Determining acceleration of free fall by of a simple pendulum.
Introduction
In this investigation I am going to use the simple pendulum to determine
the value of acceleration of free fall.
The method involves setting pendulums of certain lengths in motion and
timing the time taken for ten complete oscillations at that length.
Then using the formula
T = 2n K(l/g)
Therefore T2 = 4n2lg
I aim to find the value of acceleration of free fall from the gradient of a
graph of T2against l. From the above equation I hope to get a straight line
graph with a positive gradient passing through the origin.
As I hope to get a straight line graph the equation of the line will be in the
form of y = mx + c
But since the graph will pass through the origin my c = 0 If the above
formula is then re-arranged as follows
T2 = 4n2l
g
but T2 = gradient for the graph of T2against l l
Therefore it follows that gradient = 4n2 g
Thus the acceleration of free fall (g) will be given by the equation below
g=
4n2 gradient
My variables are length of pendulum (l), time for oscillations (T), height
from which pendulum is displaced and number of oscillations. My
dependent variable is the time T for the oscillations as it depends on the
pendulum length which is the independent variable. The number of
oscillations and the height from which the pendulum is displaced are my
constant and will not change.
The apparatus I use in the experiment are as follows:
 Pendulum bob
 Meter rule
 Stopwatch
 Stand and clamp
Method
After obtaining the above apparatus I started my investigation by hanging
the pendulum bob on a clamp as shown below, such that its length is
0.95m and set it into motion. I then measured the time for 10 complete
oscillations and record my results. I repeated the above procedure with
varying lengths of the pendulum bob of 0.9, 0.85, 0.80, 0.75, 0.70 and 0.65
m
Data collection and processing
After doing the above process I entered my data in the raw data table
below.
Raw data
Raw
data
measur
1e
2
3
4
5
6
7
Time for 10 Time for 1 T2 (T/s)
oscillations
oscillations
Uncertaint
(T/s)
(T/s)
y
19.7
1.97
Uncertainty
± Uncertainty
± 3.89
± 0.002s
19.2
1.92
3.69
2
2
0.001s
0.001s
18.7
1.87
3.53
2
2
18.1
1.81
3.30
8
8
17.6
1.76
3.12
6
6
17.0
1.70
2.90
6
6
16.5
1.65
2.73
3
3
3
I then plotted a graph of time2 against
length using3 the values shown in
the table above using graphing software. The uncertainty for the length
and time are relatively small hence I will ignore them in my graph.Graph
of time squared (T2/s) against length of pendulum (l/m)
Pendulum
length (L/m).
Uncertainty
0.95m
± 0.0005
0.90
0.85
0.80
0.75
0.70
0.65
I then proceeded to find the gradient of my graph as
follows Gradient (a) = 6y
6x
= 3.89 – 2.73
0.95 - 0.65
= 3.87
I proceeded to find the find the value of acceleration of free fall as
follows, using the equation.
g=
4n2 gradient
4n2 3.87
g =10.21 ms-1
g=
However as the straight line-graph does not pass through the origin, I
obtained the y- intercept of the graph using the graphing software to be
0.18. If this is taken into account then the new value for acceleration of
free fall will be more accurate as it has been corrected for the systematic
error that resulted in the entire line shifting upwards. Thus the new value
of acceleration of free fall will be:
10.21 - 0.18 = 10.03 ms-1
Conclusion and evaluation
The method used was sufficient enough to give an percentage
discrepancy of 4.01 % as shown below, the value being positive means it
was higher than the actual value of acceleration of free fall of 9.81 ms-;
this gives an accuracy of 96.2% as shown below Percentage discrepancy =
10.21 – 9.81 x 100 = 4.01 %
9.81
ccuracya gePercenta = 100 - 4.01 = 95.99
However when corrected for the systematic error that led to all the data
being skewed, and as result not passing through the origin, the value of
free fall determined was only 2.24% higher than the actual value and
improves the accuracy to 97.76% as shown below.
Percentage discrepancy = 10.03 – 9.81 x 100 = 2.24 %
9.81
gePercenta ccuracya = 100 - 2.24 = 97.76 %
In my experiment I expected the time of the oscillations to decrease as
the pendulum length is decreased, my data supported this. As I drew a
graph for the data it showed that the change of time for the oscillations
as the pendulum length is reduced, is proportional. Thus I was effectively
able to calculate the value of free fall of acceleration using the gradient of
the graph drawn.
My largest problem was the small scatter of data on the graph and the
data being skewed to one side of the graph due to a systematic error. This
may have been caused by my reaction time in starting and stopping the
stop watch, another reason for this could be that the string of the
pendulum may have extended when the pendulum was in motion, also
the stand holding the pendulum could be dangling from side to side as the
pendulum oscillates. Therefore I would suggest the following ways to
improve the data that will obtain better results.
● I would like to carry out the procedure of timing the oscillation at
different pendulum lengths, for a greater number of times so as to
obtain more data whose average I can use in the calculations to
obtain a more accurate answer.
● I would also like to time a greater number of complete oscillations
when the pendulum bob is set in motion so as to reduce errors that
may have arisen due to my reaction time.
● I would also like to carry out the procedure again with a different
set of apparatus, so as to eliminate the systematic error that arose
in this experiment.
● Finally I would recommend that a heavy mass be placed on the
stand holding the pendulum bob, so that it does not dangle when
the pendulum bob oscillates thus leading to a more accurate
answer.