Chapter 1:Oscillations - CTAPS · ©Dr. N. Ershaidat Phys. 207 Chapter 1: Oscillations Lecture 1 3...
Transcript of Chapter 1:Oscillations - CTAPS · ©Dr. N. Ershaidat Phys. 207 Chapter 1: Oscillations Lecture 1 3...
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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
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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
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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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 5
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
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© 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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations8
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.
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© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations9
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations10
A periodic motion could be continuous
Periodic Motion – Continuity
Examples
or discontinuous.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations11
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations12
The sine function:
is periodic of period 2ππππ
Continuous Periodic Function
(((( )))) (((( ))))xxf sin====
Fig. 1: The sine function
3
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations13
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations14
For more details see:
Further Reading
http://ctaps.yu.edu.jo/physics/Courses/Phys201/Chap6.htm
Harmonic Motion
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 16
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 17
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 19
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 20
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 22
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 23
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 27
Displacement at time t
(((( ))))}
{(((( ))))
48476φφφφ++++ωωωω==== txtx m cos
AmplitudeAngular
Frequency
Time
Phase
Phase angle*
* Phase constant
5
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 28
x(t) = xm cos (ωωωω t + φφφφ)
x(
t)
Fig. 6: Displacement vs. t
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 29
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 30
Examples of Phase Difference
Fig. 7: x/xm vs. ωωωωt
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© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 31
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 32
x, v and a
txx m ωωωω==== cos
tvv m ωωωω−−−−==== sin
taa m ωωωω==== cos
Fig. 8: x, v and a
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 33
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 34
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 35
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 37
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 38
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 39
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 40
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 43
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 45
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 46
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?
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 47
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:
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 48
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 49
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.
Sample Problem (16-2)
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 50
Sample Problem (16-2)
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
,
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 51
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 54
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 55
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 56
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 58
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 59
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 60
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 62
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.
Sample Problem (16-4)
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 63
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 ====⇒⇒⇒⇒
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 64
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−−−−====&&
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations65
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 67
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 69
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 71
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 73
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 74
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 75
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 77
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 78
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 79
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 ====++++
====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 80
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 82
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 83
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 84
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 85
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 87
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 88
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 89
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 91
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 92
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 93
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 94
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 95
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 −−−−====
→→→→
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 96
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 97
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 98
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≈≈≈≈⇒⇒⇒⇒
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 99
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 100
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 101
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 103
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⇒⇒⇒⇒
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 104
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 ?
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 105
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?
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 106
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?
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 107
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 1: Oscillations 109
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
Phys. 103: Waves and Light
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 103 Waves and Light
© Dr. Nidal M. Ershaidat
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 7
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 8
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 9
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 11
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).
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 12
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 13
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 14
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 16
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 17
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 18
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 20
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 22
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 23
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 24
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 25
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 26
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 28
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 29
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 30
Forms of the Wavefunction
ωωωω
−−−−==== tk
xkym sin
−−−−
λλλλππππ====
T
txym 2sin
(((( ))))(((( ))))tvxkym −−−−==== sin
(((( ))))txkyy m ωωωω−−−−==== sin
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 31
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 33
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 34
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 35
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’
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 36
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′
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 37
� 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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 38
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 40
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 44
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 45
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 46
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 48
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 49
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 51
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 52
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 53
(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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 55
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 57
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 59
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 60
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 61
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 62
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 64
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 66
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 67
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•
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 68
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 70
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 71
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 72
)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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 74
Case φφφφ = 0
(((( ))))txkyy m ωωωω−−−−==== sin1
(((( ))))txkyy m ωωωω−−−−==== sin2
a)
d)
)sin(2 txkyy m ωωωω−−−−====′′′′
)sin(cos222
φφφφφφφφ++++ωωωω−−−−====′′′′ txkyy m
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
φφφφφφφφ++++ωωωω−−−−====′′′′ txkyy m
ππππ====φφφφ
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 76
Case φφφφ = 2ππππ/3
f)
(((( ))))txkyy m ωωωω−−−−==== sin1
(((( ))))ππππ++++ωωωω−−−−====3
22 sin txkyy m
d)
)sin(cos233ππππππππ ++++ωωωω−−−−====′′′′ txkyy m
Superposition applet
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 77
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?
Example 2 - Problem 28
13
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 78
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
Problem 28 – Solution
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 80
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 82
Figure c shows the resultant y’=y1+y2
y’ = y1 + y2
)sin(cos222
φφφφφφφφ++++ωωωω−−−−====′′′′ txkyy m
c)Fig. 11-c
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 83
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 84
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 ++++====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 85
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 86
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.
Example 3 - Problem 31
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 87
- Using trigonometry
Let
Problem 31 – Solution
(((( ))))
(((( ))))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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 88
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 ≈≈≈≈++++++++++++====°°°°××××++++++++++++====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 89
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 90
“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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 92
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 93
Standing Wave (Definition)1
(((( )))) (((( ))))txkyytxkyy mm ωωωω−−−−====ωωωω++++==== sin,sin 21
Fig. 12
Let the two waves be represent by:
xktyyyy m sincos221 ωωωω====++++====′′′′
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 94
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 95
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 96
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)
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 97
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.
Example 4 - Problem 46
(((( )))) (((( )))) txmy ππππππππ==== 12sin2sin1.0
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 98
(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
Problem 46 - Solution
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 102
Problems
18, 24, 25, 31, 39 46 and 50
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 103
See Example 1
Problem 18
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 104
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)
Problem 24 – Solution
a)
y
gLt 2====⇒⇒⇒⇒
ygdt
dyv
dt
dy====⇒⇒⇒⇒====
tLtL
tgytdgy
dy
0000
2 ====⇒⇒⇒⇒==== ∫∫∫∫∫∫∫∫
µµµµ
ττττ====v
gyv ====⇒⇒⇒⇒
dtgy
dy====⇒⇒⇒⇒
(((( ))))µµµµ
µµµµ====
gyyv
mass
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 106
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 107
(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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 108
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 109
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 110
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 111
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≈≈≈≈
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 112
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 113
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 114
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 115
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 116
Displacement of the string particle x at any
instant t is thus given by:
Problem 50 – Solution 2
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 117
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?
Problem 50 – Solution 3
(((( )))) (((( )))) (((( ))))
(((( )))) (((( )))) (((( )))) (((( ))))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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 2: Waves I 118
(e) Sketch the standing wave at t = 0.50 s for the range x = 0. to x = 0.40 m.
Problem 50 – Solution 4
(((( )))) (((( )))) (((( ))))cmxsins.t,xy 10450 ππππ========
Chapter 3: Waves IIChapter 3: Waves IIChapter 3: Waves IIChapter 3: Waves II
Next Lecture
1
Phys. 103: Waves and Light
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 103 Waves and Light
© Dr. Nidal M. Ershaidat
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 5
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 7
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 8
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
∆∆∆∆
∆∆∆∆−−−−====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 10
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 13
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 14
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 15
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 16
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 19
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====∆∆∆∆
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 21
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 23
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 24
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.
Fully constructive interference
5
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 25
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
this occurs when the ratio ∆∆∆∆L/λλλλ verifies:
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 27
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 28
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 30
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 31
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 32
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 33
Hearing threshold
Rustle of leaves
Conversation
Rock concert
Pain threshold
Jet engine
10
Some Sound Levels (dB)
0
60
110
120
130
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 34
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 35
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
Example 1 - Problem 28
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
====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 36
(b) the power intercepted by the microphone.
Example 1 - Problem 28
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 39
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 42
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:
Example 2 - Problem 38
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 45
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 46
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 47
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====′′′′
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 48
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 49
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
Example 3 - Problem 43
212 ωωωω−−−−ωωωω====ωωωω′′′′====ωωωωbeat
3871 ====f
3384121 ====−−−−====−−−−==== ffffbeat
3-8 Doppler Effect
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 51
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 55
Stationary Source Moving Detector
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 56
Stationary Source Moving Detector
The frequency f ’ detected by D is the rate at which D intercepts wavefronts. Obviously f ’ > f.
10
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 57
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:
Moving Source Stationary Detector
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 59
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
Stationary Source Moving Detector
fv
vvf D−−−−
====′′′′
fv
vvf D±±±±
====′′′′Detector moves towards to the
sourceDetector moves away from the
source
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 60
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.
Moving Source Stationary Detector
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:
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 61
Moving Source Stationary Detector
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 62
Moving Source Stationary Detector
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 63
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 65
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 66
Problem 6 - Solution
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 70
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 ====
∆∆∆∆====⇒⇒⇒⇒
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 71
Problems 28, 38 and 43
See Examples 1, 2 and 3 respectively.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 72
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?
Problem 51 – Solution
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?
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 3: Waves II 74
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
Phys. 103: Waves and Light
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 103 Waves and Light
© Dr. Nidal M. Ershaidat
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 7
Electric and Magnetic Energy Oscillations
(((( ))))φφφφ++++ωωωω======== tC
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 8
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 10
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 12
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 13
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 14
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 15
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−−−−====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 16
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)
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 17
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 ++++====++++====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 18
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 4: Electromagnetic Oscillations and Alternating Current 19
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
Phys. 103: Waves and Light
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 103 Waves and Light
© Dr. Nidal M. Ershaidat
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
Maxwell’s Equations
0εεεε====∫∫∫∫
qAd.Evr
0====∫∫∫∫ Ad.Brr
dt
dld.E BΦΦΦΦ
−−−−====∫∫∫∫rr
====∫∫∫∫ ld.Brr
dt
di E����� 000
Ampere-Maxwell Law
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 4
James Clerk Maxwell showed that electromagnetic energy propagates in vacuum with the speed of light (c).
Maxwell’s Equations
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 6
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 7
Range of electromagnetic waves
1 nm = 10-9 m
1 Å = 10-10 m
Fig. 1
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 8
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 9
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 11
� Any accelerating charge will emit
electromagnetic waves.
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 12
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 13
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 14
• 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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 15
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
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 18
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 19
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:
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 20
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 21
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 23
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 24
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
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 25
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:
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 26
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 27
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 28
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−−−−====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 29
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 31
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 33
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 34
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
====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 35
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====⇒⇒⇒⇒
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 36
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 37
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:
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 38
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 39
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====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 40
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:
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 41
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 42
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 43
(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 ====
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 46
•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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 47
Natural Light is Unpolarized• Light from the sunWe can polarize light using special materialCrystals, Polymers with aligned atoms
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 48
Natural Light is Unpolarized
9
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 49
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 57
Light incident on an object
• Reflects (bounces)
• Refraction (bends)
•Absorbed
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 58
Angle between light beam and normal Angle of incidence = Angle of reflection
Law of Reflection
θθθθi = θθθθf
θθθθi
θθθθf
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 59
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!
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 60
Refractive index n and wavelength
11
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 61
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 62
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 63
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 65
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 66
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
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 67
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.
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 71
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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 72
Chromatic Dispersion
12
12 sinsin
n
n
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 73
Chromatic Dispersion
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 75
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:
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 76
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
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 77
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
•••••••
•••
•
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 78
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
••••
©Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 79
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:
Problem 39 – Solution
(((( )))) 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
© Dr. Nidal M. Ershaidat Phys. 103 Chapter 5: Electromagnetic Waves 81
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
Phys. 103: Waves and Light
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 103 Waves and Light
© Dr. Nidal M. Ershaidat
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
4444© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference
....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
5555© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference
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������
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 8
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،
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 9
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 10
، 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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 11
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 12
� 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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 13
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 14
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 16
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 17
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<<<<
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 18
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 21
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 22
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 23
Fully constructive interference occurs when φφφφis 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:
i.e. when the path length difference is zero or an integer multiple of the wavelength λλλλ.
Fully constructive interference
Constructive Interference
(((( )))) ππππ====φφφφ 2m
K,2,1,0====λλλλ
∆∆∆∆ L
λλλλ====∆∆∆∆ mL
or:
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 24
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
this occurs when the ratio ∆∆∆∆L/λλλλ verifies:
K,25,
23,
21
====λλλλ
∆∆∆∆ L
Fully destructive interference
λλλλ
++++====∆∆∆∆21
mL
5
6-3 Diffraction
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 26
� 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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 27
Figure below shows the diffraction of water waves traveling across the surface of water in a shallow tank.
Diffraction – Water Waves
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 28
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 29
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.
Diffraction – Schematic View2
Fig. b Fig. c
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 30
Diffraction – Schematic View3
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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 33
- 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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 35
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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 36
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 37
Source is sunlightpassing through a narrow slit.
Young – Coherent Light
Huygens' principle
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 38
•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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 39
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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 40
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 −−−−====∆∆∆∆,
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 41
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 43
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 !!!
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 44
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 46
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 47
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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 48
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 49
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 50
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 51
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 53
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 55
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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 56
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 57
To have destructive interference ∆∆∆∆D = (m + 0.5) λλλλ
Problem 20 - Solution
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 58
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?)
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 59
Problem 27 – Solution
mxxmxxd ====−−−−++++⇒⇒⇒⇒λλλλ====−−−−++++222 16
xmmmxx 21616 22 ++++====⇒⇒⇒⇒++++====++++
m
mx
216 2−−−−
====
01.16
3
7.5
1
3x(m)
42m
x
Path difference:
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 60
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 61
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 62
Problem 28
Fig. (36-8)
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 63
Problem 28 – Solution
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∆∆∆∆⇒⇒⇒⇒
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 64
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.
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 65
Problem 5 - Solution
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 66
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 67
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 69
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
Problem 10 - Solution
(((( ))))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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 78
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 81
(((( ))))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
(π−βπ−βπ−βπ−β)
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 82
φφφφ====φφφφ++++φφφφ
φφφφφφφφ++++φφφφ====
φφφφ++++φφφφ++++
φφφφ++++φφφφ====ββββ tan
cos2coscossin2sin
2coscos2sinsin
tan 2000
00
EEE
EE
(π−βπ−βπ−βπ−β)
Eθθθθ
E0
E0
E0
Calculating βy
xtωωωω
•
φφφφ
ββββ
ββββ
(((( ))))ββββ++++ωωωωθθθθ tE sin
(π−βπ−βπ−βπ−β)
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 83
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
© Dr. Nidal M. Ershaidat Phys.103 Chapter 6: Interference 85
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
Phys. 103: Waves and Light
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 103 Waves and Light
© Dr. Nidal M. Ershaidat
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 4
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.
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 5
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 8
∆∆∆∆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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 9
•
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
•
••
•
•••
•
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 10
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)
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 11
1- Phase diff. between r1 and r2 is:
2- Phase diff. between r1 and r3 is:
Phase Differences
δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ
ππππ====φφφφ sin
212 y
δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ
ππππ====φφφφ 2sin2
213 y
y∆∆∆∆
y∆∆∆∆
y∆∆∆∆
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 12
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
1- Phase diff. between r1 and r2 is:
2- Phase diff. between r1 and r3 is:
3- Phase diff. between r1 and r4 is:
δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ
ππππ====φφφφ sin
212 y
δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ
ππππ====φφφφ 2sin2
213 y
δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ
ππππ====φφφφ 3sin3
214 y
δφδφδφδφ≡≡≡≡θθθθ∆∆∆∆××××λλλλ
ππππ====φφφφ ++++ NyNN sin
211
3
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 13
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 14
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 15
Fig. below shows the resultant Eθθθθ = E(P)
using phasors.
Resultant E(P)
δφδφδφδφ====φφφφ Nββββ
16
Resultant E(P)
δφδφδφδφ====φφφφ Nββββ
2φφφφ2φφφφ
17
Resultant E(P)
We complete the isosceles triangle OAB. We start from point D by tracing a line perpendicular to OB.
δφδφδφδφ====φφφφ Nββββ
2φφφφ2φφφφ
A
O
B
D
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 18
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 19
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 21
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
αααα====θθθθ
a
a
III
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 22
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
ΙΙΙΙ====αααα
αααα====θθθθ
a
a
III
Note that the intensity is minimum when :
KK3,2,1,sinsin
====λλλλ====θθθθ⇒⇒⇒⇒ππππ====λλλλ
θθθθππππmmam
a
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 23
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
ΙΙΙΙ====αααα
αααα====θθθθ
a
a
III
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 24
“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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 25
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 26
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 32
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.): 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
λλλλθθθθππππ====αααα sina
6
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 36
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 37
The single slit pattern forms a kind of envelope for the double-slit fringes.
Diffraction Envelope2
2 sincos
αααα
ααααββββ==== mII
Diffraction envelope
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 38
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 39
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 40
Diffraction GratingDiffraction GratingDiffraction GratingDiffraction Grating����ز ا�� �د
7
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 44
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 45
λλλλ====θθθθ md sin
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 46
Intensity
The bright fringes seen on the screen are called lines because the width of each one is very small.
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 47
∆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.
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 48
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 50
Circular Aperture Diffraction
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 51
Circular Aperture Diffraction
Resolvability Resolvability Resolvability Resolvability –––– The The The The Rayleigh CriterionRayleigh CriterionRayleigh CriterionRayleigh Criterion
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 53
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”
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 54
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
© Dr. N. Ershaidat Phys. 207 Chapter 7: Diffraction Lecture 25 56
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
Phys. 103 Waves and Light
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)
Physics Department, Yarmouk University, Irbid Jordan
Phys. 103 Waves and Light
Dr. Nidal M. Ershaidat Doc. 1A
Author’s email: [email protected],
Address: Physics Department, Yarmouk University 21163 Irbid Jordan
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
Physics Department, Yarmouk University, Irbid Jordan
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
© Nidal M. Ershaidat 2012 2
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:
When the mass m is pulled to the right it suffers two forces:
,ˆ11 ixkF −−−−====
→→→→
ixkF ˆ22 −−−−====
→→→→
(2)
r and l refer respectively to the spring on the right and the spring on the left. The total
external force is:
(((( )))) ixkkFextˆ
21 ++++−−−−====→→→→
(3)
Newton’s 2nd law for the mass m (The motion is one dimensional):
(((( ))))(((( ))))
02121 ====
++++++++⇒⇒⇒⇒++++−−−−==== x
m
kkaxkkam (4)
Equation 4 can be rewritten as:
02
====ωωωω++++ xx�� (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:
(((( ))))(((( )))) (((( ))))2
22
122
121
212122 ff
m
k
m
k
m
kk++++ππππ====ωωωω++++ωωωω====++++====
++++====ωωωω (5)
22
21
2fff ++++====
ππππ
ωωωω====∴∴∴∴
© Nidal M. Ershaidat 2012 3
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:
When the mass m is pulled to the right it suffers two forces:
,ˆ11 ixkF −−−−====
→→→→
ixkF ˆ22 −−−−====
→→→→
(1
r and l refer respectively to the spring on the right and the spring on the left. The total
external force is:
(((( )))) ixkkFextˆ
21 ++++−−−−====→→→→
(2
Newton’s 2nd law for the mass m (The motion is one dimensional):
(((( ))))21 xxkam ++++−−−−==== (3)
Equation 3 can be rewritten as:
02
====ωωωω++++ xx�� (4)
This is the equation of motion of a simple harmonic motion. The angular frequency ω
and frequency f of this motion are given by:
(((( ))))(((( )))) (((( ))))2
22
122
121
212122 ff
m
k
m
k
m
kk++++ππππ====ωωωω++++ωωωω====++++====
++++====ωωωω (5)
22
21
2fff ++++====
ππππ
ωωωω====∴∴∴∴
Physics Department, Yarmouk University, Irbid Jordan
Waves and Light103 . Phys
Dr. Nidal M. Ershaidat Doc. 2A
Author’s email: [email protected],
Address: Physics Department, Yarmouk University 21163 Irbid Jordan
© 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
Physics Department, Yarmouk University, Irbid Jordan
Phys. 103 Waves and Light
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.