Essay Color Key

Free Essays
Unrated Essays
Better Essays
Stronger Essays
Powerful Essays
Term Papers
Research Papers





Factors Affecting The Time Period of a Simple Pendulum

Rate This Paper:

Length: 3183 words (9.1 double-spaced pages)
Rating: Red (FREE)      
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Factors Affecting The Time Period of a Simple Pendulum


I plan to investigate the different characteristics of a simple
pendulum and the factors affecting its time period.

Any motion that repeats itself in equal intervals of time is called
periodic motion.

Simple harmonic motion (SHM) also is a periodic motion, but has some
features:

· It is periodic

· Motion is to and fro along a straight line about an equilibrium
(mean) position

· Acceleration is directly proportional to displacement

· Acceleration is directed towards the equilibrium position

Therefore, simple harmonic motion is a periodic motion where
acceleration is directly proportional to displacement, and is always
directed towards the equilibrium position.

Simple Pendulum-

What we call a simple pendulum is a string with a mass tied to the end
of it. The mass is called a 'bob'. If we had kept the string with the
bob attached to it at a fixed point and then we displace it from this
equilibrium position, the movement of the bob we would see would be a
to and fro movement where the bob would move in a straight arced line
from left to right. The distance of the movement of the pendulum away
from the equilibrium position is called its displacement.

(Diagram 1 shows you the basic diagram of what was explained above)

The three different types of oscillations are free, damped and fixed
oscillations.

· Free oscillations occur while the pendulum, is set to its
displacement and is moving in its to and fro motion it doesn't
experience any force that prevents it from continuing this motion.
Such forces that prevent free oscillation are air resistance, etc.

· Damped oscillations occur while the pendulum, is set to its
displacement and is moving in its to and fro motion, experiences a
force, or a medium that affects its motion. For example air
resistance, water, etc.

· A forced oscillation occurs while an object is used to force one or
more pendulums into motion. An example of this is using a driving
pendulum to control the displacement of a set of 4 pendulums, which
move as a result of the driving pendulum being displaced. Another
example is using a vibrating tuning fork to force a stretched string
to vibrate and set the pendulum into motion.

Time period-

The Time Period is the time that is required for one complete
oscillation. (T)

Frequency is the number of oscillations per second, which is measured
in hertz (Hz).



Angular Velocity- the angle at which the radius vector subtends at the
center in one second is called the angular velocity of the particle.
Its unit is radians per second. That is, (rad s-1)
======================================================================

ω = ∆ (theta) / ∆t = 2π / t

Key:

ω = omega

∆ (theta) = the angle which is covered

∆t = the time taken

(Diagram 2 illustrates the simple pendulum using the formulae
specified above)

Suppose that you have a vector quantity, V. Theta is the angle dropped
to the horizontal component. Any force can be resolved into a vertical
component and a horizontal component.

Cosine Theta = Adjacent / Hypotenuse Sine Theta = Opposite /
Hypotenuse

Cosine Theta = Adjacent / V Sine Theta = Opposite / V

Adjacent = V (Cosine Theta) Opposite = V (Sine Theta)

In the below diagram, the weight, 'Mg' (mass into gravity), can be
resolved into two components. The first component 'Mg cosine'
compensates for the tension in the string. 'Mg sine' is the component
that provides the restoring force. The restoring force is required
because it brings the mass back towards the equilibrium position.

Initially it is a forced oscillation, but after two or three
oscillations, the pendulum is in simple harmonic motion. We take a
large number of oscillations, 10, because it reduces errors, as any
errors that occur will be minimized.


Factors Affecting Time Period Of A Simple Pendulum
--------------------------------------------------

* Length- For the length of the pendulum, we consider the length
from the suspension point of the string to the bottom of the bob,
and then subtract the radius. While investigating the length, all
other factors must be maintained constant for it to be a fair
test. Therefore, the mass, amplitude and the acceleration due to
gravity have to be constant. Since the amplitude and the mass will
be constant, the bob will have to displaced by 5cm for the first
test, and then so on accordingly for the other tests:

Considering the law of energy that is, energy cannot be created or
destroyed, we can say that all the potential energy at the ends will
be transferred to the maximum kinetic energy that exists at the
equilibrium position.

Max P.E at extreme ends = Max K.E at the equilibrium position

Mgh = ½ mv2; cancel out mass and multiply by 2 on both sides

2gh = v2

√2gh = v

(Below is a diagram showing how the increase in length of string)

Velocity is independent of the mass of the bob. If you use the same
linear amplitude as before, when the length increases, the height of
displacement reduces, the potential energy decreases, so kinetic
energy decreases, velocity reduces and as the distance remains the
same, Time Period increases.

* Mass of the bob- The mass of the bob could affect the time period,
but the velocity is independent of mass, since we are maintaining
the same linear amplitude, which means that it has the same speed
and same distance, and so the time taken for an oscillation is the
same. Therefore, a change in mass won't affect the time period of
a simple pendulum. This is also proven by the fact that in the
equation v = 2gh, the mass on both sides of the equation cancel
each other out.

* Amplitude- The other factor that could affect the time period of
the bob is its amplitude. To investigate this factor all other
factors must remain constant. When the amplitude increases, the
height of displacement increases, the potential energy increases,
so kinetic energy increases, velocity increases and as the
distance increases, but the Time Period remains the same.

· Gravitational field strength- the other factor that might affect the
time period of the bob is gravitational field strength. To investigate
this factor all other factors must remain constant. When the
gravitational field strength increases, the potential energy
increases, so kinetic energy increases, velocity increases and as the
distance remains the same, but the Time Period decreases.

Procedure:

List of apparatus:

* Different lengths of string

* Digital stopwatch

* Clamp stand

* Meter ruler

* Bobs (weights) of different masses

The length of the pendulum is the length of the string from the
suspension point of the string to the bottom of the bob, and then the
radius of the bob has to be subtracted from this amount. Keeping the
string at a longer length than that which is actually required, and
then decreasing the length for each test can adjust length.

For the investigations, I am planning to investigate a minimum of 5
length of string (40cm, 50cm, 60cm, 70cm, 80cm, 80cm, 90cm, etc.). I
am planning to investigate a minimum of 3 different masses for the bob
as well (400g, 800g, 1200g, etc.). And finally I am planning to
investigate a minimum of 3 different amplitudes (6cm, 9cm, 12cm,).

After the bob is set into motion, we must allow a few free
oscillations because initially they are forced oscillations.

If the starting point is taken as the extreme right an oscillation is
said to be complete when the pendulum comes back to that specific side
(right).

(Below is a diagram illustrating this)

The Bob moves from the position on the right, past the equilibrium
then to the left, then back again past the equilibrium and then to the
right side again. This is one complete oscillation.

This is what the table for the different lengths would look like:

Mass of bob = g

Linear amplitude = cm

Length (cm)

Time for 10 oscillations


Time period

= t / 10

t1 (s)

t2 (s)

Avg. t = t1 + t2 / 2

40

50

60

70

80

90

100

110

This is what the table for the different masses would look like:

Length = cm

Linear amplitude = cm

Mass (g)

Time for 10 oscillations


Time period


= t / 10

t1 (s)

t2 (s)

Avg. t = t1 + t2 / 2

400

800

1200

1600

2000

2400

This is what the table for the different amplitudes would look like:

Mass of bob = g

Length = cm

Amplitude (cm)

Time for 10 oscillations


Time period


= t / 10

t1 (s)

t2 (s)

Avg. t = t1 + t2 / 2

6

9

12

15

18

Prior Test

First I carried out a prior test using the same procedure as the
procedure explained to see exactly how the experiment works, except
for the fact that I was provided with different masses which were not
the ones that I had predicted and I used different linear amplitudes
and not the ones that I had predicted, and I had carried out three
tests when investigating each factor, not 2 as I had planned.

I used;

· A string of length 50cm

· A bob with mass 18g

· A bob with radius 0.8cm

The following is the table of results that I ended up with after
carrying out the experiment:

Amplitude (cm)

Time for 10 oscillations


Time period

= t / 10

t1 (s)

t2 (s)

t3 (s)

Avg. t = t1 + t2 / 3

5

14.16

14.20

14.14

14.17

1.417

10

14.26

14.12

14.13

14.17

1.417

15

14.17

14.16

14.22

14.18

1.418

Hence By looking at the above results, I can conclude that my
predictions were correct because the amplitude doesn't affect the time
period of the pendulum, when the mass and the length are kept
constant.

(Following is the graphical representation of these results on a
graph.)



ASHLEY. FERNANDES
=================

2. 3. 2004

SR 5 'B'

PHYSICS CW- FACTORS AFFECTING THE TIME PERIOD OF A SIMPLE PENDELUM


OBTAINING EVIDENCE
------------------

After the prior test, I decided to carry out the experiments as
described in the procedure, to analyse the length, mass and linear
amplitude factors.

BOB

RADIUS (cm)

MASS (g)

BIG

1.3

65

MEDIUM

1

29

SMALL

0.8

18

These are the results that I had obtained when considering the length
factor, where the mass was kept constant at 65g, the radius was 1.3cm
and the linear amplitude was kept constant at 15cm:

LENGTH (cm)

TESTS (s)

T = t/10 (s)

First

Second

Thrid

Average

30

11.22

11.22

11.28

11.24

1.124

40

12.9

13.04

12.92

12.95333

1.29533333

50

14.44

14.5

14.41

14.45

1.445

60

15.75

15.83

15.75

15.77667

1.57766667

70

17.22

17.03

17.25

17.16667

1.71666667

80

18.91

18.68

18.69

18.76

1.876

90

19.43

19.37

19.37

19.39

1.939

100

20.38

20.62

20.29

20.43

2.043

110

21.63

21.62

21.5

21.58333

2.15833333

120

22.03

22.19

22.31

22.17667

2.21766667

These are the results that I had obtained when considering the linear
amplitude factor, where the mass was kept constant at 65g, the radius
was 1.3cm and the length was also kept constant at 50cm:

AMPLITUDE (cm)

TESTS (s)

T = t/10 (s)

First

Second

Thrid

Average

5

14.31

14.25

14.24

14.26667

1.42666667

10

14.34

14.36

14.32

14.34

1.434

15

14.44

14.5

14.41

14.45

1.445

These are the results that I had obtained when considering the mass
factor, where the linear amplitude was kept constant at 15cm and the
length was also kept constant at 50cm. The radius for each mass was
0.8cm (small bob), 1cm (medium bob) and 1.3cm (big bob) respectively:

MASS (g)

TESTS (s)

T = t/10 (s)

First

Second

Thrid

Average

Big 65

14.44

14.5

14.41

14.45

1.445

Medium 29

14.2

14.46

14.22

14.29333

1.42933333

Small 18

14.44

14.5

14.41

14.45

1.445

(Below are the graph done on plotted sheets to show length against
time, amplitude against time and mass against time)



ASHLEY. FERNANDES
=================

2. 3. 2004

SR 5 'B'

PHYSICS CW- FACTORS AFFECTING THE TIME PERIOD OF A SIMPLE PENDELUM


ANALYSIS
--------

Conclusions:

1. For the graph time period against length we suggest that it is
not a linear graph, instead it is a curve. This is because as
length increases, time period increases. For the same linear
amplitude, as length increases the height through which the bob
moves decreases. Hence when height decreases, kinetic energy will
decrease, so also causing potential energy to decrease. Velocity
therefore decreases. As velocity decreases, time period will
increases. Therefore the prediction that I had made in planning is
proven right.

2. The graph time period against mass does not affect time period.

Mgh at extreme ends = ½ mv at extreme ends

V = √2gh

Velocity doesn't depend on mass. Distance remains the same, due to the
linear amplitude being constant. So time period remains the same and
therefore mass is not affected by time period.

3. Time period is not affected by change in amplitude. When linear
amplitude increases, height through which the bob moves increases.
When height increases potential energy and kinetic energy
increases. As amplitude increases, the velocity and distance
increases. Therefore time period is not affected.

(Below are the specified graphs for the conclusions)

We use the theory of Simple Harmonic Motion to find the relationship
between the length and time squared. The relationship proves to be
directly proportional.

F = ma

Ma = mg sine θ

A = g sine θ

Therefore, F = mg sine θ

The formula for angle in a sector, a = g θ, this is used to find θ
incase of it being small.

a = g x A(1enght) / radius

= g x BB1 / radius

= g x - (x / length)

In the above equation, 'BB1' represents the displacement and 'length'
represents radius. Since 'a' (acceleration) opposes 'x', x will have a
negative charge. Since 'g' and 'length' are constants, we can conclude
that a is directly proportional to -x. (feature of a simple harmonic
motion)

Therefore, a = -ω2x (where ω is Omega)

So, ω2 = a / x (where ω2 is a positive constant)

So, ω2 = a / x

Therefore, the time period, T, of a simple harmonic motion is:

T = 2 π / ω

T = 2π ÷ a / x

T = 2Ï€ x x/a

From the above equation, g x (- x / l) ; a = g -x / l

x / a = l / g

Therefore, T = 2Ï€ l / g

T2 = 4Ï€2 x l /g

T2 = 4Ï€2 / g (l)

Hence 4Ï€2 / g, is a constant

Thus T2 is directly proportional to l

L is directly proportional to T2

Therefore L directly proportional to T2 graph is needed.

Following is a table of the length against time squared. This table is
used to find the value for gravity.

Length (m)

T^2 (s)

Gravitational val.

0.3

1.263376

9.37

0.4

1.677888444

9.41

0.5

2.088025

9.45

0.6

2.489032111

9.52

0.7

2.946944444

9.38

0.8

3.519376

9

0.9

3.759721

9.45

1

4.173849

9.46

1.1

4.658402778

9.32

1.2

4.918045444

9.63

9.399

Therefore the average gravity value obtained is, 9.399 ≈ 9.40

I will hence plot a graph for length against time squared for the
following values.

Length (cm)

T^2 (s)

30

1.263376

40

1.677888444

50

2.088025

60

2.489032111

70

2.946944444

80

3.519376

90

3.759721

100

4.173849

110

4.658402778

120

4.918045444

I am going to use the following formula to see if I obtain the right
value for gravity from my graph.

Gradient = T2 ' L

g = 4Ï€2 x l / T2

Gravitational strength = 4Ï€2 x 1 / Gradient

Gradient = (2.50 / 0.6) - (1.26 / 0.3) = 1.24 / 0.3 = 4.133

Gravitational strength = 4Ï€2 / 4.133

= 9.5520

'G' value from table- 9.40

'G' value from graph- 9.55

From the above statements, both values are around the value, 9.8 so I
can conclude that my results were accurate.



Quantitative Conclusions
========================

To confirm the accuracy of my results I will investigate further using
quantitative tests.

Sine θ = A / L

θ = sin-1 A / L

Cosine = L1 / L = L - h / L

(L Cosine) = L - h

H = L - (L Cosine)

V = ω x λ, where 'V' is the velocity, 'ω' is Omega and 'λ' is the
amplitude

ω = 2π / T

Kinetic energy = ½ mv2, where 'm' is the mass and 'v' is the velocity.

Gravitational potential energy = M x g x H, where 'M' is the mass, 'g'
is the gravity that is found from the table and the graph and 'H' is
the height.

Using the above formulas, I will find the height of displacement,
velocity, kinetic energy, and potential energy. I will compare the
values for kinetic energy and potential energy, and see if they are
almost the same. If they are, then the experiment that I carried out
was accurate.

Following is the calculation as done for the 0.3m strings:

= Sine-1 A / L

= Sine-1 15 / 30

= 30Ëš

H = L - (L Cosine)

H = 30 - (30 x cos30)

H = 4.019cm = 0. 0401m

V = ω x λ

V = 2Ï€ / T x 0.15m

V = 0.838 m/s

Kinetic energy = ½ mv2

= ½ x 0.065 x 0.8382

= 2.282293 x 10-2 J

Gravitational potential energy = Mgh

= 0.065 x 9.8x 0.0401

= 2.55437 x 10-2 J

Length (L)

(m)

Height

H = L - L Cosine (m)

Velocity

V = ω x λ (m/s)

Kinetic Energy

K.E. = ½ mv2 (J)

Potential Energy

P.E. = Mgh (J)

0.3

0.0401

0.838

0.02282293

0.0255437

0.6

0.9

From the above results I find that the values for kinetic energy and
potential energy are quite similar, therefore I can conclude that my
results were accurate. Looking at the table we see that as length
increases, velocity decreases therefore the time period is increasing.



ASHLEY. FERNANDES
=================

4. 3. 2004

SR 5 'B'

PHYSICS CW- FACTORS AFFECTING THE TIME PERIOD OF A SIMPLE PENDELUM


EVALUATION
----------

After analyzing the results that were obtained, I concluded that my
investigation on the factors that effect the time period of a simple
pendulum were correct. This can be established using the various
values that were calculated from the results, including the value for
the gravitational force acting on the bob. Moreover, once a pendulum
is in simple harmonic motion, the potential energy it has at the
extreme ends will be equal to the kinetic energy at its equilibrium
position, due to none of its energy is lost; only instead changed.
This was also indicated in the results obtained from the readings for
the Potential energy that was calculated was similar to the value for
the kinetic energy of the bob. From the tables in obtaining and
analysis where we see the observed time period and calculated time
period, we notice that there are slight differences in the values for
the Time Period that were obtained and those that were calculated
using the formula. (T = 2∠x / (L / g) [where g = 9.81 N/kg]) However,
there is merely a difference of 0.01 of a second in most of the
readings. Furthermore, both results show a general trend that as the
length of the pendulum increases, the time period also increases.
Taking this in consideration, I can conclude that my results, and the
procedure itself, were accurate.

I can conclude that though the method is reliable for investigation
purposes, it is not very suitable for obtaining reliable results for
calculation. This is because there is too much scope for human error,
as in the recording of the time taken for ten oscillations (parallax
error), in the measurement of the amplitude of the oscillations, and
even in taking the length of the string for the pendulums created. To
curtail this margin of error, perhaps pendulums of standard lengths
could be used during the experiment, instead of measuring out
different lengths. Also, for measuring the times for the various
oscillations, Infra- red sensors connected to computers could be used,
so that the times taken for ten oscillations could be monitored and
recorded precisely.

Extended Investigation

mg

x

I could investigate the oscillations of a spring mass system. This
system is made up of a suspended spring that has a mass kept at its
end. Pulling on this string increases the tension of the string hence
a force that is countered by the weight of the mass, and the pulling
force acting in the opposite direction. This is shown below:

[IMAGE]

[IMAGE]


[IMAGE][IMAGE][IMAGE]

[IMAGE]

[IMAGE]

[IMAGE]


When the mass is pulled down, the tension in the string will increase.
So, when the mass is released, it will move continuously up and down,
oscillate, about its equilibrium position. The time period of this
oscillation can then be found out using a calculation:

In Figure 2,

The force acting downwards = Mass of weight x Force of gravity on
object

Therefore, F = M x A

Since the mass is at rest in Figure 2, the upward force (or tension)
of the spring will be equal to the downward force (mg). This force can
be given by the formula, F = k x e.

'K is the Spring Constant and 'E' is extension of the spring under the
authority of mass. This force should be equal to the weight of the
mass acting downwards.

Therefore, Mg = K.E

As shown in Figure 3, when the spring is pulled down, it is extended
more by the distance (x). Considering this, the tension of the spring
acting upwards once it is pulled further can be given by the formula,
F = k (e + x)

As the spring is being pulled downwards, there will be a resultant
downward force (causing the spring to move downwards). This resultant
downward force can be calculated using the formula:

Resultant force 'F' (downwards) = Mg - k (e + x)

Since Mg = K.E, the above equation can be formed as,

F = K.E - k (e + x)

Once lengthened, this equation will look as follows:

F = K.E - K.E - KX

F = - KX

Therefore, F = - KX

Since k is a constant, the relationship between the resultant
downwards force and the extension can be written as, F α - x

This states that as the resultant downward force increases, the
extension of the spring (x) will also increase.

Furthermore, since 'F = -KX' and F = MA (Newton's 2nd Law), a new
equation can be formed:

MA = -KX

A = (-k / m) x

'K' and 'M' are both constants, and so they can be replaced by 'ω2,'
which is a positive constant. Hence, ω2 = k / m

Therefore we get, ω = √ (k / m)

For a SHM the time period = 2∠/ ω

Considering the previous equation, I can thus say,

T = 2∠/ √ (k / m) otherwise, T = 2∠x √ (m / k)

In this fashion, the Time Period of the spring oscillations can be
calculated. Also, the effect of the varying factors, such as using an
object of greater mass, increasing the length of the spring, or even
increasing the pulling force on the spring, on the Time Period of the
oscillation, may perhaps be studied.

How to Cite this Page

MLA Citation:
"Factors Affecting The Time Period of a Simple Pendulum." 123HelpMe.com. 21 Apr 2014
    <http://www.123HelpMe.com/view.asp?id=121478>.




Related Searches





Important Note: If you'd like to save a copy of the paper on your computer, you can COPY and PASTE it into your word processor. Please, follow these steps to do that in Windows:

1. Select the text of the paper with the mouse and press Ctrl+C.
2. Open your word processor and press Ctrl+V.

Company's Liability

123HelpMe.com (the "Web Site") is produced by the "Company". The contents of this Web Site, such as text, graphics, images, audio, video and all other material ("Material"), are protected by copyright under both United States and foreign laws. The Company makes no representations about the accuracy, reliability, completeness, or timeliness of the Material or about the results to be obtained from using the Material. You expressly agree that any use of the Material is entirely at your own risk. Most of the Material on the Web Site is provided and maintained by third parties. This third party Material may not be screened by the Company prior to its inclusion on the Web Site. You expressly agree that the Company is not liable or responsible for any defamatory, offensive, or illegal conduct of other subscribers or third parties.

The Materials are provided on an as-is basis without warranty express or implied. The Company and its suppliers and affiliates disclaim all warranties, including the warranty of non-infringement of proprietary or third party rights, and the warranty of fitness for a particular purpose. The Company and its suppliers make no warranties as to the accuracy, reliability, completeness, or timeliness of the material, services, text, graphics and links.

For a complete statement of the Terms of Service, please see our website. By obtaining these materials you agree to abide by the terms herein, by our Terms of Service as posted on the website and any and all alterations, revisions and amendments thereto.



Return to 123HelpMe.com

Copyright © 2000-2013 123HelpMe.com. All rights reserved. Terms of Service