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Carry out an experiment of simple harmonic motion using a simple pendulum and determine the acceleration due to gravity, to verify the equation T = 2P&amp;Ouml;(l/g) and show the relationship between time period and length.

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Introduction

Page  of

SIMPLE HARMONIC MOTION AND THE SIMPLE PENDULUM

Aim

To carry out an experiment of simple harmonic motion using a simple pendulum and determine the acceleration due to gravity, to verify the equation T = 2Π√(l/g) and show the relationship between time period and length.

Method

The apparatus is set up, as above, the string must be measured carefully with a ruler to minimise any error; the length should start off fairly short at about 0.2m for example.  The bob should be secure.

A table should be made for results, this should include length, time (twice), average time and time period (= average time/20).

The pendulum is set into motion by a gentle push, some practice in doing this and also counting and timing the oscillations beforehand may help to achieve greater accuracy.

The angle of amplitude should be kept similar if not the same for each experiment, to make sure the forces acting on it are the same.

It is easier to count the oscillations from the equilibrium position as this is where the pendulum has no energy.

The number of oscillations counted should be about 20 as it is easier to count a larger number.  The reaction time needed to count one single oscillation would be so small that errors could easily occur.

Middle

22.08

1.10

0.55

0.4

25.1

25.31

25.21

1.26

0.63

0.5

27.94

28.00

27.97

1.40

0.71

0.6

30.44

30.47

30.455

1.52

0.77

0.7

32.44

32.56

32.5

1.63

0.84

0.8

34.91

34.88

34.89

1.74

0.90

Analysis of Results

The gradient of the graph can be calculated from   ΔΥ    and is equal to  2ΔΧg

ΔΥ  = 1.18    = 1.98

ΔΧ     0.595

1.98 = 2  = 6.2

g      g

g =  6.2

1.98

= 3.13

g = 3.13²

g = 9.8ms²

Conclusion

The value I obtained for the acceleration of the simple pendulum due to gravity is 9.8ms², which is good.  This shows that the experiment was accurate and verifies the equation for the time period.  The value is slightly less than the accepted value of 9.81ms² for the acceleration of gravity which could be due to a number of uncertainties and errors that may have occurred throughout the experiment despite efforts to keep them at a minimum.

Timing for the 20 oscillations of each length was repeated because although it should take the same time, as amplitude does not affect the time period, we knew errors would occur and calculating an average time from 2 sets would be more efficient.  The main uncertainty in the experiment was human reaction time for starting and stopping the stopwatch which must be fairly fast because it is does not take long at all for one oscillation and errors could occur very easily.

Conclusion

Grandfather clocks have a time period of 2 seconds.

The length of the pendulum needed, can be obtained from the following equation

Where T = time period, L = length, g = gravity

T = 2L/g

2 = 2L/9.81

2  = L/9.81

2

(0.32)² = L/9.81

1. = L/9.81

L = 0.1 x 9.81

L = 0.981 m

The acceleration of the spring must be less than or equal to the gravitational force for the mass to not lose contact with the tray.  a < g

Therefore the maximum acceleration when contact is not lost must be 9.8ms².  With this information it is then possible to calculate the maximum amplitude the spring can have without losing contact, by first finding the time period.

where T = time period, m = mass, k = spring constant, f = frequency, a = acceleration and x = displacement.

T = 2Π√(m/k)

T = 2Π√(0.13/15)

T = 0.58s

Then the frequency needs to be calculated

f = 1/T

f = 1/0.58

f = 1.72Hz

And finally the displacement can be calculated from the equation for SHM.

a = - (2Πf)² x

x = a/(2Πf)²

x = 9.81/(2Π1.72)²

x = 9.81/116.8

x = 0.084

x = 8.4cm

If the mass were to lose contact with the tray, it would happen when the acceleration is greater than the force of gravity, and would happen at the highest point of displacement when the tray is at its maximum acceleration on the way back down.

A > g.

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