Final Year Thesis

PROJECT

CLASSICAL HARMONIC

OSCILLATOR

Group Members

SADAF SHAFIQ HIRA KHAN

( ROLL NO 2) ( ROLL NO 21)

LARAIB SALEEM SYEDA FAIZA RUBAB SHERAZI

( ROLL NO 4) (ROLL NO 34)

SUPERVISED BY:

SIR ASSAD ABBAS MALIK

I

S H

P

Y

CS

YEAR

2010

DEPARTMENT OF PHYSICS

UNIVERSITY OF SARGODHA

2010

2

DEPARTMENT OF PHYSICS

UNIVERSITY OF SARGODHA

3

DEDICATED TO,

Our respected teacher

SIR ASSAD ABBAS MALIK

4

Acknowledgements

First of all we thank Allah Almighty who provided us opportunity and gave the strength to

undertake this work of great importance.

Several persons helped us directly or indirectly in our work. Laraib and Faiza have done the

calculations sensibly. While Sadaf and Sabeen not only typed and edited the manuscript with

admirable competence but always went through the innumerable revision with patience, care

and understanding. The responsibility for any remaining errors or shortcomings is, of course,

ours.

At the end, we would like to thank our respected teacher, Sir Assad Abbas Malik,

University of Sargodha for his valuable criticism, suggestions comments and he most

willingly, come forward to review our work and gave very valuable suggestions. Also we

especially wish to thank Professor Khalid Naseer, University of Sargodha for helping us in

using mathematica while making graphs.

Laraib, Faiza, Sabeen and Sadaf.

5

CONTENTS

Chapter-1 Classical Simple Harmonic Motion (7-28)

1-1 Introduction ……….…………………………………………………………….. 7

1-2 Horizontal Oscillation ……………………………………………………......…. 9

1-2-1 Comparison b/w displacement, velocity and acceleration ………………… .. 15

1-3 Vertical Oscillation …………………………………………………………..… 15

1-4 SHM In Two Dimensions ………………………………………..…………….. 17

1-5 SHM In Three Dimensions ……………………………………..……………… 20

1-6 Simple Pendulum …………………………………………………………...…. 21

1-7 Physical Pendulum …………………………………..………………………… 23

1-8 Spring In Series and Parallel ………………………………………………...… 23

1-8-1-1 Parallel Configuration ................................................................................... 24

1-8-1-2 Variation In Parallel configuration ………………………………………………….. 25

1-8-2 Series Configuration ….………………………………………….............................….. 26

1-9 Floating Objects ………..………………………....………….…….………....……... 27

1-10 Tunnel In Earth …………...……………………………………………...…… 28

6

Chapter-2 Simple Harmonic Motion In Lagrangian (28-36)

2-1 Introduction …..……………………………………………...………………… 30

2-2 Horizontal Oscillation………………………………………………...………... 30

2-3 Vertical Oscillation ……………………….……………………………………. 31

2-4 Simple Harmonic Motion in 2- D…………….………………….……………... 31

2-5 Simple Harmonic Motion In 3- D ……………………………………………... 31

2-6 Simple Pendulum …………………………………………………………….... 32

2-7 Physical Pendulum ……..……..………………………………………………. 33

2-8-1 Parallel configuration ………………..…………………………………….… 34

2-8-2 Series Configuration ………………….…………………………………….. 35

2-9 Floating Objects ……………………………………………………...……….. 36

2-10 Tunnel in Earth …………………………...………………………………...… 36

References ………………………………………………………………………...… 37

7

CLASSICAL SIMPLE HARMONIC MOTION

1-1 Introduction

Our study of physics started with the discussion of bodies at rest. We then consider the

simple case of motion along a straight line. Motion can take any path. Linear motion is the

easiest motion to discuss and we know how equations of motion and Newton's laws of

motion can help us to determine velocities, displacements, accelerations, force, inertia, etc.

Oscillatory motion is periodic in nature. That is after a regular time interval, the motion

repeats itself. The object returns to the given position after a fixed time interval unlike linear

or circular motion, the force responsible for the motion of a body undergoing Simple

Harmonic Motion, varies in magnitude and direction, the variation is also periodic in nature

and hence the simple harmonic is also periodic in nature. Let us now try and understand

how SHM occurs and what conditions are necessary for an SHM in real life.

We heard the one about the opera singer breaking glass of water with their voice or it can

be the specs worn by the audience too. Later we will discuss the breaking of the glass with a

loud speaker but first we will try to work out some rules for the things which vibrate, or

1

8

oscillate. There are many forms of oscillations in the real world- oscillations determine the

sound of the musical instrument, the color of rainbow, the ticking the clock, needle of a

sewing machine goes up and down at a certain regular frequency, a child on a swing in a

park goes back and forth repeatedly and even the temperature of the cup of tea. A classic

example of SHM is the motion of a leaf in a pond. If waves are generated in the pond, the

leaf will bob up and down in a repeated manner. When a bat hits a baseball, the bat may

oscillate enough to stings the batter’s hands or even to break a part. When wind blows past a

power line, the line may oscillate so severely that it rips apart, shutting off the power supply

to a community. When a train travel around a curve, its wheel oscillate horizontally as they

are forced to turn in new direction (you can hear the oscillation). A similar example of SHM

is a cork bobbing up and down, when placed in a beaker containing water. A mechanical

oscillation is a repeating movement –an electrical oscillation is a repeating change in voltage

and current.

Mechanical oscillations are probably easiest to think about, so it is better to start from there.

Let us think about the movement of an object which is moving back and forth over a fixed

range of positions, such as the motion of the swing in the park, or the bouncing up and down

of the weight on the end of the spring.

When initially the body is in equilibrium position then forces on it balance. Now let us

displace it from equilibrium by applying force on it to a distance x, when it is released a

restoring force comes into play and returns it to its original position which is zero. Inertia on

the other hand tries to oppose any change in its velocity. The elasticity returns body to its

original position imparting to it an appropriate negative velocity 𝒅𝒙

𝒅𝒕 , when body reaches its

equilibrium position again at that point the negative velocity is maximum, which produces

negative displacement. The body overshoots its equilibrium position. The restoring force now

becomes positive (helps increase x) and it must now overcome inertia of negative velocity.

Consequently velocity keeps on decreasing until it is zero by that time displacement has

become large and negative and has become large and negative and process is reversed. This

process of restoring force trying to bring x to zero by imparting a velocity and inertia

preserving the velocity and making x, to overshoot, repeats itself and that is the way body

oscillate.

As when we displace a system work has to be done on it .The restoring force F obviously

depends on work done to give a displacement x. Thus F is some general function of x. For

systems oscillating violently (large x) the dependence of F on x is very complex. We will

discuss only those systems in which moving part always stays close to its mean position

(small x).Now we'll define Hook's Law which states that:

Restoring force is directly proportional to the displacement and opposes its increase.

F = − k x (1.1)

Negative sign indicates that F is opposite to increase in x .The constant of proportionality k

is called the force constant .The SI unit of k is Nm−2

and its magnitude depends upon the

elastic properties of the system under study. Equation (1.1) is statement of Hook's Law for

elastic forces.

9

1-2 Horizontal Oscillations

The motion in which the restoring force is proportional to displacement from mean

position and opposes its increase is called the simple harmonic motion. This happens when

the force acting on the body is elastic restoring force or simply we can say return force. We

can define linear oscillator or an oscillator for which the restoring force is proportional to

displacement.

A body of mass m attached to a spring of spring constant k and is free to move over a

frictionless horizontal surface as shown

m

m

m

mg

mg

mg

Figure1-1: Model for the horizontal periodic motion.

Case 1: When spring is stretched the net force acting on the body will be given below

𝑭𝒏𝒆𝒕= 𝑭𝒓+ 𝑭𝒈 + 𝑭𝑵 , (1.2)

where 𝑭𝒏𝒆𝒕 is net force and is given by 𝐹𝑥𝒙 +𝐹𝑦 𝒚 , 𝑭𝒓 is restoring force is acting along

negative x-axis an its magnitude is 𝐹𝑟 , gravitational force 𝑭𝒈 is equal to weight 𝑚𝑔 and its

direction is along negative y-axis similarly 𝑭𝑵 is given by 𝐹𝑁 (𝒚 ). By substituting the values

we get

𝐹𝑥𝒙 +𝐹𝑦 𝒚 = −𝑘𝑥𝒙 + (𝑁 −𝑚𝑔) 𝒚 , (1.3)

10

By comparing the components of force and magnitude of components of normal and

gravitational force balanced each other so y-component is zero. And force will be only along

x- axis that is 𝐹𝑥 equal to −𝑘𝑥 and then substituting 𝐹𝑥 may be written equal to 𝑚𝑑2𝑥

𝑑𝑡2 . The

final result will be

𝑑2𝑥

𝑑𝑡2 + 𝜔2𝑥 = 0. (1.4)

Here, 𝜔 is angular frequency given by 𝜔2 = 𝑘

𝑚 .

Case 2: When the spring is compressed then the expression of net force is given by equation

(1.2). Here the difference will be that the restoring force will act towards the positive x-axis

and will tend to take the body towards the mean position substituting the values an then

comparing the components of the force and substituting the values 𝐹 = −𝑚𝑎 and 𝜔2 = 𝑘

𝑚

we get

𝑑2𝑥

𝑑𝑡2 + 𝜔2𝑥 = 0. (1.5)

For a linear oscillator 𝐹 ∝ −𝑥, after resolving the proportionality sign we get the constant

of proportionality k which is spring constant or stiffness factor. It value depends upon the

nature of the spring. It is large for the hard spring and small for soft spring. so using 𝐹 =

−𝑘𝑥, and then substituting 𝐹 = 𝑚𝑥 , we get the equation of simple harmonic oscillator

𝑥 + 𝜔2𝑥 = 0. (1.6)

Here, 𝜔 is the angular velocity given by 𝑘

𝑚.

The Physical Statement corresponding to this equation is: The acceleration of the body in

simple harmonic motion is proportional (and opposite in sign) to displacement. In order to

determine that what type of motion is represented by above equation, we need to solve this

differential equation i.e. obtain an expression of displacement x as function of time t. The

equation of motion is homogenous second order ordinary differential equation. We can solve

this second order differential equation using trigonometric solution, power series solution and

exponential solution. We will discuss only one of them here (differential solution).

Multiply by 𝑥 on both sides of equation (1.5) and by substituting the values 𝑥 𝑥 = 𝑑

𝑑𝑡(

1

2𝑥2 )

and 𝑥𝑥 =𝑑

𝑑𝑡𝑥2, we get

𝑑

𝑑𝑡(

1

2𝑥 2 +

1

2𝑥2𝜔2) = 0, (1.7)

Integrating on both sides and twice and using B and C as constant of integration and then

simplifying we will get

11

𝑠𝑖𝑛−1 𝑥

2𝐵

𝜔2

= 𝜔𝑡 + 𝐶, (1.8)

where C is constant of integration, 𝜔 is the angular velocity and 𝑥 is the displacement. By

applying the boundary conditions: 𝑡 = 0 , 𝑥 = 𝑥0, and simplifying we get

𝑥 = 𝑥𝑚 sin(𝜔𝑡 + ∅). (1.9)

Above equation can also be written as

𝑥 = 𝑥𝑚 cos(𝜔𝑡 + ∅). (1.10)

The smallest time period during which oscillation repeats itself is called time period or

simply period, is given by

𝑇 = 2𝜋√𝑚

𝑘. (1.11)

From this note that time period does not depend upon amplitude or displacement. If we

change time by 2𝜋

𝜔 then value of 𝑥 remains unchanged .In other words displacement repeats

itself in a time interval of 2𝜋

𝜔 .

The maximum displacement from the mean position is called amplitude. Since maximum

and minimum values of x are respectively +1 and −1 so the maximum and minimum values

are respectively −𝑥𝑚 and +𝑥𝑚 .

𝑥𝑚 = 2𝐵

𝜔2. (1.12)

This is possible if sin(𝜔𝑡 + ∅) = 1 or cos(𝜔𝑡 + ∅) =1. The number of vibrations completed

in one second is called frequency given by

𝑓 = 1

2𝜋 𝑘

𝑚. (1.13)

We can get the Velocity of the harmonic oscillator by differentiating the displacement,𝑣 =dx

dt substituting value of 𝑥 and then differentiating we get the equation for velocity given by

𝑣 = −𝑥𝑚 𝜔 sin(𝜔𝑡 + ∅). (1.14)

Substituting the value of sin(𝜔𝑡 + ∅) we can get a more precise result for the velocity of

harmonic oscillator

𝑣 = ± 𝑥𝑚 𝜔 1 −𝑥2

𝑥𝑚 2. (1.15)

12

For the relation of acceleration we differentiate the displacement twice and then putting the

value of cos(𝜔𝑡 + ∅) =𝑥

𝑥𝑚 we can get more précised result for the acceleration of the

harmonic oscillator

𝑥 = −𝜔2𝑥. (1.16)

The argument 𝜔𝑡 + ∅ of cosine function is called the phase of the motion. The constant

∅ is called the Initial phase. The knowledge of phase constant enables us to find out how far

from mean position the system was at time t=0 e.g. If ∅ = 0 then

𝑥 𝑡 = 𝑥𝑚𝑐𝑜𝑠𝜔𝑡. (1.17)

Figure1-2: Graph of displacement versus time period for SHM.

0 4 8 12 16 20 241

0

1

time

dis

pla

cem

ent

13

Figure 1-3: Graph of velocity versus time period for SHM.

Figure 1-4: Graph of acceleration versus time period for SHM.

During simple harmonic motion the total energy of oscillator remains conserved.

As the total energy is the sum of its kinetic energy and the potential energy. But at mean

position the potential energy is equal to zero so we are left only with the kinetic energy and

so the total energy at the mean position is only the kinetic energy. While at the extreme position the kinetic energy of the oscillator is equal to zero so the total energy will be only its

potential energy.

The total mechanical energy of the harmonic oscillator at any instant other then the mean

and the extreme position is partly kinetic and partly potential.

0 4 8 12 16 20 241

0

1

time

vel

oci

ty

0 4 8 12 16 20 241

0

1

time

acce

lera

tion

14

Potential energy at any instant is given by U which is given by 𝐹𝑑𝑥𝒙

𝟎. Integrating and then

substituting the value of the displacement 𝑥 we get the relation for the energy of the harmonic

oscillator at any instant

U=1

2𝑘[𝑥𝑚

2 𝑐𝑜𝑠2 (𝜔𝑡 + ∅)]. (1.18)

The maximum value of the potential energy varies with time and as cos(𝜔𝑡 + ∅) = 1

So has maximum value of potential energy will be

U=1

2𝑘𝑥𝑚

2. (1.19)

The kinetic energy at any instant is given by K.E=1

2𝑚𝑥 2, substituting the value of "𝑥" and

then simplifying the above equation we can get the precised value for the maximum value of

the kinetic energy at any instant

K.Emax = 1

2𝑘 𝑥𝑚

2. (1.20)

The total mechanical energy at any instant will be given by adding the maximum value of

kinetic and potential energy at any instant

T.E= 1

2𝑘𝑥𝑚

2 . (1.21)

From equations (1.19) ,(1.20) and (1.21) we find that

P.Emax = K.Emax =T.E.

Figure 1-5: Graph of energy for SHM. The bold curve represents P.E while the dotted

represents the K.E, where K.E and P.E. are the functions of position for SHM.

15

1-2-1 Comparison Between Displacement, Velocity And Acceleration

1) When displacement is maximum (−𝑥𝑚 𝑎𝑛𝑑 + 𝑥𝑚 ) , the velocity 𝑣 =0, because the

oscillator has to return and velocity changes its direction.

2) If displacement is maximum acceleration is also maximum (− 𝜔2𝑥𝑚 𝑜𝑟 + 𝜔2𝑥𝑚) and is

directed opposite to displacement.

3) When x =0, i.e. when cos(𝜔𝑡 + ∅) = 0 the velocity is maximum (𝜔𝑥 𝑜𝑟 − 𝜔𝑥) and

acceleration is zero.

Figure 1-6: Comparison b/w graph of displacement, velocity and acceleration versus time

period.

1-3 Vertical Oscillations

Suppose we have a spring with force constant k and suspended from it a body with a mass m.

Case 1: The body hangs at rest in equilibrium. In this position the spring is stretched to an

amount 𝑙𝑜 just great enough that the spring's upward vertical force 𝑘𝑙𝑜 on the body balances

its weight mg,

𝑘𝑙𝑜 = 𝑚𝑔 (1.22)

Case 2: When the body is displaced below the equilibrium position, and then stretched to an

amount 𝑦 then the total extension in the spring will be 𝑙𝑜 + 𝑦 and the net force will be:

𝑭𝒏𝒆𝒕 = 𝑭𝒓 + 𝑭𝒈, (1.23)

0 4 8 12 161

0

1

time

dis

pla

cem

ent

16

m

m

m

Figure 1-7: Model for the vertical periodic motion.

Here, 𝑭𝒏𝒆𝒕 is the net force which is only along 𝑦-axis and is equal to𝐹𝑦𝒚 , 𝑭𝒓 is the

restoring force of the spring its magnitude will be equal to 𝑘 𝑙𝑜 + 𝑦 𝒚 and 𝑭𝒈 is

gravitational force of the earth equal to the weight of the body 𝑚𝑔(−𝒚 ), by substituting the

values in the above equation and then solving we will get

𝑦 +𝜔2𝑦 = 0, (1.23)

Here, 𝜔 = 𝑘

𝑚.

Case 3: When the spring is compressed and moved upwards then the extension in the string

is (𝑦 − 𝑙𝑜) . 𝑭𝒓 is the restoring force of the spring it is equal to 𝑘 𝑦 − 𝑙𝑜 (−𝒚 ) and 𝑭𝒈 is

gravitational force of the earth equal to the weight of the body 𝑚𝑔(−𝒚 ), by substituting the

values in equation(1.23) then solving we will get 𝑭𝒏𝒆𝒕 = 𝑘 (-𝒚 ) which then yields

𝑚𝑦 + 𝑘𝑦 = 0. (1.24)

1-4 Harmonic Oscillator In 2 Dimensions

Consider the motion of the particle of mass m at any point P as shown in the figure moving

in accordance with the following equation of motion. Net force can be written as

𝑭𝒏𝒆𝒕 = 𝑭𝒓, (1.25)

17

Here, 𝑭𝒏𝒆𝒕 is equal to 𝐹𝑥𝒙 +𝐹𝑦 𝒚 and similarly 𝑭𝒓 can be written as 𝐹𝑟𝑐𝑜𝑠𝜃𝒙 + 𝐹𝑟𝑠𝑖𝑛𝜃𝒚 and

then we may also write 𝐹𝑟𝑐𝑜𝑠𝜃 equal to 𝑘𝑥𝑟𝑐𝑜𝑠𝜃𝒙 similarly 𝐹𝑟𝑠𝑖𝑛𝜃 equal to 𝑘𝑦𝑟𝑠𝑖𝑛𝜃𝒚 , then

after comparing coefficients of 𝒙 and 𝒚 on both sides we get two equations

P(x,y)

Figure 1-8: Motion in 2-D.

𝑥 + 𝜔𝑥2𝑥 = 0, (1.26)

𝑦 + 𝜔𝑦2𝑦 = 0, (1.27)

We know that 𝜔𝑥 has the value equal to 𝑘𝑥

𝑚 and 𝜔𝑦 has the value

𝑘𝑦

𝑚. Then the solutions

of equations (1.26) and (1.27) are respectively given by

𝑥 = 𝑥𝑚cos(𝜔𝑥 𝑡 + ∅𝑥). (1.28)

𝑦 = 𝑦𝑚cos(𝜔𝑦 𝑡 + ∅𝑦). (1.29)

1-5 Harmonic Oscillator In 3 Dimensions

Consider a motion of particle of mass m in space. The net force is given by equation (1.25)

but here, 𝑭𝒏𝒆𝒕 is equal to 𝐹𝑥𝒙 +𝐹𝑦 𝒚 + 𝐹𝑧 𝒛 and similarly 𝑭𝒓 can be written as 𝐹𝑟𝑐𝑜𝑠𝜃𝒙 +

𝐹𝑟𝑠𝑖𝑛𝜃𝒚 +𝐹𝑟𝒛 and then we may also write 𝐹𝑟𝑐𝑜𝑠𝜃 equal to 𝑘𝑥𝑟𝑐𝑜𝑠𝜃𝒙 similarly 𝐹𝑟𝑠𝑖𝑛𝜃 equal

to 𝑘𝑦𝑟𝑠𝑖𝑛𝜃𝒚 and 𝐹𝑟𝒛 equal to𝑘𝑧𝑧𝒛 then after comparing coefficients of 𝒙 and 𝒚 on both sides

we get three equations

𝑥 + 𝜔𝑥2𝑥 = 0, (1.30)

𝑦 + 𝜔𝑦2𝑥 = 0, (1.31)

𝑧 + 𝜔𝑧2𝑧 = 0, (1.32)

18

We know that 𝜔𝑥 has the value equal to 𝑘𝑥

𝑚 and 𝜔𝑦 has the value

𝑘𝑦

𝑚 and in the same way

𝜔𝑧 has the value 𝑘𝑧

𝑚 after solving the above equations the solutions are given by the

following equations respectively

𝑥 = 𝑥𝑚cos(𝜔𝑥 𝑡 + ∅𝑥). (1.33)

𝑦 = 𝑦𝑚cos(𝜔𝑦 𝑡 + ∅𝑦). (1.34)

𝑧 = 𝑧𝑚cos(𝜔𝑧 𝑡 + ∅𝑧). (1.35)

As 𝑘𝑥 = 𝑘𝑦 = 𝑘𝑧 then the period of motion is written by

𝑇 =2𝜋𝑛𝑥

𝜔𝑥=

2𝜋𝑛𝑦

𝜔𝑦=

2𝜋𝑛𝑧

𝜔𝑧. (1.36)

If angular frequencies are such that for some set of integers 𝑛𝑥 ,𝑛𝑦 𝑎𝑛𝑑 𝑛𝑧 the angular

frequencies are given by

𝜔𝑥

𝑛𝑥 =

𝜔𝑦

𝑛𝑦=

𝜔𝑧

𝑛𝑧. (1.37)

The graphs relating the displacement, velocity and acceleration are given below, one point

which we had noticed while plotting these graphs is that one can get confused with the

characteristics of the equations as graphs are made using the equations (1.30), (1.31), (1.32),

(1.33), (1.34) and (1.35) and the thing which is common in these equations is that all of them

are the equations with one variable only but still we can plot and represent them in 3

dimensions. It means one variable equations can be represented in 3 dimensions

Figure 1-9: The displacement of the simple harmonic oscillator represented in 3 Dimensions.

19

The above graphic is representing equations (1.33), (1.34) and (1.35) and it shows the

displacement of the simple harmonic oscillator in 3 dimensions. The difference between the

fig (1-2) and fig (1-9) is that the graphic in two dimensions can show us only the two sides of

any graphic, the front or the back while a graphic in 3 dimensions can illustrate us the shape

as well as the area under the curve.

Figure 1-10: The velocity of the simple harmonic oscillator represented in 3 Dimensions.

The above graphic shows the velocity of the simple harmonic oscillator in 3 dimensions

which is given by the equation, −𝑥cos(𝜔 𝑡 + ∅). The difference between the fig (1-3) and

fig (1-10) is that the graphic in two dimensions can show us only the two sides of any

graphic, the front or the back while a graphic in 3 dimensions can illustrate us the shape as

well as the area under the curve.

20

Figure 1-11: The acceleration of the simple harmonic oscillator represented in 3 Dimensions.

The above graphic is representing equations (1.30), (1.31) and (1.32) and it shows the

acceleration of the simple harmonic oscillator in 3 dimensions. The difference between the

fig (1-4) and fig (1-11) is that the graphic in two dimensions can show us only the two sides

of any graphic, the front or the back while a graphic in 3 dimensions can illustrate us the

shape as well as the area under the curve.

Figure 1-12: The energy of the simple harmonic oscillator represented in 3 Dimensions.

1-6 Simple Pendulum

A simple pendulum has a point mass suspended from a frictionless support by a light and

inextensible string.When we pull the simple pendulum to one side of equilibrium position and

release it, the pendulum swing in vertical plane under the influence of gravity. The motion is

periodic and oscillatory.

Now let us consider a pendulum of length l and mass m. When pendulum is displaced from

mean to extreme position, let x and 𝜃 be the linear and angular displacements then the forces

acting on mass m are weight mg and tension in the string. The motion will be along the arc of

the circle with radius l. In equation (1.23) we may write 𝑭𝒈 equal to

𝑚𝑔𝑐𝑜𝑠𝜃(−𝒚 )+𝑚𝑔𝑠𝑖𝑛𝜃(−𝒙 ) and the tension in the string 𝑭𝑻 equal to 𝑇(𝒚 ), also we know

that the net force is equal to 𝐹𝑥𝒙 +𝐹𝑦 𝒚 . Substituting the values and comparing the

coefficients we get

𝑚𝑑2𝑥

𝑑𝑡2 = −𝑚𝑔𝑠𝑖𝑛𝜃, (1.38)

𝑇 = 𝑚𝑔𝑐𝑜𝑠𝜃. (1.39)

21

Since along y-axis there is no motion so acceleration is zero. The two components of

restoring force 𝐹𝜃 are 𝐹𝜃1= mg cosθ and 𝐹𝜃2=mgsin𝜃, the component 𝐹𝜃1 is balanced by the

tension T in the string so the tangential component of the net force is 𝐹𝜃

=−mg sinθ, here

negative sign indicate fo noitcerid ot etisoppo si F taht 𝜃, rof simple harmonic motion 𝜃

should be very small. So the tangential component of restoring force 𝐹𝜃 will become equal to

–𝑚𝑔θ. Also we have knowledge that for small displacements, the restoring force is

proportional to the displacement and is oppositely directed. Substituting 𝑚𝑔

𝑙= 𝑘(𝑐𝑜𝑛𝑠𝑡. )

we reach at the final relation for the time period of the simple pendulum

T = 2𝜋 𝑙

𝑔. (1.40)

Or

T = 2𝜋 𝑚

𝑘 . (1.41)

From this expression we might expect that a larger mass would lead a longer period.

However increasing the mass also increases the forces that act on the pendulum: gravity an

tension in the string. This increases k as well as m so the time period of the pendulum is

independent of m.

l

-

O

mgsin

mgcos

T

Figure 1-13: The force on the bob of simple pendulum.

1-7 Physical Pendulum

A rigid body mounted so that it can swings in a vertical plane about some axis passing

through it is called physical pendulum.

A body of irregular shape is pivoted about a frictionless horizontal axis through P the

position is that in which centre of mass C of the body lies vertically below P. The distance

from the pivot to centre of mass is d. The rotational inertia of the body about an axis through

22

the pivot is I and mass of the body is m. In the displaced position weight mg acts vertically

downward at an angle 𝜃 with 𝜏 = (𝑓𝑜rce)× (𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚) written as

𝜏 = 𝐹 × 𝑟. (1.42)

Here, F is along y-axis and its magnitude is 𝑚𝑔𝑠𝑖𝑛𝜃 while the displacement is along x-axis

having magnitude 𝑑 so 𝜏 will be along z-axis and is written as 𝜏𝑧 and is equal to −𝑚𝑔𝑑𝑠𝑖𝑛𝜃.

As the restoring torque is proportional to the 𝑠𝑖𝑛𝜃. So motion is simple harmonic motion.

Hence, 𝑠𝑖𝑛𝜃 ≈ 𝜃 and then we get 𝜏 = −𝑚𝑔𝑑𝜃, negative sign indicates that restoring force is

clockwise when the displacement is counterclockwise and vice versa. The equation of motion

is

𝜏𝑧 = 𝐼𝛼𝑍 . (1.43)

dcos

dsin

d

O z

mgsin

mgcos

mg

Figure 1-14: Dynamics of physical pendulum.

Here, 𝛼𝑍 is the angular acceleration along the 𝜏𝑧 and is given by 𝑑2𝜃

𝑑𝑡2, 𝜃 is the angular

displacement substituting these values and then by simplifying we get

𝜔 = 𝑚𝑔𝑑

𝐼𝜃, (1.44)

where 𝜔 is the angular frequency, d is the distance, 𝜃 is the angular displacement and 𝐼 is

the moment of inertia.Also we get the values of the frequency and time period of the physical

pendulum respectively given by

23

𝑓 =1

2𝜋 𝑚𝑔𝑑

𝐼. (1.45)

T= 2𝜋 𝐼

𝑚𝑔𝑑. (1.46)

1-8 Springs In Parallel Or Series

When springs are organized in a parallel configuration as shown in figure (1.11) or in

series as shown in figure (1.12) then the springs act as a single spring with an effective spring

constant.

m

Figure 1.11: Springs in parallel.

m

Figure 1-15: Springs in series.

1-8-1-1 Parallel Configuration

When the spring m in fig (1.16) is displaced by amount x, each spring is displaced by

corresponding amounts 𝑥1 = 𝑥2 = 𝑥, the net restoring force applied by the spring is

𝑭𝒏𝒆𝒕 = 𝑭𝒓 + 𝑭𝒈 + 𝑭𝑵, (1.47)

𝑭𝒏𝒆𝒕 is net force an is equal to𝐹𝑥𝒙 +𝐹𝑦 𝒚 , 𝑭𝒓 is the restoring force here it is equal to kx and

for parallel configuration k is equal to 𝑘1 + 𝑘2, 𝑭𝒈 is the gravitational force equal to 𝑚𝑔𝒚 ,

𝑭𝑵 is the normal force equal to 𝑁𝒚 . Substituting the values in equation (1.47) and then

putting the value of 𝐹𝑦 = 0 and 𝐹𝑥 = −𝑘𝑒𝑓𝑓 𝑥 we get equation (1.6). This equation leads us

24

to the result that the two springs act as a single spring having 𝑘𝑒𝑓𝑓 = 𝑘1 + 𝑘2, where 𝑘𝑒𝑓𝑓 is

the effective spring constant for parallel configuration of the springs.

m

m

m

Relaxed

Stretched

Compressed

Figure 1-16: Spring in parallel.

1-8-1-2 Variation Of The Parallel Configuration

If 𝑥 is positive then first spring is stretched and the second is compressed by the same

amount of 𝑥, then the net force is given by equation (1.47) where, 𝐹𝑘 is equal to 𝑭𝒌𝟏 + 𝑭𝒌𝟐

and 𝑭𝒌𝟏 is given by 𝑘1𝑥1(−𝒙 ) and while 𝑭𝒌𝟐 is equal to 𝑘2𝑥2(−𝒙 ), 𝑭𝒈is the gravitational

force and is equal to the weight of the body so is given by mg(-𝒚 ) and 𝑭𝑵 is the normal

force which is given by N(𝒚 ) so we can write after simplification

𝑭𝒙 = −𝑘𝑒𝑓𝑓 𝑥. (1.48)

And finally we get equation of motion that is equation (1.6), where, 𝜔2 is the angular velocity

and here its value is 𝑘𝑒𝑓𝑓

𝑚.

25

When 𝑥 is negative then first spring is compressed and the second is stretched by the same

amount 𝑥 . Thus in such case the net force on the mass is 𝐹𝑥𝒙 +𝐹𝑦 𝒚 , , 𝑭𝒓 is the restoring

force here it is equal to kx and for parallel configuration k is equal to 𝑘1 + 𝑘2, 𝑭𝒈 is the

gravitational force equal to 𝑚𝑔(−𝒚 ), 𝑭𝑵 is the normal force equal to 𝑁𝒚 also from figure it

is clear that there is no force along the y-axis so it will be equal to zero and we are left with

the net force acting only along the x-axis, after substituting the values and solving we get

𝑥 +𝑘𝑒𝑓𝑓

𝑚= 0, (1.49)

It is clear from here that we get the same result that is equation (1.6) with the value of angular

velocity equal to

𝜔 = 𝑘𝑒𝑓𝑓

𝑚 (1.50)

m

m

m

Figure 1-17: A variation of a parallel spring mass-system.

1-8-2 Series Configuration

For this case the springs are arranged in series as shown. If a force F is applied so as to

displace a mass m by an amount x then the first spring is displaced by an amount 𝑥1 and the

second spring by an amount 𝑥2 but the total displacement will be, 𝑥 = 𝑥1 + 𝑥2,

26

Relaxed

Stretched

Compressed

m

m

m

Figure 1-18: Spring in series.

As we know that for the series configuration

1

𝑘𝑒𝑓𝑓=

1

𝑘1+

1

𝑘2. (1.51)

From here we find the value of 𝑘𝑒𝑓𝑓 equal to 𝑘1 𝑘2

𝑘1+𝑘2 , with the understanding that the hook's

law still applies we can surely write equqtion (1.1) as 𝐹 = −𝑘𝑒𝑓𝑓 𝑥, Force can be written

equal to 𝑚𝑥 , and after solving the equation after substitution we get equaton (1.6) with

𝜔 = 𝑘𝑒𝑓𝑓

𝑚 (1.52)

1-9 Floating Objects (A Bottle In A Bucket)

Case 1: We considered a bottle floating upright in a large bucket of water as shown. In

equilibrium it is submerged to a depth 𝑑𝑜 below the surface of water.

The two forces on the bottle are its weight acting downward mg and the upward buoyant

force 𝜚𝑔𝐴𝑑 where 𝜚 is the density of water and A is the cross sectional area of the bottle and

by archidemes principle we know that the buoyant force is 𝜚𝑔 times the volume submerged

which is just Ad.The equilibrium depth 𝑑𝑜 is determined by the condition 𝑚𝑔 = 𝜚𝑔𝐴𝑑. The

net force on the bottle in equilibrium condition is given by

𝑭𝒏𝒆𝒕= 𝑭𝒃 + 𝑭𝒈 , (1.53)

27

Figure 1-19: (a) Bottle in equilibrium position. (b) Bottle lying below the equilibrium

position. (c) Bottle lying above the equilibrium position.

Here, 𝑭𝒏𝒆𝒕 is the net force and is equal to 𝐹𝑥𝒙 +𝐹𝑦 𝒚 and 𝑭𝒃 is the buoyant force and it acts

along y-axis and its magnitude is equal to 𝜚𝑔𝐴 𝑑𝑜 + 𝑦 also 𝑭𝒈 that is gravitational force

which is equal to the weight of the body is acting in the downward direction given

by 𝑚𝑔(−𝒚 ). After substituting the values and then comparing the coefficients of x and y on

both sides we arrive at the result that as there is no force along x-axis so it will become equal

to zero while comparing the forces along y-axis we get

𝑚𝑔 = 𝜚𝑔𝐴𝑑𝑜 . (1.54)

Case 2: When the bottle is more dipped in the water up to the depth 𝑑𝑜 + 𝑦 so that it lies

below the equilibrium position, in this case the net force will be given by equation (1.52)

Here, is the net force acting on the bottle in the compressed position and is equal to the 𝐹𝑥𝒙

+𝐹𝑦 𝒚 , while 𝑭𝒃 is the buoyant's force and is equal to 𝜚𝑔𝐴 𝑦 − 𝑑𝑜 (𝒚 ) and 𝑭𝒈 is the

gravitational force which will be equal to the weight of the bottle and is given by 𝑚𝑔(− 𝒚 )

put these values in equation (1.52). As there is no force acting along x-axis so it is equal to

zero and we are left only with the y-component of the force given by

𝐹𝑦 = 𝜚𝑔𝐴𝑦.

Substituting the value of 𝐹𝑦 = −𝑚𝑦 and then putting 𝑑𝑜 =𝑚𝑔

𝜚𝐴 we get the more précised

result

𝑦 + 𝜔2𝑦 = 0, (1.55)

𝜔 = 𝑔

𝑑𝑜 . (1.56)

F

mg

F

mg(a) (b) (c)

28

Case 3:

In this case the large portion of the bottle is outside or above the surface of water as shown

Or we can say this is similar to the compression case, So we have the net force acting on the

bottle in the compressed position and is equal to 𝐹𝑥𝒙 +𝐹𝑦 𝒚 , while 𝑭𝒃 is the buoyant's force

and is equal to 𝜚𝑔𝐴 𝑦 − 𝑑𝑜 (𝒚 ) and 𝑭𝒈 is the gravitational force which will be equal to the

weight of the bottle and is given by 𝑚𝑔(− 𝒚 ) putting these values in the above

equation.Here there is no force acting along x-axis so it is equal to zero and we are left only

with the y-component of the force given by

𝐹𝑦 𝒚 = −𝜚𝑔𝐴𝑦(𝒚 ) (1.57)

Substituting the value of 𝐹𝑦 = 𝑚𝑦 and then putting 𝑑𝑜 =𝑚𝑔

𝜚𝐴 we get equation (1.54) with

value of the angular velocity equal to

𝜔 = 𝑔

𝑑𝑜 . (1.58)

1-10 Tunnel In Earth

We drill a hole through the earth having the radius R and mass M along a diameter and

drop a small ball of mass m down the hole. Here, earth assumed to be a uniform sphere (a

poor assumption).

O

m

R

M

Figure 1-20: A hole through the centre of the earth (assume to be uniform). When an object

is at varying distance y from the centre.

When y> 0 then the net force is 𝑭𝒏𝒆𝒕 acting on the ball which is along x-axis as well as y-

axis and is given by 𝐹𝑥𝒙 +𝐹𝑦 𝒚 and 𝑭𝒈is the gravitational force and its value is given

by𝑭𝒈(−𝒚 ), so from above relation

𝑭𝒏𝒆𝒕 = 𝑭𝒈. (1.59)

29

Comparing the x and y components on both sides we get 𝐹𝑦 = −𝐹𝑔 but as there is no force

along the x-axis so it is equal to zero. Also,

Fraction of mass = 4

3𝜋𝑦3

4

3𝜋𝑅3

. (1.60)

So we get the fraction of the mass equal to 𝑦3

𝑅3 , gravitational force becomes 𝐹𝑔 = −

𝐺𝑚𝑀𝑦

𝑅3.

Putting the value of 𝐹𝑔 in the above equation and then substituting the value of 𝐹𝑦 equal to

𝑚𝑑2𝑦

𝑑𝑡2 and simplifying we can get the précised result,that is given by

𝑑2𝑦

𝑑𝑡2+ 𝜔2𝑦 = 0, (1.61)

Here,

𝜔 = 𝐺𝑀

𝑅2 . (1.62)

The equation (1.61) shows that the motion of the bob in tunnel of earth is an example of

simple harmonic motion.

30

REFORMULATIONS OF PROBLEMS USING

LAGRANGIAN TECHNIQUES

2-1 Introduction

Lagrangian mechanics is used as the first alternative to the newtonian mechanics. Using

lagrangian mechanics is sometimes useful instead of newtonian mechanics in the problems

where the equations of newtonian mechanics becomes difficult to solve. In lagrangian

mechanics we begin by defining a quantity called the LAGRANGIAN (L) which is defined

as: The difference between kinetic energy kinetic energy T and potential energy V.

2-2 Horizontal Oscillations

As we know that kinetic energy of the system is

𝑇 =1

2 𝑚𝑥 2. (2.1)

The potential energy is

𝑉 =1

2 𝑘𝑥2. (2.2)

The lagrangian L of the system is given by

𝐿 = 𝑇 − 𝑉. (2.3)

Substituting the values from equqtions (2.1) and (2.2) in equqtion (2.3) we can get the value

of lagrangian L. Now since the lagrangian equation of the dynamical system is

𝑑

𝑑𝑡 𝜕𝐿

𝜕𝑥 −

𝜕𝐿

𝜕𝑥= 0. (2.4)

2

31

By substituting the values from equation (2.1) and (2.2) in (2.4) we can get the equation of

motion of simple harmonic oscillator (1.6).

2-3 Vertical Oscillations

To solve the vertical oscillations using lagrangian , first of all we consider the kinetic

energy of the system which can be written as

𝑇 =1

2 m𝑦 2. (2.5)

According to the law of conservation of energy the potential energy is

𝑉 =1

2𝑘(𝑦 + 𝑙𝑜)2 −𝑚𝑔𝑦. (2.6)

We can get the lagrangian L by usimg equations (2.5) and (2.6) in equation (2.3), the

lagrange's equation for the vertical oscillations will be

𝑑

𝑑𝑡 𝜕𝐿

𝜕𝑦 −

𝜕𝐿

𝜕𝑦= 0. (2.7)

After putting the values and then solving we get

𝑦 +𝜔2𝑦 = 0. (2.8)

2-4 Simple Harmonic Motion In 2 Dimensions

Kinetic energy of system in 2-Dimensions is given by

T=1

2𝑚 𝑟 2 + 𝑟2𝜃 2 (2.9)

Here, 𝑟 cos𝜃-r sin𝜃 is equal to 𝑥 similarly 𝑟 sin𝜃+ r cos𝜃 is equal to 𝑦 , Potential energy is

given by the relation V = 1

2𝑘𝑥2 +

1

2𝑘𝑦2 and then we can write it as V =

1

2𝑘𝑟2. The

lagrangian formula after substitution of the values is given by

L=1

2𝑚(𝑟 2+𝑟2𝜃 2) −

1

2𝑘𝑟2. (2.10)

The lagrangian equation is given by the equation (2.7).After substitution of the values the

final result of the equation of motion in 2-Dimensions

𝑎𝑟 + 𝜔2𝑟 = 0. (2.11)

2-5 Simple Harmonic Motion In 3 Dimensions

T = 1

2𝑚 𝑥 2 + 𝑦 2 + 𝑧 2 . (2.12)

32

V = 1

2𝑘(𝑥2 + 𝑦2+𝑧2). (2.13)

Along x-axis the lagrangian equation is given by equation (2.4), after substitution it will

become

𝑚𝑥 + 𝑘𝑥=0. (2.14)

Along y-axis the lagrangian equation is given by equation (2.7), after substitution it will

become

𝑚𝑦 + 𝑘𝑦=0. (2.15)

Along z-axis the lagrangian equation is written as 𝑑

𝑑𝑡(𝜕𝐿

𝜕𝑧 )−

𝜕𝐿

𝜕𝑧= 0, after substitution it will

become

𝑚𝑧 + 𝑘𝑧=0. (2.16)

2-6 Simple Pendulum

m

O

x

A

CB

l

Figure 2-1: Simple Pendulum.

We choose the angle 𝜃 made by string of the pendulum with the vertical axis as generalized

coordinates. The kinetic energy of the pendulum bob is

T= 1

2𝑚(𝑙𝜃) 2, (2.17)

where m is the mass of the bob and l is the length of the pendulum. The potential energy of

the mass is

33

𝑉 = 𝑚𝑔𝑙(1−cos𝜃). (2.18)

Where we take a horizontal plane as the reference level which is passing through the lowest

point of the bob, the lagrangian is given equation by (2.3) and after substituting the values we

get

𝐿 =1

2m𝑙2𝜃 2 − 𝑚𝑔𝑙(1 −cos𝜃). (2.19)

The equation of motion of the mass is given by the lagrang's equation

𝑑

𝑑𝑡 𝜕𝐿

𝜕𝜃 −

𝜕𝐿

𝜕𝜃= 0, (2.20)

By substituting the values an then differentiating and solving the above equation

𝜃 +𝜔2𝜃 = 0, (2.21)

Where

𝜔 = 𝑔

𝑙. (2.22)

2-7 Physical Pendulum

The rotational inertia of the body about an axis through the pivot is I and mass of the body

is M.The kinetic energy of physical pendulum is given by

𝑇 =1

2𝐼𝜃 2. (2.23)

dcos

dsin

d

O z

mgsin

mgcos

mg

Figure 2-2: Physical pendulum.

34

The potential energy of physical pendulum is given by

𝑉 = 𝑚𝑔𝑑(1− 𝑐𝑜𝑠𝜃). (2.24)

The lagrangian L is given by equation (2.3) and after substituting the values we get

𝐿 =1

2𝐼𝜃 2 − 𝑚𝑔𝑑(1− 𝑐𝑜𝑠𝜃) (2.25)

Since lagrangian equation of motion is given by equation (2.20) so after substituting the

values and then solving we get the equation of motion of physical pendulum.

𝜃 +𝜔2𝜃 = 0, (2.26)

where

𝜔 = 𝑚𝑔𝑑

𝐼. (2.27)

2-8-1 Parallel Configuration

m

m

m

Relaxed

Stretched

Compressed

Figure 2.3: Springs in parallel.

When the spring m in fig (2.3) is displaced by amount x, each spring is displaced by

corresponding amounts 𝑥1 = 𝑥2 = 𝑥 , the kinetic energy of the spring is

35

And because we have 𝑥1 = 𝑥2 = 𝑥 so kinetic energy of the system is given by the equqtion

(2.1). The potential energy is given by 𝑉 =1

2𝑘1𝑥1

2 +1

2𝑘2𝑥2

2, after simplification we get

𝑉 =1

2𝑘𝑒𝑓𝑓 𝑥

2 (2.28)

The lagrange's equation is given by equqtion (2.3) and then after putting the values we get

𝐿 =1

2m𝑥2 −

1

2𝑘𝑒𝑓𝑓 𝑥

2. (2.29)

The lagrange's equation is given by equation (2.7) after solving we get

𝑚𝑥 + 𝑘𝑒𝑓𝑓 𝑥 = 0. (2.30)

2-8-2 Series Configuration

Relaxed

Stretched

Compressed

m

m

m

Figure 2.4: Springs in series.

For this case the springs are arranged in series as shown. If a force F is applied so as to

displace a mass m by an amount x then the first spring is displaced by an amount 𝑥1 and the

second spring by an amount 𝑥2 but the displacement will be 𝑥 = 𝑥1 + 𝑥2 Where 𝑥1 =𝐹

𝑘1

and𝑥2 =𝐹

𝑘2 the kinetic energy will be

𝑇 =1

2𝑚𝑥 2. (2.31)

While the potential energy is given by

V= 𝐹2

2 𝑘𝑒𝑓𝑓 =

𝑥2𝑘𝑒𝑓𝑓2

2 𝑘𝑒𝑓𝑓=

1

2𝑘𝑒𝑓𝑓 𝑥

2 (2.32)

36

As the lagrangian is given by the equation (2.3) and the lagrange's equation is given by

equation (2.7). If 𝜔 is the angular velocity given by 𝑘𝑒𝑓𝑓

𝑚 then finally we will get the

equqtion of motion of simple harmonic oscillator that is equation (1.6).

2-9 Floating Object

In order to find the lagrangian for the floating object we have to find the kinetic as well as

the potential energy of the object. The kinetic energy of the floating object is given by

𝑇 =1

2𝑚𝑦 2. (2.33)

According to the law of conservation of energy

V=1

2𝜌𝐴𝑔 𝑦 + 𝑑𝑜

2 −𝑚𝑔𝑦. (2.34)

We can substitute these values in the formula of lagrangian and will get

𝐿 =1

2𝑚𝑦 2 −

1

2𝜌𝐴𝑔 𝑦 + 𝑑𝑜

2 +𝑚𝑔𝑦. (2.35)

The lagrangian equation is given by the equation (2.7). By substituting all the above values

and then simplifying we can get equation of motion of the floating object

𝑦 +𝑔

𝑑𝑜𝑦 = 0. (2.36)

2-10 Tunnel In Earth

In order to find the lagrangian for the earth, we must know the kinetic and potential energy

of the earth. The kinetic energy of the earth is given by equation (2.5). The potential energy

of the earth can be given by

𝑉 =𝐺𝑚 (

𝑦3

𝑅3𝑀)

2𝑦. (2.37)

And then after simplification we can get 𝐺𝑚𝑀

2𝑅3 𝑦2.We get after substitution of the values in the

formula of lagrangian we can get

L=1

2𝑚𝑦 2 −

1

2

𝐺𝑚𝑀

𝑅3 𝑦2. (2.38)

The lagrangian equation is given by the equation (2.7).After substituting the values and then

after simplification we can equation (2.8) that is the equation of motion.

37

REFERANCES:

[1] Javier E. Hasbun, Classical Mechanics, by Jones and Bartlett publishers, USA, 2009.

[2] Hugh D. Young & Roger A. Freedman, University physics with modern physics (12th

edition), Sears And Zemansky’s edition, by Addison-Wesley publishing company, USA,

2007.

[3] Raymond A. Serway & John W. Jewett, Physics for scientists and engineers (6th

edition),

By Thomson-Brooks/Cole Custom Publishing company, 2003.

[4] Francis W. Sears, University physics (3rd

edition), by Addison - Wesley Press Inc, USA,

2000.

[5] John R. Taylor, Classical Mechanics, by university science books, B. Jane Ellis

South Orange, New Jersey, 2005.

[6] Tai L. Chows, Classical Mechanics, by john wiley & sons Inc. USA, 1995.

[7] Murray R. Spiegel, Schaum's outline of theoretical mechanics with lagrange's equation,

by Hill Education, 1977.

[8] Herbert G. , Charles P. Poole & John L. Safco, Classical Mechanics (3rd

Edition) ,

Addison Wisley, USA, 2002.

[9] David M. , Introduction to classical mechanics edition, Cambridge University Press ,

2008.

[10]Charles K. & Walter D. Knight, Mechanics (2nd

edition), by Tata Mgraw Hill, 1973.

[11] Vernon B. & Marton O. , Classical Mechanics a modern perspective (2nd

edition),

McGraw-Hill College, 1995.

[12] Ronald L. Reese, 1st

edition, University Physics, Brooks Cole, 1999.

38

39

40