Y
Y  m X b
1
0.5
1
0
0
1
2
3
4
5
6
0
0
7
0
X
1
2
3
4
5
6
0
7
Y
0.5
8
1.1
1.25
y = 1/x
Y  12
X
Y  m 12  b
X
6
X
y = 1/x2
0.75
4
X
1.25
1
2
Y 1
X
1
0.75
Y
Y 1
X
1
0.9
Y  m 1 b
X
y = 1/x1/2
Y  m 1 b
X
0.8
Y
0.7
0.5
0.6
0.25
0.25
0.5
0
0
1
2
3
4
X
5
6
7
0.4
0
0
0
1
2
3
4
X
5
6
1
2
3
4
7
X
5
6
7
Plot a new graph where time squared
is the independent variable:
Distance, d versus Time Squared,
2
t
Revised Data Table
time-t,s time2-t2, s2 distance-d,m
0
5
0
1
9.9
1
2
24.6
4
3
49.1
9
4
83.4
16
5
127.5
25
6
181.4
36
7
245.1
49
8
318.6
64
9
401.9
81
10
495
100


Distance versus Time Squared
500
450
400
distance, m
350
300
250
200
150
100
50
0
0
10 20 30 40 50 60 70 80 90 100 110
time squared
Analysis of Graph
Y  mX b
distance
time
squared
d
2
t
2
d  mt  b
Slope Calculation :
m = Y  d2
X t
Points Chosen :
2
2
(4s , 24.6m) and (64s ,318.6m)
24.6m  294m
m = 318.6m
2
2
2
64s  4s
60s
m  4.9 m2
s
2
m
d  4.9 2 t  b
s
With units of m/s2 the slope represents the acceleration of the
object.
Distance versus Time Squared
500
450
400
distance, m
350
300
250
200
intercept,b
150
It will be difficult to
determine the intercept from
the graph!
100
50
0
0
10 20 30 40 50 60 70 80 90 100 110
time squared
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.
2
t  0  b  5m
2
m
 4.9 2  t  b
s
2
m
b  d  4.9 2  t
s
2. Start with the partial equation: d
Solve for “b”:
Choose any data pair and substitute the values of “d” and “t2”
into the equation for “b”:
(25s2, 127.5m)
2
m
b  127.5m  4.9 2  25s
s
b  5m
Final Equation
2
m
d  4.9 2 t  5m
s
Dependence of Radiation
Intensity on Distance
from Source
PURPOSE:
The purpose of this laboratory exercise is to investigate the
transfer of energy by radiation and the dependence of intensity
on distance.
INTRODUCTION:
Radiation is the mechanism of heat (and other energy) transfer
by electromagnetic waves. Electromagnetic radiation can be
classified according to its frequency (f or u) and the energy
transferred. The energy transported by an electromagnetic wave
is directly proportional to its frequency. Typically EM radiation is
divided into eight categories called the EM spectrum. Listed in
order of increasing frequency (increasing energy) the components
are: 1) Radio, 2) Television, 3) Microwaves, 4) Infrared, 5) Visible
Light,6) Ultraviolet, 7) X-Rays, 8) Gamma Rays. Infrared
radiation is what we sense as heat. The visible light component
can be further broken down into the visible light spectrum where
different frequencies appear as different colors. Listed in order of
increasing frequency: Red, Orange, Yellow, Green, Blue, Indigo,
Violet
As radiation travels from a source it spreads
spherically.
As it spreads the intensity (brightness) decreases.
Source
Distance
from
Source
How does the intensity depend on the distance?
Experimental Apparatus
123.5
Selector
Turn Selector counterclockwise
to 200 m
The radiation detector converts radiation emitted by the light bulb
into an electrical voltage which is measured by the multimeter.
Procedure
Step 1. Set the Radiation Detector next to the meter stick. Align the
detector opening with the height of the light bulb and 20 cm away.
Step 2. Turn on the light to maximum brightness. Slide the ring on
the radiation detector forward to uncover the aperture. Turn
multimeter selector counterclockwise to 200m (V).
Step 3. Record the detector's voltage output in the data table.
Step 4. Move the detector to 30 cm from the bulb and record the
detector's output.
Step 5. Repeat Step 4 for distances to 100 cm in 10 cm increments.
Data Table
Distance from Source, cm Radiation, mV
20
65.3
Data Table
Distance from Source, cm Radiation, mV
20
65.3
30
52.8
40
45.0
50
37.3
60
28.1
70
14.6
80
8.3
90
5.5
100
0.8
Complete the extended data table.
Plot graphs of Radiation, R versus:
1) distance, d
2) reciprocal of distance, 1
d
3) reciprocal of the square root of distance,
4) reciprocal of distance squared,
1
d
1
2
d
Each group member do a different graph.
R, mV
Conclusion
Put a 4 by the proportion that gives the relationship
between radiation intensity, R and distance from the source, d.
__R versus d
__R versus 1
d
__R versus 1
__R versus 12
d
d
Suppose the radiation intensity was proportional to the distance
2
squared, R  d .
If the distance from the source was doubled (multiplied by 2) the
radiation intensity would be multiplied by 22 = 4. if the distance
was multiplied by 3 (tripled) the intensity would be multiplied by
32 = 9.
If the distance we cut in half (multiplied by 1/2) the intensity would
be multiplied by (1/2)2 = 1/4.
Suppose that at 50cm from the light bulb the light intensity was
100 mV, according to your graphical analysis, at a distance of
100cm the intensity would be ______mV.
Suppose that at 50cm from the light bulb the light intensity was
100 mV, according to your graphical analysis, at a distance of
25cm the intensity would be ______mV.