To Investigate The Relationship Between The Period And The Length Of A Simple
Pendulum And From This To Calculate Acceleration Due To Gravity (g).
Split cork
l
Timer
Bob
20:30
Equipment Needed:
 A small spherical mass to which thread can be tied (ie. a pendulum bob)
 A split cork, some thread, a retort stand and a clamp.
 A timer or a stopwatch and a metre stick.
Procedure
 Measure length from the bottom of a split cork to the centre of the bob

Measure length in meters

Use a split cork to hold the thread for definite suspension in one plane only

Use an angle of less than 5 degrees

Count for 30 oscillations:

Remember to divide by 30 to get the time for ONE oscillation

Too few oscillations will give error in time

Too many oscillations will result in more air friction and loss of energy

Count in seconds

When calculating T2, ensure to divide the time by 30, before squaring

Put length into meters

Draw graph of length versus time squared

Use an in extendable string to get a fixed length

Ensure length is greater than 40 cm to ensure little % error in the length.

Repeat experiment by increments of 10cm

Use a small heavy spherical bob:


You know where the centre of gravity is

Too big a bob will increase the air friction and lose energy

A heavy bob will ensure a fixed length
Include the formula in your answer…..
g
=
42l/t2
Which becomes
g
=
42 (slope)
Precautions:
1. The bob must swing in one plane.
2. The angle must be less than 5
3. Count 30 oscillations
4. Use a medium ball
5. Use a split cork
SAMPLE QUESTIONS
1. Draw a suitable graph to investigate the relationship between the period of the
simple pendulum and its length. What is this relationship? Justify your answer.
Graph should be L / m against T2 for 1 oscillation / seconds
From the graph, we see a straight line through the origin.
This shows that L is proportional to T2
Hence T2 is proportional to L
Hence T is proportional to Square Root of L
Note, that there are a lot of relationships that can be formed by the formula
2. Describe how the pendulum was set up so that it swung freely about a fixed
point?
Through a split cork on a clamp of retort stand. (as seen in the diagram)
3. Give a precaution taken when allowing the pendulum to swing?
The bob must swing on the one plane.
4. What is the advantage of having a large number of oscillations?
It means that we should make less of an error when finding the time for one
oscillation and hence reduce time percentage error.
5. Why do you use a medium sized bob?
Because if the bob is too big the air resistance will make the experiment inaccurate.
But if the bob is too small it will not keep the string taut and this may also effect the
experiment.
6. Why should the length of the pendulum not be less than 40cm?
It shouldn’t be less than 40cm because percentage errors in measuring its length
becomes significant.
7. How does the number of oscillations timed affect the accuracy of the
experiment?
If there is too many oscillations air friction becomes a factor but if there is only a few
oscillations time percentage error increases.
8. Why must the angle that the pendulum makes with the vertical be small?
Because the formula for simple harmonic motion only works accurately with an angle
less than 5.
9. Why is the time for one oscillation not measured directly?
If you measured one oscillation the time would be a significant % error. .
10. How do you measure the length?
From the bottom of the split cork to the centre of gravity of the bob. Using a ruler.
11. Why do we use a bob?
Because we know where its centre of gravity is and we measure length to the centre of
gravity.
12. What effect would the period experience if the same measurements were taken
on the moon where gravity is a sixth that of earth when
L=1m g=1.63ms-2
and T1 for one oscillation =1.98s? (Data is taken from a previous question)
g = 42 _L_
T2 = 42 _L_
T2 ,
g
If the denominator gets smaller, then T2 gets bigger, hence T gets bigger.
Proof
T2 = 42 _L_
G
T2 = 42(1)
1.63
T2 = 24.22
T = 4.92 which is bigger than 1.98s
13. A pendulum executes 40 oscillations in 30 seconds. What is the length (l) of the
pendulum (g =9.8ms-2)
= 42__L__
T2
g
40 oscillations in 30 sec
We know g =9.8ms-2
L
=
g T2
42
L
=
(9.8)(0.5625)
42
L
=
0.14m
=
0.75 for T1
=
T2
=
0.5625
14. From YOUR graph, determine the time it would have taken for 30 oscillations,
had the length been 65cm?





GO to your graph and draw a horizontal line from 0.65 m to the line and down to
the x-axis.
In this case we find the reading on the x-axis to be 2.56 (Taken from a booklet!!!)
Hence T2 is 2.56 s
Hence T is 1.6 s
Hence, with 30 oscillations the time is 30 X 1.6 = 48 s
13. This pendulum was placed on the moon where gravity is 1/6 of earth, find
the number of oscillations in one minute.
What do we know:
Gearth = 9.8 ms-2
Gmoon = 1.63 ms-2
Length of THIS pendulum is 0.14m
If we find the time for one oscillation, then we can find the total number of oscillations in
1 min (60 seconds) by 60/ T1 oscillation
g = 42__L__
T2
T2 = 42 L
g
T2 = (42)(0.14)
1.63
T2 = 3.39
T = 1.84
If time for one oscillation is 3.11s, then in one min (60s) we have:
60
1.84 =
32.6 oscillations