Hooke’s Law Experiment
for Computer Applications (Geng0015)
Submitted by: Omar El
Hossieny, oeh1g14
§
Overview: I have carried out a Hooke’s Law experiment
that investigates the behaviour
of three materials, giving the following set of results. The first two results,
Y1 andY2, are for two different elastic materials, which are both still in
their linear regions. The results, Z, describe the behaviour of a material which has gone
past its elastic region (in the plastic region). X is the force applied (in Newtons)
and y1, y2 and z are the deformation (in mm).
§
Objective: I aim to tabulate my results of
the 3 experiments and plot them in a matter that would help me observe the
behavior and trend shown by each of the three materials Y1,Y2 and Z, which will
help me analyze the correspondence and relationship between them and come up
with some conclusions about the experiment.
Ø What is Hooke’s Law?
"It’s the law of elasticity
discovered by the English scientist Robert Hooke
in 1660, which states that, for relatively small deformations
of an object, the displacement
or size of the deformation is directly proportional to the deforming force or
load. Under these conditions the object returns to its original shape and size
upon removal of the load. Elastic behavior of solids according to Hooke’s law
can be explained by the fact that small displacements of their constituent molecules,
atoms,
or ions
from normal positions is also proportional to the force that causes the
displacement."
It’s
satisfied by the equation: F= kΔL
F is the force applied
K is the springs constant
ΔL is the change in length
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| Picture 1 |
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| Picture 2 |
§ Data
1. Equations
Y1
|
ax+b
|
Y2
|
(a+0.5)*x+0.2
|
Z
|
X**3+b
|
2. Table
I used the equations shown in
Figure 1 to tabulate the values using an excel spreadsheet.
Table Of Values
|
||||
X
|
Y1
|
Y2
|
Z
|
|
1.00
|
3.00
|
2.26
|
2.38
|
|
2.00
|
4.50
|
4.32
|
9.38
|
|
3.00
|
6.00
|
6.37
|
28.38
|
|
4.00
|
7.50
|
8.43
|
65.38
|
|
5.00
|
9.00
|
10.49
|
126.38
|
|
6.00
|
10.50
|
12.55
|
217.38
|
|
7.00
|
13.00
|
14.61
|
344.38
|
|
8.00
|
14.00
|
16.67
|
513.38
|
|
9.00
|
15.00
|
18.72
|
730.38
|
|
(Figure 2)
3. Data
Representation
The initial values tabulated
where the ones for X and Y1 in order to calculate the a and b to help us derive
the equations for Y2 and Z. Figure 3 below shows
the relationship between X which is force in Newtons and Y1 which is the
deformation in mm.
 |
Figure 3 |
After plotting the graph of X
against Y1 and using a trendline to show the relationship between them, I was
able to work out an equation which helped me figure out the two unknowns a
and b. From the equation on top of the line in Figure 3 I
now know that a=1.5583 and b=1.375. Using those values for the unknowns I
was able to write the functions for Y2 and Z providing me with new equations
which I typed into the fx
tab in excel to help me calculate the values, and are clear in
Figure 4.
Y1
|
1.5583x+1.375
|
Y2
|
(1.5583+0.5)*x+0.2
|
Z
|
X**3+1.375
|
Having found the Y2 data ,I
plot on the same graph for Y1 the graph for Y2. On the graph the
equation for line Y2 is shown.
 |
Figure 5 |
As we can see in Figure 5 the
two linear equations for Y1 and Y2 clearly intersect with each other at one point.
Initially I estimated the intersection
point to be around (2.4,5) but in order to check my estimation I solved the two
equations simultaneously through the next steps
Y1=Y2
2.0583x + 0.2 = 1.5583x + 1.375
1.375- 0.2 = 2.0583x- 1.5583x
1.175 = 0.5x
x=2.35
And by
replacing 2.35 as x in the equation we should know Y:
y= (2.0583 * 2.35) + 0.2= 5.0370
X= 2.35 Y= 5.037
|
The point of intersection
between Y1 and Y2 is (2.35,5.037) so I could say that my estimate was almost
correct as it was close to the exact values.
After calculating the values
for Z I was able to draw a graph of X( Force applied) against Z
 |
Figure 6 |
We can see from this diagram that the shape of the line is not linear like Y1 and Y2 but is an exponential trendline proving that Z is not in it’s elastic region and doesn’t obey Hooke’s law anymore.
§
Data Analysis
Referring to Hooke’s law, it
states that the extension of a spring is directly proportional to the
stretching force up to the elastic limit, providing us with the equation F=Kx
where K is the springs constant.
·
For Y1 things are not straight forward, there is
direct proportionality but not obeying
Hooke’s Law that well which shows that there is already a slight deformation in
the spring. We can see that in Figure 3 as line Y1 never passes through origin which
is an imperative for a direct relationship to occur, instead the line starts at
(0,1.375).
·
For Y2 if we look back at Figure 5 we can see that the trend line it provided us matches Hooke’s law much better though not a 100% still as
it produces a directly proportional relationship with the force through a line
passing through (0,0.2) which is really close to the origin, that’s due to the
extra constant added which helped provide this linearity. So we can conclude
that Y2 was still in its elastic region with no deformation to the spring and
material Y2 had much better elasticity than material 1.
If there was no deformation in
the spring used for Y1 the graph should’ve looked like Figure 7, with no intersections.
 |
Figure 7 |
- For Z, the graph in Figure 6 proves that Z has passed its elastic point and is not linear anymore but more of an exponential trend line with varying responses to force added. Z extends more with weaker force, so as more force is added the linearity decreases and more deformation occurs. The spring Z faces irreversible deformation and will never go back to its original shape. Moreover if any other forces are added the spring might reach its fracture point and break.
§ Conclusion
Every material has a different elasticity, and as long as more
weights are added and a bigger force is applied but the spring remains in its
elastic region it will always return to its original shape as in the case for
Y1 and Y2. On the other hand if the material passes its elastic limit it will
be deformed and can never return to its original shape as what happened to
material Z. Furthermore if you keep adding weights after the material surpasses
its elastic limit it might reach a fracture point and break.
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| Picture 3 |
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| Picture 4 |
§ Errors
Of course
no experiment is flawless, so my experiment might have had some
inaccuracy or miscalculation. I didn’t do the experiment myself but when handed
the graphs I noticed some irregularities.
For
the graph of Y1 shown in Figure 3 you can notice that there is a point that
doesn’t fit the line which is the point at x=7.00 N. If I would take a guess I
would say measurements were taken using a ruler which in some cases might be a
little bit inaccurate, as there is always the uncertainty of ± the smallest measurable unit which in our
case is a millimeter. Or simply that could be a parallax error when taking
measurement due not proper angle of measurement. We should take into account
that the error didn’t affect the graph that much as a best fit line was done
and not a point-to-point line. Other than that the experiment and graphs were
done pretty well with no visible errors that might affect the outcome of the
experiment.
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| Picture 5 |
- References:
http://www.britannica.com/EBchecked/topic/271336/Hookes-law (Picture 2,4)
http://bernetwpcaroline.blogspot.co.uk/2010/05/parallax-error.html (Picture 5)
http://bernetwpcaroline.blogspot.co.uk/2010/05/parallax-error.html (Picture 5)
