COMPOUND PENDULUM LAB
APPLIED PHYSICS, LAB 7
GROUP 9
A. GROUP MEMBERS
Following are the group members of 9
B. TABLE OF CONTENTS
#
EXPERIMENT
Pg. #
1.
THEORY ......................................................................................................................................2
2.
ABSTRACT ..................................................................................................................................2
3.
PROCEDURE ...............................................................................................................................3
4.
GRAPHS AND TABLES ..................................................................................................................3
5.
DATA ANALYSIS ..........................................................................................................................4
6.
CONCLUSIONS AND RESULTS ......................................................................................................4
LAB INSTRUCTORS:
APPLIED PHYSICS LAB
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COMPOUND PENDULUM LAB
EXP
FINDING THE VALUE OF “g” USING A
COMPOUND PENDULUM
1. Theory
In case of a simple pendulum a weight is suspended from a pivot so that it can swing freely. On the
other hand, a compound pendulum, also known as physical pendulum, is the type of pendulum in which a
rigid body is free to rotate about a fixed horizontal axis. The mass of a simple pendulum is concentrated
within the bob attached to the inelastic and almost massless string. On the contrary, in a compound
pendulum the mass is uniformly distributed over its entire body. The appropriate length for calculating the
period of such pendulums is the distance from the pivot to the center of oscillation. This is the point that
is exactly located under the center of gravity at a distance from the pivot traditionally called the radius of
oscillation. It is related to the length of the pendulum, the moment of inertia of the pendulum about the
pivot point and its mass. When the rigid bar is pivoted at its center the distance is zero as it is suspended
at its center of gravity, and it does not act as a pendulum i.e.it fails to exhibit oscillatory motion. The
relationship between the distance from the center of gravity and time period of the compound pendulum
can be explained as:



Initially when the suspension point is at the end of the bar the time period has a greater value.
When the suspension point is brought closer to the center of gravity it becomes shorter.
However, when the distance is small, and the suspension point is very close to the center of gravity
the time period starts increasing again.
The explanation of parabolic curve between the graph of distance between the point of suspension and
center of gravity and time period of the compound pendulum can also be analyzed using the formula for
time period of the pendulum, which is:
𝑙
𝑇 = 2𝜋√
𝑔
Where, l is the distance length and g is the gravitational acceleration.
This formula can also be expressed as:
𝑇 2 = 4𝜋 2
𝑙
𝑔
Here, 𝑇 2 ∝ 𝑙, which shows a parabolic relationship.
2. Abstract
In this experiment we will try to find the value of gravitational acceleration (g) by using compound
pendulum. First of all, we take measurements of period of oscillations and length of a compound pendulum
using a stopwatch and a meter rule. Then, by using our observations, calculations, formulae (theoretical)
we calculated calculate the value of gravitational acceleration along with its error.
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COMPOUND PENDULUM LAB
3. Procedure
i.
The center of gravity(C.O.G) of the pendulum was found by balancing the pendulum on a
wedge. The point at which the pendulum was stable was marked as C.O.G.
ii.
The C.O.G was considered to divide the pendulum into 2 sides, namely A and B.
iii.
The distance bewtween the center of gravity and each hole on side A was measured and
recorded.
iv.
The distance bewtween the center of gravity and each hole on side B was measured and
recorded.
v.
Side A:
a. The pendulum was hung on the stand from the first hole on side A (numbering of holes
begins from end of pendulum).
b. The pendulum was slightly displaced and then released.
c. The time taken for 5 vibrations was noted. The avg time period(T) for one vibration was
then calculated.
d. The pendulum was then hung from the stand on the next hole.
e. Steps (ii) and (iii) were repeated to get avg time period at this hole.
f. Steps (ii) to (v) were repeated for all holes on side A.
vi.
Side B:
a. The pendulum was hung on the stand from the first hole on side B (numbering of holes
begins from end of pendulum).
b. The pendulum was slightly displaced and then released.
c. The time taken for 5 vibrations was noted. The avg time period(T) for one vibration was
then calculated.
d. The pendulum was then hung from the stand on the next hole.
e. Steps (ii) and (iii) were repeated to get avg time period.
f. Steps (ii) to (v) were repeated for all holes on side B.
vii.
Graph was then plotted using data gathered in previous steps.
viii.
The time period(T) was taken along the y-axis(drawn in the middle of graph paper. Distances of
holes from C.O.G. were taken along the x-axis.
ix.
Points were plotted on the graph to reveal two smooth curves on either side of the y-axis.
x.
A line, ABCDE, was drawn parallel to x-axis intercepting both curves but not near the bends of
the curves.
xi.
The distance AD was taken as l1 and BE as l2. The length, l, was calculated from l1 and l2. The
value of gravitational acceleration (g) was determined from l and T.
4. Graphs and Tables
a. Table
From end 𝐴
𝑑
𝑡5 𝑣𝑖𝑏
(cm)
(s)
48.3
8.1
43.2
7.8
38.0
7.5
32.9
7.7
27.9
7.8
𝑇
(s)
1.62
1.36
1.50
1.54
1.54
𝑑
(cm)
48.5
43.5
38.5
33.2
28.1
From end 𝐵
𝑡5 𝑣𝑖𝑏
(s)
8.1
7.8
7.7
7.6
7.8
𝑇
(s)
1.62
1.56
1.54
1.52
1.56
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22.9
17.9
12.7
7.6
2.7
7.8
8.0
8.8
10.9
16.0
1.57
1.61
1.76
2.18
3.2
23.2
17.9
13.0
8.0
3.0
7.9
8.0
8.8
10.9
16.0
1.58
1.61
1.76
2.18
3.2
b. Graph
Graph (x-axis values) not to scale.
5. Data Analysis
From Graph,
Distances:
AC=46.5 cm
EC=48.5 cm
CD=16.5 cm
BC=16.5 cm
𝒍𝒂𝒗𝒈 = 𝑙1 + 𝑙2 =
63+65
2
𝑙1 = 𝐴𝐵 + 𝐵𝐶 = 46.5+16.5 = 63cm
𝑙2 = 𝐸𝐶 + 𝐵𝐶 = 48.5+16.5= 65cm
= 𝟔𝟒cm
T = 1.625s
As we know,
𝒍
T=𝟐𝝅√𝒈 ⇒ 𝒈 =
g=
4×3,14152 ×64
1.6252
𝟒𝝅𝟐 𝒍𝒂𝒗𝒈
𝑻𝟐
= 𝟗𝟓𝟔𝒄𝒎𝒔−𝟐
6. Conclusions and Results
a. Error Evaluation
For error in g, we know the theoretical value to be 𝑔 = 980 𝑐𝑚𝑠 −2 so
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%𝑬𝒓𝒓𝒐𝒓 =
980−956
980
× 100 = 𝟐. 𝟒𝟒%
b. Results
Finally, we can combine our results as follows:
𝒈 = 𝟗𝟓𝟔 𝒄𝒎𝒔−𝟐 ± 𝟐. 𝟒𝟒%
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