Chapter 1:Oscillations - CTAPS · ©Dr. N. Ershaidat Phys. 207 Chapter 1: Oscillations Lecture 1 3 Oscillations are motions that repeat themselves. ... /Phys201/Chap6.htm Harmonic

1

Phys. 103: Waves and Light

Physics DepartmentYarmouk University 21163 Irbid Jordan

Phys. 103 Waves and Light

© Dr. Nidal M. Ershaidat

Chapter 1: Oscillations

http://ctaps.yu.edu.jo/physics/Courses/Phys103/Chapter1© Dr. N. Ershaidat Phys. 207 Chapter 1: Oscillations Lecture 1

2

Oscillations are motions that repeat themselves.We are surrounded by oscillations:

16-1 Oscillations

• There are swinging clock pendulums .

• There are oscillating guitar strings, drums, bells.

• Diaphragms in telephones and speaker systems .

• Quartz crystals in wrist watches.

© Dr. N. Ershaidat Phys. 207 Chapter 1: Oscillations Lecture 1

3

Oscillations are motions that repeat themselves.We are surrounded by oscillations:

16-1 Oscillations

• There are swinging clock pendulums .

• There are oscillating guitar strings, drums, bells.

• Diaphragms in telephones and speaker systems .

• Quartz crystals in wrist watches.

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 4

Less evident are the oscillations of:

More Oscillations

• Air molecules that transmit the sensation of sound,

• Oscillations of the atoms in a solid that convey the sensation of temperature

and

• The oscillations of the electrons in the antennas of radio and TV transmitters that convey information.


Damped and Undamped Oscillations

Oscillations in the real world are usually damped; that is, the motion dies out gradually, transferring mechanical energy to thermal energy by the action of frictional forces.

In this course we shall concentrate on undamped oscillations.

Earthquakes for example

Periodic Motion

Harmonic Motion

2

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations7

dt

pdF ext

→→→→→→→→

====∑∑∑∑ .

Dynamics is the study of motion taking into consideration what caused it, i.e. forces.

Mechanics is the branch of physics which deals with motions

Kinematics is the study of motion regardless of what caused it.

Newton’s second law which relates the external forces on a system and its acceleration is the general principle of dynamics.

Mechanics, Motion , Forces

→→→→→→→→====∑∑∑∑ amF ext . 1


The goal of mechanics is to solve the equation of motion, i.e. to find the relation between the space coordinates of the mechanical system in motion and time, i.e.

• Motion can be translational (the system translates in space).

Motions, Equation of Motion

(((( ))))tfr ====→→→→

The linear motion is an example.

• Motion can be rotational (the system rotates around a fixed point (axis) in space.

2


Periodic Motion – Definition

A periodic motion is one that

repeats itself in successive equal

intervals of time.

PeriodThe period is the time required for one

complete repetition of the motion.


A periodic motion could be continuous

Periodic Motion – Continuity

Examples

or discontinuous.


We also say that the function f(t) repeats itself every T interval.

Periodic Motion - MathsMotion is said to be periodic if the

displacement in space as a function of time

is a periodic function.

A continuous or discontinuous function f(t)

is said to be periodic of period T if we have

f(t) = f(t + T)

Equivalently Motion is said to be periodic if the

solution of its equation of motion is a periodic

function.


The sine function:

is periodic of period 2ππππ

Continuous Periodic Function

(((( )))) (((( ))))xxf sin====

Fig. 1: The sine function

3


0 L-L 2L-2L 3L

The Heaviside step function defined by:

is periodic of period 2L

Discontinuous Periodic Function

(((( ))))

<<<<<<<<

<<<<<<<<−−−−====

Lx

xLxf

01

00

Fig. 2: Heaviside step function


For more details see:

Further Reading

http://ctaps.yu.edu.jo/physics/Courses/Phys201/Chap6.htm

Harmonic Motion


A periodic motion is said to be harmonic if the system passes by an equilibrium position after each quarter of a period.

Dynamically this motion is caused by a force which can restore the system to a given equilibrium position.

The following figure shows a harmonic motion.

Harmonic Motion

x=0t=0

t=T/4

t=T/2

Fig. 3: Harmonic motion


The net force on the particle in the above periodic motion is such that the magnitude of the force is proportional to the displacement of the particle but the direction of the force is always opposite to that of the displacement (the force is always directed toward the midpoint).

A Simple Harmonic Motion The simplest periodic motion is a particle moving back and forth between two fixed points along a straight line (See figure 3 below).

To undergo such a motion, the particle must be subject to a "restoring" force that is opposite to the displacement at least part of the time. Simple Harmonic

Motion

4


A harmonic motion is said to be simple harmonic motion (SHM) if the magnitude of the restoring force acting on the system is proportional to the displacement from the equilibrium position of the system.

Typical examples are:

Hooke’s Force: Restoring Force of a spring F= - k x

The pendulum (angular displacement is small)

Simple Harmonic Motion


Simple Harmonic Motion

The most common object that obeys Hooke's law on large length scale is a spring. Therefore, the motion of a particle on a spring is a classical example of simple harmonic motion.

The motion is periodic and can be described as that of a sine function (or equivalently a cosine function), with constant amplitude. It is characterized by its amplitude, its period and its phase.

Simple Harmonic

Oscillator (SHO)

A system which undergoes a SHM is

called a simple harmonic oscillator


One important property of oscillatory

motion is its frequency, or number of

oscillations that are completed each

second. The symbol for frequency is f,

and its SI unit is the hertz (abbreviated

Hz),

Period T is defined by:

Frequency

1 hertz = 1 Hz = 1 Oscillation per second = 1 s-1

f

1T ==== 3


The angular frequency (symbol ωωωω) ωωωω = 2 ππππ f is also used in the study of

oscillations.

Period T is defined by:ωωωω

ππππ====

2T

Its SI unit is the rad s-1

Angular Frequency

x=0

4

Equation of Motion

5

25

Simple Harmonic Motion - Animation

x=0

Fig. 4: “Sine” behaviour © Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 26

This motion can be described by

In general the motion can be described by:

Equation of Motion

txx m ωωωω==== cos

Fig. 5: Displacement vs. t

5


Displacement at time t

(((( ))))}

{(((( ))))

48476φφφφ++++ωωωω==== txtx m cos

AmplitudeAngular

Frequency

Time

Phase

Phase angle*

* Phase constant

5


x(t) = xm cos (ωωωω t + φφφφ)

x(

t)

Fig. 6: Displacement vs. t


The phase is defined by the initial conditions of the motion

Phase

x = xm cos (ωωωω t + φφφφ )

x(t=0) = 0 = cos (φφφφ ) ⇒⇒⇒⇒ φφφφ = -ππππ/2 ⇒⇒⇒⇒ x = xm sin ωωωω t

cos

( ωω ωωt

+ φφ φφ

)

ωωωω t


Examples of Phase Difference

Fig. 7: x/xm vs. ωωωωt

6


Velocity is defined as:

Acceleration is defined as:

Velocity and Acceleration

)tsin(xdt

dxv m φφφφ++++ωωωωωωωω−−−−========

)cos(2 φφφφ++++ωωωωωωωω−−−−======== txdt

dva m

mmm xv),t(sinvv ωωωω====φφφφ++++ωωωω−−−−====

mmm xataa 2),cos( ωωωω====φφφφ++++ωωωω−−−−====

6

7

8

9


x, v and a

txx m ωωωω==== cos

tvv m ωωωω−−−−==== sin

taa m ωωωω==== cos

Fig. 8: x, v and a


a is proportional to x

(((( )))) xtxdt

dva m

22 cos ωωωω−−−−====φφφφ++++ωωωωωωωω−−−−========

02 ====ωωωω++++ xa

02

2

2

====ωωωω++++ xdt

xd

02 ====ωωωω++++ xx&&

or

13

10

11

12


The general solution of a homogeneous 2nd order differential equation of the form:

General solution of a linear 2nd order DE

14

15

02 ====ωωωω++++ yy&&

is given by:

)cos( φφφφ++++ωωωω==== tyy m


General solution of a linear 2nd order DE

The general solution ofof the 2nd order DE:

16

17

02 ====ωωωω++++ yy&&

Where αααα and ββββ are the roots of the

auxiliary equation r2 + a r + b = 0

tt eBeAy βα ++++====

Here we have αααα= i ωωωω and ββββ = - i ωωωω = αααα*

titi eBeAy ωω −−−−++++====

15)cos( φφφφ++++ωωωω==== tyy m

is given by:

Which we can rewrite as:

1-3

Force Law for a Simple

Harmonic Motion

7


which means that the restoring force is proportional to the displacement x and acts in an opposite direction to the displacement.

The restoring force of a spring displaced a distance x from its equilibrium point is given by Hooke's law:

Hooke’s Law

xkF −−−−====

Fig. 9: Spring Restoring Force


Hooke’s Law – Elasticity Constant

Hooke’s law holds (approximately) as long as x

remains below the spring's elastic limit. Even

when the elastic limit is not exceeded, pulling

a coil spring far enough to uncoil it results in a

much larger spring constant than the spring's

"coiled" value.

xkF −−−−==== 18

The proportionality constant k is the spring constant or elasticity constant. Its unit in SI is

N/m.


We retrieve here the 2nd order differential

equation, we have seen, which defines a

simple harmonic motion.

Equation of motion for a SHM

xkF −−−−====

0=+⇒−= xm

kaxkam

Physlet\contents\oscillations_waves\periodic_motion\illustration16_4.html

00 2 ====ωωωω++++⇒⇒⇒⇒====++++ xxxm

kx &&&&

The angular frequency of the SHM is:

m

k=ω


Another way to define a simple harmonic

motion is to say that if the displacement x

of a system is governed by a differential

equation of the type:

Equation of motion for a SHM

02 =ω+ xx&&

then the motion is a simple harmonic one.

13

Simple Harmonic

Oscillator (SHO)الهز�از التوافقي البسيطالهز�از التوافقي البسيطالهز�از التوافقي البسيطالهز�از التوافقي البسيط© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 42

The solution of the equation of motion is a

cosine function and we have

Displacement for a SHM

(((( )))) (((( ))))φφφφ++++ωωωω==== txtx m cos

Where xm and φφφφ are integration constants

which we determine using the initial

conditions of the motion (x(0)=xm ).

The system oscillates around its equilibrium

position.

A system which undergoes a SHM (eq. 5) is

called a simple harmonic oscillatorsimple harmonic oscillator

5

8


The solution of the equation of motion is a

cosine function and we have

Displacement as a function of time for a SHM

( ) ( )φ+ω= txtx m cos

Where xm and φφφφ are integration constants

which we determine using the initial

conditions of the motion (x(0)=xm ).

Examples


c

Sample Problem (16-1)

2.What is the maximum speed vm of the oscillating block, and where is the block when it occurs?

1.What are the angular frequency, the frequency, and the period of the resulting motion?

3.What is the magnitude am of the maximum acceleration of the block?

A block whose mass m is 680 g is fastened to a

spring whose spring constant k is 65 N/m. The

block is pulled a distance x = 11 cm from its

equilibrium position at x = 0 on a frictionless

surface and released from rest at t = 0.


Solution1

11

78.968.0

65 −−−−−−−−

≈≈≈≈========ωωωω sradkg

mN

m

k

Hzf 56.12

≈≈≈≈ππππ

ωωωω====

mssf

T 64064.01

====≈≈≈≈====

mxmNkkgm m 11.0,65,68.0 ============

Angular frequency:

Frequency:

Period of the resulting motion?


Solution2

11 1.1)11.0()(78.9 −−−−−−−− ≈≈≈≈====ωωωω==== smmsradxv mm

mxmNkkgm m 11.0,65,68.0 ============Computing vm

0cos0cos)( ====ωωωω⇒⇒⇒⇒====ωωωω==== mmmm ttxtx

)2(mod ππππππππ====ωωωω⇒⇒⇒⇒ mt

mssT

tm 16016.042

============ωωωω

ππππ====

(((( )))) 2212 5.10)11.0()(78.9 −−−−−−−− ≈≈≈≈====ωωωω==== smmsradxa mm

Computing am

This occurs when the block passes by x = 0, i.e.

at tm given by:


Solution3 – Phase ConstantWhat is the phase constant φφφφ for the motion?

The displacement function x(t) for the spring–block system:

[[[[ ]]]]0)8.9(cos)11.0(

)(cos)(

1 ++++====

φφφφ++++ωωωω====

−−−− tsradm

txtx m

)(cos)( φφφφ++++ωωωω==== txtx m mxx ====)0(

(((( )))) (((( ))))φφφφ++++××××ωωωω======== 0cos0 mm xxx

[[[[ ]]]]tsradmtx )8.9(cos)11.0()( 1−−−−====

with

φφφφ==== cosmm xx 01cos ====φφφφ⇒⇒⇒⇒====φφφφ⇒⇒⇒⇒

9


At t = 0, the displacement x(0) of the block in a linear oscillator like that of fig. 4 is -8.50 cm. (Read x(0) as “x at time zero.”). The block's velocity v(0) then is -0.920 m/s, and its acceleration a(0)is +47.0 m/s2.




What is the angular frequency ωωωω of this system?

Solution: x(0) = -8.5 cm, v(0) = 0.92m/s, a(0) = 47 m/s2

(((( )))) φφφφ==== cos0 mxx (((( )))) φφφφωωωω−−−−====φφφφ−−−−==== sinsin0 mm xvv

(((( )))) φφφφωωωω====φφφφ==== coscos0 2mm xaa

(((( ))))(((( ))))

12

5.23085.0

0.47

0

0 −−−−−−−−

====−−−−

−−−−========−−−−====ωωωω⇒⇒⇒⇒ sradm

sm

x

a

,


What are the phase constant φφφφ and amplitude xm?

Computing xm

φφφφ And xm (16-2)

x(0) = -8.5 cm, v(0) = 0.92 m/s, a(0) = 47 m/s2

(((( ))))(((( ))))

(((( ))))(((( )))) φφφφωωωω−−−−====

φφφφ

φφφφωωωω−−−−==== tan

cos0

sin0

00

x

x

xv

(((( ))))(((( ))))

°°°°≈≈≈≈====−−−−××××

−−−−−−−−====

ωωωω−−−−====φφφφ⇒⇒⇒⇒ −−−−−−−− 257.24

)085.0(5.23

92.0tan

0

0tan 11

x

v

(((( ))))cm

cmxxm 4.9

25cos

5.8

cos

0−−−−≈≈≈≈

°°°°

−−−−====

φφφφ====

1-4 Energy in SHM

53

The potential energy of “SHM” is obtained by:

Potential Energy in SHM

( ) ∫∫ =−= dxxkdxxFtU )(

)(cos2

1 22 φ+ω= txk m

( )( )2

2

1txk=

)(cos2

1 222 φ+ωω= txm m

*The dependence of U on t is implicit in the integral. It is in x

*

19


The kinetic energy of “SHM” is given by:

Kinetic Energy in SHM

( )( ) ( )[ ]22sin

2

1

2

1)( φ+ωω−== txmtvmtK m

)(sin2

1 22 φ+ω= txk m

)(sin2

1 222 φ+ωω= txm m

*The dependence of K on t is explicit.

*

20

10


The total mechanical energy of “SHM” is given by:

E does not depend on t. It is constant. This is

an illustration of the law of conservation of energy, since we considered the SHO as an isolated system.

Total Energy in SHM

)()()( tKtUtE +=

22

2

1mxm ω=

)(sin2

1)(cos

2

1 222222 φ+ωω+φ+ωω= txmtxm mm

2

2

1mxk= 21


The figures below show the change of energy of the SHO with time.

Fig. 10: SHM. Energy vs. t

Energy vs. t

1-5 Angular Simple

Harmonic Oscillator


Figure 11 shows an angular version of a simple

harmonic oscillator;

The element of springiness or elasticity is associated with the twisting of a suspension wire rather than the extension and compression of a spring as we previously had.

The device is called a torsion pendulum, with torsionreferring to the twisting.

An Angular Simple Harmonic Oscillator

Fig. 11: Torsion pendulum


An angular simple harmonic oscillator, or torsion pendulum, is an angular version of the linear simple harmonic oscillator.

The disk oscillates in a horizontal plane; the reference line oscillates with angular amplitude θθθθm.

The twist in the suspension wire stores potential energy as a spring does and provides the restoring torque.

Restoring Torque


Restoring Torque

Newton’s 2nd law in rotational motion is written as:

θθθθ====ττττ &&I

0, ====θθθθκκκκ

++++θθθθ⇒⇒⇒⇒θθθθ====θθθθκκκκ−−−−I

I &&&&

θθθθκκκκ−−−−====ττττ

ττττ Replaces F, I (moment of inertia) replaces mand is the angular acceleration. With θθθθ&&

02 ====θθθθωωωω++++θθθθ&&

κκκκππππ====⇒⇒⇒⇒

κκκκ====ωωωω

IT

I2

)cos( φφφφ++++ωωωωθθθθ====θθθθ⇒⇒⇒⇒ tm

11

See Sample

Problem 16-4


Figure 12-a shows a thin rod whose length L is 2.4 cm and whose mass m is 135 g. suspended at its midpoint from a long wire. Its period Ta of angular SHM is measured to be 2.53 s.


An irregularly shaped object (X) is then hung from the same wire as in Fig. 12-b and its period Tb

is found to be 4.76 s. What is the rotational inertia of object X about its suspension axis?

Fig. 12


Sample Problem (16-4) Solution

24

2

24

2

2

1012.6)53.2(

)76.4(1073.1 mkg

T

TII

a

bab

−−−−−−−− ××××====××××========

L=12.4 cm, m=135 g, Ta=2.53 s, Tb=4.76 s.

κκκκππππ====

κκκκππππ==== b

ba

a

IT

IT 2,2

2

12

1LmIa ==== mkg

42 1073.1)124.0(135.012

1 −−−−××××====××××××××====

2

2

a

bab

T

TII ====⇒⇒⇒⇒


The simple harmonic motion can be defined by either of the following:

"motion in which the acceleration of the oscillator is proportional to, and opposite in direction to the displacement from its equilibrium position".

“SHM is a periodic motion where the restoring force is proportional to the displacement from the equilibrium position and opposite in direction to this displacement”.

Simple Harmonic Motion (Review)

xx 2ωωωω−−−−====&&


The simple harmonic oscillator is a mechanical system which is in simple harmonic motion

“A simple harmonic oscillator is a system whose potential energy is proportional to the square of its displacement from an equilibrium position in space (U = ½ k x2)

Another way to define the SHO is the following:

Simple Harmonic Oscillator

1-6

Applications Applications Applications Applications –––– PendulumsPendulumsPendulumsPendulums

12


A simple pendulum is a mechanical system composed of an unstretchable, massless rod of length L fixed at one end and to the another end is attached a mass m.

The Simple Pendulum

m is called the bob of the pendulum.

m swings to the right and back to the left.

Here the motion is associated with the gravitational force. We shall see that this is a SHM. →→→→→→→→

==== gmFgFig. 13: Simple pendulum© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 68

- Tension from the string

m is subject to 2 forces:

- Gravitational force

Forces acting on a Simple Pendulum

→→→→→→→→==== gmFg

→→→→T

Fig. 13 shows the resolution

of these 2 forces in a system

of coordinate where m is the

origin.

One can easily see that the

rotation of m is caused by the

torque:

(((( ))))θθθθ−−−−====ττττ singFL


Pendulum - Equation of Motion

(((( ))))θθθθ−−−−====ττττ singFL

Newton’s 2nd law:mgFLmI g ======== ,2

θθθθ−−−−====θθθθ sinLFI g&&

For small θθθθ: θθθθ≈≈≈≈θθθθsin

00 2 ====θθθθωωωω++++θθθθ⇒⇒⇒⇒====θθθθ++++θθθθ &&&&L

g

θθθθ−−−−====θθθθ sin2 LgmLm &&

0sin ====θθθθ++++θθθθL

g&&

gLT

L

gππππ========ωωωω 2, Animaton1 Pendulum

where

The Physical Pendulum


For small θθθθ, sin θθθθ ≈≈≈≈ θθθθ

The mass in a real physical pendulum is no more a point mass. C is the center of mass and h is the distance between C and the pivot point.

The Physical Pendulum - Equation of Motion

0====θθθθ++++θθθθ⇒⇒⇒⇒θθθθ−−−−====θθθθI

hgmhgmI &&&&

θ−=θ=τ sinhgmI &&

hgmIT ππππ==== 2

I

hgm====ωωωω====θθθθωωωω++++θθθθ ,02&&

Note that I here is the moment of inertia about O

Fig. 14: Physical pendulum

Example 4 Example 4

(Problem 16(Problem 16--5)5)

13


Example 4 - Problem 16-5

In Fig. 15(a) A meter stick

suspended from one end

at a distance h from its

center of mass, point C, as

a physical pendulum.

(b) A simple pendulum

whose length L0 is chosen

so that the periods of the

two pendulums are equal.

Point P on the pendulum of

(a) marks the center of

oscillation.Fig. 15


If I/cm is the moment of inertia of a system about

its center of mass then the moment of inertia about an axis parallel to the axis of rotation is given by:

Where h is the distance between the two axes, i.e.

between the 2 points C and O.

Theorem of Parallel Axis

I/O = I/cm + M h2

C

Oh


Parallel Axis Theoremنظري�ة المحور الموازي نظري�ة المحور الموازي نظري�ة المحور الموازي نظري�ة المحور الموازي II//Axis//Axis يعطى بالعالقة يعطى بالعالقة يعطى بالعالقة يعطى بالعالقة hالكتلة ويبعد عنه مسافة الكتلة ويبعد عنه مسافة الكتلة ويبعد عنه مسافة الكتلة ويبعد عنه مسافة حول محور مواز للمحور المار بمركز حول محور مواز للمحور المار بمركز حول محور مواز للمحور المار بمركز حول محور مواز للمحور المار بمركز mعزم القصور الذاتي لجسم كتلته عزم القصور الذاتي لجسم كتلته عزم القصور الذاتي لجسم كتلته عزم القصور الذاتي لجسم كتلته ““““ = I= I/cm/cm + m h+ m h22 ””””C

Oh

))))Cالنقطة النقطة النقطة النقطة ((((مركز كتلة الجسم مركز كتلة الجسم مركز كتلة الجسم مركز كتلة الجسم OOOO الكتلةالكتلةالكتلةالكتلة هي النقطة التي يمر فيها محور مواز للمحور المار بمركز هي النقطة التي يمر فيها محور مواز للمحور المار بمركز هي النقطة التي يمر فيها محور مواز للمحور المار بمركز هي النقطة التي يمر فيها محور مواز للمحور المار بمركز....© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 76

(a) What is its period of oscillation T ?

Period

hgm

IT O/2 ππππ====

2Lh ====

g

L

Lg

LT

3

22

3

22

2

ππππ====ππππ====

sT 64.18.93

122 ====

××××

××××ππππ====

2// hMII CO ++++==== 2

22

3

1

212

1Lm

LmLm ====

++++====

hg

L

hgm

Lm

32

32

22

ππππ====ππππ====


L0

(b) What is the distance Lo

between the pivot point O of

the stick and the center of

oscillation of the stick?

Center of Oscillation: is defined

as the point P, which if we let

the physical pendulum hang

from it and oscillate, its period

of oscillation is TP is equivalent

to the period of oscillation

around point O, T0. This period

is also equivalent to that of a

simple pendulum of length Lo.

Lo


Period(b) The period of a simple

pendulum of Length L0 is

given by:

g

LTsp

02 ππππ====

g

LT

3

220 ππππ====

mL

LTTsp3

2

3

200 ========⇒⇒⇒⇒====

14


O

P

C

Lo

L

S

Sgm

IT P

P ππππ==== 2

2SmII cmP ++++====

22

121 SmLmIP ++++====

g

L

Sgm

I

TT

P

OP

3

222 ππππ====ππππ

====

mSLSmLm

mSLI P

32

121

32

22 ====++++

====


mSLIP 32====

O

P

C

Lo

L

S

SL

L ++++====2

0

mSLSmLm32

121 22 ====++++

012

2

322 ====++++−−−− LSLS

6LS ====

LLL

L3

2

620 ====++++====

81

1111----7: Simple Harmonic Motion 7: Simple Harmonic Motion 7: Simple Harmonic Motion 7: Simple Harmonic Motion

and Uniform Circular Motionand Uniform Circular Motionand Uniform Circular Motionand Uniform Circular Motion


Fig. 16-a shows a reference particle P´ moving in uniform circular motion with (constant) angular speed ωωωω in a reference circle.

Uniform Circular Motion

The radius xm of the circle is the magnitude of the particle's position vector.

Fig. 16-a: UCM

At any time t, the angular position of the particle is

ωωωωωωωωtt + + φφφφφφφφ, where φφφφ is its angular position at t = 0.


From earlier courses (Phys. 101) we know:

Kinematics of a UCM

1. The linear velocity v is given by:

2. The linear acceleration aR is given by:

22

ωωωω======== mm

R xx

va

ωωωω==== mxv

We also know that a central force is responsible of this kind of motion and we have:

mm

RR xx

vmamF

2

−−−−========rr


The projection of particle P´ onto the x-axis is a

point P, which we take to be a second particle. The

projection of the position vector of particle P´ onto

the x axis gives the location x(t) of P. Thus, we find:

Projection on the x-axis

(((( )))) )(cos φφφφ++++ωωωω==== txtx m

Which is nothing else but the equation of a motion of a simple harmonic oscillator.

15


Similarly, the projection of particle P´ onto the y-axis

is a point Q, which we take to be a second particle. The projection of the position vector of particle P´

onto the y-axis gives the location y(t) of Q. Thus, we find:

Projection on the y-axis

It is easy to see that φφφφ + φφφφ’ = ππππ

(((( )))) )(cos)(sin φφφφ′′′′++++ωωωω====φφφφ++++ωωωω==== tytyty mm

Which is also the equation of motion of a SHO on the y axis.

Fig. 16-b© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 86

UCM = Linear combination of 2 SHM

The position vector of the particle P is given by:

(((( ))))

(((( )))) (((( ))))

jtyitx

jtyitx

rrtr

mmˆ)(sinˆ)(cos

ˆˆ

ˆ

φφφφ++++ωωωω++++φφφφ++++ωωωω====

++++====

====→→→→

Which is a linear combination, we say a superposition of two simple harmonic oscillators.

One on the x-axis and the other on the y-axis.


UCM = Linear combination of 2 SHM

Simple harmonic motion is the projection of uniform circular motion on a diameter of the circle in which the latter motion occurs.


Figure 17 shows the velocity v(t) of the reference particle.

Velocity

)(sin)( φφφφ++++ωωωω−−−−==== tvtv m

The magnitude of the velocity vector is ωωωωxm; its projection on the x

axis is

)(sin)( φφφφ++++ωωωωωωωω−−−−==== txtv m

Fig. 17


Acceleration

Fig. 18 shows the radial acceleration of the

reference particle:

aR = ωωωω2xm ;

Its projection on

the x-axis is :

)(cos)( 2 φφφφ++++ωωωωωωωω−−−−==== txta m

Fig. 18

Problems

16


In the figure below, two identical springs of

spring constant k are attached to a block of

mass m and to fixed supports. Show that

the block's frequency of oscillation on the

frictionless surface is:

Problem 24

Fig. 19

m

kf

2

2

1

ππππ====

x


Let’s write Newton’s 2nd law for the mass m: When the mass is pulled to the right it suffers two forces:

Problem 24 – Solution

m

kf

m

k 2

2

122

ππππ====⇒⇒⇒⇒====ωωωω

02

2 ====++++⇒⇒⇒⇒−−−−==== xm

kaxkam

xkF 2−−−−====

ixkFrˆ−−−−====

→→→→

ixkFlˆ−−−−====

→→→→,

ixkFextˆ2−−−−====

→→→→

K

mT

22 ππππ====⇒⇒⇒⇒

x

m→→→→

rF→→→→

lF

i


Problem 25

Suppose that the two springs in fig. 16 have

different spring constants k1 and k2. Show that

the frequency f of oscillation of the block is

then given by:

where f1 and f2 are the frequencies at which the

block would oscillate if connected only to

spring 1 or only to spring 2.

x

22

21 fff ++++==== 1 2


Problem 25 – SolutionSolution: Newton’s 2nd law for the mass m. When the mass is pulled to the right it suffers two forces:

22

21

22

21

212 fffm

k

m

k++++====⇒⇒⇒⇒ωωωω++++ωωωω====++++====ωωωω

(((( ))))(((( ))))

02121 ====

++++++++⇒⇒⇒⇒++++−−−−==== x

m

kkaxkkam

(((( )))) xkkF 21 ++++−−−−====

m→→→→

1F→→→→

2F

ixkF ˆ11 −−−−====

→→→→ixkF ˆ

22 −−−−====→→→→,

i

(((( )))) ixkkFextˆ

21 ++++−−−−====→→→→

x


In figure 20, two springs are joined and

connected to a block of mass m. The surface is frictionless. If both of the springs have spring constant k find find ωωωωωωωω.

Solution: Newton’s 2nd law for the mass m. When the mass is pulled to the right it suffers the forces:

Problem 27 (Two springs in Series)

m

Fig. 20

x

x1

x2

ixkF ˆ111 −−−−====

→→→→ixkF ˆ& 222 −−−−====

→→→→


Problem 27 - Solution

222111 xkFxkFxkF eff ====================

21 xxx ++++====

021

21 ====++++

++++ xkk

kkxm &&

xkk

kx

21

12

++++====

21

212 x

k

kxxxx −−−−====−−−−====,

48476 2

21

12

x

xkk

kkF

++++−−−−====⇒⇒⇒⇒

++++====ωωωω⇒⇒⇒⇒

21

211

kk

kk

m

xk

kx ====

++++⇒⇒⇒⇒

1

22 1

21

11

1

kk

keff

++++====

m

keff====ωωωωIf we define then

17


A 5.00 kg object on a horizontal frictionless

surface is attached to a spring with spring

constant 1000 N/m. The object is displaced from

equilibrium 50.0 cm horizontally and given an

initial velocity of 10.0 m/s back toward the

equilibrium position. (a) What is the frequencyof the motion? What are (b) the initial potential energy of the block–spring system, (c) the initial kinetic energy, and (d) the amplitude of the oscillation?

Problem 33


Problem 33 - SolutionInformation: m=5 kg, k=1000 N/m, x(0)=0.5 m and v(0)=-10 m/s

,2

200

5

1000

2

1

2

1

ππππ====

ππππ====

ππππ====

m

kf

JxkU 1255.0100021)0( 22

21 ====××××××××========

JvmK 2501055.02

1)0( 22

0 ====××××××××========

a)

b)

c)

d) 2

2

1)0()0( mxkKUE ====++++====

mxm500

375====⇒⇒⇒⇒

22 5002

1250125 mm xxkE ========++++==== cmxm 87≈≈≈≈⇒⇒⇒⇒


Problem 52A stick with length L oscillates as a physical pendulum, pivoted about point O as shown in the Figure.

a) Derive an expression for the period of the pendulum.

hgm

IT 02 ππππ====

xhxmLmIo ====++++==== ,12

1 22

g

x

xg

L

xgm

xmLmT ++++ππππ====

++++ππππ====

122

122

222

Fig. 21


Problem 52 – Solution1b) For what value of χ χ χ χ = x/L is the period a minimum?

χχχχ++++χχχχ

ππππ====12

12

g

L

T is minimum if and only if

g

x

xg

LT ++++ππππ====

122

2

<<<<

χχχχ

====χχχχ

χχχχ====χχχχ

χχχχ====χχχχ

0

0

min

min

2

2

d

Tdsign

d

dT


Problem 52 – Solution2

0112

10

12

1

112

1

2

2

====−−−−χχχχ

⇔⇔⇔⇔====

χχχχ++++χχχχ

−−−−χχχχ

ππππ====χχχχ g

L

d

dT

29.012

1min ≈≈≈≈====χχχχ

T

x/L© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 102

A wheel is free to rotate about its fixed axle. A spring is attached to one of its spokes a distance r from the axle, as shown in the Figure. (a) Assuming that the wheel is a hoop of mass m and radius R, obtain the angular frequency of small oscillations of this system in terms of m, R, r and the spring constant.

Problem 58

Fig. 19

θ r

x

How does the result change if (b) r = R and (c) r = 0?

18


Problem 58 - Solutiona) Computing ωωωω

b) If r = R then

c) If r = 0 then

θθθθκκκκ−−−−====κκκκ−−−−====ττττ 2rrx

02 ====θθθθκκκκ++++θθθθ rI &&

02

====θθθθκκκκ

++++θθθθI

r&&

m

κκκκ====ωωωω

0====ωωωω

θ r

x

2

22

Rm

r

I

r κκκκ====

κκκκ====ωωωω⇒⇒⇒⇒


QUESTION 1

Which of the following relationships

between the acceleration a and the

displacement x of a particle involves SHM:

(a) a = 0.5 x

(b) a = 400 x2

(c) a = -20 x

(d) a = -3 x2 ?


QUESTION 2

Given x(t) = (2.0 m) cos(5 t) for SHM and

needing to find the velocity at t = 2 s, should

you substitute for t and then differentiate

with respect to t or vice versa?


QUESTION 3

The acceleration a(t) of a particle undergoing SHM

is graphed in Fig. 16-18.

(a) Which of the labeled points corresponds to the

particle at -xm?

(b) At point 4, is the velocity of the particle

positive, negative, or zero?

(c) At point 5, is the particle

at -xm, at +xm , at 0,

between -xm and 0,

or between 0 and +xm?


QUESTION 4

Which of the following describes φφφφ for the

SHM of Fig. 16-19-a?

(a) -ππππ < φφφφ <

(b) ππππ < φφφφ <

(c) < φφφφ < - ππππ ?2

3 ππππ−−−−

2

ππππ−−−−

2

3 ππππ

Fig. 16-19-a

108

QUESTION 5

The velocity v(t) of a particle undergoing SHM is graphed in Fig. 16-19 b.

Is the particle momentarily stationary, headed toward

-xm, or headed toward +xm at (a) point A on the graph

and (b) point B?

Is the particle at -xm, at +xm,

at 0, between -xm and 0,

or between 0 and +xm

when its velocity is represented by

(c) point A and (d) point B?Is the speed of the particle increasing or

decreasing at (e) point A

and (f) point B?-xm xm

Fig. 16-19-b

19


QUESTION 6

Figure 16-20 gives, for three

situations, the displacements

x(t) of a pair of simple harmonic

oscillators (A and B) that are

identical except for phase. For

each pair, what phase shift (in

radians and in degrees) is

needed to shift the curve for A

to coincide with the curve for B?

Of the many possible answers,

choose the shift with the

smallest absolute magnitude.

Fig. 16-20110

Figures 16-21 a and b show the positions of four linear

oscillators with identical masses and spring constants, in snapshots at the same instant. What is the phasedifference of the two linear oscillators in (a) Fig. 16-21a and (b) Fig. 16-21 b?

QUESTION 7

Fig. 16-21

R

G

R

G

(c) What is the phase difference between the oscillator

R in Fig. 16-21 a and the oscillator G in Fig. 16-21 b?

Next Lecture

Chapter 2 : Waves I

1





Chapter 2 : Waves Chapter 2 : Waves Chapter 2 : Waves Chapter 2 : Waves I

http://ctaps.yu.edu.jo/physics/Courses/Phys103/Chapter2© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 2

When a beetle moves along the sand within a few tens of centimeters of this sand scorpion, the scorpion immediately turns toward the beetle and dashes to it (for lunch). The scorpion can do this without seeing (it is nocturnal) or hearing the beetle.

How can the scorpion so precisely locate its prey?

Waves I

The answer is in this chapter.

What is a Wave?

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 5

A wave is a disturbance of a medium

which transports energy through the

medium without permanently transporting

matter.

What is a Wave?

In a wave, particles of the medium are

temporarily displaced and then return to

their original position.

2-2

Types of Waves


There are 3 main types of waves:

1.1. Mechanical waves. Mechanical waves.

1. Mechanical Waves

These waves are most familiar because we encounter them almost constantly;

Common examples include water waves,

sound waves, and seismic waves.

All these waves have certain central features: They are governed by Newton's laws, and they can exist only within a

material medium, such as water, air, and rock.

2


These waves are less familiar, but you use them constantly; common examples

include visible and ultraviolet light, radio and television waves, microwaves, X rays,

and radar waves.

These waves require no material medium to exist. Light waves from stars, for

example, travel through the vacuum of space to reach us. All electromagnetic waves travel through vacuum at the same speed c.

2. Electromagnetic Waves


Although these waves are commonly used in modern technology, their type is probably very unfamiliar to you.

Much of what we discuss in this chapter applies to waves of all kinds. However, for specific examples we shall refer to mechanical waves.

These waves are associated with electrons, protons, and other fundamental particles, and even atoms and molecules.

Because we commonly think of these things as constituting matter, such waves are called matter waves (See Phys. 251).

3. Matter Waves

2-3 Transverse and

Longitudinal Waves


Waves, where the parameter of interest (displacement, mechanical stress, etc.) oscillates along the axis of the wave propagation, are called longitudinal waves.

Oscillation vs. Propagation

If oscillation occurs perpendicularly to the direction of the wave propagation, then such a wave is called transversewave (electro-magnetic waves, for example, are transverse ones).


Displacement

In a transverse wave, particles of the In a transverse wave, particles of the

medium are displaced in a direction medium are displaced in a direction

perpendicular to the direction of energy perpendicular to the direction of energy

transport. transport.

In a longitudinal wave, particles of the In a longitudinal wave, particles of the

medium are displaced in a direction medium are displaced in a direction

parallel to energy transport. parallel to energy transport.


1.Single pulse on a stretched string.

2.Continuous pulse on a stretched string

Pulse on a Stretched String

Applet Transverse and Longitudinal

waves.

Fig. 1

3


One way to study the waves of Fig. 17-1a is to

monitor the wave forms (shapes of the waves)

as they move to the right. Alternatively, we

could monitor the motion of an element of the

string as the element oscillates up and down

while a wave passes through it.

Waveform

We would find that the displacement of every

such oscillating string element is perpendicular

to the direction of travel of the wave, as

indicated in Fig. 17-1a. This motion is said to be

transverse, and the wave is said to be a

transverse wave. © Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 15

Figure 2 (17-2) shows how a sound wave can be

produced by a piston in a long, air-filled pipe.

If you suddenly move the piston rightward

and then leftward, you can send a pulse of

sound along the pipe.

Production of Sound waves

Fig. 2


The rightward motion of the piston moves the

elements of air next to it rightward, changing

the air pressure there. The increased air

pressure then pushes rightward on the

elements of air somewhat farther along the

pipe.

Changes in Air Pressure


Moving the piston leftward reduces the air

pressure next to it. Once they have moved

rightward, the nearest elements, and then farther

elements, move back leftward. Thus, the motion of

the air and the change in air pressure travel

rightward along the pipe as a pulse.

Sound waves – Longitudinal Waves


If you push and pull on the piston in simple

harmonic motion, as is being done in Fig. 17-2, a

sinusoidal wave travels along the pipe.

Sound waves – Longitudinal Waves

Because the motion of the elements of air is

parallel to the direction of the wave's travel, the

motion is said to be longitudinal, and the wave is

said to be

a longitudinal wave.

Transverse and

Longitudinal Waves are

Traveling Waves

4


Both a transverse wave and a longitudinalwave are said to be traveling waves because they both travel from one point to another, as from one end of the string to the other end in Fig. 17-1 or from one end of the pipe to the other end in Fig. 17-2 .

Traveling Waves

Animation of transverse and longitudinal waves

2-4 Wavefunction


دا�� ا��

To completely describe a wave on a string

(and the motion of any element along its

length), we need a function that gives the

shape of the wave.

This means that we need a relation in the

form y = h(x, t), in which y is the transverse

displacement of any string element as a

function h of the time t and the position x of

the element along the string.

Wavefunction


We shall start by the simplest form of

disturbance, i.e. that which causes a

simple harmonic motion of the

particles of the medium.

This means that the disturbance of the

medium’s particles can be represented

by a sinusoidal function.

The resultant wave is called sinusoidal

wave.

Simplest Disturbance - SHM


In general, a sinusoidal shape like the wave in Fig. 17-1b can be described with h being either a

sine function or a cosine function; both give the same general shape for the wave. In this chapter we use the sine function.

Sinusoidal Shape

xkyy sin0====

At t = 0


Sinusoidal Shape

xyxkyyλλλλ

ππππ========

2sinsin 00

λλλλ

ππππ====

2kwhere

As time goes on each point will oscillate up and down with angular frequency ωωωωsuch that :

)(sin0 txkyy ωωωω−−−−====λλλλ

5


The crest of a wave is the point on the medium which

exhibits the maximum amount of positive or upwardsdisplacement from the rest position (Points and b).

Crest and Trough of a Wave

y

x

The trough of a wave is

the point on the

medium which exhibits the maximum amount

of negative or downwards

displacement from the

rest position (Points c and d).

a b

c d

Fig. 6© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 27

A = amplitude of the wave = maximum value of the displacement.

λ= wavelength of the wave = distance between 2 crests or 2 troughs.

Amplitude and Wavelength of a Wave

y

x

A

λλλλ

graph for t = constant

λλλλ-A

A


Period and Wavenumber

T = period of the wave (in seconds) = time it takes the wave to repeat itself

= time it takes the wave to travel a distance of one wavelength

k = wavenumber of the

wave = number of “sinus”

in 2 ππππ

λλλλλλλλ = = v Tv T

λλλλ

ππππ====⇒⇒⇒⇒

2k


vT

λλλλ====

Dimensions

λλλλ

ππππ====

2k

Period T s

Amplitude L m

Wavenumber (k) L-1 m-1

Wavelength (λλλλ) L m

Parameter Dimension Unit in SI


Forms of the Wavefunction

ωωωω

−−−−==== tk

xkym sin

−−−−

λλλλππππ====

T

txym 2sin

(((( ))))(((( ))))tvxkym −−−−==== sin

(((( ))))txkyy m ωωωω−−−−==== sin


2-5 Speed of a Traveling Wave

wavetheofphasetxk ====ωωωω−−−−

0====∆∆∆∆ωωωω−−−−∆∆∆∆ txk ff

kt

xv λλλλ====

λλλλππππ

ππππ====

ωωωω====

∆∆∆∆

∆∆∆∆====⇒⇒⇒⇒

/2

2

(((( )))) (((( ))))(((( ))))tvxkytxy m −−−−==== sin,We, thus, can write:

Fig. 4

6

2-5 Traveling Wave


Consider a wave pulse on a string, moving from left to right (along x-direction) with speed = v

Traveling Waves

x

y

x

yv vvt

A

Pulse at t = 0

This is a transverse wave i.e. the displacement

of the string (the medium) is in the y-direction

Pulse at time t

Fig. 5


The maximum displacement A is called the amplitude of the wave.

At t = 0, peak of pulse is at x = 0.

At a later time, t, peak of pulse is at x = v t.

Amplitude

x

y

x

yv vvt

A

Pulse at t = 0 Pulse at time t


The shape of the pulse is given as a function of x’ at t=t

by the function:y(x’) with respect to origin O’.

The shape of the pulse is given as a function of x at t=0

by the function: y = y(x)

with respect to origin O.

Phase

Pulse at t = 0

x

y

v

x’

yv

Note that both y(x) and y(x’)describe the pulse such that y(x) = y(x’) = y(x - vt) Pulse at t = t

OO’ = x - vt.

O

O’


Phase x - vt = constant

Note that the quantity (x – vt) = constant does not

change.

(x – vt) is called the phasephase of the wave.

It changes in such a way that as t changes x

changes to keep the quantity (x – vt) = constant.

x

y

x

yv vvt

A

x'O′


� A mathematical function that describes a (propagating) wave is called a wavefunction.

Wavefunction

� We can describe the wave pulse that we have been considering by a function of the form

)(),( tvxftxyy −−−−========

displacement

(along y-axis)

position of pulse (along x-axis)

� y is a function of the quantity (x - vt).

� y is a “function of two variables”.

7


A wave traveling to the right with speed v

can be described by a wavefunction of the

form:

Traveling Wave

A wave traveling to the left with speed v

can be described by a wavefunction of the form:

)(),( tvxftxyy −−−−========

)(),( tvxftxyy ++++========

Sinusoidal Waves


y = f(x , t)

(((( ))))

−−−−

λλλλ

ππππ==== tvxAy

2sin

This sinusoidal wave is described by the

expression:

(((( ))))tvxfy −−−−====

TvTv

λλλλ====⇒⇒⇒⇒====λλλλ

(((( ))))

−−−−

λλλλππππ====⇒⇒⇒⇒

T

txAtxy 2sin,

Recall that


Angular Frequency and Wavenumber

fT

v ππππ====ππππ

====λλλλ

ππππ≡≡≡≡ωωωω 2

22

�We can define two “new” quantities:

λλλλ

ππππ≡≡≡≡

2k

(((( )))) (((( ))))txkAtxy ωωωω−−−−==== sin,

Thus we can write:

= Angular frequency

= Wavenumber


f = frequency of the wave= number of times that a crest passes a fixed point each second

f is measured in s-1 or Hertz (symbol Hz)

Frequency

====λλλλ====

λλλλ====⇒⇒⇒⇒====λλλλ

fv

TvTv

Tf

1====

speed of a sinusoidal wave

m/s s-1 = Hz m


Velocity, Frequency and Wavelength

(((( ))))smk

fvωωωω

====λλλλ====

Tf

ππππ====ππππ====ωωωω

22

λλλλ

ππππ====

2k

Angular frequency

Wavenumber

Thus the velocity of the wave is given by:

8

Wave speed on a

stretched string


Consider a single symmetrical pulse such as that in the figure, moving from left to right along a string with speed v.

Pulse along a string with speed v

θ

Fig. 7


Geometry

θ

Consider a small string element within the pulse, of length ∆l, forming an arc of a circle of radius R and subtending an angle 2θ at the center of that circle.

The linear mass density is µµµµ.© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 50

Force on the element ∆∆∆∆l

A force with a magnitude equal to the tension in the string pulls tangentially on this element at each end.

The horizontal components of these forces cancel, but the vertical components add to form a radial restoring force. At any instant t this force (2 ττττ sin θθθθ) is central

The horizontal components of these forces cancel, but the vertical components add to form a radial restoring force. At any instant t this force (2 ττττ sin θθθθ) is central


v

θθθθθττττ==== sin2F

θθθθττττ≈≈≈≈θθθθττττ==== 2sin2am

,lm ∆∆∆∆µµµµ==== θθθθ====∆∆∆∆ Rl 2

θθθθµµµµ==== Rm 2

θθθθττττ====θθθθµµµµ 22 aR

RR

v

µµµµ

ττττ====

2

The speed of a wavealong a stretched ideal string depends only on the tension and linear density of the string and not on the frequency of the wave.

RaaR R

µµµµ

ττττ====⇒⇒⇒⇒ττττ====µµµµ⇒⇒⇒⇒

µµµµ

ττττ====⇒⇒⇒⇒ v


A sinusoidal wave is traveling on a string with speed 40 cm/s. The displacement of the particles of the string at x = 10 cm is found to vary with time according to the equation y = (5.0 cm) sin[1.0 - (4.0 s-1)t]. The linear density of the string is 4.0 g/cm.

(a) Find the frequency of the wave.

(b) Find the wavelength of the wave

Example 1 - Problem 18

ππππ====⇒⇒⇒⇒ππππ

====ππππ

ωωωω==== /2

2

4

2ff

11.0,1.02 ====××××====⇒⇒⇒⇒λλλλππππ==== kmxatxxk

mmk ππππ====λλλλ⇒⇒⇒⇒====λλλλππππ====⇒⇒⇒⇒ −−−− 20.0102,10 1

Information: y(x=0.1m, t) = (5.0 cm) sin[1.0 - (4.0 s-1)t ]

9


(c) Write the general equation giving the transverse displacement of the particles of the string as a function of position and time

(d) Calculate the tension in the string

Problem 18 – c and d

[[[[ ]]]]txmy 410sin)05.0( −−−−====

smkfv 4.0104 ========ωωωω====λλλλ====

(((( ))))212 4.04 smcmgv ××××====µµµµ====ττττ −−−−

Nsmmkg 064.04.04.0 2221 ====××××==== −−−−−−−−

2-7 Energy and Power of a

Traveling String Wave


When we set up a wave on a stretched string, we provide energy for the motion of the string.

As the wave moves away from us, it transports that energy as both kinetic energy and elastic potential energy.

Energy and Power of a Traveling String Wave

Let us consider each form in turn.© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 56

When a wave travels along a string with velocity vto the right as shown in the figure, an element of mass dm oscillates along the y direction with

transverse velocity u=dy/dt. Then the element dm

has kinetic energy associated with it transverse velocity u.

The Kinetic energy is given

by:

Kinetic Energy

2

2

1udmdK ====

2

2

1

====

dt

dydm

(((( ))))txkydm m ωωωω−−−−ωωωω==== 222 cos2

1

Fig. 8


As a string element of length dx oscillates transversely, its length must increase and decrease in a periodic way if the string element is to fit the sinusoidal wave form.

Elastic potential energy is associated with these length changes, just as for a spring.

(Elastic) Potential Energy


The oscillating string element thus has both its maximum kinetic energy and its maximum elastic potential energy at y = 0.

In the snapshot of the Figure, the regions of the string at maximum displacement have no energy, and the regions at zero displacement have maximum energy. As the wave travels along the string, forces due to the tension in the string continuously do work to transfer energy from regions with energy to regions with no energy.

Energy Transport

10


The “average rate of change of kinetic energy” is given by:

Average Rate of Change of K

(((( ))))(((( ))))avgmavg

dttxkydt

dx

dt

dKωωωω−−−−ωωωωµµµµ====

222

cos2

1

(((( )))) 22222

4

1)(cos

2

1mavgm yvtkxyv ωωωωµµµµ====ωωωω−−−−ωωωωµµµµ====

(((( ))))(((( ))))2

1cos2 ====ωωωω−−−− avgdttxk

22

4

1m

avg

yvdt

dKωωωωµµµµ====

∴∴∴∴

22

4

1m

avg

yvdt

dU, ωωωωµµµµ====


cos2(k x - ωωωω t)avg is the average over an integer number of periods. It is defined by:

Calculating the cos2(k x - ωωωω t)avg

(((( ))))

(((( ))))∫∫∫∫

∫∫∫∫

ωωωω−−−−++++====

ωωωω−−−−====ωωωω−−−−

T

T

avg

dt)tkxcos(T

dt)tkx(cosT

)tkx(cos

0

0

22

212

1

1

[[[[ ]]]]

[[[[ ]]]]0

)sin()sin(

)sin()cos(2 0

0

====

−−−−ωωωω−−−−αααα====

ωωωω−−−−αααα====ωωωω−−−−∫∫∫∫

kxTkx

txkdttxkT

T

2

1====


The average rate of change of total energy is, thus, given by:

Average Rate of Change of E

22

2

12 m

avgavg

yvdt

dK

dt

dEωωωωµµµµ====

====

Note the proportionality with all parameters µ µ µ µ, v, ω ω ω ω2 and ym.

Due to the conservation of energy and considering the system isolated, the average rate of change of total energy is given by:

22

4

1m

avgavg

yvdt

dK

dt

dUωωωωµµµµ====

====

2-8

Superposition Principle


It often happens that two or more waves pass

simultaneously through the same region. We say

that these waves interfere “with each other”

When we listen to a concert, for example, sound

waves from many instruments fall simultaneously

on our eardrums. You still can distinguish between

the violin or the piano.

The Superposition Principle for Waves

Although waves could have different amplitudes

and frequencies and travel independently from

each other, we are still able to distinguish their

sources.© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 65

Suppose that two waves travel simultaneously

along the same stretched string.

Waves Along a Stretched String

),(),(),( 21 txytxytxy ++++====′′′′

Let y1(x,t) and y2(x,t) be the displacements

that the string would experience if each

wave traveled alone. The displacement of

the string when the waves overlap is then

the algebraic sum :

11


This summation of displacements along the string means that:

Overlapping waves algebraically add to produce a resultant wave (or net wave).

That is an example of what is called the That is an example of what is called the

superposition principle.superposition principle.

The superposition principle in physics is, in The superposition principle in physics is, in

general, the sum of two ore more effects.general, the sum of two ore more effects.

The Superposition Principle


الميكانيكية في الوسط الناقل والكهرومغناطيسية الميكانيكية في الوسط الناقل والكهرومغناطيسية الميكانيكية في الوسط الناقل والكهرومغناطيسية الميكانيكية في الوسط الناقل والكهرومغناطيسية ((((تنتشر األمواج تنتشر األمواج تنتشر األمواج تنتشر األمواج • . . . . نستطيع التمييز بين البيانو والطبل في اوركسترا موسيقيةنستطيع التمييز بين البيانو والطبل في اوركسترا موسيقيةنستطيع التمييز بين البيانو والطبل في اوركسترا موسيقيةنستطيع التمييز بين البيانو والطبل في اوركسترا موسيقيةلهذا السبب لهذا السبب لهذا السبب لهذا السبب بشكل مستق+ل الواحدة عن األخرى، وبشكل مستق+ل الواحدة عن األخرى، وبشكل مستق+ل الواحدة عن األخرى، وبشكل مستق+ل الواحدة عن األخرى، و) ) ) ) في الفراغفي الفراغفي الفراغفي الفراغ....والصوت في حياتناوالصوت في حياتناوالصوت في حياتناوالصوت في حياتنا نعيش يوميا وفي كل+ لحظة تراكب أمواج الضوء نعيش يوميا وفي كل+ لحظة تراكب أمواج الضوء نعيش يوميا وفي كل+ لحظة تراكب أمواج الضوء نعيش يوميا وفي كل+ لحظة تراكب أمواج الضوء ونحنونحنونحنونحنالمرونة المرونة المرونة المرونة مبدأ التراكب صالح طالما أن: اإلزعاجات المحص+لة ال تتعدى حد: مبدأ التراكب صالح طالما أن: اإلزعاجات المحص+لة ال تتعدى حد: مبدأ التراكب صالح طالما أن: اإلزعاجات المحص+لة ال تتعدى حد: مبدأ التراكب صالح طالما أن: اإلزعاجات المحص+لة ال تتعدى حد: •

مبدأ التراكبمبدأ التراكبمبدأ التراكبمبدأ التراكب

....مساوية لمجموع اإلزعاجات التي سببتها األمواج كل[ على حدةمساوية لمجموع اإلزعاجات التي سببتها األمواج كل[ على حدةمساوية لمجموع اإلزعاجات التي سببتها األمواج كل[ على حدةمساوية لمجموع اإلزعاجات التي سببتها األمواج كل[ على حدةالميكانيكية وفي الفراغ في حالة األمواج الكهرومغناطيسية، الميكانيكية وفي الفراغ في حالة األمواج الكهرومغناطيسية، الميكانيكية وفي الفراغ في حالة األمواج الكهرومغناطيسية، الميكانيكية وفي الفراغ في حالة األمواج الكهرومغناطيسية، ، لجزيئات الوسط في حالة األمواج ، لجزيئات الوسط في حالة األمواج ، لجزيئات الوسط في حالة األمواج ، لجزيئات الوسط في حالة األمواج disturbancedisturbancedisturbancedisturbanceاإلزعاج اإلزعاج اإلزعاج اإلزعاج إذا حدث وصادفت هذه األمواج عائقا فإن:ها تتداخل وتكون محص+لة إذا حدث وصادفت هذه األمواج عائقا فإن:ها تتداخل وتكون محص+لة إذا حدث وصادفت هذه األمواج عائقا فإن:ها تتداخل وتكون محص+لة إذا حدث وصادفت هذه األمواج عائقا فإن:ها تتداخل وتكون محص+لة . . . . مبدأ التراكب ليس إال عملية جمع تأثيراتمبدأ التراكب ليس إال عملية جمع تأثيراتمبدأ التراكب ليس إال عملية جمع تأثيراتمبدأ التراكب ليس إال عملية جمع تأثيرات•


Figure 9 below shows a

sequence of snapshots of two pulses traveling in opposite directions on the same stretched string. When the pulses overlap, the resultant pulse is their sum. Moreover, each pulse moves through the other, as if the other were not present:

Overlapping Waves

Overlapping waves do not in

any way alter the travel of

each other.

Applet Two pulses travel in opposite direction. Fig. 9

2-9

Interference of Waves


Suppose we send two sinusoidal waves of

the same wavelength and amplitude in the

same direction along a stretched string. The

superposition principle applies. What

resultant wave does it predict for the

string?

Resultant Wave


Consider a wave traveling along the stretched

string:

and another, shifted from the first, by a phase

factor φφφφ:

These waves have the same angular frequency ωωωω,

the same angular wave number k and the same

amplitude ym. They both travel in the positive

direction of the x axis, with the same speed.

Superimposing 2 Waves in a Stretched String

(((( ))))txkyy m ωωωω−−−−==== sin1

(((( ))))φφφφ++++ωωωω−−−−==== txkyy m sin2

12


)sin(cos222

φφφφφφφφ++++ωωωω−−−−====′′′′ txkyy m

Using the trigonometric identity

We get

The amplitude of the resultant wave is

and its phase is

y’ = y1 + y2

(((( )))) (((( ))))φφφφ++++ωωωω−−−−++++ωωωω−−−−====++++====′′′′ txkytxkyyyy mm sinsin21

++++

−−−−====++++

2sin

2cos2sinsin

BABABA

2cos2

φφφφ====′′′′ mm yy

2

φφφφ

)sin(2

φφφφ++++ωωωω−−−−′′′′==== txkym

73

Examples of y1 and y2

(((( ))))txkyy m ωωωω−−−−==== sin1 (((( ))))φφφφ++++ωωωω−−−−==== txkyy m sin2

a) b) c)

d) e) f)

21 yyy ++++====′′′′

Fig. 10


Case φφφφ = 0

(((( ))))txkyy m ωωωω−−−−==== sin1

(((( ))))txkyy m ωωωω−−−−==== sin2

a)

d)

)sin(2 txkyy m ωωωω−−−−====′′′′

)sin(cos222


0====φφφφ© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 75

Case φφφφ = π π π π

e)

(((( ))))txkyy m ωωωω−−−−==== sin1

(((( ))))ππππ++++ωωωω−−−−==== txkyy m sin2

c)

0====′′′′y

)sin(cos222


ππππ====φφφφ


Case φφφφ = 2ππππ/3

f)

(((( ))))txkyy m ωωωω−−−−==== sin1

(((( ))))ππππ++++ωωωω−−−−====3

22 sin txkyy m

d)

)sin(cos233ππππππππ ++++ωωωω−−−−====′′′′ txkyy m

Superposition applet


Two sinusoidal waves, identical except for

phase, travel in the same direction along a

string and interfere to produce a resultant

wave given by:

y´(x, t) = (3.0 mm) sin(20 x - 4.0 t + 0.820 rad),

with x in meters and t in seconds.

What are

a) the wavelength λλλλ of the two waves,

b) the phase difference between them and

c) their amplitude ym?


13


y´(x, t) = (3.0 mm) sin(20 x - 4.0 t + 0.820 rad),

a) Wavelength of the two waves:

b) Phase difference between the two waves

c) Amplitude of each of the two waves

Information


mk 314.010/220 ====ππππ====λλλλ⇒⇒⇒⇒λλλλππππ========

°≈=⇒= 9464.182.02

radrad φφ

(((( ))))mm

radyyy mmm 2.2

82.0cos

5.13cos2

2========⇒⇒⇒⇒========′′′′ φφφφ

)sin(cos222

φφφφφφφφ++++ωωωω−−−−====′′′′ txkyy m )sin(

2

φφφφ++++ωωωω−−−−′′′′==== txkym

2-10 Phasors


We can represent a string wave (or any other type of wave) vectorially with a phasor. In essence, a phasor is a vector that has a magnitude equal to the amplitude of the wave and that rotates around an origin; the angular speed of the phasor is equal to the angular frequency ωωωω of the wave. For example, the wave:

is represented by the phasor shown in Fig. 11-a

Phasor = Vector

)sin(11 txkyy m ωωωω−−−−====

txk ωωωω−−−−

Fig. 11-a© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 81

Interference using Phasors

)sin(22 φφφφ++++ωωωω−−−−==== txkyy m


Figure c shows the resultant y’=y1+y2

y’ = y1 + y2

)sin(cos222


c)Fig. 11-c


The Resultant y’ can be written as :

Where y’, is its amplitude and ββββ its phase :

Defining y’ using Phasors

c)

(((( )))) 22

221 )sin(cos φφφφ++++φφφφ++++====′′′′ mmmm yyyy

φφφφ++++

φφφφ====ββββ

cos

sintan

21

2

mm

m

yy

y

)sin( ββββ++++ωωωω−−−−′′′′====′′′′ txkyy m

φφφφ++++++++==== cos2 212

22

1 mmmm yyyy

14


Defining y’ using Phasors

And equivalently from the “cosines law”

The rectangular triangle OAB

O

(((( )))) 22

221 )siny(cosyyy mmmm φφφφ++++φφφφ++++====′′′′

φφφφ++++

φφφφ========ββββ

cosyy

siny

OB

ABtan

mm

m

21

2

φφφφ++++++++====′′′′ cosyyyyy mmmmm 212

22

1 2

A

B

C

(((( )))) (((( )))) (((( ))))222BAOBOA ++++====


In general we can plot the phasors with the one of zero phase along the x-axis and plot the others on the same figure with their phases as if it is the angle they make with the x-axis.

Phase of the First phasor

b)φφφφ++++++++====′′′′ cos2 212

22

1 mmmmm yyyyy


Three sinusoidal waves of the same frequency

travel along a string in the positive direction of

an x axis. Their amplitudes are y1, y2, and y3,

and their phase constants are 0, ππππ/2, and ππππ,

respectively.

What are (a) the amplitude and (b) the phase

constant of the resultant wave?

(c) Plot the wave form of the resultant wave at

t = 0, and discuss its behavior as t increases.



- Using trigonometry

Let


(((( ))))

(((( ))))ππππ++++ωωωω−−−−====

ππππ

++++ωωωω−−−−====

ωωωω−−−−====

txky

Y

txky

Y

txkyY

sin3

2sin

2

sin

13

12

11

(((( )))) (((( ))))

ππππ++++ωωωω−−−−++++

ππππ++++ωωωω−−−−++++ωωωω−−−−====

++++++++====′′′′⇒⇒⇒⇒

txksintxksintxksiny

YYYY

3

1

22

11

321


Problem 31 – Resultant Amplitude

(((( ))))121221 sin φφφφ++++ωωωω−−−−====++++====′′′′ txkyyyy m

1

212

22

2121

22

2112 tan,)2/cos(2

y

yyyyyyyym ====φφφφ++++====ππππ++++++++====

6.262

1tan,

2

54

1

121

2

1

2

112≈==+= −φyyyym

)sin()sin( 31212

3

ππππ++++ωωωω−−−−++++φφφφ++++ωωωω−−−−====

++++′′′′====′′′′′′′′

txkytxky

yyy

m

12331223

212123 cos2 φφφφ++++++++==== yyyyy mmm

1121

23

22

21 37.15.0

9

1

4

116.26cos

3

1

4

52 yyyyyy ≈≈≈≈++++++++++++====°°°°××××++++++++++++====


Phase of the Resultant

)cos(

)sin(

)cos(

)sin(

2tan

12312

123

12312

123123

φφφφ−−−−

φφφφ====

φφφφ−−−−ππππ++++

φφφφ−−−−ππππ====

φφφφ

yy

y

yy

y

m

m

3

1

9.0

45

2

1

3

1

3

1

cos2

5

tan3

1

2tan

12

12123

−−−−

××××====

−−−−φφφφ

φφφφ====

φφφφ

°°°°====φφφφ 6.20123

1

212tan

y

y====φφφφ

15


“Phasors” represent respectively

the 3 waves. The resultant is the sum of these

phasors.

Using Phasors→→→→→→→→→→→→

BCandABOA ,

B C

O A D

(((( )))) 2222ABCBOADCODOC ++++++++====++++====

( ) ( )1

2

1

2

114.123 yyyy =++=

8

3

3

2

tan

11

1 ====++++

====

++++========φφφφ

yy

y

CBOA

AB

OD

CD

°°°°≈≈≈≈ 6.20

φφφφ

2-11

Standing Waves


If two sinusoidal waves of the same amplitude and wavelength travel in opposite directions along a stretched string, their interference with each other produces a standing wave.

Two Sinusoidal Waves Traveling in opposite directions


Standing Wave (Definition)1

(((( )))) (((( ))))txkyytxkyy mm ωωωω−−−−====ωωωω++++==== sin,sin 21

Fig. 12

Let the two waves be represent by:

xktyyyy m sincos221 ωωωω====++++====′′′′


Standing Wave (Definition)2The resultant wave of the form:

xktyy m sincos2 ωωωω====′′′′

does not have the form of a traveling wave, i.e. y’=f(x - vt).

This is what we call a standing wave.

(((( )))) (((( ))))3214434421

876

termgOscillatintatAmplitude

m

ntDisplaceme

txkytxy ωωωω====′′′′

====

cossin2,

0


Nodes and Antinodes

If we examine the resultant figure, we see points presented by dots that never move, those are called nodes.

Half way between any pair of nodes there exists a point which oscillate with maximum amplitude it is called antinode

Fig. shows a standing wave at different times.

16


Nodes and Antinodes

The nodes occur at points x where the

amplitude of y’ is Zero.

22

0

λλλλ====

λλλλππππ

ππππ====

ππππ====

ππππ========

n/

nk

nx

nkxofvaluesatkxsin

Nodes occur at every half wavelength(λλλλ/2)


A rope, under a tension of 200 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by:

What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

where x = 0 at one end of the rope, x is in meters, and t is in seconds.


(((( )))) (((( )))) txmy ππππππππ==== 12sin2sin1.0


(a) the length of the rope. For the 2nd harmonic we have:

(c) the mass of the rope?

(b) the speed of the waves on the rope


mk

LL

0.42

22

2

222 ====

ππππ

ππππ====

ππππ====λλλλ============λλλλ

122

2

2 242

12

2

−−−−====ππππ

ππππ====

ππππ

ωωωω========⇒⇒⇒⇒====

λλλλ==== smLLfLv

L

vvf

kgv

LM

M

Lv 38.1

24

420022

====××××

====ττττ

====⇒⇒⇒⇒ττττ

====µµµµ

ττττ====

Standing Wave and

Resonance

100

Three simplest patterns where nodes appear are shown in Fig. 14.

Patterns

22

λλλλ====λλλλ====L

Fig. 14

2

λλλλ====L

23

λλλλ====L

n

LornL n

2

2====λλλλ

λλλλ====

101

A standing wave in a string can be set up on a string of length L if the resonant frequencies verify the relation:

The collection of all possible oscillation modes is called the harmonic series. n is called the harmonic number of the nth harmonic.

Harmonic Series

L

vn

vf

nn

2====

λλλλ====

f1 is called fundamental frequency

or 1st harmonic

f2 is called the 2nd harmonic

f3 is called the 3rd harmonic

f4 is called the 4th harmonic

17


Problems

18, 24, 25, 31, 39 46 and 50


See Example 1

Problem 18


Problem 24

(b) Show that the time a

transverse wave takes to

travel the length of the rope is

given by

A uniform rope of mass m and

length L hangs from a ceiling.

(a) Show that the speed of a

transverse wave on the rope

is a function of y, the distance

from the lower end, and is

given by

y

ygv ====

gLt 2====© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 105

(b)


a)

y

gLt 2====⇒⇒⇒⇒

ygdt

dyv

dt

dy====⇒⇒⇒⇒====

tLtL

tgytdgy

dy

0000

2 ====⇒⇒⇒⇒==== ∫∫∫∫∫∫∫∫

µµµµ

ττττ====v

gyv ====⇒⇒⇒⇒

dtgy

dy====⇒⇒⇒⇒

(((( ))))µµµµ

µµµµ====

gyyv

mass


Problem 25A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of 1.00 cm. The motion is continuous and is repeated regularly 120 times per second. The string has linear density120 g/m and is kept under a tension of 90.0 N.

(a)Find the maximum value of the transverse speed u

Solution: Available information:

A = ym= 0.005 m, f = 120 Hz, µµµµ=0.12 kg/m and ττττ=90 N

sradf ππππ====××××ππππ====ππππ====ωωωω 24012022

smyfyu mm /77.3005.02402max ====××××ππππ====ππππ====ωωωω====


(b)Find the maximum value of the transverse component of the tension ττττ.

(Hint: The transverse component is ττττ sin θθθθ, where θθθθis the angle the string makes with the horizontal. You will need to relate angle θθθθ to dy/dx.)

θθθθx

Problem 25 - b

(((( ))))txkyy m ωωωω−−−−==== siny

max

tandx

dy====θθθθ

18


Problem 25 – Solution b

smv 66.8120.090 ========µµµµττττ====

N405.23sin90max

sin ====°°°°====θθθθττττ

(((( )))) mmmaxm ykytxkcoskytanλλλλ

ππππ========ωωωω−−−−====θθθθ

2

°°°°====θθθθ⇒⇒⇒⇒====λλλλ

ππππ====θθθθ⇒⇒⇒⇒ 5234350

2..my

tan

fv====λλλλ

120077.3

240 −−−−≈≈≈≈ππππ

====ωωωω

==== msm

srad

uk

m.07210====


c)Show that the two maximum values calculated above occur at the same phasevalues for the wave. What is the transversedisplacement y of the string at these phases?

occurs, also, for kx – ωωωωt = n ππππ.

At this value of the phase y = ym sin (n ππππ ) = 0

The maximum speed occurs for kx – ωωωωt = n ππππ.

Problem 25 - c

(((( )))) ππππωωωω====ωωωω−−−−ωωωω======== ncosytxkcosydt

dyu mmaxm

max

max

maxsin θθθθττττ


d)What is the maximum rate of energy transfer along the string?

e)What is the transverse displacement y when this maximum transfer occurs?

This maximum transfer occurs when kx – ωωωωt = n ππππ,i.e.

Problem 25 – d, e

(((( ))))txkyvdt

dKm ωωωω−−−−ωωωωµµµµ==== 222

cos2

1

ππππωωωωµµµµ====⇒⇒⇒⇒ nyvdt

dKm

222

max

cos2

1

(((( )))) (((( )))) 0sin ====ππππ====ππππ nyny m


f)What is the minimum rate of energy transfer along the string?

Problem 25 – f

2)12( ππππ++++====ωωωω−−−− ntxk

ofvaluesatoccuretransfereenergyminimumThe

(((( ))))(((( ))))ππππ++++ωωωωµµµµ==== 12cos2

1 222 nyvdt

dKm

f

(((( )))) (((( ))))22322005.024077.310120

2

1

2

1ππππ××××××××××××××××====ωωωωµµµµ==== −−−−

myv

J2.3≈≈≈≈


Problem 25 – gg)What is the transverse displacement y when

this minimum transfer occurs?

(((( ))))(((( )))) mmf ynyy ±±±±====ππππ++++==== 12sin

This minimum transfer occurs when k x – ωωωω t = (2n+1) ππππ, i.e.


A string that is stretched between fixed

supports separated by 75.0 cm has resonant

frequencies of 420 and 315 Hz, with no

intermediate resonant frequencies. What are

(a) the lowest resonant frequency and (b) the

wave speed?

a)

b)

Problem 39

HzfnL

vnvnfn 315

21

1

============λλλλ

====

(((( )))) (((( )))) (((( )))) HzfnL

vnvnfn 4201

2

111

11 ====++++====

++++====

λλλλ

++++====++++

Hzf 1051 ====⇒⇒⇒⇒

11 5.1572 −−−−======== smfLv

19


For a certain transverse standing wave on a long string, an antinode is at x = 0 and a node is at x = 0.10 m. The displacement y(t) of the string particle at x = 0 is shown in Fig. 17-34.

When t = 0.50 s, what are the displacements of the string particles at (a) x = 0.20 m and (b) x =

0.30 m?

Problem 50


From the figure we have:

Problem 50 – Solution 1

Hz.T

fs.T 501

02 ========→→→→====

Node number 1 occurs at x1 = 10 cm and since:

cmm.ym 4040 ========

111432

−−−−−−−− ≈≈≈≈ππππ====ππππ====ωωωω srad.sradf

(((( )))) 11 5612040040 −−−−−−−− ====ππππ××××≈≈≈≈××××ππππ====ωωωω==== scm.sm..yv m

13140

20

2

20

2 −−−−====ππππ

====ππππ

====⇒⇒⇒⇒ cm.k

n

xn2====λλλλ then cmm.

x2020

1

2 1 ============λλλλ


Displacement of the string particle x at any

instant t is thus given by:


xktyy m sincos2 ωωωω====′′′′ (((( )))) (((( )))) (((( )))) (((( ))))cmtcosxsint,xy ππππππππ====⇒⇒⇒⇒ 104

Displacements of the string particles at

(a) x = 0.20 m and (b) x = 0.30 m? when t = 0.5 s.

(((( )))) (((( )))) (((( )))) 050245020 ====××××ππππππππ============ .cossins.t,cmxy

(((( )))) (((( )))) (((( )))) 050345030 ====××××ππππππππ============ .cossins.t,cmxy


At x = 0.20 m, what are the transverse velocities of the string particles at (c) t = 0.50 s and (d) t =

1.0 s?


(((( )))) (((( )))) (((( ))))

(((( )))) (((( )))) (((( )))) (((( ))))1105612

−−−−ππππππππ−−−−====⇒⇒⇒⇒

ωωωωωωωω−−−−====

scmtsinxsin.t,xv

tsinxksinyt,xv m

Velocity ∀ ∀ ∀ ∀(x,t):

For x = 20 cm:

(((( )))) (((( )))) 13450342

205020 −−−−−−−−====××××ππππ−−−−

============ scm..sin.

Tt,cmvs.t,cmv

(((( )))) (((( )))) 01342

20120 ====××××ππππ−−−−====

============ .sin.

Tt,cmvst,cmv

(((( )))) (((( )))) (((( )))) (((( )))) 13420561220

−−−−ωωωω−−−−≈≈≈≈ωωωω−−−−==== scmtsin.tsinsin.t,m.v


(e) Sketch the standing wave at t = 0.50 s for the range x = 0. to x = 0.40 m.


(((( )))) (((( )))) (((( ))))cmxsins.t,xy 10450 ππππ========

Chapter 3: Waves IIChapter 3: Waves IIChapter 3: Waves IIChapter 3: Waves II

Next Lecture

1





http://ctaps.yu.edu.jo/physics/Courses/Phys103/Chapter3

Chapter 3: Waves II18 Waves—II

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 3

Sound waves in modern life:� Seismic prospecting teams use such waves to probe Earth's crust for oil.� Ships carry sound-ranging gear (sonar) to detect underwater obstacles .� Submarines use sound waves to stalk other submarines.� Explore the soft tissues of the human body.

3-1 Sound WavesA sound wave is defined roughly as any longitudinal wave

4

S: a point source that emits sound wave in all

directions.

Definitions

Wavefronts are surfaces over which the oscillations of the air due to the sound wavehave the same value; such surfaces are represented by whole or partial circles in a two-dimensional drawing for a point source.

Rays are directed lines perpendicular to the

wavefronts that indicate the direction of travel of the wavefronts.

Fig. 1


Fig. 1: A sound wave travels from a point source S

through a 3D medium. The wavefronts form spheres centered on S; the rays are radial to S. The short, double-headed arrows indicate that elements of the medium oscillate parallel to the rays.

Wavefronts - Rays

3-2 The Speed of

Sound

2


The speed of any mechanical wave, transverse or longitudinal, depends on two important properties of the medium:

• The inertial property of the medium (to store kinetic energy).

• The elastic property of the medium (to store potential energy).

Speed of a Mechanical Wave

µµµµττττ==== /vIn the case of a stretched string

ττττ represents the elasticity property

µµµµ represents the inertial property


If the medium is air for example and the

wave is longitudinal as for a sound wave

then the inertial property is simply the

volume density ρρρρ of air.

The elasticity property here is the bulk

modulus B defined by:

Where ∆∆∆∆V/V is the fractional change in volume

produced by a change in pressure ∆∆∆∆p.

Inertial and Elasticity Properties

VV

pB

∆∆∆∆

∆∆∆∆−−−−====

9

Potential energy is associated with periodic compressions and expansions of small elements of the air

The property that determines the extent to which an element of a medium changes in volume when the pressure (force per unit area) on it changes is the bulk modulus B

The minus sign illustrates the fact that when we increase the pressure on air the volume decreases and vice-versa.

The Bulk Modulus

VV

pB

∆∆∆∆

∆∆∆∆−−−−====


B measures the substance's resistance to

uniform compression.

It has the dimension of a pressure:

Its units in the MKS is the pascal = N.m-2

Bulk Modulus

[[[[ ]]]] [[[[ ]]]]pB ∆∆∆∆====

Air 1.42×105 Pa (adiabatic bulk modulus)

Air 1.01×105 Pa (constant temperature bulk modulus)

Water 2.2×109 Pa

Methanol 8.23×108 Pa (at 20°C and 1 Atm)

11

The Speed of Sound

aAt 0°C and 1 atm pressure, except

where noted.

1284Hydrogen

965 Helium

343 Air (20°C)

331 Air (0°C)

Gases

Speed (m/s) Medium

6000 Granite

5941 Steel

6420 Aluminum

Solids

1522 Seawaterb

1482 Water (20°C)

1402 Water (0°C)

Liquids

TABLE 2-1 The Speed of

Sounda

bAt 20°C and 3.5% salinity.

The speed of sound wave in any medium is given by:

ρρρρ====

Bv

Derivation of v

3


Figure 2a shows a pulse sent down a long air-filled tube. We choose a frame of reference such that the pulse be at rest.

A slice of air of width ∆∆∆∆x moves towards the

pulse with velocity v.

Frame of Reference

Fig. 2a

Fig. 2b


Figure 2a shows a pulse sent down a long air-filled tube. We choose a frame of reference such that the pulse be at rest.

A slice of air of width ∆∆∆∆x moves towards the pulse with velocity v.

Frame of reference

Fig. 2a

Fig. 2b


Time

The mass of the slice is ∆∆∆∆m.If ρρρρ is the density of the medium then ∆∆∆∆m = ρρρρ A ∆∆∆∆x.

The leading face of the slice enters the pulse where the pressure is p + ∆∆∆∆p. It slows down and its velocity becomes v +∆∆∆∆ v where ∆∆∆∆v is negative.The “slowing” is complete when the rear face of the slice has reached the pulse. This operation requires a time interval:

v

xt

∆∆∆∆====∆∆∆∆

0>>>>∆∆∆∆p


Formal Derivation of vThe leading face of the slice enters the pulse (Figure b).

The average forces acting on the leading and the trailing faces are respectively:

The average net force on the system (the slice) is:

(((( )))) AppFApF lt ∆∆∆∆++++−−−−======== ,

(((( )))) ApApppAFext ∆∆∆∆−−−−====∆∆∆∆++++−−−−====∑∑∑∑ . am∆∆∆∆====

17

Newton's 2nd Law

t

v

tv

p

x

pa

∆∆∆∆

∆∆∆∆====

∆∆∆∆ρρρρ

∆∆∆∆−−−−====

∆∆∆∆ρρρρ

∆∆∆∆−−−−====⇒⇒⇒⇒

v

v

tvA

tvA

V

V ∆∆∆∆====

∆∆∆∆

∆∆∆∆∆∆∆∆====

∆∆∆∆

axAAp ∆∆∆∆ρρρρ====∆∆∆∆−−−−∴∴∴∴

(((( )))) ApApppAF .ext ∆∆∆∆−−−−====∆∆∆∆++++−−−−====∑∑∑∑

VV

P

vv

∆∆∆∆

∆∆∆∆−−−−====ρρρρ

1

ρρρρ====⇒⇒⇒⇒

Bv

am∆∆∆∆====

xAVm ∆∆∆∆ρρρρ====∆∆∆∆ρρρρ====∆∆∆∆

v

Pv

∆∆∆∆

∆∆∆∆−−−−====ρρρρ⇒⇒⇒⇒

V

Vvv

∆∆∆∆====∆∆∆∆⇒⇒⇒⇒

BVV

Pv ====

∆∆∆∆

∆∆∆∆−−−−====ρρρρ⇒⇒⇒⇒ 2

3-3 Traveling Sound

Waves

4


Displacement and Pressure Variation

Fig. 3

20

Displacement and Pressure Variation

(((( )))) (((( ))))txkstxs m ωωωω−−−−==== cos,876

Amplitude

(((( )))) (((( ))))txkptxp m ωωωω−−−−∆∆∆∆====∆∆∆∆ sin,43421

Oscillating term

Displacement

Pressure variation

The pressure variation will be derived in the next section.

(((( )))) mm svp ωωωωρρρρ====∆∆∆∆


Pressure Variation

∆s,A∆V ====

V

VBp

∆∆∆∆−−−−====∆∆∆∆

x

sB

x

sBp

∂∂∂∂

∂∂∂∂−−−−====

∆∆∆∆

∆∆∆∆−−−−====∆∆∆∆

(((( ))))[[[[ ]]]] (((( ))))txksktxksxx

smm ωωωω−−−−−−−−====ωωωω−−−−

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂sincos

(((( ))))txkskBp m ωωωω−−−−====∆∆∆∆ sin

(((( )))) mmm skvskBp ρρρρ========∆∆∆∆ 2

∆xAV ====

Fig. 3b 3-4 Interference


Interference

Path length difference

∆∆∆∆L = |L2 – L1|.

To relate phase difference φφφφ to path length

difference ∆∆∆∆L, we recall (from Section 17-4).

that a phase difference of 2ππππ rad

corresponds to one wavelength. Thus, we

can write the proportion

λλλλ

∆∆∆∆====

ππππ

φφφφ L

2

Fig. 4


Fully constructive interference occurs when φφφφ is zero, 2ππππ, or any integer multiple of 2ππππ. We can write this condition as:

Fully constructive interference

this occurs when the ratio ∆∆∆∆L/λλλλ verifies:

Constructive Interference

(((( )))) ππππ====φφφφ 2m

K,2,1,0====λλλλ

∆∆∆∆ L

i.e. when the path length difference is zero

or an integer multiple of the wavelength λλλλ.


5


Destructive Interference

Fully destructive interference occurs when φφφφis an odd integer multiple of ππππ. A condition we can write as:

Fully destructive interferenceFully destructive interference(((( )))) ππππ++++====φφφφ 212 m


K,2

5,

2

3,

2

1====

λλλλ

∆∆∆∆ L

i.e. when the path length difference is half an integer multiple of the wavelength λλλλ.

Fully destructive interferenceFully destructive interference

3-5 Intensity and

Sound Level


The intensity I of a sound wave at a surface is the average rate per unit area at which energy is transferred by the wave through or onto the surface. We can write this as:

Intensity and Sound Level

A

PI ====

We will show that the intensity I is related to the displacement amplitude Sm by the relation:

, where P is the average power

22

2

1mSvI ωωωωρρρρ====


Fig. 18-10 A point source S emits sound waves uniformly in all directions. The waves pass through an imaginary sphere of radius r that is centered on S.

Variation of Intensity with Distance

The intensity of sound at any point on the surface of a sphere of radius r is given by:

24 r

PI s

ππππ====

Fig. 4

The Decibel Scale


What kind of sounds humans can hear?!!

I = f (v, f, Sm)

We suggested earlier that the intensity

of sound is given by:

It is clear from the above equation that

the intensity I depends on the speed of

sound wave v and on the frequency f of

sound wave or the wave length λλλλ, and

the displacement amplitude Sm.

22

2

1mSvI ωωωωρρρρ====

6


The Decibel Scale

The displacement amplitude which the human ear can hear ranges from about 10-5

m for the loudest tolerable sound to about 10-11 m for the faintest detectable sound, a ratio of 106.

We shall speak of the sound level ββββ, instead of speaking of the intensities of sound wave at this large limit or ratio 1012.

The ratio of the intensities at both limits is 1012. It is a huge range to consider


The unit of sound level is dB. It is the abbreviation for decibel, a name that was chosen in recognition of the work of Alexander Graham Bell.

Sound Level

I0 is a standard reference intensity

(= 10-12 W/m2), chosen because it is near the

lower limit of the human range of hearing.

For I = I0 , the above equation gives

ββββ = 10 log 1 = 0

0

log10I

I====ββββ


Hearing threshold

Rustle of leaves

Conversation

Rock concert

Pain threshold

Jet engine

10

Some Sound Levels (dB)

0

60

110

120

130


Derivation of

2

2

1svdmdK ====

(((( )))),sin txkst

sv ms ωωωω−−−−ωωωω−−−−====

∂∂∂∂

∂∂∂∂====

dxAdm ρρρρ====

(((( ))))⇒⇒⇒⇒ωωωω−−−−ωωωωρρρρ==== txksdxAdK m222 sin

2

1 (((( ))))txksvAdt

dKm ωωωω−−−−ωωωωρρρρ==== 222 sin

2

1

22

4

1m

avg

svAdt

dKωωωωρρρρ====

22

2

12

mavgavg

svA

dt

dK

A

dt

dE

I ωωωωρρρρ====

====

====

22

2

1msvI ωωωωρρρρ====


A point source emits 30.0 W of sound

isotropically. A small microphone intercepts

the sound in an area of 0.750 cm2, 200 m from

the source. Calculate (a) the sound intensitythere and (b) the power intercepted by the microphone.

S


Mic 0.75 cm2

200 m

Solution: (Pavg = 30 W)

a) I at the mic. is:

24 r

PI

avg

ππππ====

25

221096.5

2004

30mW

m

WI

−−−−××××====××××ππππ

====


(b) the power intercepted by the microphone.


S

Mic 0.75 cm2

200 m

WcmmWArIrP 9225 1047.475.01096.5)()( −−−−−−−− ××××====××××××××====××××====

dB9.5210

1096.5log10

12

5

====××××

====ββββ−−−−

−−−−

Sound level:

7

3-6 Sources of

Musical Sound

38

a) The simplest standing wave pattern of displacement

for (longitudinal) sound waves in a pipe with both both

ends openends open has an antinode (A) across each end and a

node (N) across the middle*.

Standing Waves Patterns

Fig. 5-a

*The longitudinal displacements represented by the

double arrows are greatly exaggerated.

Fig. 5-b

b) The corresponding standing wave pattern for (transverse) string waves.


Standing Waves Patterns

- As the compression region of the wave

exits the open end of the pipe, the

constraint of the pipe is removed and the

compressed air is free to expand into the

atmosphere

The open end of a pipe is a displacement

antinode in the standing wave

The open end corresponds with a pressure

node.

It is a point of no pressure variation

40

Standing wave patterns for string waves superimposed on pipes to represent standing sound wave patterns in the pipes.

Pipe with both ends open

Fig. 6-a

With both ends of the pipe open, any harmonic can be set up in the pipe

22ndnd harmonicharmonic

33rdrd harmonicharmonic

44thth harmonicharmonic

41

Pipe with both ends open

When both ends are open this leads to:

L

vnf

n

Lnn

2

2====⇒⇒⇒⇒====λλλλ


With only one end open, only odd harmonics can be set up.

Pipe with one ends closed

With only one end open this leads to:

K,7,5,3,1,4

========λλλλ nn

Ln ),5,3,1(,

4K========⇒⇒⇒⇒ noddn

L

vnfn

Fig. 6-b

8

43

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.330

m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

Solution:

For the tube:

For the string:


7,5,3,1,4

======== nL

vnfn Hz

L

vf 5.71

2.14

343

4, 1 ≈≈≈≈

××××========

N

fLfv

7.64)5.7133.02(33.0

0096.0

)2()(

2

21

22

====××××××××====

××××µµµµ====λλλλµµµµ====µµµµ====ττττ

3-7 Beats


If we listen, a few minutes apart, to two sounds whose frequencies are, say, 552 and 564 Hz, most of us cannot tell one from the other.

Beats - 1

Fig. 7

564 Hz

552 Hz


Beats - 2

However, if the sounds reach our ears simultaneously, what we hear is a soundwhose frequency is 558 Hz, the average of the two combining frequencies.

We also hear a striking variation in the intensity of this sound—it increases and decreases in slow, wavering beats that repeat

at a frequency of 12 Hz, the difference between the two combining frequencies.


Let wave 1 be given by:

and wave 2 be given by:

The resultant wave s’ is given by:

Using the identity

we have

How do we hear the beats?

tss m 22 cos ωωωω====

tss m 11 cos ωωωω====

(((( ))))ttssss m 2121 coscos ωωωω++++ωωωω====++++====′′′′

2cos

2cos2coscos

BABABA

++++−−−−====++++

(((( )))) (((( ))))ttss m 21212

1cos

2

1cos2 ωωωω++++ωωωωωωωω−−−−ωωωω====′′′′


If the two frequencies are nearly equal, i.e.

if ωωωω1 ≈≈≈≈ ωωωω2 = ω then

Note that ωωωω’ is very small. We thus have:

Which is a “decaying”sine wave.

Time

ω1≈ω

2

ttss m ωωωω′′′′ωωωω====′′′′ coscos2

ωωωω′′′′====ωωωω−−−−ωωωω

ωωωω====ωωωω++++ωωωω

2,

2

2121

ωωωω′′′′ππππ====′′′′ 2T

c

9


A tuning fork of unknown frequency makes three beats per second with a standard fork of

frequency 384 Hz. The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork?

Solution: The beat frequency is ωωωω1 - ωωωω2


212 ωωωω−−−−ωωωω====ωωωω′′′′====ωωωωbeat

3871 ====f

3384121 ====−−−−====−−−−==== ffffbeat

3-8 Doppler Effect


This is true for waves of all types. We shall discuss here the Doppler effect for mechanical (sound) waves.

Doppler Effect is a general phenomenon which occurs each time there is a relative motion between a source and a detector.

What is Doppler Effect?

Johann Christian Doppler

54

Fig. 18-19 A stationary source of sound S emits

spherical wavefronts, shown one wavelengthapart, that expand outward at speed v.

A sound detector D, represented by an ear, moves with velocity vD toward the source. The detector senses a higher frequency because of its motion.

Stationary Source Moving Detector





The frequency f ’ detected by D is the rate at which D intercepts wavefronts. Obviously f ’ > f.

10


Let us now consider the situation in which Dis moving opposite the wavefronts.

Moving Source Stationary Detector

vt

v

Dv

vD

S

In time t, the wavefronts move to the right a distance vt but here the detector D moves to the left a distance vDt.

λλλλ λλλλ

58

Thus in time t the distance moved by the wavefronts relative to D is vt + vDt. The number of wavelengths in this relative distance vt +vDt is the number of wavelengths intercepted by D in time t, and this number is:


The rate at which D intercepts wavelengths in this situation is the frequency f ’ given by:

(((( ))))λλλλ

++++====

λλλλ++++====′′′′ DD vv

t

tvtvf f

v

vv

fv

vvf DD ++++

====++++

====′′′′⇒⇒⇒⇒

(((( ))))λλλλ

++++====

λλλλ++++ DD vv

t

tvtv


Similarly, if D is moving away from the source, then the wavefronts move a distance vt – vDt relative to D in time t and we have:

Summary: Stationary source moving detector


fv

vvf D−−−−

====′′′′

fv

vvf D±±±±

====′′′′Detector moves towards to the

sourceDetector moves away from the

source


Now Let S move at speed vS toward D, which is stationary with respect to the body of air. Consider two waves emitted by the source: W1 and W2. During the period T, the wavefront W1 moves a distance vt and the source moves a distance vSt. At the end of T wavefront W2 is emitted.


fvv

v

fvfv

v

TvTv

vvf

SSS −−−−====

−−−−====

−−−−====

λλλλ′′′′====′′′′

In the direction of motion of S, the distance between the 2 wavefronts (which is the wavelength detecteddetected by D is λλλλ’ = vt – vS t

The frequency f ’ detected by the detector is:



ffvv

v

tvtv

vvf

ss

>>>>−−−−

====−−−−

====λλλλ′′′′

====′′′′

N vttvNtvN ========λλλλ

(((( )))) (((( ))))tvvNtvvN ss −−−−====−−−−====λλλλ′′′′

N(v-vS)t

N vS t

When both are stationary,

the frequency f = v/λλλλ



ffvv

v

s

>>>>−−−−

====

N vt

N(v-vS)t

N vS

t

When both are stationary,

the frequency f = v/λλλλ

tvtv

vvf

s−−−−====

λλλλ′′′′====′′′′

11


We can summarize all the previous results in one single equation, namely:

General Doppler Effect Equation

fvv

vvf

S

D

m

±±±±====′′′′

When the motion of detector or source is toward the other, the sign on its speed must give an upward shiftupward shift in frequency.

When the motion of detector or source is away from the other, the sign on its speed must give a downward shiftdownward shift in frequency.

Waves II - Problems


The speed of sound in a certain metal is V. One end of a long pipe of that metal of length L is struck a hard blow. A listener at the other end hears two sounds, one from the wave that travels along the pipe and the other from the wave that travels through the air. (a) If v is the speed of sound in air, what time interval t elapses between the arrivals of the two sounds? (b) Suppose that t = 1.00 s and the metal is steel. Find the length L

Problem 6



a) Let v and V be respectively the speed of sound in air and in the metal:

b)

−−−−====−−−−====∆∆∆∆Vv

Lttt mA11

smvsmVst Steel 343,5941,1 ============∆∆∆∆

mvV

vVtL 364

3435941

34359411 ====

−−−−

××××××××====

−−−−∆∆∆∆====

∴∴∴∴tm= L/V, tA = L/v

67

Problem 11

The pressure in a traveling sound waveis given by the equation (φφφφ = 0)

Find (a) the pressure amplitude, (b) the frequency, (c) the wavelength, and (d) the speed of the wave.

Papm 5.1====∆∆∆∆

Hzf 5.15723152 ====ππππππππ====ππππωωωω====

smk

v 3509.0

315====

ππππ

ππππ====

ωωωω====

mk 22.29.022 ====ππππππππ====ππππ====λλλλ

(((( )))) ][ )()( 11315900.0sin50.1 tsxmPap

−−−−−−−− −−−−ππππ====∆∆∆∆

a)

b)

c)68

Problem 13

In Fig. 18-33, two loudspeakers, separated by

a distance of 2.00 m, are in phase.

Assume the amplitudes of the sound from the speakers are approximately the same at

the position of a listener, who is 3.75 mdirectly in front of one of the speakers.

12

69

Problem 13(a) For what frequencies in the audible

range (20 Hz to 20 kHz) does the listener hear a minimum signal?

nnL λλλλ++++====∆∆∆∆ )(21

Solution: Minimum signal occurs when the interference of the sound waves from the two sources is destructive, i.e.

mL 5.075.3275.322 =−+=∆

5.0

343

21

21 )()( ××××++++====

∆∆∆∆++++====⇒⇒⇒⇒ n

L

vnfn

nf

vn )(

21++++====

∆∆∆∆L is the path difference between the waves of the two sources


Problem 13 – Solution b

20000343maxmin 0 <<<<========⇒⇒⇒⇒ nn fHzff

28,4,3,2,1,068621 )( K====++++==== ×××× nfornfn

282000068621

maxmax )( ====⇒⇒⇒⇒≤≤≤≤++++⇒⇒⇒⇒ ×××× nn

5.0

343

21)( ++++==== nfn

(b) For what frequencies is the signal a maximum?

nn

f

vnnL ====λλλλ====∆∆∆∆

20000686maxmin

<<<<==== nn fHzf

2920000686 maxmax ====⇒⇒⇒⇒≤≤≤≤××××⇒⇒⇒⇒ nn

5.0

343n

L

vnfn ====

∆∆∆∆====⇒⇒⇒⇒


Problems 28, 38 and 43

See Examples 1, 2 and 3 respectively.


A French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic

(Fig. 18-37). The French sub moves at 50.0

km/h, and the U.S. sub at 70.0 km/h. The French sub sends out a sonar signal (sound

wave in water) at 1000 Hz. Sonar waves

travel at 5470 km/h.

Problem 51 – Doppler Effect

73

Solution:

(a) What is the signal's frequency as detected by the U.S. sub?


Hzvv

vvff

s

DFUS 1022

5420

55401000

505470

7054701000 ====××××====

−−−−

++++××××====

−−−−

++++====

Hzvv

vvff

s

DUSFref

10445400

55201022

705470

5054701022 ====××××====

−−−−

++++××××====

−−−−

++++====

(b) What frequency is detected by the French sub in the signal reflected back to it by the U.S. sub?


A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound

emission frequency of the bat is 39000 Hz. During one fast swoop directly toward a flat

wall surface, the bat is moving at 0.025 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

Problem 54 – Doppler Effect

Hzvv

vf

vv

vvff

ss

Dw 40000

025.0343343

3433900000 ====

××××−−−−====

−−−−====

±±±±

±±±±====

Hzvv

vvff

s

Dwbatref

41000343

025.034334340000 ====

××××++++====

±±±±

±±±±====

13

Chapter 4 Chapter 4 Chapter 4 Chapter 4 –––– Electromagnetic Electromagnetic Electromagnetic Electromagnetic

Oscillations and Alternating CurrentsOscillations and Alternating CurrentsOscillations and Alternating CurrentsOscillations and Alternating Currents

Next Lecture

1





Chapter 4: Electromagnetic Oscillations and Alternating Current

http://ctaps.yu.edu.jo/physics/Courses/Phys103/Chapter4

4-1 New Physics

—Old Mathematics

3

LC Circuit

Fig. 1

In this chapter you will see how the electric charge q varies with time in a circuit made up of an inductor L, and a capacitor C.

From another point of view, we shall discuss how energy shuttles back and forth between the magnetic field of the inductor and the electric field of the capacitor.

CL

4

The energy stored in the capacitor is called the electric energy because it is associated with the energy stored between the capacitor plates as electric field. Which is equal to:

The energy stored in the inductors is called the magnetic energy because it is associated with the energy stored in the inductor as magnetic field Which is equal to:

C

qU E

2

2

1====2

2

1iLUB ====

LC Oscillation - Qualitative

Fig. 2-a

5

Consider the LC circuit. According to

Kirchhoff’s second law: the (algebraic) sum of potential differences equals zero, i.e.

Thus we get a homogeneous linear 2nd order DE:

LC Oscillator

0====++++ CL VV

0====++++C

q

dt

dIL

2

2

dt

qd

dt

dI

dt

dqI ====⇒⇒⇒⇒====

C

qVC ====

dt

dILVL ====

⇒⇒⇒⇒====++++ 01

2

2

qCdt

qdL ⇒⇒⇒⇒====++++ 0

12

2

qCLdt

qd

02 ====ωωωω++++ qq&&

LC

12 ====ωωωωwhere

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 6

We know that the solution is of the form:

LC Oscillator

(((( ))))φφφφ++++ωωωω==== tQq cos (((( ))))φφφφ++++ωωωω==== txx cos0

(((( )))) (((( ))))φφφφ++++ωωωω−−−−====φφφφ++++ωωωωωωωω−−−−==== tItQi sinsin

which is similar to

The current in this circuit is given by:

(((( ))))φφφφ++++ωωωω−−−−==== tvv sin0which is similar to

2


Electric and Magnetic Energy Oscillations

(((( ))))φφφφ++++ωωωω======== tC

Qq

CU E

22

2cos

22

1

(((( ))))φφφφ++++ωωωωωωωω======== tQLiLUB2222

sin2

1

2

1

CL

CL

11 22 ====ωωωω⇒⇒⇒⇒====ωωωω

(((( ))))φφφφ++++ωωωω==== tQC

UB22

sin2

1

tC

QUUU BE ∀∀∀∀====++++====

2

2


The inductor and capacitor transfer energy from one to the other as shown below:

LC Oscillation - Energy

Fig. 3

4-3 The Electrical–

Mechanical Analogy


LC Oscillation - Energy

InductorBlock

CapacitorSpring

EnergyElementEnergyElement

LC Oscillator Block–Spring System

The Energy in Two Oscillating Systems Compared

2

2

1xk

2

2

1vm

21

2

1q

C

2

2

1iL

dt

dqi ====

dt

dxv ====

Problemson Electromagnetic Oscillations

and Alternating Current


The energy in an oscillating LC circuit containing a 1.25 H inductor is 5.70 µµµµJ. The maximum charge on the capacitor is 175.0 µµµµC. Find

The mass m corresponds to the inductance, i.e.

m = 1.25 kg.

The spring constant k corresponds to the reciprocal of the capacitance. Since the total energy is given by U = Q2/2C, where Q is the maximum charge on the capacitor and C is the capacitance then:

Problem 33-7

(a) the mass:

b) the spring constant:

( )F

J

C

U

QC

3

6

2262

1069.21070.52

10175

2

−

−

−

×=××

×== mNk 372

1069.2

13

====××××

====∴∴∴∴−−−−

3


The maximum displacement xm corresponds to the maximum charge, thus

xm

= 175 × × × × 10-6m= 175.0 µµµµm

and (d) the maximum speed for a mechanical system with the same period.

The maximum speed vm corresponds to the maximum current. The maximum current is:

c) the maximum displacement,

Problem 33-7

Thus v = 3.02 ×××× 10-3 m/s

(((( ))))(((( ))))A

FH

C

CL

QQI 3

3

6

1002.3

1069.225.1

10175 −−−−

−−−−

−−−−

××××====××××

××××========ωωωω====


The frequency of oscillation of a certain LC

circuit is 200 kHz. At time t = 0, plate A of the capacitor has maximum positive charge. At what times t > 0 will (a) plate A again have maximum positive charge, (b) the other plate of the capacitor have maximum positive charge, and (c) the inductor have maximum magnetic field?

Problem 33-5

CL


Problem 33-5 – Solution

a) Charge on the capacitor

c) The Magnetic field is maximum when the current is maximum and that occurs at t = T/4

when q=0, at t=1.25 µµµµs not that:

b)

(((( )))) (((( ))))⇒⇒⇒⇒========

φφφφ++++ωωωω====

Qqtat

tQtq

0

cos

0cos ====φφφφ⇒⇒⇒⇒φφφφ==== QQ

tQq ωωωω==== cos

sfTtatQq µµµµ====================⇒⇒⇒⇒ 5200000/1/1

sTtat µµµµ======== 5.22/

4/sin TtatIitIi ====−−−−====⇒⇒⇒⇒ωωωω−−−−====


In the circuit shown in Fig. 33-23 the switch is kept in position a for a long time. It is then thrown to position b. (a) Calculate the frequency of the resulting oscillating current. (b) What is the amplitude of the current oscillations?

Solution

Problem 33-13

a)CL

fππππ

====2

1

Hzf 275

102.610542

1

63====

××××××××××××ππππ====

−−−−−−−−

QfQI ππππ====ωωωω==== 2

AI 364.0108.21027526

====××××××××××××ππππ====−−−−

CVFVCQ µµµµ====××××µµµµ======== 8.2100.342.6

b)


The total energy U is the sum of the energies in the inductor and capacitor. If q is the charge on the capacitor, C is the capacitance, i is the current, and L is the inductance, then:

In an oscillating LC circuit, L = 25.0 mH and C = 7.80 mF. At time t = 0 the current is 9.20 mA, the charge on the capacitor is 3.80 µ µ µ µC, and the capacitor is charging.

(a) What is the total energy in the circuit?

Problem 33-17

(((( ))))(((( ))))

(((( )))) (((( ))))2

100.25102.9

1080.72

1080.3323

6

26HA

F

C−−−−−−−−

−−−−

−−−− ××××××××++++

××××××××

××××====

JJ µµµµ====××××====−−−− 98.11098.1

6

22

22 Li

C

qUUU BE ++++====++++====


Problem 33-17, b, c and d

(b) What is the maximum charge on the capacitor?

Solve U = Q2/2C for the maximum charge Q:

(c) What is the maximum current?

(d) If the charge on the capacitor is given by q = Q cos(ωωωωt + φφφφ), what is the phase angle φφφφ?

If q0 is the charge on the capacitor at time t = 0, then :

CJFUCQ 6661056.5)1098.1()1080.7(22

−−−−−−−−−−−− ××××====××××××××××××========

mAH

F

L

UI 6.12

100.25

)1080.7(223

6

====××××

××××××××========

−−−−

−−−−

φφφφ==== cos0 Qq °°°°±±±±====

µµµµ

µµµµ====

====φφφφ⇒⇒⇒⇒ −−−−−−−− 6.49

56.5

80.3coscos

101

C

C

Q

q

4


Problem 33-17, e

For φφφφ = +46.9°, the charge on the capacitor is

decreasing; for φφφφ = - 46.9°, it is increasing. To

check this, calculate the derivative of q with

respect to time, evaluated for t = 0. You should

get - ωωωω Q sinφφφφ. You want this to be positive.

Since sin(+46.9°) is positive and sin(-46.9°) is

negative, the correct value for increasing

charge is φφφφ = -46.9 ° ° ° °

(e) Suppose the data are the same, except that the capacitor is discharging at t = 0. What then is φφφφ?

Now you want the derivative to be negative and sinφφφφ to be positive. Take φφφφ = +46.9°.

Chapter 5 –

Electromagnetic Waves

Next Lecture

1





Chapter 5Chapter 5Chapter 5Chapter 5Electromagnetic WavesElectromagnetic WavesElectromagnetic WavesElectromagnetic Waves

© Dr. Nidal Ershaidat

http://ctaps.yu.edu.jo/physics/Courses/Phys103/Chapter5© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 2

In 1870, James Clerk Maxwell established a mathematical frame based on four equations in electricity and magnetism, in which all phenomena electricity and magnetism can be explained.

Maxwell’s Equations

Electricity and magnetism were thus unified in what we call now Electromagnetism.

©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 3

Gauss’ Law for electricity

Gauss’ Law for magnetism

Faraday’s Law


0εεεε====∫∫∫∫

qAd.Evr

0====∫∫∫∫ Ad.Brr

dt

dld.E BΦΦΦΦ

−−−−====∫∫∫∫rr

====∫∫∫∫ ld.Brr

dt

di E�� 000

Ampere-Maxwell Law


James Clerk Maxwell showed that electromagnetic energy propagates in vacuum with the speed of light (c).


His major conclusion was that light is nothing else but em waves, i.e. a beam of light is a traveling wave of electric and magnetic fields—an electromagnetic wave

Optics, the study of visible light, became a branch of electromagnetism.

We shall concentrate on strictly electric and magnetic phenomena, and we build a foundation for optics.

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 5

c and MKS

00

1

µµµµεεεε====c

The MKS system (SI) has for base this equation.

All other units are defined starting from the definition of c.

1200 ====µµµµεεεε cor


In Maxwell's time, the visible, infrared, and ultraviolet forms of light were the only electromagnetic waves known.

Heinrich Hertz discovered what we now call radio waves and verified that they move through the laboratory at the same speed as visible light.

H. Hertz

2


Range of electromagnetic waves

1 nm = 10-9 m

1 Å = 10-10 m

Fig. 1


The Electromagnetic Spectrum• Note the overlap between types of waves

•Types are distinguished by frequency or wavelength

Fig. 2

•Visible light is a small portion of the spectrum


Visible Light

Figure 3 shows the relative sensitivity of the

average human eye to electromagnetic waves at different wavelengths.

Fig. 3

This portion of the electromagnetic spectrumto which the eye is sensitive is called visible light.

5-2 The Traveling

Electromagnetic

Wave, Qualitatively


� Any accelerating charge will emit

electromagnetic waves.


�An electromagnetic wave is a wave that

combines the electric wave “electric field E”

and magnetic wave “magnetic field B’.

� “Oscillating charges” is an example

accelerating charges.


An LC circuit is used to produce emoscillations. An energy source is used to compensate for the energy loss in the wires (represented by the resistance R)

Oscillations are transmitted to an antenna which emits the em waves in space.

Generation of electromagnetic waves

Fig. 4

3


D e s c r i p t i o n o f t h e e l e c t r i c a n d

m a g n e t i c f i e l d s a s t h e y p a s s

p o i n t P o n t h e F i g . 4

P r o p a g a t i o n o f a n e m w a v e

F i g . 5

P r o p a g a t i o n d i r e c t i o n o f

t h e w a v e


• Two rods are connected to an ac source, charges oscillate between the rods (a)• As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b)

• The charges and field reverse (c)

• The oscillations continue (d)

EM Waves from an Antenna

Fig. 5


Propagating Oscillations

+

-

xz

y

Animation of E & B fields “polarization”

Current (up and down) creates B field into and out of the page!

Properties of the electric

and magnetic fields in

electromagnetic waves.

17

© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves


xz

y

• Transverse

• E perpendicular to B and in phase.

• Can travel in empty space

f = v/λλλλ

v = c = 3 x 108 m/s (186,000 miles/second!)

c =1

ε0µ0

E

B= c


1. EM waves are transverse waves.

2.The electric field is always perpendicular to the

magnetic field.

EM Waves are transverse waves

The electric and magnetic fields and are always perpendicular to the direction of travel of the wave., as discussed in Chapter 17.

Er

Br

3.The cross product always gives the

direction of travel of the wave.

BErr

××××

4


Properties of the electric and magnetic field in electromagnetic waves

)txksin(B)t,x(B

)txksin(E)t,x(E

m

m

ωωωω−−−−====

ωωωω−−−−====

4.The fields always vary sinusoidally, just like the transverse waves discussed in Chapter 17 . Moreover, the fields vary with the same frequency and in phase (in step) with each other.

5.The E and B field can be described by:


6. All electromagnetic waves, including visible

light, have the same speed c in vacuum

Where c is given by:

Speed of Propagation

00

1εεεεµµµµ

============m

m

B

E

B

Ec


Where n is a parameter which characterizes the medium and called the index of refraction of the medium.

7. Speed of light in dense medium is less than c:

Speed in Material Media

n

cv ====

Electromagnetic FieldElectromagnetic FieldElectromagnetic FieldElectromagnetic Field


1. EM waves are transverse waves.

2.The electric field is always perpendicular to the

magnetic field.

EM Waves are transverse waves

The electric and magnetic fields and are always perpendicular to the direction of travel of the wave., as discussed in Chapter 17.

Er

B•

3.The cross product always gives the

direction of travel of the wave.

BE

II

I


Properties of the electric and magnetic field in electromagnetic waves

)txksin(B)t,x(B

)txksin(E)t,x(E

m

m

ωωωω−−−−====

ωωωω−−−−====

4.The fields always vary sinusoidally, just like the transverse waves discussed in Chapter 17 . Moreover, the fields vary with the same frequency and in phase (in step) with each other.

5.The E and B field can be described by:

5


The previous equations define what we call a plane wave

Plane Electromagnetic waves

• • • )))) (((( )))) (((( ))))txkBtxBtxkEtxE mm ωωωω−−−−====ωωωω−−−−==== sin,,sin,

(((( )))) (((( )))) (((( )))) (((( ))))txki

m

txki

m eBtxBeEtxE ωωωω−−−−ωωωω−−−− ======== ,,,

Which we can write in a complex form as:


Consider a representative plane of area A.

Fig. 9 shows a wavefront at some reference instant t0. After time dt, the wavefront has traveled cdt.

propagates in the y+

direction and in the z+

direction

Em/Bm= c , Induced Electric Field

Fig. 9

Er

Br


Let us apply Faraday’s law of induction along the dashed ………… blue path:

Em/Bm= c, Induced Electric Field

dt

dldE BΦΦΦΦ

−−−−====→→→→→→→→

∫∫∫∫ .

d

x

h

(((( )))) dEhhEhdEEldE ====−−−−++++====→→→→→→→→

∫∫∫∫ .

dt

ABd

dt

d B )(====

ΦΦΦΦ

dt

dBdxhdEh −−−−====

dt

dB

dx

dE−−−−====

dt

dBA====

dt

dBdxh====

E

E + dE


Em/Bm= c

(((( )))) (((( ))))txkBtxkkE mm ωωωω−−−−ωωωω++++====ωωωω−−−−⇒⇒⇒⇒ coscos

ckBE mm ====ωωωω====⇒⇒⇒⇒

(((( )))) (((( ))))txkEtxE m ωωωω−−−−==== sin,Oscillations are described by:

(((( )))) (((( ))))txkBtxB m ωωωω−−−−==== sin,

(((( )))) (((( ))))txkEkx

txEm ωωωω−−−−====

∂∂∂∂

∂∂∂∂⇒⇒⇒⇒ cos,

( ) ( )txkBt

txBm ωω −−=

∂

∂cos

,

dt

dB

dx

dE−−−−====


Let us apply Ampere-Maxwell law of induction along the dashed ………… blue path:

Em/Bm= c, Induced Magnetic Field

dt

dldB EΦΦΦΦ

µµµµεεεε−−−−====→→→→→→→→

∫∫∫∫ 00.

dx

h(((( )))) dBhhBhdBBldB −−−−====−−−−++++−−−−====→→→→→→→→

∫∫∫∫ .

dt

AEd

dt

d E )(====

ΦΦΦΦ

dt

dEdxhdBh 00

dt

dE

dx

dB00••••

dt

dEA•

dt

dEdxh•

••• BdB

•B

→→→→E

30

Em/Bm= c

(((( )))) (((( ))))txkEtxkkB mm ωωωω−−−−ωωωωµµµµεεεε++++====ωωωω−−−−⇒⇒⇒⇒ coscos 00

(((( )))) ckB

E

m

m

0000

11µµµµεεεε

====ωωωωµµµµεεεε

====⇒⇒⇒⇒

(((( )))) (((( ))))txkEtxE m ωωωω−−−−==== sin,Oscillations are described by:

(((( )))) (((( ))))txkBtxB m ωωωω−−−−==== sin,

(((( )))) (((( ))))txkEt

txEm ωωωω−−−−ωωωω−−−−====

∂∂∂∂

∂∂∂∂⇒⇒⇒⇒ cos,

(((( )))) (((( ))))txkBkx

txBm ωωωω−−−−====

∂∂∂∂

∂∂∂∂ sin,

dt

dE

dx

dB00 µµµµεεεε−−−−====

6


We have:

ε0 µ0 c2 = 1

cB

E

m

m ====

cB

E

m

m

00

1µµµµεεεε

====

1200 ====µµµµεεεε c

00

1

µµµµεεεε====c

Which gives:

The speed of em waves is:

227270

229

0

104104

10361

−−−−−−−−−−−−−−−−−−−−

−−−−

××××ππππ====××××ππππ====µµµµ

××××ππππ====εεεε

sCNAN

CmN

187

9103

104

1036 −−−−−−−−

××××====××××ππππ

××××ππππ≈≈≈≈ sm

5-4 Energy Transport

and the Poynting Vector


From sun light, We all know that an electromagnetic wave can transport energy and deliver it to a body on which it falls.

The rate of energy transport per unit area in such a wave is described by a vector , called the Poynting vector.

Sr

It is called after physicist John Henry Poynting(1852–1914), who first discussed its properties.

Poynting Vector


The direction of the Poynting vector of an electromagnetic wave at any point gives the wave's direction of travel and the direction of energy transport at that point.

Sr

Poynting Vector

The Poynting vector is defined by:

→→→→→→→→→→→→

××××µµµµ

==== BES0

1

====

µµµµ====

====→→→→→→→→→→→→

→→→→→→→→→→→→

shrSandBE

BES

..,

1

0

The magnitude of the Poynting vector is given by:

BES0

1µµµµ

====


The units of S is power per unit area or W/m2

Its magnitude S is related to the rate at which energy is transported by a wave across a unit area at any instant.

Remember that for a plane wave:

Power emitted and Poynting

[[[[ ]]]] [[[[ ]]]] [[[[ ]]]] [[[[ ]]]]][][ Area

Power

Area

TimeEnergyS ========

cB

E

B

E

m

m ========

2

0

20

0

2B

cSorEc

c

ES

µµµµ====εεεε====

µµµµ====⇒⇒⇒⇒


The intensity is defined as:

or:

Intensity Emitted

c

EE

cSI m

avgavg0

22

0 21

µµµµ====

µµµµ========

20

0

2

21

2 mm Ec

BcI εεεε====

µµµµ====

c

Erms

0

2

µµµµ====

c

Erms

0

2

µµµµ====

7


Intensity Emitted2

24 r

PI s

ππππ====

Fig. 34-8 A point source S emits

electromagnetic waves uniformly in

all directions. The spherical wave

fronts pass through an imaginary

sphere of radius r that is centered

on S.

BE uBBcEu ====µµµµ

====εεεε====εεεε==== 2

0

220

20 2

121

21

The energy density “energy per unit volume” stored in the electric

field and in the magnetic field is given by (respectively):

Where uB is the energy density stored in the magnetic field.

The intensity at any point r is given

by:


The energy of electromagnetic waves is related to its momentum p by the following relation:

AA

If an electromagnetic wave is incident normally on

an area A as shown in Fig. 10 in a time ∆∆∆∆t and it is totally absorbed then there is momentum change given by:

Radiation Pressure

cpU ====

c

Up

∆∆∆∆====∆∆∆∆

Fig. 10

Where this momentum change is totally transferred to the area A


Radiation Pressure

The net force on the wall is:

The radiation pressure Pr is:

(((( ))))t

U

c

or

t

pF

∆∆∆∆

∆∆∆∆====

∆∆∆∆

∆∆∆∆====

21

(((( ))))c

Ior

A

FPr 21========

If the electromagnetic wave is totally reflected then the change in momentum is:

c

Up

∆∆∆∆====∆∆∆∆

2

(((( ))))t

tAI

c

or

∆∆∆∆

∆∆∆∆====

21 (((( ))))AI

c

or 21====


The maximum electric field at a distance of 10m from an isotropic point light source is 2.0 V/m. What are:

a) the maximum value of the magnetic field andb) the average intensity of the light there?c) What is the power of the source?

Problem 17

a)

b)

c)

Teslac

EB 8

810666.0

103

2 −−−−××××====××××

========

278

0

2

/0106.0104103

4mW

c

EI m ====

××××ππππ××××××××====

µµµµ====

−−−−

WIrP 32.130106.010044 2 ====××××××××ππππ====ππππ====

Solution:


A plane electromagnetic wave, with wavelength 3.0 m, travels in vacuum in the positive x direction with its electric field , of amplitude 300 V/m, directed along the y axis

(a) What is the frequency f of the wave?

(b)What are the direction and amplitude of the magnetic field associated with the wave?

Problem 25

Hzcf 88 103103 ====××××====λλλλ====

axizzthealongcEB −−−−====××××======== −−−−68 10103300

→→→→E


Problem 25(c)What are the values of k and ωωωω if

E = Em sin(k x – ωωωω t)?

m.k 12322 ====ππππ====λλλλππππ====

188 102861022 −−−−××××====××××ππππ====ππππ====ωωωω srad.f

,

(d)What is the time-averaged rate of energy flow in watts per square meter associated with this wave?

(((( ))))

(((( )))) (((( )))) 2782

02

31191041032300

2

mW.

cESI mavg

====××××ππππ××××××××××××====

µµµµ========

−−−−

8


(e)If the wave falls on a perfectly absorbing sheet of area 2.0 m2, at what rate is momentum delivered to the sheet and what is the radiationpressure exerted on the sheet?

Problem 25

Nc

AI

dt

dp 88

105.79103

23.119 −−−−××××====××××

××××========

291075.39 mNc

Ipr

−−−−××××========

5-2 Polarization

45

Orientation of E field matters when the EM wave traverses matter

Polarization

n

cv ====


•Transverse waves have a polarization

(Direction of oscillation of the electric field E for

light)

Polarization

Types of Polarization

• Linear (Direction of E is constant)

• Circular (Direction of E rotates with time)

• Unpolarized (Direction of E changes randomly)

xz

y


Natural Light is Unpolarized• Light from the sunWe can polarize light using special materialCrystals, Polymers with aligned atoms


Natural Light is Unpolarized

9


Linear Polarizers absorb all electric fields perpendicular to their transmission axis.

Linear Polarizers

50

Unpolarized Light on Linear Polarizers

202

1EcI εεεε====

•Most light comes from electrons accelerating in random directions and is unpolarized.

•Averaging over all directions, intensity of

transmitted light reduces due to reduction in E

51

Linearly Polarized Light on Linear Polarizer

TATATATA

θθθθθθθθ is the angle between the incoming light’s polarization, and the transmission axis

θθθθ

Transmission axisTransmission axisTransmission axisTransmission axis

Incident E E Transmitted

EEEEabsorbed

absorbed

absorbed

absorbed

=Eincidentcos(θθθθ)

Etransmitted = Eincident cos(θθθθ)Itransmitted = Iincident cos2(θθθθ)

52

Example - 1• Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges is:

– Zero

– 1/2– 1/3– 1/4– Need more information

21

is value average Therefore,

cycle fullyover randomly varying is

22

θθθθ

θθθθ∝∝∝∝⇒⇒⇒⇒∝∝∝∝ cosIEI

53

Example - 2

• Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges is:

– Zero

–1/2

–1/3

–1/4

–Need more information

090 is o

22

====⇒⇒⇒⇒θθθθ

θθθθ∝∝∝∝⇒⇒⇒⇒∝∝∝∝

ΙΙΙΙ

χοσχοσχοσχοσΙΙΙΙΕΕΕΕΙΙΙΙ

54

Example - 3

unpolarized

light

E1

45°

I = I0

TA

TA

90°

TA

E0

I3

B1

unpolarized

light

E1

45°

I = I0

TA

TA

90°

TA

E0

I3

B1

1) Intensity of unpolarized light incident on a

linear polarizer is reduced by half . I1= I0 / 2

I = I0I1

I2

2) Light transmitted through first polarizer is vertically 2) Light transmitted through first polarizer is vertically 2) Light transmitted through first polarizer is vertically 2) Light transmitted through first polarizer is vertically

polarized. Angle between it and second polarizer is polarized. Angle between it and second polarizer is polarized. Angle between it and second polarizer is polarized. Angle between it and second polarizer is θθθθ=90=90=90=90ºººº . . . .

I2= I1 cos2(90º) = 0

10

55

Example - 3

3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third

polarizer is q=45º. I3= I2 cos2(45º)

unpolarized

light

E1

45°

I = I0

TA

TA

90°

TA

E0

I3

B1

unpolarized

light

E1

45°

I = I0

TA

TA

90°

TA

E0

I3

B1

2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is

θθθθ=45°°°°. I2= I1 cos2 (45º)= 0.5 I1= 0.25 I0

I2= I1cos2(45)

= 0.125 I0

I1= 0.5 I0 Light Incident on

an Object


Light incident on an object

• Reflects (bounces)

• Refraction (bends)

•Absorbed


Angle between light beam and normal Angle of incidence = Angle of reflection

Law of Reflection

θθθθi = θθθθf

θθθθi

θθθθf


Index of Refraction

v =

c

n

Speed of light

in medium Index of refraction

Speed of light in

vacuum

n > 1v < c so

300 000 km/second: it’s not just a

good idea, it’s the law!


Refractive index n and wavelength

11


When light travels from one medium to another the speed changes v=c/n, but the frequency is constant.

Snell’s Law

θθθθ1

So the light bends: n1 sin(θθθθ1)= n2 sin(θθθθ2)

n1n2

θθθθ2


Snell’s Law – Example 1Which is true? 1) n1 > n2

2) n1 = n2

3) n1 < n2

nnnn1111

nnnn2222

θθθθ1111

θθθθ2222

θθθθ1 < θθθθ2

sin θθθθ1 < sin θθθθ2

n1 > n2


A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle θθθθr = 60°°°°. What is θθθθ2?

Snell’s Law – Example 2

1θ

2θ

sin(60°°°°) = 1.33 sin(θθθθ2)

θθθθ2 = 40.6 degrees

rθθθθθ1 =θθθθr =60°°°°

n2 sin(θθθθ2) = n2 sin(θθθθ2)n2=1.33

n1=1.0

normal

64

Snell’s Law: n1 sin(θθθθ1)= n2 sin(θθθθ2)when n1 > n2 , θθθθ2 > θθθθ1

Total Internal Reflection

When θθθθ1 = sin-1(n2/n1) θθθθ2 = 90°°°°This is a critical angle!

Light incident at a larger angle will be completely

reflected θθθθi = θθθθr

normal

θθθθ2

θθθθ1

n2

n1

θθθθc

θθθθiθθθθr


Total Internal ReflectionTotal internal reflection can occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction.

Ray 5 suffers internal reflection


A particular angle of incidence will result in an angle of refraction of 90°

Critical Angle

This angle of incidence is called the critical angle

211

2sin nnforn

n>>>>====θθθθ

12


Critical Angle

• For angles of incidence greater than the

critical angle, the beam is entirely reflected

at the boundary

–This beam obeys the Law of Reflection at

the boundary.

• Total internal reflection occurs only when

light attempts to move from a medium of

higher index of refraction to a medium of

lower index of refraction.


Chromatic DispersionThe index of refraction n encountered by light in any medium except vacuum depends on the wavelength of the light.

The dependence of n on wavelength implies that when a light beam consists of rays of different wavelengths, the rays will be refracted at different angles by a surface; that is, the light will be spread out by the refraction.

This spreading of light is called chromatic dispersion, in which “chromatic” refers to the colors associated with the individual wavelengths and “dispersion” refers to the spreading of the light according to its wavelengths or colors


Chromatic Dispersion

12

12 sinsin

n

n


Chromatic Dispersion


The maximum electric field at a distance of 10m from an isotropic point lightsource is 2.0 V/m. What are:a) the maximum value of the magnetic field andb) the average intensity of the light there?c) What is the power of the source?

P r o b l e m 1 7

a)

b)

c)

Tesla

c

E

B

8

8

10666.0

103

2I

II

I

II

2

78

0

2

/0106.0

104103

4

mW

c

E

I

m

I

IIII

I

I

I

I

WIrP 32.130106.010044

2

IIIIIII

Solution:


P r o b l e m 2 5

A plane electromagnetic wave, with wavelength 3.0 m, travels in vacuum in the positive x direction

with its electric field , of amplitude 300 V/m,

directed along the yaxis

(a) What is the frequency f of the wave?

(b)What are the direction and amplitude of the magnetic field associated with the wave?

Hzcf

88

103103 IIIII

axizzthealongcEB IIIII

I 68

10103300

I

E

1 3


P ro b le m 2 5 – c , d

(c)W hat are the values of kand • if

E= Emsin(k x–•t)?

mk 1.2322 ••••••

1881028.61022

••••••••• sradf

,

(d)W hat is the tim e-averaged rate of energy flow in w atts per square m eter associated w ith this w ave?

• •

• • • •2782

02

3.1191041032300

2

mW

cESI mavg

•••••••

•••

•


P ro b le m 2 5 – e

(e)If the w ave falls on a perfectly absorbing sheet of area 2.0 m2, at w hat rate is m om entum delivered to the sheet and w hat is the radiationpressure exerted on the sheet?

Nc

AI

dt

dp 88

105.79103

23.119 •••

•

•••

291075.39 mN

c

Ipr

••••


A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through 360° while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of 5.0 during the rotation, what fraction of the intensity of the original beam is associated with the beam's polarized light?

Problem 39

80

Let I0 be the intensity of the incident beam and f

the fraction of it which is polarized.Then the intensity of the polarized portion is f I0:


(((( )))) θθθθ++++−−−−====′′′′ 200 cos21 IfIfI

and the intensity of the unpolarized portion is (1-f) I0:Initially the intensity of transmitted portion before rotating the polariod is:

The minimum intensity is :

The maximum intensity is:

(((( )))) (((( )))) {{{{ }}}} (((( )))) 21211cos 0002

max IfIfIfII ++++====++++−−−−========θθθθ′′′′====′′′′

(((( )))) (((( )))) {{{{ }}}} (((( )))) 210210cos 002

min IfIfII −−−−====++++−−−−========θθθθ′′′′====′′′′

325

11

min

max ====⇒⇒⇒⇒====−−−−

++++==== f

f

f

I

I


In Fig. 34-52 , light enters a 90°°°° triangular prism at point P with incident angle θθθθ and then some of it refracts at point Q with an angle of refraction of 90°°°°(a) What is the index of refraction of the prism in terms of θθθθ? (b) What, numerically, is the maximum value that the index of refraction can have? Explain what happens to the light at Q if the incident angle at Q is (c) increased slightly and (d) decreased slightly.

Problem 59

Chapter 6: InterferenceChapter 6: InterferenceChapter 6: InterferenceChapter 6: Interference

Next Lecture

1





InterferenceInterferenceInterferenceInterference

http://ctaps.yu.edu.jo/physics/Courses/Phys207/Chapter6© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 2

نضال محمد الرشيدات جامعة اليرموك–قسم الفيزياء إربد األردن21163

ا��ء

http://ctaps.yu.edu.jo/physics/NErshaidat/Light

3333© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference

....األجسام تشع ضوءا كما كان يظن اليونانيوناألجسام تشع ضوءا كما كان يظن اليونانيوناألجسام تشع ضوءا كما كان يظن اليونانيوناألجسام تشع ضوءا كما كان يظن اليونانيوناستقبال العين للضوء المنعكس عن األجسام وليس ألن! استقبال العين للضوء المنعكس عن األجسام وليس ألن! استقبال العين للضوء المنعكس عن األجسام وليس ألن! استقبال العين للضوء المنعكس عن األجسام وليس ألن! أن!نا نبصر األشياء بفضل أن!نا نبصر األشياء بفضل أن!نا نبصر األشياء بفضل أن!نا نبصر األشياء بفضل ) ) ) ) كتاب المناظركتاب المناظركتاب المناظركتاب المناظر((((عرف ابن الهيثم عرف ابن الهيثم عرف ابن الهيثم عرف ابن الهيثم كيف ينتشر، كيف نراه وكيف نرى األشياء حولنا؟كيف ينتشر، كيف نراه وكيف نرى األشياء حولنا؟كيف ينتشر، كيف نراه وكيف نرى األشياء حولنا؟كيف ينتشر، كيف نراه وكيف نرى األشياء حولنا؟مم! يتكون، مم! يتكون، مم! يتكون، مم! يتكون، . . . . منذ األزل واإلنسان يبحث في طبيعة الضوءمنذ األزل واإلنسان يبحث في طبيعة الضوءمنذ األزل واإلنسان يبحث في طبيعة الضوءمنذ األزل واإلنسان يبحث في طبيعة الضوءالضوءالضوءالضوءالضوء


....يرمى في الماءيرمى في الماءيرمى في الماءيرمى في الماءموجي!ة أي أن!ه عبارة عن موجات كتلك التي يحدثها حجر موجي!ة أي أن!ه عبارة عن موجات كتلك التي يحدثها حجر موجي!ة أي أن!ه عبارة عن موجات كتلك التي يحدثها حجر موجي!ة أي أن!ه عبارة عن موجات كتلك التي يحدثها حجر ه مكون من جسيمات وطبيعة ه مكون من جسيمات وطبيعة ه مكون من جسيمات وطبيعة ه مكون من جسيمات وطبيعة طبيعة جسيمي!ة للضوء، أي أن!طبيعة جسيمي!ة للضوء، أي أن!طبيعة جسيمي!ة للضوء، أي أن!طبيعة جسيمي!ة للضوء، أي أن!منذ القرن السابع عشر وآراء الفيزيائيين تتأرجح بين منذ القرن السابع عشر وآراء الفيزيائيين تتأرجح بين منذ القرن السابع عشر وآراء الفيزيائيين تتأرجح بين منذ القرن السابع عشر وآراء الفيزيائيين تتأرجح بين جسيمات أم موجات؟ جسيمات أم موجات؟ جسيمات أم موجات؟ جسيمات أم موجات؟::::الضوءالضوءالضوءالضوء


نيوتن

الضوء عبارة عن الضوء عبارة عن الضوء عبارة عن الضوء عبارة عن الضوء عبارة عن الضوء عبارة عن الضوء عبارة عن الضوء عبارة عن جسيمات جسيمات جسيمات جسيمات جسيمات جسيمات جسيمات جسيمات

CorpusclesCorpusclesCorpusclesCorpusclesCorpusclesCorpusclesCorpusclesCorpuscles ��

2

© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 7

....نقطة أخرى على شكل موجاتنقطة أخرى على شكل موجاتنقطة أخرى على شكل موجاتنقطة أخرى على شكل موجاتتنتقل الطاقة بجميع أشكالها من نقطة في الفراغ إلى تنتقل الطاقة بجميع أشكالها من نقطة في الفراغ إلى تنتقل الطاقة بجميع أشكالها من نقطة في الفراغ إلى تنتقل الطاقة بجميع أشكالها من نقطة في الفراغ إلى وتمث)ل كل وتمث)ل كل وتمث)ل كل وتمث)ل كل . . . . موجات دائري>ة مركزها نقطة سقوط الحجرموجات دائري>ة مركزها نقطة سقوط الحجرموجات دائري>ة مركزها نقطة سقوط الحجرموجات دائري>ة مركزها نقطة سقوط الحجركمثال تنتشر الطاقة الناجمة عن إلقاء حجر في بحيرة كمثال تنتشر الطاقة الناجمة عن إلقاء حجر في بحيرة كمثال تنتشر الطاقة الناجمة عن إلقاء حجر في بحيرة كمثال تنتشر الطاقة الناجمة عن إلقاء حجر في بحيرة ....الناتج عن انتقال الطاقةالناتج عن انتقال الطاقةالناتج عن انتقال الطاقةالناتج عن انتقال الطاقة” ” ” ” االزعاج االزعاج االزعاج االزعاج ””””وتمث)ل الموجة وتمث)ل الموجة وتمث)ل الموجة وتمث)ل الموجة والذي ينتقل خالل والذي ينتقل خالل والذي ينتقل خالل والذي ينتقل خالل ) ) ) ) الماء هناالماء هناالماء هناالماء هنا((((الوسط الوسط الوسط الوسط ” ” ” ” ازعاجازعاجازعاجازعاج””””موجة موجة موجة موجة ....السطحالسطحالسطحالسطح��


الموجة المستوية والموجة الكروي�ة....لى شكلها في لحظة سابقةلى شكلها في لحظة سابقةلى شكلها في لحظة سابقةلى شكلها في لحظة سابقةإإإإالموجة استنادا الموجة استنادا الموجة استنادا الموجة استنادا وضع هويغنز طريقة هندسي ة تسمح بإيجاد مقدمة وضع هويغنز طريقة هندسي ة تسمح بإيجاد مقدمة وضع هويغنز طريقة هندسي ة تسمح بإيجاد مقدمة وضع هويغنز طريقة هندسي ة تسمح بإيجاد مقدمة . . . . الموجة التي تكون مقدمتها عبارة عن سطح كرويالموجة التي تكون مقدمتها عبارة عن سطح كرويالموجة التي تكون مقدمتها عبارة عن سطح كرويالموجة التي تكون مقدمتها عبارة عن سطح كرويعبارة عن سطح مستو والموجة الكروي ة هي عبارة عن سطح مستو والموجة الكروي ة هي عبارة عن سطح مستو والموجة الكروي ة هي عبارة عن سطح مستو والموجة الكروي ة هي الموجة التي تكون مقدمتها الموجة التي تكون مقدمتها الموجة التي تكون مقدمتها الموجة التي تكون مقدمتها فالموجة المستوية هي فالموجة المستوية هي فالموجة المستوية هي فالموجة المستوية هي يمكن تصنيف الموجات حسب مقدمة الموجة، يمكن تصنيف الموجات حسب مقدمة الموجة، يمكن تصنيف الموجات حسب مقدمة الموجة، يمكن تصنيف الموجات حسب مقدمة الموجة،


الموجة المستويةالموجة المستويةالموجة المستويةالموجة المستوية. . . . سطح ماسطح ماسطح ماسطح ماالحقيقي�ة، فهي في الواقع تقاطع الموجة الكروي�ة مع الحقيقي�ة، فهي في الواقع تقاطع الموجة الكروي�ة مع الحقيقي�ة، فهي في الواقع تقاطع الموجة الكروي�ة مع الحقيقي�ة، فهي في الواقع تقاطع الموجة الكروي�ة مع الموجة المستوية هي تقريب للموجة الكروي�ة الموجة المستوية هي تقريب للموجة الكروي�ة الموجة المستوية هي تقريب للموجة الكروي�ة الموجة المستوية هي تقريب للموجة الكروي�ة


، في عام ، في عام ، في عام ، في عام هيغنزهيغنزهيغنزهيغنزبرهن الفيزيائي الهولندي كريستيان برهن الفيزيائي الهولندي كريستيان برهن الفيزيائي الهولندي كريستيان برهن الفيزيائي الهولندي كريستيان ه يمكن تفسير انعكاس الضوء وانكساره ه يمكن تفسير انعكاس الضوء وانكساره ه يمكن تفسير انعكاس الضوء وانكساره ه يمكن تفسير انعكاس الضوء وانكساره ، أن�، أن�، أن�، أن�1678167816781678 ....الموجي�ةالموجي�ةالموجي�ةالموجي�ةلى النظري�ة لى النظري�ة لى النظري�ة لى النظري�ة إإإإاستنادا استنادا استنادا استنادا مبدأ هيغنزمبدأ هيغنزمبدأ هيغنزمبدأ هيغنز


مبدأ هيغنزمبدأ هيغنزمبدأ هيغنزمبدأ هيغنزملخص هذه الطريقة، والتي تسم�ى مبدأ ملخص هذه الطريقة، والتي تسم�ى مبدأ ملخص هذه الطريقة، والتي تسم�ى مبدأ ملخص هذه الطريقة، والتي تسم�ى مبدأ هيغنز، هو أنه يمكن اعتبار كل� نقطة من هيغنز، هو أنه يمكن اعتبار كل� نقطة من هيغنز، هو أنه يمكن اعتبار كل� نقطة من هيغنز، هو أنه يمكن اعتبار كل� نقطة من مقدمة الموجة وكأن�ها مصدر لمويجات ثانوي�ة مقدمة الموجة وكأن�ها مصدر لمويجات ثانوي�ة مقدمة الموجة وكأن�ها مصدر لمويجات ثانوي�ة مقدمة الموجة وكأن�ها مصدر لمويجات ثانوي�ة

. . . . تنتشر بنفس سرعة انتشار األمواج نفسهاتنتشر بنفس سرعة انتشار األمواج نفسهاتنتشر بنفس سرعة انتشار األمواج نفسهاتنتشر بنفس سرعة انتشار األمواج نفسهاويمكن استخدام السطح المماس للمويجات ويمكن استخدام السطح المماس للمويجات ويمكن استخدام السطح المماس للمويجات ويمكن استخدام السطح المماس للمويجات !!!!الثانوي�ة لتحديد مكان الموجة في لحظة تاليةالثانوي�ة لتحديد مكان الموجة في لحظة تاليةالثانوي�ة لتحديد مكان الموجة في لحظة تاليةالثانوي�ة لتحديد مكان الموجة في لحظة تالية


� The first person to advance a convincing wave theory for light was Dutch physicist Christian Huygens, in 1678 .

Light as a Wave: Huygens' Principle

� Its great advantages are that it accounts for the laws of reflection and refraction in terms of waves and gives physical meaning to the index of refraction.

Huygens' wave theory is based on a geometrical construction that allows us to tell where a given wavefront will be at any time in the future if we know its present position.

3


This geometric construction is based on Huygens' principle, which is:

All points on a wavefront serve as point

sources of spherical secondary wavelets.

After a time t, the new position of the

wavefront will be that of a surface tangent

to these secondary wavelets.

Huygens' principle


Huygens' principle The Figure shows , the present location of a

wavefront of a plane wave traveling to the right in vacuum represented by plane ab, perpendicular to the

page.

The wavefronts of all the spherical waves reach the

plane ed after a time t.

All the spherical waves have a radius ct when they reach the

plane ed after a time t.

The points dots represent secondary sources which emit

spherical waves.

Refraction


Refraction

tvtvbFigFrom 2211 ,)( ====λλλλ====λλλλ

1

2

1

2

v

v====

λλλλ

λλλλ⇒⇒⇒⇒

2211 sin&sin λλλλ====θθθθλλλλ====θθθθ chch

1

2

1

2

sinsin

θθθθ

θθθθ====

λλλλ

λλλλ


The ratio of the speed of light in vacuum to that in a material is defined as (n the refractive index)

Thus if the wavelength in vacuum is λλλλ then the wavelength (λλλλn) in a medium of refractive index n is:

Huygens' Principle & Snell’s Law

v

cn ====

1

2

2

1

1

2

1

2

1

2

sinsin

θθθθ

θθθθ================

λλλλ

λλλλ⇒⇒⇒⇒

n

n

nc

nc

v

v2211 sinsin θθθθ====θθθθ⇒⇒⇒⇒ nn

2

1

1

2

n

n====

λλλλ

λλλλ

nn

λλλλ====λλλλ λλλλ<<<<


Huygens' Principle

Note that the wavelength in dense medium is

smaller than that in vacuum

4

6-4 Interference

© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference

It often happens that two or more waves

pass simultaneously through the same

region. We say that these waves interfereinterfere

“with each other”

When we listen to a concert, for example,

sound waves from many instruments fall

simultaneously on our eardrums. You still

can distinguish between the violin or the

piano.

The Superposition Principle for Waves


Consider two sources of waves S1 and S2 (Fig. 4).

L1 and L2 represent two waves originating from the two sources S1 and S2, respectively

L1 and L2 reach point P (on some screen)

Interference

Fig. 4


Interference

Path length difference ا��ق �� ا��ر

∆∆∆∆L = |L2 – L1|.

To relate phase difference φφφφ to path length difference ∆∆∆∆L, we recall (from Section 17-4) that a phase difference of 2ππππ rad corresponds to one wavelength. Thus, we can write the proportion

λλλλ

∆∆∆∆====

ππππ

φφφφ L

2

Fig. 4


Fully constructive interference occurs when φφφφis zero, 2ππππ, or any integer multiple of 2ππππ. We can write this condition as:



i.e. when the path length difference is zero or an integer multiple of the wavelength λλλλ.


Constructive Interference

(((( )))) ππππ====φφφφ 2m

K,2,1,0====λλλλ

∆∆∆∆ L

λλλλ====∆∆∆∆ mL

or:


or:

Fully destructive interference occurs when φφφφ is an odd integer multiple of ππππ. A condition we can write as:

Fully destructive interference

i.e. when the path length difference is half an integer multiple of the wavelength λλλλ.

Destructive Interference

(((( ))))ππππ++++====φφφφ 12 m


K,25,

23,

21

====λλλλ

∆∆∆∆ L

Fully destructive interference

λλλλ

++++====∆∆∆∆21

mL

5

6-3 Diffraction


� If a wave encounters a barrier that has an opening of dimensions similar to the wavelength, the part of the wave that passes through the opening will diffract(flare (spread) out ) —into the region beyond the barrier.

� The flaring is consistent with the spreading of wavelets in the Huygens construction of Fig. 36-1.

� Diffraction occurs for waves of all types, not just light waves.

Diffraction


Figure below shows the diffraction of water waves traveling across the surface of water in a shallow tank.

Diffraction – Water Waves


Diffraction – Schematic View1

The situation for an incident plane wave of wavelength λλλλ encountering a slit that has width a = 6.0 λλλλ and extends into and out of the page is shown in fig. a.

Fig. a


The wave flares out on the far side of the slit. Figures 36-5b (with a = 3.0 λλλλ ) and 36-5c (a = 1.5 λλλλ ) illustrate the main feature of diffraction: the narrower the slit, the greater the diffraction.


Fig. b Fig. c



Fig. a Fig. cFig. b

6

6-3 Interference

32

Coherence is a necessary condition for two or more

wave sources to maintain a stable interference pattern.

Coherence

)sin(,)sin( 2211 trkEEtrkEE mm ωωωω−−−−====ωωωω−−−−====

(((( )))) (((( ))))

ωωωω−−−−

++++

−−−−====++++==== t

rrkrrkEEEE m 2

sin2

cos2 121221

For coherent sources, the phase difference at P only depends on the path difference ∆∆∆∆r = (r2 – r1) →→→→ ∆φ∆φ∆φ∆φ = k ∆∆∆∆r

= 2ππππ ∆∆∆∆r /λλλλ

Two sources S1 and S2 are coherent if their phase

difference is independent of time, otherwise they are incoherent.


- same frequency (necessary, but not sufficient)- constant or zero phase difference

- come from the same source, or synchronized

sources

Examples of some sources of light:

• Light bulb thermal radiation, incoherent

• Sunlight (emitted by independent, hot atoms)

• Laser coherent (quantum effect)

Requirements for coherent waves

Double Slit Interference


Coherence is a necessary condition for

two or more wave sources to maintain a

stable interference pattern.

Coherence

Two sources S1 and S2 are coherent if

their phase difference is independent of

time, otherwise they are incoherent.


Young’s Interference Experiment (1801)– observed a stable interference pattern from 2

slits- This was the first conclusive proof that light is a

wave→→→→ How did he obtain coherent light without a laser?

Young’s Double Slit Experiment

7


Source is sunlightpassing through a narrow slit.

Young – Coherent Light

Huygens' principle


•Coherent incident light enters slits S1 and S2 with the same phase.

Double Slit Experiment - Analysis

•Phase difference at point P is k ∆∆∆∆r = 2ππππ (r2-r1) / λλλλ

•Path difference for L >> d is ∆∆∆∆∆∆∆∆rr = (= (rr22--rr11) = ) = dd sinsinθθθθθθθθ

Source is sunlightpassing through a

narrow slit.


Double Slit Experiment - Fringes

- Constructive interference for ∆∆∆∆r = m λλλλ (m = 0, 1, 2, …)

(((( ))))Lyd

md

====

λλλλ====θθθθsin

- Destructive interference for ∆∆∆∆r = (m + ½ ) λλλλ

(((( ))))Lyd

md

====

λλλλ

++++====θθθθ21sin

- y represents the vertical position of point P.


For an EM wave:

Intensity is:

Applying the superposition principle:

Intensity pattern (central part)

(((( ))))txkEE m ωωωω−−−−==== sin

(((( )))) (((( ))))222

0

sin2

1mWtxkE

cI m ωωωω−−−−

µµµµ====

(((( )))) (((( ))))trkEtrkE

EEE

ωωωω−−−−++++ωωωω−−−−====

++++====

2010

21

sinsin

221 rr

r++++

====

ωωωω−−−−

++++

−−−−==== t

rrk

rrkE

2sin

2cos2 2121

0

φφφφ∆∆∆∆====θθθθ====∆∆∆∆⇒⇒⇒⇒ sindkrkLet:

θθθθ++++==== sin12 drr

21 rrr −−−−====∆∆∆∆,


Intensity pattern (central part)

where I0 is the intensity with only one slit open,

We get

Intensity is:

(((( ))))trkEE ωωωω−−−−

φφφφ∆∆∆∆==== sin

2cos2 0

Amplitude:

(((( )))) (((( ))))ββββ====µµµµ

φφφφ∆∆∆∆==== 2

00

220 cos4

22cos4

Ic

EI

ββββ====

φφφφ∆∆∆∆==== cos2

2cos2 00 EE

and ββββ = ½ ∆φ∆φ∆φ∆φ = ½ k ∆∆∆∆r = ππππ d sinθθθθ/λλλλ

42

Intensity pattern

(((( )))) (((( )))) (((( ))))λλλλθθθθππππ====ββββ====θθθθ sincos4cos4 20

20 dIII

I(θθθθ) is max when cos (ππππ d sinθθθθ/λλλλ) = ±±±± 1

i.e. (ππππ d sinθθθθ/λλλλ) = mππππ →→→→ m λλλλ = d sinθθθθ

I(θθθθ) is min when cos (ππππ d sinθθθθ/λλλλ) = 0

i.e. (ππππ d sinθθθθ/λλλλ) = (m + ½) ππππ →→→→ (m + ½) λλλλ = d sinθθθθ

8


I

Shape of the Intensity pattern

The width of each band is the distance between successive minima or maxima:

For minima: (m + ½ ) λλλλ = d sin θθθθ.

Assuming small angle for central “fringe”:

Separation is ∆∆∆∆m = 1; λλλλ = d ∆∆∆∆ (sinθθθθ) ≈≈≈≈ d ∆θ∆θ∆θ∆θFringe spacing ∆θ∆θ∆θ∆θ goes as 1/d !!!


Narrow and Wide Slits

Assuming small angle for central “fringe”: Separation is ∆∆∆∆m = 1; λλλλ = d ∆∆∆∆ (sinθθθθ) ≈≈≈≈ d ∆θ∆θ∆θ∆θFringe spacing ∆θ∆θ∆θ∆θ goes as 1/d !!!

Interference and CoherenceInterference and CoherenceInterference and CoherenceInterference and Coherence


Difference between Coherent and Incoherent light

-Coherent: phase difference is stable in time →→→→stable interference pattern- Incoherent: phase difference is random in time →→→→ no interference

For coherent sources, the phase difference at P

only depends on the path difference ∆∆∆∆r = (r2 – r1) →→→→∆φ∆φ∆φ∆φ = k ∆∆∆∆r = 2ππππ ∆∆∆∆r /λλλλ


Coherence and Interference Pattern

For 2 slits, the average intensity over the screen is the same in both cases, but only the coherent light shows an interference pattern.


What is the minimum slit separation dd to produce

at least one maximum other than the central maximum?

•Second minimum for positive θθθθ occurs at:

1.5 1.5 λλλλλλλλ = = dd sinsinθθθθθθθθ = = dd sin (sin (ππππππππ/2) = /2) = dd

Example 1

9


Example 1For larger d, the second minimum will occur

at a smaller value of θθθθ, so more maxima will

appear on the screen. Therefore, d ≥≥≥≥ 1.5 λλλλ

NB: to see interference effects, the central

peak must be defined: d ≥≥≥≥ 0.5 λλλλ

d ≥≥≥≥ λλλλ just makes it to the second peak, not to

the next minimum


A double-slit arrangement produces interference

fringes for sodium light (λλλλ = 589 nm = 5890 Å) that

are 0.20°°°° apart. What is the angular fringe

separation if the entire arrangement is immersed

in water (n=4/3)?

Example 2


Example 2 – Solution

∆∆∆∆m λλλλ = d ∆∆∆∆ (sinθθθθ) ≈≈≈≈ d ∆θ∆θ∆θ∆θ

In water, the wavelength changes to λλλλλλλλ’’ = = λλλλλλλλ/n/n

75.0431

============λλλλ

λλλλ′′′′====

θθθθ∆∆∆∆

θθθθ′′′′∆∆∆∆

n

°°°°====°°°°××××====θθθθ∆∆∆∆====θθθθ′′′′∆∆∆∆⇒⇒⇒⇒ 15.020.043

43

The The The The Michelson InterferometerMichelson InterferometerMichelson InterferometerMichelson Interferometer


Principle

d1

d2

• light from a source is split into 2 beams,reflected from 2 mirrors, and recombined.

• path difference ∆r = 2(d2 – d1)

• recombined light shows an interferencepattern at the detector

• if one mirror is moved a distance λ/2then the path difference changes by λand exactly one fringe moves across thedetector

Michelson, 1907 Nobel prize:

1 meter = 1,553,163.5 wavelengthsof red Cadmium light!

This allows a precise distance measurementby counting fringes!

Double Slit Interference - Review

10


λλλλ

Phase

difference:

∆φ∆φ∆φ∆φ = k ∆∆∆∆r

= 2ππππ d sinθθθθ / λλλλ

d

y

0

∆∆∆∆r = d sinθθθθ

•

θθθθ

intensity on the screen at angle θθθθ:

Double Slit

(((( ))))λλλλθθθθππππ====θθθθ sincos4)( 2 dII o

)2(cos~)(~ 2221 φφφφ∆∆∆∆++++ EEI

I0 = intensity for

one slit alone.


In Fig. 36-29 , S1 and S2 are identical radiators of

waves that are in phase and of the same wavelength λλλλ. The radiators are separated by distance d = 3.00 λλλλ. Find the greatest distance from S1, along the x axis, for which fully destructive interference occurs. Express this distance in wavelengths.

Problem 20


To have destructive interference ∆∆∆∆D = (m + 0.5) λλλλ


λλλλ

++++====−−−−++++2122

mxxd λλλλ

++++====−−−−++++λλλλ⇒⇒⇒⇒219 22

mxx

xmmxx λλλλ

++++++++λλλλ

++++++++====++++λλλλ212

219 2

2222

xmm λλλλ

++++++++λλλλ

++++====λλλλ212

219 2

22 λλλλ

++++−−−−

++++

λλλλ====⇒⇒⇒⇒

21

21

212

9m

m

x

-0.464?0.55 λ

2

8.75λ

0

2.25 λx

31m


Problem 27

S1 and S2 in Fig. 36-29 are point sources of

electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power. (a) If a detector is moved to the right along the x axis from source S1, at what distances from S1 are the first three interferencemaxima detected? (b) Is the intensity of the nearest minimum exactly zero? (Hint: Does the intensity of a wave from a point source remain constant with an increase in distance from the source?)



mxxmxxd ====−−−−++++⇒⇒⇒⇒λλλλ====−−−−++++222 16

xmmmxx 21616 22 ++++====⇒⇒⇒⇒++++====++++

m

mx

216 2−−−−

====

01.16

3

7.5

1

3x(m)

42m

x

Path difference:


Problem 28The double horizontal arrow in Fig. 36-9 marks the

points on the intensity curve where the intensity of the central fringe is half the maximum intensity.

Show that the angular separation ∆θ∆θ∆θ∆θ between the

corresponding points on the viewing screen is:

∆θ∆θ∆θ∆θ = λλλλ / 2d if θθθθ in Fig. 36-8 is small enough so that

sin θθθθ = θθθθ.

11


Problem 28

2 1 0 1 2 m for maxima

2 1 0 0 1 2 m for minima

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5 ∆∆∆∆L/λλλλ


Problem 28

Fig. (36-8)



02

02

0 2sin

cos42

cos4 Id

Irk

II ====

λλλλ

θθθθππππ====

∆∆∆∆====′′′′

4sin ππππ

====λλλλ

θθθθππππ⇒⇒⇒⇒

d

21sin

cos2 ====

λλλλ

θθθθππππ⇒⇒⇒⇒

d

d4sin λλλλ

====θθθθ⇒⇒⇒⇒

d2λλλλ

====θθθθ∆∆∆∆⇒⇒⇒⇒


Problem 5Ocean waves moving at a speed of 4.0 m/s are approaching a beach at an angle of 30°°°° to the normal, as shown from above in Fig. 36-26. Suppose the water depth changes abruptly at a certain distance from the beach and the wave speed there drops to 3.0 m/s. Close to the beach, what is the angle θθθθ between the direction of wave motion and the normal? (Assume the same law of refraction as for light.)

Explain why most waves come in normal to a shore even though at large distances they approach at a variety of angles.



1

2

2

1

sinsin

θθθθ

θθθθ====

n

n

2

1

1

2

2

1

sinsin

θθθθ

θθθθ========

n

n

v

v

22

11 sinsin θθθθ====θθθθ

v

v

°°°°====θθθθ⇒⇒⇒⇒====°°°°==== 22375.030sin43


In Fig. 36-3 , assume that the two light waves, of

wavelength 620 nm in air, are initially out of phase

by ππππ rad. The indexes of refraction of the media are

n1 = 1.45 and n2 = 1.65. (a) What is the least

thickness L that will put the waves exactly in phase once they pass through the two media? (b) What is the next greater L that will do this?

Problem 10

Fig. 36-3 Two light rays travel

through two media having different indexes of refraction.

12


If N2 - N1 = integer, the two waves will be in phase and

will interfere constructively

The phase difference between two light waves can change if

the waves travel through different materials having different indexes of refraction.

Interference in Different Media

Number of wavelengths in medium 1:

λλλλ====

λλλλ====

LnLN

n

11

1

λλλλ====

λλλλ====

LnLN

n

22

2

Number of wavelengths in medium 2:

If N2 - N1 = half integer, the two waves will be out of

phase and will interfere destructively


The optical path difference ∆∆∆∆∆∆∆∆DD=|=|NN22--NN11| | λλλλλλλλ = (= (nn22--nn11))LLminmin

The phase difference ∆φ∆φ∆φ∆φ = k ∆∆∆∆D = 2ππππ ∆∆∆∆D/λλλλ

To be in phase when they come out: ∆φ∆φ∆φ∆φ=ππππ

The next larger L that do that is when ∆φ∆φ∆φ∆φ = 3ππππ


(((( ))))ππππ====

λλλλ

−−−−ππππ====

λλλλ

∆∆∆∆ππππ====φφφφ∆∆∆∆ min1222

LnnD

(((( ))))nm

nm

nnL 1550

2.02620

2 12min ====

××××====

−−−−

λλλλ====⇒⇒⇒⇒

(((( ))))nmL

nnL 46503

23

min12

========−−−−

λλλλ====

Three Slits Interference


P

θθθθ

r2

r1

r3

I(θθθθ) = ?

What if there are more than 2 slits?

Three Slits Interference

Let’s consider 3 slits and find I(θθθθ)!

•

2d

2d sin θθθθ

79

Phase difference due to path difference:

φφφφ = k ∆∆∆∆r = 2ππππ d sinθθθθ/ λλλλ between 1-2 and 2-3

Phase Difference

∴∴∴∴Phase difference between 1-3 is = 2φφφφ

)2sin(),sin(),sin( 321 φφφφ++++ωωωω====φφφφ++++ωωωω====ωωωω==== tEEtEEtEE ooo

Represent three waves originating from 1,2 and 3.

P

θθθθ

r2

r1

r3

•

2d

2d sin θθθθ

I(θθθθ) = ?

80

Resultant Wave

P

θθθθ

r2

r1

r3

I(θθθθ) = ?•

2d

2d sin θθθθ

We need to know the amplitude of the E field to calculate the intensity at P

)tsin(EEEE)(E ββββ++++ωωωω====++++++++====θθθθ θθθθ321

(((( )))) 2

21

θθθθµµµµ

====θθθθ⇒⇒⇒⇒ Ec

Io

13


(((( ))))φφφφ++++ωωωωθθθθ tE sin

tE ωωωωθθθθ sin

(((( ))))φφφφ++++ωωωωθθθθ 2sin tE

follow angles around a full circle starting here:

Eθθθθ

E0

E0

E0

Calculating I(θ)sin( ) sin( ) sin( 2 )sin( )o o o

E t E t EE ttθ ω β ω ω φ ω φ+ ++ += +

y

xtωωωω

φφφφ

φφφφ

ββββ

ββββ

(π−βπ−βπ−βπ−β)

)sin( ββββ++++ωωωωθθθθ tE

(((( )))) φφφφ====ββββ→→→→ππππ====ββββ−−−−ππππ++++φφφφ 222



φφφφ====φφφφ++++φφφφ

φφφφφφφφ++++φφφφ====

φφφφ++++φφφφ++++

φφφφ++++φφφφ====ββββ tan

cos2coscossin2sin

2coscos2sinsin

tan 2000

00

EEE

EE


Eθθθθ

E0

E0

E0

Calculating βy

xtωωωω

•

φφφφ

ββββ

ββββ

(((( ))))ββββ++++ωωωωθθθθ tE sin



But we showed that ββββ = φφφφ = 2 ππππ d sinθθθθ/λλλλ

E0 cosββββ E0

E0

E0

Eθθθθ

ββββ

ββββ

[[[[ ]]]] (((( ))))20

2

0

2

0

2

1cos21cos222

)( 0 ++++φφφφ====++++φφφφµµµµ

====µµµµ

====θθθθ θθθθ Ic

E

c

EI

θ

(((( ))))1cos2cos2 000 ++++ββββ====++++ββββ====θθθθ EEEE

(((( ))))1cos20 ++++φφφφ====θθθθ EE

Intensity:

Now find the amplitude Eθθθθ

Three Slits Interference (Intensity) Pattern


3 Slit Intensity Pattern

Zero when cosφφφφ = - ½ or φφφφ = (2ππππ/3, 4ππππ/3) + 2ππππ n

I(((( )))) (((( ))))2

0 1cos2 ++++φφφφ====θθθθ II

Max when cosφφφφ = 1φφφφ = 2ππππ n, or

d sin θθθθ = n λλλλ

λλλλθθθθππππ====φφφφ sin2 d

θθθθ or φφφφ

9I0

86

Phasor ApproachDivide the slit into N equal pieces and add the electric field

phasors for each piece, in the limit of large N (see previous

lectures)

( )( ) sin( ) sin( ) sin( ) ... sin( )2o

E P E t Nt t tδφδ ω ω δφωφω δ= + + + + + +

Light at point P, angle θθθθ to center of slit:

Eθθθθ

y

x

β

(As the number of stepsgets very large, the string of component phasors traces a circular arc from the origin to the tip of Eθθθθ .)

Next step: what is the value of Eθθθθ , the amplitudeof the resultant wave at P ?

String of phasors, each of length δδδδEoand rotated by angle δφδφδφδφ. Total rotation: N δφδφδφδφ = k a sinθθθθ

( ) sin( )E P E tθ ω β= +

14

Chapter 7: DiffractionChapter 7: DiffractionChapter 7: DiffractionChapter 7: Diffraction

Next Lecture

1





Chapter 7: Diffraction

http://ctaps.yu.edu.jo/physics/Courses/Phys207/Chapter7© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 2

λλλλ

θθθθ

Question: If light waves from 2 slits can interfere with each other, can light from different parts of the same slit interfere also?

Introduction

© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 3

Answer:

Diffraction by a Single Slit

This is an example of diffraction.

Every point in the slit acts like a source of

secondary Huygens’ wavelets, and the

phase difference between the wavelets

gives rise to an interference pattern.

Yes!

λλλλ

θθθθ


The interference pattern in this case is called diffraction pattern.

When light passes through a narrow slit the diffraction pattern is shown in fig. 37-1

Diffraction Pattern

This phenomenon is related to the wave nature of light.

If light behaves like being formed of straight rays the light resulting should be a sharp bright rendition of the slit on the screen.


1- Diffraction of light by a razor blade:

View on a screen behind the object shows interference fringes around the sharp edges.

Other Examples of Diffraction

6

2- Diffraction of light by an opaque diskopaque disk:View on a screen behind the object shows a surprising bright spot in the center of the image.

Other Examples of Diffraction

See last paragraph page 891.

2

Diffraction by a Diffraction by a Diffraction by a Diffraction by a Single Single Single Single SlitSlitSlitSlit


∆∆∆∆r = ∆∆∆∆y sin θθθθ

PD

a ∆∆∆∆y

Geometry: D >> a

Let’s consider 2 wavelets originating from two secondary sources distant by ∆∆∆∆y.

The path length difference between these 2 waves is:

Analysis of Single Slit Diffraction

•

•

θ

•

∆∆∆∆r = ∆∆∆∆y sin θθθθ

Phase difference between any two spherical wavelets at P: δφδφδφδφ = k ∆∆∆∆r = k ∆∆∆∆y sinθθθθ


•

PD

a

Geometry: D >> a

Techniques: In order to compute the total electric field at P we Add up electric field contributions from an infinite string of point sources across the width of the slit, keeping track of the relative phase at P.

Superposition Principle

•

••

•

•••

•


Each point source contribution to E(P):

with a different value of r, since the points are distributed across the slit of width a.

Contribution to E

(((( ))))trkE ωωωω−−−−δδδδ sin0

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))(((( ))))δφδφδφδφ++++ωωωω++++++++δφδφδφδφ++++ωωωω++++δφδφδφδφ++++ωωωω++++ωωωωδδδδ==== NttttEPE sin2sinsinsin0 K

θθθθ====δφδφδφδφ sinakNwith (full phase shift across the slit)


1- Phase diff. between r1 and r2 is:


Phase Differences

δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ

ππππ====φφφφ sin

212 y


ππππ====φφφφ 2sin2

213 y

y∆∆∆∆

y∆∆∆∆

y∆∆∆∆


Phase Differences

…………N - Phase diff. between r1 and rN+1 is:

Note that N∆∆∆∆y=a where a is the slit width, N δφδφδφδφ = φφφφ





ππππ====φφφφ sin

212 y



213 y



214 y


ππππ====φφφφ ++++ NyNN sin

211

3


1.The incident electric field has an amplitude of Em

Contribution to E

2.At the slit each point source has amplitude δδδδEm.

3. Em = N δδδδEm

The wave from the mth source reaches point P on the screen with electric field :

(((( )))) (((( ))))δφδφδφδφ++++ωωωω−−−−δδδδ====δφδφδφδφ++++ωωωω−−−−δδδδ mtrkEmtrkE mm sinsin

Em

δδδδEm


P

θθθθ

The total electric field at point P is:

Total Electric Field at P

(((( ))))

(((( )))) (((( ))))(((( )))) (((( ))))(((( ))))

(((( ))))

δφδφδφδφ++++ωωωω−−−−++++

δφδφδφδφ++++ωωωω−−−−

++++δφδφδφδφ++++ωωωω−−−−++++δφδφδφδφ++++ωωωω−−−−

++++δφδφδφδφ++++ωωωω−−−−++++ωωωω−−−−

δδδδ====θθθθ

Ntrk

trk

trktrk

trktrk

EE m

sin

4sin

3sin2sin

sinsin

KK

KK


Fig. below shows the resultant Eθθθθ = E(P)

using phasors.

Resultant E(P)

δφδφδφδφ====φφφφ Nββββ

16

Resultant E(P)


2φφφφ2φφφφ

17

Resultant E(P)

We complete the isosceles triangle OAB. We start from point D by tracing a line perpendicular to OB.


2φφφφ2φφφφ

A

O

B

D


amplitude of E field at P:

The electric field phasors add up to form a circular arc of radius R

Arc length Em = N δδδδEm

Analysis

φφφφ====⇒⇒⇒⇒====δφδφδφδφ====φφφφ mm ERREN

(((( ))))2sin2 φφφφ====θθθθ RE

total rotation of all N phasors: φφφφ = N δφδφδφδφ

4


Analysis

Note that Em = N δδδδE0 is the amplitude of the total E field inside the slit itself, i.e. it is what we started with!

φφφφ==== mER

(((( ))))2sin2 φφφφ====θθθθ RE

(((( ))))2

2sin

φφφφ

φφφφ====⇒⇒⇒⇒ θθθθ mEE

αααα

αααα====θθθθ

sinmEE


The intensity at point P is:

The electric field at point P is:

Intensity

(((( )))) (((( )))) (((( ))))trkEtrkEE mm ωωωω−−−−αααα

αααα====ωωωω−−−−

φφφφ

φφφφ

====θθθθ sinsin

sin

2

2sin

(((( ))))2

2

02

2

02

2

0sin

sinsin

2

2sin

sin

λλλλ

θθθθππππ

λλλλ

θθθθππππ

====

φφφφ

φφφφ

ΙΙΙΙ====αααα


a

a

III


Note that the intensity is maximum when :

And when θθθθ = 0 too!

Maximum & Minimum Intensity

KK3,2,1,0,2

1sin

2

1sin====λλλλ

++++====θθθθ⇒⇒⇒⇒ππππ

++++====λλλλ

θθθθππππmmam

a

(((( ))))2

2

02

2

02

2

0sin

sinsin

2

2sin

sin

λλλλ

θθθθππππ

λλλλ

θθθθππππ

====

φφφφ

φφφφ



a

a

III

Note that the intensity is minimum when :

KK3,2,1,sinsin

====λλλλ====θθθθ⇒⇒⇒⇒ππππ====λλλλ

θθθθππππmmam

a


Intensity

====λλλλ++++

====λλλλ====θθθθ

3210)21(

4321sin

,,,mmaximumm

,,,mminimumma

(((( ))))2

2

02

2

02

2

0sin

sinsin

2

2sin

sin

λλλλ

θθθθππππ

λλλλ

θθθθππππ

====

φφφφ

φφφφ



a

a

III


“Straight through” light travels to the screen

with intensity

(same intensity as the incoming wave at the center of the screen.)

Let αααα ≡≡≡≡ φφφφ/2 where φφφφ = k a sinθθθθ = 2ππππ a sinθθθθ / λλλλ is the

full phase shift across the slit

c

EI

o

m

µµµµ====

2

2

0

Diffraction Pattern

2

0

222sinsin

22)(

αααα

αααα≡≡≡≡

αααα

αααα

µµµµ====

µµµµ====θθθθ θθθθ I

c

E

c

EI

o

m

o


0II central maximum

Very bright central max: width ~ 1/a

Diffraction Pattern2

0

sin)(

αααα

αααα====θθθθ II

λλλλθθθθππππ====αααα sina

1sin

lim0

====

αααα

αααα

→→→→αααα

ππππθθθθ

K,3,2,0sin ππππππππππππ====αααα⇒⇒⇒⇒====αααα

λλλλ====αααα⇒⇒⇒⇒ ma sin

minima:K,23,21sin ππππππππ====αααα⇒⇒⇒⇒±±±±====αααα

λλλλ

++++====αααα⇒⇒⇒⇒2

1sin ma

other maxima (approx.):

5


Case a << λλλλNote that ∆θ∆θ∆θ∆θ is very large nearly 2ππππ if a<<λλλλ

So that the central maximum covers the entire screen.

∆θ∆θ∆θ∆θ

Single Slit Diffraction - Review


0II central maximum

Very bright central max: width ~ 1/a

Diffraction Pattern2

0

sin)(

αααα

αααα====θθθθ II


1sin

lim0

====

αααα

αααα

→→→→αααα

ππππθθθθ

K,3,2,0sin ππππππππππππ====αααα⇒⇒⇒⇒====αααα

λλλλ====αααα⇒⇒⇒⇒ ma sin

minima:K,23,21sin ππππππππ====αααα⇒⇒⇒⇒±±±±====αααα

λλλλ

++++====αααα⇒⇒⇒⇒2

1sin ma

other maxima (approx.): Diffraction Pattern by Diffraction Pattern by Diffraction Pattern by Diffraction Pattern by

Double Narrow SlitDouble Narrow SlitDouble Narrow SlitDouble Narrow Slit

34

The two slits will give an interference pattern like this one:

Introduction

where I0 is the intensity with only one slit open, and ββββ = ½ ∆φ∆φ∆φ∆φ = ½ k ∆∆∆∆r = ππππ d sinθθθθ/λλλλ

Note the condition for interference is a<<λλλλ, a is the width of the

slit

ββββ====

φφφφ∆∆∆∆

µµµµ====

µµµµ====θθθθ θθθθ 2

02

0

20

2

cos42

cos2

4

2)( I

c

E

c

EI

o

35

With the condition that the slit width a << λλλλ, each slit will give us a diffraction pattern like the following one:

a << λλλλ

Note that the central maximum covers a large part of the screen!

22

0

22sinsin

22)(

αααα

αααα≡≡≡≡

αααα

αααα

µµµµ====

µµµµ====θθθθ θθθθ

om

o

Ic

E

c

EI


6


θθθθλλλλ

ππππ====ββββ sin

d

What we will see on the screen is both intensities combined together. (Fig. )

Combined intensities

22 sin

cos

αααα

ααααββββ==== mII

θθθθλλλλ

ππππ====αααα sin

a


The single slit pattern forms a kind of envelope for the double-slit fringes.

Diffraction Envelope2

2 sincos

αααα

ααααββββ==== mII

Diffraction envelope


In a double-slit experiment, the wavelength λλλλ of the light source is 405 nm, the slit separation d is 19.44 µµµµm

and the slit width a is 4.05 µµµµm.

Example 1– Sample Problem 37-4

λλλλ====θθθθsina

Consider the interference of the light from the two slits and also the diffraction of the light through each slit.

Solution:

The limits of the central peak are the first minima in the diffraction pattern due to either slit, individually.

These first minima are defined by:

How many bright fringes are within the central peak of the diffraction pattern


The number of bright fringes in the envelope of the central peak = m2 (2 refers to double-slit)

Example 1– Sample Problem 37-4

8.42 ========⇒⇒⇒⇒a

dmλλλλ====θθθθ 2sin md

This means that there are 4 bright fringes in the envelope of the central peak.

a

λλλλ====θθθθsin


Diffraction GratingDiffraction GratingDiffraction GratingDiffraction Grating��ز ا�� د

7


This device is somewhat like the double-slitarrangement of Fig. 36-8 but has a much greater number N of slits, often called rulings, perhaps as many as several thousands per millimeter.

Diffraction Grating


λλλλ====θθθθ md sin


Intensity

The bright fringes seen on the screen are called lines because the width of each one is very small.


∆θ∆θ∆θ∆θhw is the angle between the central maximum and the first minimum. Thus ∆θ∆θ∆θ∆θhw is the width of the first minimum

Width of the Lines

∆θ∆θ∆θ∆θhw is called half width the line.


maximumcentralthefordN

dN hwhwλλλλ

====θθθθ⇒⇒⇒⇒λλλλ====θθθθsin

maximummthefordN

thhw

θθθθ

λλλλ====θθθθ

cos

We know from the diffraction of single slit that the first maximum occurs when the path difference between the top ray and the bottom ray is mλλλλ

Circular Aperture Circular Aperture Circular Aperture Circular Aperture DiffractionDiffractionDiffractionDiffraction

8


Circular Aperture Diffraction


Circular Aperture Diffraction

Resolvability Resolvability Resolvability Resolvability –––– The The The The Rayleigh CriterionRayleigh CriterionRayleigh CriterionRayleigh Criterion


Diffraction is used to measure unknown wavelengths.

Resolvability

Is many cases, such as atomic spectra, we have to distinguish very close wavelengths. The diffraction “obstacle” used is chosen so that one is able to “see” these λλλλ and λλλλ’ distinctly.

The ability of the diffracting device is called “resolvability”


The Rayleigh criterion is the generally accepted criterion for the minimum resolvable detail.

Rayleigh Criterion

The imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.

Resolved

UnresolvedRayleigh

Criterion

Fresnel and Fresnel and Fresnel and Fresnel and Fraunhofer DiffractionFraunhofer DiffractionFraunhofer DiffractionFraunhofer Diffraction

9


The Fraunhofer diffraction deals with the limiting cases where the light approaching the diffracting object is parallel and monochromatic, and where the image plane is at a distance large compared to the size of the diffracting object. The more general case where these restrictions are relaxed is called Fresnel diffraction

Fresnel and Fraunhofer

Fresnel diffraction refers to the general case. This makes it much more complex mathematically. Some cases can be treated in a reasonable empirical and graphical manner to explain some observed phenomena.

��א�� א��

Last Lecture

Physics Department, Yarmouk University, Irbid Jordan


Dr. Nidal M. Ershaidat Doc. 1A

Author’s email: [email protected],

Address: Physics Department, Yarmouk University 21163 Irbid Jordan

© Nidal M. Ershaidat 2012

1

Mass on Spring Resonance

A mass on a spring has a single resonant frequency determined by

its spring constant k and the mass m. Using Hooke's law and

neglecting damping and the mass of the spring, Newton's second

law gives the equation of motion:

2

2

dt

xdmamxkgm ========−−−−

The solution to this differential equation is of the form:

(((( )))) (((( )))) BtAtx ++++φφφφ++++ωωωω==== cos

which when substituted into the motion equation gives:

(((( ))))(((( )))) (((( ))))(((( ))))BtAmBtAkgm ++++φφφφ++++ωωωωωωωω−−−−====++++φφφφ++++ωωωω−−−− coscos2

Collecting terms gives B=mg/k, which is just the stretch of the

spring by the weight, and the expression for the resonant

vibrational frequency:

m

k====ωωωω

2

This kind of motion is called simple harmonic motion and the

system a simple harmonic oscillator.

© Nidal M. Ershaidat 2012 2

Energy in Mass on Spring

The simple harmonic motion of a mass on a spring is an example

of an energy transformation between potential energy and kinetic

energy. In the example below, it is assumed that 2 joules of work

has been done to set the mass in motion.

Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html (Accessed June 2007)






1

A. Springs - Two Springs and a Mass

Consider a mass m with a spring on either end, each attached to a wall.

Let k1 and k2 be the spring constants of the springs. A displacement of the

mass by a distance x results in the first spring lengthening by a distance x

(and pulling in the x− direction), while the second spring is compressed by

a distance x (and pushes in the same x− direction). The equation of

motion then becomes

( ) xkkxkxkxm 2121 +−=−−=••

(A-1)

or

021 =

++

••x

m

kkxm (A-2)

Equation (A-2) can be written in the form:

0=+••

xm

kxm

eff (A-3)

21 kkkeff += is the effective spring constant of the system, and the angular

oscillation frequency ω is

m

kk 21 +=ω (A-4)

B. Springs - Two Springs in Parallel

The force exerted by two springs attached in parallel to a wall on a mass m

is given by: xkxkxkF eff−−−−====−−−−−−−−==== 21 (B-1)

2

21 kkkeff += is the effective spring constant of the system. The equation of

motion of the system is thus:

0=+••

xm

kxm

eff (B-2)

and the angular oscillation frequency ω is

m

kk 21 +=ω (B-3)

C. Springs - Two Springs in Series

Consider two springs placed in series with a mass m on the bottom of the

second. The force is the same on each of the two springs. Therefore

2211 xkxkF −=−= (C-1)

Solving for x1 in terms of x2, we have:

2

1

21 x

k

kx =

(C-2)

The force exerted on the mass can also be written as: ( )21 xxkF eff +−=

Where effk is the effective spring constant of the system. The total

displacement of the mass is 21 xxx += .

Equating equation (C-3) and (C-1) we have:

+= 22

1

222 xx

k

kkxk eff

(C-3)

Dividing both side of the previous equation by x2 we get:

+= 1

1

22

k

kkk eff

(C-4)

or

1

21

11−

+=

kkkeff

(C-5)

and the angular oscillation frequency ω is

( )mkk

kk

21

21

+=ω (C-6)

Reference: http://scienceworld.wolfram.com/physics/SpringsTwoSpringsandaMass.html

© 1996-2007 Eric W. Weisstein


Waves and Light103 . Phys

Dr. Nidal M. Ershaidat C1. Doc

Author’s email: [email protected], Address: Physics Department, Yarmouk University 21163 Irbid Jordan © Nidal M. Ershaidat 2012

1

Oscillations - Problems

Problem 16-24

In the Fig. below, two identical springs of spring constant k are attached to a block of

mass m and to fixed supports. Show that the block's frequency of oscillation on the

frictionless surface is :

m

kf

2

2

1

ππππ====

(1)

Solution:

When the mass m is pulled to the right it suffers two forces:

,ixkFr −−−−====→→→→

ixkFlˆ−−−−====

→→→→

(2)

r and l refer respectively to the spring on the right and the spring on the left. The total

external force is:

ixkFextˆ2−−−−====

→→→→

(3)

Newton’s 2nd law for the mass m (The motion is one dimensional):

02

2 ====++++⇒⇒⇒⇒−−−−==== xm

kaxkam (4)

Equation 4 can be rewritten as:

02

====++++ xm

kx�� (5)

This is the equation of motion of a simple harmonic motion. The angular frequency ω

and frequency f of this motion are given by:

m

kf

m

k 2

2

122

ππππ====⇒⇒⇒⇒====ωωωω


Problem 16-25

Suppose that the two springs in the figure below have different spring constants k1

and k2. Show that the frequency f of oscillation of the block is then given by:

22

21 fff ++++==== (1)

where f1 and f2 are the frequencies at which the block would oscillate if connected

only to spring 1 or only to spring 2.

Solution:


,ˆ11 ixkF −−−−====

→→→→

ixkF ˆ22 −−−−====

→→→→

(2)


external force is:


21 ++++−−−−====→→→→

(3)


(((( ))))(((( ))))

02121 ====

++++++++⇒⇒⇒⇒++++−−−−==== x

m

kkaxkkam (4)


02

====ωωωω++++ xx�� (5)



(((( ))))(((( )))) (((( ))))2

22

122

121

212122 ff

m

k

m

k

m

kk++++ππππ====ωωωω++++ωωωω====++++====

++++====ωωωω (5)

22

21

2fff ++++====

ππππ

ωωωω====∴∴∴∴


Problem 16-27

In the figure below, two springs are joined and connected to a block of mass m. The

surface is frictionless. If both of the springs have spring constant k find ω.

Solution:


,ˆ11 ixkF −−−−====

→→→→

ixkF ˆ22 −−−−====

→→→→

(1


external force is:


21 ++++−−−−====→→→→

(2


(((( ))))21 xxkam ++++−−−−==== (3)


02

====ωωωω++++ xx�� (4)



(((( ))))(((( )))) (((( ))))2

22

122

121

212122 ff

m

k

m

k

m

kk++++ππππ====ωωωω++++ωωωω====++++====

++++====ωωωω (5)

22

21

2fff ++++====

ππππ

ωωωω====∴∴∴∴


Waves and Light103 . Phys




© Nidal M. Ershaidat 2012

1

Waves تعريف الموجة )أ

. للتعبير عن انتقال الطاقة" كائن رياضي"يستخدم مفهوم الموجة وهي

تنقل الطاقة خالل هذا مادي في وسط أو إزعاج " خلخلة"هي ،كموجة الصوت، الموجة الميكانيكي)ةمن مواقعها " مؤقت"في الموجة تنزاح جزيئات الوسط بشكل . المادة بشكل دائم" نقل"الوسط دون

.ثم) تعود إلى وضعها األصلي ميكانيكي لوسط مادي، تعبLر عن كيفية "إزعاج"الناتجة عن الموجة الميكانيكي)ة وبعبارة أخرى نقول أن)

.انتشار الطاقة الميكانيكي)ة في هذا الوسط .تنتشر في الفراغ وسط ناقل بل إلىال تحتاج : electromagnetic wave الموجة الكهرومغناطيسية

ما يحدث هنا هو انتقال االهتزازات الكهرومغناطيسي)ة الناتجة عن مجال كهربائي متذبذب ومجال .متذبذب بشكل عمودي على كال المجالين) متعامد مع المجال الكهربائي(مغناطيسي

األمواجتصنيف )ب :من حيث الوسط الناقل1..I الموجة الميكانيكية mechanical wave : الماءوأمواج الصوت كأمواج وسط ناقل إلىوتحتاج . .II وسط ناقل بل تنتشر في الفراغإلىال تحتاج التي الموجة الكهرومغناطيسية . . غير دوريةأودورية : من حيث اعتمادها على الزمكان وانتشارها2. :الوسط الناقل" إزعاج"من حيث 3..I لموجة المستعرضة اtransverse wave : جزيئات الوسط عن وضع االتزان إزاحةوفيها تكون

. مثل الموجة الناشئة عن اهتزاز سلك-عمودية على اتجاه انتشار الموجة.II الموجة الطوليةlongitudinal wave : جزيئات الوسط عن وضع االتزان في إزاحةوفيها تكون

. الصوتأمواج مثل -اتجاه انتشار الموجة : االنتشارأثناء wavefrontث شكل مقدمة الموجة من حي 4..I الموجة المستويةplane wave :مقدمة الموجة سطح مستو. .II الموجة الكرويةspherical wave :مقدمة الموجة سطح كروي. : االنتشارأثناء wavefrontمن حيث شكل مقدمة الموجة 5. .I الموجة المستويةplane wave : الموجة سطح مستومقدمة. .II الموجة الكرويةspherical wave :مقدمة الموجة سطح كروي.

© Nidal M. Ershaidat 2012 – Phys. 103 Supplements Doc 2

2

Harmonic plane waves التوافقيةية المستواألمواج 3.

هي الفئة التي يمكن وصف حركة المصدر المنتج لها بحركة توافقية األمواج من األهملعل) الفئة تنتشر الموجة في هذه الحالة على شكل دالة . simple harmonic motion) ب.ت.ح(بسيطة

: مما يليأهميتها وتأتيجيبية بداللة المكان والزمن البسيطة عن اإلزاحاتلتي تحوي تغيرات دورية منتظمة يمكن، وفي حالة افي الظواهر الفيزيائية -

.وضع االتزان، تقريب هذه التغيرات بحركة توافقية بسيطة- kدالة جيب التمامأو الرياضي للحركة التوافقية البسيطة هو الدالة الجيبية الحل . ن)ها أ كان نوعها وكأي)ا حركة توافقية أي)ةبفضل تحليل فورييه يمكن اعتبار ) األهموهذه هي النقطة ( -

. لحركات توافقية بسيطةseries سلسالت أو) linear superposition(تراكب خطLي

Superposition Principle التراكبمبدأ 4.

بشكل مستقLل الواحدة ) الميكانيكية في الوسط الناقل والكهرومغناطيسية في الفراغ (األمواجتنتشر �� ا�� وا�� اورآ ��ا ��، واألخرىعن �� ا�� حدث إذا. �"!ا ا�

، لجزيئات الوسط في disturbance اإلزعاجتتداخل وتكون محصLلة ا فإن)ه عائقا األمواجوصادفت هذه اإلزعاجات الكهرومغناطيسية، مساوية لمجموع األمواج الميكانيكية وفي الفراغ في حالة األمواجحالة

. كلr على حدةاألمواجالتي سببتها بمت)جهات فان) عملية التراكب األمواجل وعند تمثيتأثيرات عملية جمع إال التراكب ليس مبدأنذكLر بان) . أكثر تصبح عملية جمع مت)جهات ال أكثر أولموجتين

نعيش يوميا وبأننا المحصLلة ال تتعدى حد) المرونة أن) اإلزعاجات التراكب صالح طالما مبدأ بان) أيضانذكLر .ومية الضوء والصوت في حياتنا اليأمواجوفي كلL لحظة تراكب

© Nidal M. Ershaidat 2012 – Phys. 103 Supplements Doc 2

3

Wave Function دالة الموجة 5.

Ψ(x,t): على الصورةt والزمن x المكان إحداثيات بدالة تعتمد على v مثال، بسرعة مقدارها xتمث"ل سعة موجة كالسيكية تنتشر في بعد واحد، في اتجاه المحور .لكالسيكية الألمواج Ψ(x,t)يلخ>ص الجدول التالي بعض خصائص ". دالة الموجة"تدعى سرعة انتشار الموجة : تمث>لΨ(x,t)الدالة الموجة

µµµµ====

Fv

. المستعرضة عن وضع اتزاناإلزاحة المستعرضة الميكانيكي"ة موجة تنتشر في سلك مشدود: مثال

=Fقوة الشد =µكتلة وحدة الطول للسلك المشدود

ρρρρ====

Bv

. الطولية عن وضع اتزانحةاإلزا الطولية الميكانيكي"ة هي سعة تخلخل الوسطΨ الصوت التي تنتشر في وسط مادي وتكون أمواج: مثال

=B للوسطالمرونة الحجميمعامل =ρكثافة الوسط

المجال المغناطيسيأوتذبذب المجال الكهربائي الكهرومغناطيسية v = c

الكالسيكيةلألمواج Ψخصائص دالة الموجة



Dr. Nidal M. Ershaidat Doc. 2B

Author’s email: [email protected], Address: Physics Department, Yarmouk University 21163 Irbid Jordan

1

Superposition of Waves

Basic Concepts

The principle of superposition may be applied to waves whenever two (or

more) waves travel through the same medium at the same time. The

waves pass through each other without being disturbed. The net

displacement of the medium at any point in space or time, is simply the

sum of the individual wave displacements. This is true of waves which are

finite in length (wave pulses) or which are continuous sine waves.

1. Two sine waves traveling in the same direction: Constructive

and Destructive Interference

Let’s consider two waves (with the same amplitude, frequency, and

wavelength) (((( )))) (((( ))))txkytxy m ωωωω−−−−==== sin,1 and (((( )))) (((( ))))φφφφ++++ωωωω−−−−==== txkytxy m sin,2

traveling in the same direction on a string. Using the principle of

superposition, the resulting string displacement may be written as:

(((( )))) (((( )))) (((( ))))

++++−−−−

====

++++−−−−++++−−−−====

222

φω

φ

φωω

txksincosy

txksinytxksinyt,xy

m

mm

(1)

which is a traveling wave whose amplitude depends on the phase (φ).

When the two waves are in-phase (φ=0°), they interfere constructively and

the resultant wave has twice the amplitude of the individual waves. When

the two waves have opposite-phase (φ=180°), they interfere destructively

and cancel each other out.

2. Two sine waves traveling in opposite directions create a

standing wave

A traveling wave moves from one place to another, whereas a standing

wave appears to stand still, vibrating in place. Consider that two waves

2

(with the same amplitude, frequency, and wavelength) are traveling in

opposite directions on a string. Using the principle of superposition, the

resulting string displacement may be written as:

(((( )))) (((( )))) (((( ))))

(((( )))) (((( ))))txky

txkytxkytxy

m

mm

ωωωω====

ωωωω++++++++ωωωω−−−−====

cossin2

sinsin,

(2)

This wave is no longer a traveling wave because the position and time

dependence has been separated. The displacement of the string as a

function of position has an amplitude of 2ymsin kx. This amplitude does not

travel along the string, but stands still and oscillates up and down

according to cos ωt. Characteristic of standing waves are locations with

maximum displacement (antinodes) and locations with zero displacement

(nodes).

3. Two sine waves with different frequencies: Beats

Two waves of equal amplitude are traveling in the same direction. The two

waves have different frequencies and wavelengths, but they both travel

with the same wave speed. Using the principle of superposition, the

resulting particle displacement may be written as:

(((( )))) (((( )))) (((( ))))

(((( )))) (((( )))) (((( )))) (((( ))))

ωωωω++++ωωωω−−−−

++++

ωωωω−−−−ωωωω−−−−

−−−−====

ωωωω++++++++ωωωω−−−−====

txkk

txkk

y

txkytxkytxy

m

mm

22sin

22cos2

sinsin,

21212121

2211

(3)

This resulting particle motion is the product of two traveling waves. One

part is a sine wave which oscillates with the average frequency f = ½(f1 +

f2). This is the frequency which is perceived by a listener. The other part is

a cosine wave which oscillates with the difference frequency f = ½(f1 - f2).

This term controls the amplitude "envelope" of the wave and causes the

perception of "beats". The beat frequency is actually twice the difference

frequency, fbeat = (f1 - f2).

Reference: http://www.kettering.edu/~drussell/Demos/superposition/superposition.htm

*This web page contains animations of the four situations.

Chapter 1:Oscillations - CTAPS · ©Dr. N. Ershaidat Phys. 207 Chapter 1: Oscillations Lecture 1 3 Oscillations are motions that repeat themselves. ... /Phys201/Chap6.htm Harmonic

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