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Waves and Vibrations
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PHY 1114
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Contents
1.
Oscillations ............................................................................................................................. 3
1.1 Simple Harmonic Motion:................................................................................................. 3
1.2 Energy Swapping:............................................................................................................. 4
1.3 Angular Simple Harmonic Motion .................................................................................... 5
1.4 Energy of the System ........................................................................................................ 7
1.5 The moment of inertia....................................................................................................... 7
1.6 The Simple Pendulum ....................................................................................................... 9
1.7 Small Angle Approximation ........................................................................................... 10
1.8 The Real (Nonlinear) Pendulum...................................................................................... 11
1.9 The Physical Pendulum................................................................................................... 11
1.10
The superposition of simple harmonic motions ............................................................ 13
2.Wave Motion ............................................................................................................................. 17
2.1 Properties of waves. ....................................................................................................... 17
2.1 Mathematical representation of a wave ........................................................................... 18
2.3 The speed of a travelling (or progressive) wave ................................................................... 20
2.4 Standing waves ................................................................................................................... 21
2.5 Two sine waves travelling in the same direction. ..................................................................... 23
Constructive and Destructive Interference. .................................................................................... 23
2.5.1 Interference of two sound sources in a room: ................................................................ 24
2.6 Reflection of Waves from Boundaries ................................................................................. 25
Reflection from an impedance discontinuity.............................................................................. 26
2.7 Refraction of a sound wave ................................................................................................. 27
2.8 Resonance and Standing Waves .......................................................................................... 28
2.9 The Human Ear ................................................................................................................... 36
2.10 Energy and Power of a travelling wave.............................................................................. 37
2.11 Derivation of the wave speed............................................................................................. 38
2.11.1 Wave speed on a stretched string ................................................................................ 38
2.11.2Wave speed of a sound wave in air .............................................................................. 39
2.12 Travelling Sound Waves ................................................................................................... 40
3. The Doppler Effect................................................................................................................. 45
3.1 Introduction......................................................................................................................... 45
3.2 Supersonic Speeds ............................................................................................................... 50
3.3 The Doppler effect for light ................................................................................................. 52
3.3.1 The Expansion of the Universe ..................................................................................... 53
4. Ultrasound ................................................................................................................................ 53
4.1 Introduction......................................................................................................................... 53
4.1.1 Ultrasound devices: ...................................................................................................... 54
4.1.2 Medical uses of ultrasonic machines: ............................................................................ 54
5. Dispersion ................................................................................................................................. 56
5.1 Dispersion ........................................................................................................................... 56
5.2 Group Velocity and Phase Velocity ..................................................................................... 57
2
1.Oscillations
A motion that repeats itself is an oscillation.
There are many applications. Examples:
Oscillations in material objects:
Strings, Drums, Air columns, Pendulums
Diaphragms in telephones and speakers
Quartz Crystals I wrist watches
Boats bobbing at anchor
Air molecules that transmit sound
Atoms in a solid that transmit heat and convey the temperature
Electrons in an antenna that transmit radio and TV waves
Oscillations not confined to material medium:
Light, Radio waves, X-rays, Gamma rays etc.
1.1
Simple Harmonic Motion:
The force acting on the mass is directed towards a fixed point and its magnitude is proportional
to the distance to the mass from the fixed point.
In order for mechanical oscillation to occur, a system must possess two quantities: elasticity and
inertia. When the system is displaced from its equilibrium position, the elasticity provides a
restoring force such that the system tries to return to equilibrium. The inertia property causes the
system to overshoot equilibrium. This constant play between the elastic and inertia properties is
what allows oscillatory motion to occur. The natural frequency of the oscillation is related to the
elastic and inertia properties by:
--------------(1.1.1)
f 0 = frequency = number of oscillations per second, w0 = angular frequency = radians/sec.
The simplest example of an oscillating system is a mass connected to a rigid foundation by way of a
spring. The spring constant k provides the elastic restoring force, and the inertia of the mass m
provides the overshoot. By applying Newton's second law F=ma to the mass, one can obtain the
equation of motion for the system:
3
m!x! = -kx
--------------(1.1.2)
----------(1.1.3)
where w0 =
k
m
is the natural angular oscillating frequency. The general solution to this equation of
motion takes the form x( t ) = a sin w0 t + b cos w0 t .
---------(1.1.4)
x!( t ) = a w0 cos w0 t - bw0 sin w0 t and !x!( t ) = -aw02 sin w0 t - bw02 cos w0 t = -w02 x .
Special Case: If x = 0 at t = 0
Þb =0
and x( t ) = a sin w0 t .
General solution x( t ) = a sin w0 t + b cos w0 t
é
ù
a
b
= a2 + b2 ê
a sin w0 t +
cos w0 t ú .
êë a 2 + b 2
úû
a2 + b2
= xm [sin w0 t cos f + cos w0 t sin f ] where tan f = b / a .
= x m sin ( w0 t + f )
---------(1.1.5)
where xm is the amplitude of the oscillation, ( w0 t + f ) is the phase of the motion and φ is the phase
constant of the oscillation. Both xm and φ are constants determined by the initial condition (initial
displacement and velocity) at time t=0 when one begins observing the oscillatory motion.
T – period of motion = time for one complete oscillation (or cycle)
For a cycle
x = xm sin ( w0 t + f ) = xm sin ( w0 ( t + T ) + f )
This happens when w0 ( t + T ) = w0 t + 2p or w0T = 2p Þ T = 2p / w0 = 1 / f 0
T = 2p
m
k
------------(1.1.6)
Note: By selecting the constant phase factor one could write the General solution as
x( t ) = xm cos ( w0 t + f )
1.2
------------(1.1.7)
Energy Swapping:
The elastic property of the oscillating system (spring) stores potential energy
---------(1.1.8)
and the inertia property (mass) stores kinetic energy
--------(1.1.9)
4
As the system oscillates, the total mechanical energy in the system trades back and forth between
potential and kinetic energies. The total energy in the system, however, remains constant, and
depends only on the spring constant and the maximum displacement (or mass and maximum
velocity vm=ωxm)
------------1.1.10
The total mechanical energy in a simple undamped
mass-spring oscillator is traded between kinetic and
potential energies while the total energy remains
constant.
1.3
Angular Simple Harmonic Motion
Example: The torsion pendulum
Consider a disk suspended from a torsion wire attached to its center. This setup is known as a
torsion pendulum. A torsion wire is essentially inextensible, but is free to twist about its axis. Of
course, as the wire twists it also causes the disk attached to it to rotate in the horizontal plane. Let q
be the angle of rotation of the disk, and let q =0 correspond to the case in which the wire is
untwisted.
5
Any twisting of the wire is inevitably associated with mechanical deformation. The wire resists such
deformation by developing a restoring torque, t , which acts to restore the wire to its untwisted
state. For relatively small angles of twist, the magnitude of this torque is directly proportional to the
twist angle. Hence, we can write
t = -Cq ----------(1.2.1)
where C > 0 is the torque constant of the wire. The above equation is essentially a torsional
equivalent to Hooke's law. The rotational equation of motion of the system is written as
t = I!q! ----------(1.2.2)
where I is the moment of inertia of the disk (about a vertical axis through its center). The moment of
inertia of the wire is assumed to be negligible. Combining the previous two equations, we obtain
I!q! = -Cq ----------(1.2.3)
Equation (1.2.3) is clearly a simple harmonic equation (cf., Eq. (1.1.2)). Hence, we can immediately
write the standard solution [cf., Eq. (1.1.5)]
q = q m sin ( w0 t + f )--------(1.2.4)
where w0 =
C
I
and T = 2p
I
C
--------(1.2.5)
We conclude that when a torsion pendulum is perturbed from its equilibrium state (i.e., q = 0 ), it
executes torsional oscillations about this state at a fixed frequency, w0 , which depends only on the
torque constant of the wire and the moment of inertia of the disk. Note, in particular, that the
frequency is independent of the amplitude of the oscillation [provided q remains small enough].
Torsion pendulums are often used for time-keeping purposes. For instance, the balance wheel in a
mechanical wristwatch is a torsion pendulum in which the restoring torque is provided by a coiled
spring.
6
1.4
Energy of the System
Rotational Kinetic Energy =
1 !2
Iq
2
----------(1.2.6)
q
Potential Energy = Work done against the torque = - ò ( -Cq ) dq =
0
Total Energy =
1.5
1 !2 1 2 1 2 1 !2
Iq + Cq = Cq m = Iq m
2
2
2
2
1 2
Cq ---------(1.2.7)
2
--------(1.2.8)
The moment of inertia
The moment of inertia is a function not only of the mass of the object hanging from the wire but
also of how that mass is distributed about the axis of rotation. Some examples are:
Example 1:
Two point masses, m each, at a distance r from the axis of
oscillation, connected by a massless rod.
I = 2 mr2 -----------(1.2.9)
Example 2:
Example 2: A disc of mass M, radius r and thickness t oscillating
around the diameter that goes through the center of mass.
---------(1.2.10)
7
Example 3:
The same disc as before, but oscillating around the
axis perpendicular to its surface through the center.
-------(1.2.11)
Example 4:
A cylinder of mass M, radius r and length t.
----------(1.2.12)
Note: this is the same result as Example 2 above.
Example 5
A hollow cylinder of mass M, length L,
inner radius a and outer radius b.
------(1.2.13)
8
Example 6:
To the left we show the cylinders from
Examples 4 and 5 combined into a single
complex object. The total moment of inertia
is just the sum of those two examples:
Note: Deriving these examples requires knowing that the moment of inertia of a differential mass
dm rotating at a distance r from the axis of rotation has a differential moment of inertia dI = r2 dm.
Then one integrates all the differential masses over the entire body to calculate the total moment of
inertia
For the Example 3 above: Consider a ring of radius x and thickness dx
dI = x2dm
t = thickness
dm = 2px dx t r
r = density
r
r
0
0
x
I = ò 2px dx t r x 2 = 2pt r ò x 3 dx
I = 2pt r
1.6
r4
, Mass of the disk M = pr 2 tr Þ I = ½ Mr2
4
The Simple Pendulum
A simple pendulum consists of a mass m hanging from a string of length L and fixed at a pivot
point P. When displaced to an initial angle and released, the pendulum will swing back and
forth with periodic motion. By applying Newton's second law for rotational systems, the
equation of motion for the pendulum may be obtained
, ------------(1.3.1)
and rearranged as
.------------(1.3.2)
If the amplitude of angular displacement is small enough that the small angle approximation (
) holds true, then the equation of motion reduces to the equation of simple harmonic
motion
.--------(1.3.3)
9
The simple harmonic solution is
-----------(1.3.4)
1.7
Small Angle Approximation
How small q should be used?
Example: If q = 50 = 0.0873 rad, then sin q = 0.0872 and the difference is ~ 0.1%.
With the assumption of small angles, the frequency and period of
the pendulum are independent of the initial angular displacement
amplitude. All simple pendulums should have the same period
regardless of their initial angle (and regardless of their masses).
This simple approximation is illustrated in the movie at left
(shown in the class). All three pendulums cycle through one
complete oscillation in the same amount of time.
The period for a simple pendulum does not depend on the
mass or the initial angular displacement, but depends only
on the length L of the string and the value of the
gravitational field strength g, according to
-----------(1.3.5)
The movie at left shows (see in the class) two pendulums,
with different lengths.
10
1.8 The Real (Nonlinear) Pendulum
When the angular displacement amplitude of the pendulum is large enough that the small angle
approximation no longer holds, then the equation of motion must remain in its nonlinear form
.
This differential equation does not have a closed form solution, and must be solved numerically
using a computer. Mathematica numerically solves this differential equation very easily with the
built in function NDSolve[ ].
The small angle approximation discussed above is valid for small initial angular displacements. For
small initial angular displacements the error in the small angle approximation becomes evident only
after several oscillations.
When the initial angular displacement is significantly
large that the small angle approximation is no longer
valid, the error between the simple harmonic solution and
the actual solution becomes apparent almost immediately,
and grows as time progresses. In the movie at left (see in
the class), the dark blue pendulum is the simple harmonic
approximation, and the light blue pendulum (initially
hidden behind the dark blue one) shows the numerical
solution of the nonlinear differential equation of motion.
1.9
The Physical Pendulum
Hanging objects may be made to oscillate in a manner similar to a
simple pendulum. The motion can be described by "Newton's 2nd law
for rotation:
t = I sup porta = I sup portq!! -------(1.4.1)
where the torque is t = - mgLcm sin q -------(1.4.2)
and the relevant moment of inertia ( I sup port ) is that about the point of
suspension. The resulting equation of motion is:
I sup portq!! = -mgLcm sinq --------(1.4.3)
11
For small q ,
The Period
q!! = -
T = 2p
mgLcm
q = -w 02q
I sup port
I sup port
mgLcm
mgLcm
--------(1.4.4)
I sup port
where w 0 =
--------(1.4.5)
I sup port = I CG + mL2cm \ --------(1.4.6)
where I CG is the moment of inertia about a parallel axis through the center of gravity of the
pendulum, which can be written as I CG = mk 2 . Here, k is called the radius of gyration. (For a disk
of radius r, I CG = 21 mr 2 , k = a / 2 )
mk 2 + mL2cm
k 2 + L2cm
T = 2p
= 2p
= 2p
mgLcm
gLcm
k2
Lcm
+ Lcm
g
----------(1.4.7)
k2
+ Lcm -----(1.4.8)
Lcm
Lets consider the point P at a distance l from the point of suspension (O)
as shown in the figure
The equation (1.4.7) is equivalent to a simple pendulum of length l =
P- Center of oscillation & O-Center of suspension
Let LCM = h
(1.4.8)
two roots.
h 2 - lh + k 2 = 0
2
h1 + h2 = l and h1 h2 = k
k2
.
Lcm
Therefore, if the system is suspended from P the period would be the same.
Let h1 = Lcm , then h2 = l - h1 = l - Lcm =
Similarly there are infinite number of axes parallel to the normal at
the center of mass, which are passing through the two circles:
1. Circle of radius k2/Lcm centered at CM.
2. Circle of radius Lcm centered at CM.
If the body were suspended from the CM the period would be
infinite. For any other axis the period is given by equation (1.4.7)
12
T = 2p
k 2 + L2cm
gLcm
This has a minimum value when
This is minimum when Lcm
k 2 + L2cm
is minimum, which can be written as
Lcm
( Lcm - k )2 + 2 kLcm
Lcm
= k and the minimum value of the period is
Tmin = 2p
2k
g
--------(1.4.9)
The variation of the period as a function of the distance to the point of suspension is shown below.
T
Distance from center of gravity
1.10
The superposition of simple harmonic motions
The resultant motion of an object under several SHMs is given by the algebraic sum of individual
motions.
Consider the superposition of x1 = A1 sin ( w1 t + f1 ) and x 2 = A2 sin ( w 2 t + f2 )
The resultant motion is
x = x1 + x2 = A1 sin (ω1t + φ1 ) + A2 sin (ω 2t + φ2 ) --------(1.5.1)
Example 1. When the two frequencies are equal.
x = A1 sin ( wt + f1 ) + A2 sin ( wt + f2 ) ---------(1.5.2)
x = A1 (sin wt cosf1 + cos wt sinf1 ) + A2 (sin wt cosf2 + cos wt sinf2 )
13
x = sin wt [ A1 cosf1 + A2 cosf2 ] + cos wt [ A1 sinf1 + A2 sinf2 ]
x = sin wt [ B1 ] + cos wt [ B2 ]
x = B12 + B22 [sin wt
B1
B12 + B22
+ cos wt
B2
B12 + B22
]
x = A [ sin wt cos f + cos wt sin f ] = A sin ( wt + f ) ---------(1.5.3)
B
where A = B12 + B22 and tan f = 2
--------(1.5.4)
B1
Therefore, the resultant motion is a SHM of same frequency but different amplitude and phase
constant.
Example 2. When the two amplitudes are equal.
From
(1.5.1)
x = A sin ( w1 t + f1 ) + A sin ( w 2 t + f2 )
Let x1 = x 2 = 0 at t=0 then f1 = f2 = 0
x = A sin w1 t + A sin w 2 t = A(sinw1 t + sin w 2 t )
w + w2
w - w2
x = 2 A sin ( 1
)t cos( 1
)t ---(1.5.5)
2
2
First term shows the variation of the amplitude
and the second term shows the envelope. The
result is the production of beats and the beat
frequency, the number of maximas per second,
is f1 – f2.
Example 3. Two SHM s of same frequency in
perpendicular directions.
x = a sin ( wt + f )
y = b sin wt
y2
Þ sin wt = y / b & cos wt = 1 - 2
b
x = a(sin wt cos f + cos wt sin f ) = a [
y
y2
cos f + 1 - 2 sin f ]
b
b
x y
y2
[ - cos f ] 2 = ( 1 - 2 ) sin 2 f
a b
b
2
2
x
2 xy
y
cos
f
+
= sin 2 f -------(1.5.6)
2
2
ab
a
b
14
Lets look at the result for different phase factors.
Case 1: When f = 0
From (1.5.6)
x2
a
2
-
2 xy y 2
+
=0
ab b 2
or (
x y 2
- ) =0
a b
b
x
a
The resultant motion is along a straight line of slop b/a.
y=
Case 2: When f = p
x2
2 xy y 2
From (1.5.6) 2 +
+
=0
ab b 2
a
x y
b
( + )2 = 0 y = - x
a b
a
The resultant motion is along a straight line of slop -b/a.
Case 3: When f =
2
From (1.5.6)
p
2
y2
x
+
=1
a 2 b2
a=b
Case 4: When f =
p
4
x
2 xy
y
1
+
=
a2
2 ab b 2 2
2
2
a>b
3p
4
2
2
x
2 xy
y
1
+
+
=
a2
2 ab b 2 2
a<c
Case 5: When f =
The above cases considered here are the simplest possible out of many. Following figures shows the
resultant motion when the two frequencies are simple ratios.
15
Frequency ratio fx: fy = 6:4=3:2
Frequency ratio fx: fy = 6:4 = 2:3
The figures discussed here are called Lissajou’s Figures. These figures could be demonstrated using
an oscilloscope. Two appropriate signals have to be provided for horizontal and vertical oscillations
of the electrons in the beam.
16
2. Wave Motion
Waves and Particles are important concepts in Classical Physics. Almost every branch of Physics is
associated with one or the other.
Particle: Concentration of matter capable of transmitting energy.
Wave: A broad distribution of energy filling the space through which it passes.
Quantum mechanically a particle can be associated with a wave and vice versa. A particle with
h
linear momentum p is associated with a wave of wavelength l such that l =
-------(2.1),
p
where h is the Planck’s constant. The above relationship, that relates the wave particle duality,
namely the particle nature with its wave nature is called the de Broglie relationship.
2.1
Properties of waves.
What is a wave? A wave is a disturbance or variation, which travels through a medium. The
medium through which the wave travels may experience some local oscillations as the wave passes,
but the particles in the medium do not travel with the wave.
Webster's dictionary defines a wave as "a disturbance or variation that transfers energy
progressively from point to point in a medium and that may take the form of an elastic deformation
or of a variation of pressure, electric or magnetic intensity, electric potential, or temperature."
Mechanical Waves: A material medium, with elastic and inertia, is needed to propagate a
mechanical wave. Newton’s laws govern the propagation of a mechanical wave. The speed of the
wave depends on the properties of the medium.
Electromagnetic Waves: Require no material medium and the propagation is governed by
Maxwell’s equations. Speed is a constant, c=2.99x108m/s, in vacuum.
The disturbance may take any of a number of shapes, from a finite width pulse to an infinitely long
sine wave.
Finite width Pulse wave:
Infinitely long sine wave:
17
Longitudinal waves:
As the wave passes through, the particles in the air oscillate back and forth about their
equilibrium positions but it is the disturbance that travels, not the individual particles in the
medium. Sound wave is an example.
Transverse waves: The particles in the medium are displaced up and down as the wave travels
from left to right, but the individual particles do not experience any net displacement. A wave
propagating on a string is an example.
Both transverse and longitudinal waves can be set in a solid medium. Following is an example.
Sand
Scorpion
A sand scorpion can catch a beetle moving
within few cms from it without seeing or
hearing. When a beetle disturbs the sand, it sends
pulses along the surface of the sand. Two sets of
pulsus, one longitudinal and one transverse will
be set when the solid medium is disturbed. The
longitudinal wave propagates faster than the
transverse wave. In this case vl ~150 m/s and vt ~
50 m/s. Eight legs of the scorpion spread roughly
in a circle of ~5cm, intercepting the longitudinal
wave first, from which the scorpion can find the
direction of the beetle. When the slower
transverse wave arrives the scorpion the distance
to the beetle is determined instantly by sensing
the time difference between the two waves.
Dt =
d d
.
vt vl
Same principle is used in locating the epicenter
of a seismic wave.
2.1
Mathematical
representation of a wave
A wave can have many shapes but
fundamental to each shape are:
Wavelength( l ): The length at which the
shape of the wave begins to repeat.
Frequency (f): The number of times per
second that a displacement repeats or the
number of cycles per second.
18
The displacement of a wave y depends on two independent variables, the position x and the time t.
i.e. y(x,t) = h(x,t).
For a sinusoidal wave y( x , t ) = y m sin ( k x - w t ) --------(2.2.1)
Why sinusoidal form? All waveforms can be constructed by adding up sin waves with suitable
amplitudes and wavelengths.
Fourier Analysis: A French mathematician, J.B. Fourier explained how the principle of
superposition could be used to analyze nonsinusoidal waveforms.
A Fourier series is an expansion of a periodic function f(x) in terms of an infinite sum of sines and
cosines. Fourier series make use of the orthogonality relationships of the sine and cosine functions.
The computation and study of Fourier series is known as harmonic analysis and is extremely useful
as a way to break up an arbitrary periodic function into a set of simple terms that can be plugged in,
solved individually, and then recombined to obtain the solution to the original problem or an
approximation to it to whatever accuracy is desired or practical. Examples of successive
approximations to common functions using Fourier series are illustrated below.
In particular, since the superposition principle holds for solutions of a linear homogeneous ordinary
differential equation, if such an equation can be solved in the case of a single sinusoid, the solution
for an arbitrary function is immediately available by expressing the original function as a Fourier
series and then plugging in the solution for each sinusoidal component. In some special cases where
the Fourier series can be summed in closed form, this technique can even yield analytic solutions.
µ
µ
n= 1
n= 1
f ( x ) = 21 a0 + å a n cos ( nx ) + å bn sin ( nx ) ---------(2.2.2)
19
Fourier series for the sawtooth can be written as
1
1
1
1
y( t ) = - sin wt sin 2wt sin 3wt sin 4wt --------(2.2.3)
p
2p
3p
4p
Displacement as a function of x:
At a given time, t=0 for example, (2.2.1) for a sinusoidal wave reduces to
y( x , 0 ) = y m sin k x ----- (2.2.4).
By definition of l , y is same at x1 and x1+ l .
y ( x ,0 ) = y m sin k x 1 = y m sin k ( x 1 + l ) = y m sin ( kx 1 + kl )
This is true when
kl = 2p -----------(2.2.5)
k is defined as the angular wave number and the wave number
K=
1
l
=
k
-------(2.2.6)
2p
Displacement as a function of t:
At a given point, x=0 for example, (2.2.1) reduces to
y( 0 , t ) = y m sin ( -wt ) = - y m sin ( wt ) -------(2.2.7)
If T is the period, y is same at t = t1 and t = t1 + T
y( 0 , t ) = y m sin ( -wt 1 ) = - y m sin w ( t 1 + T )
The above relationship is true when wT = 2p ----(2.2.8)
Angular frequency w =
2p
T
-----(2.2.9)
and the frequency f =
1
w
-------(2.2.10)
=
T 2p
2.3 The speed of a travelling (or progressive) wave
Consider two “snap-shots”
of a wave as shown in the
figure.
The speed of the wave
Dx dx
v=
=
Dt dt
x
A give displacement y is same for all points with the same phase,
i.e. kx - wt = Cons tan t -------(2.3.1)
20
In taking the derivative of the above equation with respect to time k
of the wave:
v=
dx w
=
dt
k
dx
- w = 0 . Hence, the speed
dt
--------(2.3.2)
l
= fl ------(2.3.3)
T
For a wave propagating in the positive x direction v is positive and it should be negative for a wave
along the negative x direction.
Using equations 2.2.5 and 2.2.9 v =
y( x , t ) = y m sin ( k x - w t ) -----(2.3.4) for a wave along the positive x direction and
y( x , t ) = y m sin ( k x + w t ) ------(2.3.5) for a wave along the negative x direction.
The general form of a travelling wave can be written as y( x , t ) = h( kx ± wt ) --------(2.3.6)
2.4 Standing waves
A standing wave, also known as a stationary wave, is a wave that remains at a fixed
position. This phenomenon can occur because the medium is moving in the opposite direction to the
wave, or it can arise in a stationary medium as a result of interference between two waves travelling
in opposite directions. There is no net propagation of energy.
Case 1: As an example of the first type, under certain meteorological conditions standing waves
form in the atmosphere in the lee of mountain ranges.
Generation of Mountain Waves
(schematic drawing)
1 = Mountain
2 = Wind
3 = Rotor
4 = Lee Wave (Standing wave)
5 = typical cloud (lenticularis)
6 = typical cloud (cumulus)
From Wikipedia
Case 2: The interference of two waves travelling in opposite directions produce a standing wave in
a stationary medium.
Consider two sinusoidal waves travelling in opposite directions, y1 ( x , t ) = y m sin ( k x - w t ) and
y 2 ( x , t ) = y m sin ( k x + w t ) . The resultant wave is given by the superposition principle.
y( x , t ) = y m sin ( k x - w t ) + y m sin ( k x + w t )
----------(2.4.1)
= 2 y m sin kx cos wt
21
This is not the form (2.3.6) of a travelling wave. y as a function of x at four different t values are
shown in the following figure.
As can be seen from the above plots, the y displacements are zero at certain x values regardless of
the time. These positions are called “Nodes”. Similarly, the points that could reach the highest
amplitudes are at certain x values and theses points are called “Antinodes”.
For nodes; sin kx = 0 , kx = np ------(2.4.2)
where n = 0, 1, 2, …..
nl
From (2.2.5) & (2.4.2): x =
; n= 0,1,2,…. ---------(2.4.3)
2
The nodes are separated by
l
.
2
Similarly, for maximum amplitude, which is 2ym, sin kx = 1 . Therefore, for antinodes;
kx = ( n + 21 )p
Using (2.2.5)
x = ( n + 21 )
Transverse speed:
--------(2.4.3) where n= 0,1,2……
l
---------(2.4.5)
2
dy
dt
y( x , t ) = 2 y m sin kx cos wt
At nodes sin kx = 0
\
dy
=0
dt
dy
= ! 2 y m w sin wt
dt
At antinodes sin kx = ±1
Transverse acceleration:
Strain:
dy
= - 2 y m w sin kx sin wt --------(2.4.6)
dt
d2y
dt
2
= -2 y mw 2 sin kx cos wt ----(2.4.7)
=0
at nodes
2
= ! 2 y mw cos wt at antinodes.
dy
= 2 y m k cos kx cos wt = ± 2 y m k cos wt
dx
=0
at nodes
antinodes
-----(2.4.8)
22
2.5 Two sine waves travelling in the same
direction.
Constructive and Destructive Interference.
Consider two waves of same wavelength and frequency, but different amplitudes and phase
constants, travelling in the x-direction.
y1 = a sin ( kx - wt )
y2 = b sin ( kx - wt + f )
= b [ sin ( kx - wt ) cos f + cos ( kx - wt ) sin f ]
The resultant wave is
y = y1 + y 2
= sin ( kx - wt ) ( a + b cos f ) + cos ( kx - wt ) ( b sin f )
= sin ( kx - wt ) ( A cos q ) + cos ( kx - wt ) ( A sin q )
-----(2.5.1)
= A sin( kx - wt + q )
b sin f
where A 2 = A 2 sin 2 q + A 2 cos 2 q = a 2 + b 2 + 2ab cos f and tan q =
---(2.5.2)
a + b cos f
The result is a travelling wave with different amplitude and different phase constant.
Special case when a=b
f
f
A = 2 a cos and tan q = tan
2
2
y = 2 a cos
f
sin ( kx - wt +
f
) -------(2.5.3)
2
2
which is a travelling wave whose amplitude depends on the phase (phi). When the two waves are
in-phase (phi=0), they interfere constructively and the result has twice the amplitude of the
individual waves. When the two waves have out-of-phase (phi=180), they interfere destructively
and cancel each other out. Note that f is the phase difference between the two waves. The above
conditions can be generalized as follows. The interference at a given point is determined by the
phase difference between the two waves reaching that point.
For constructive interference the phase difference
----------(2.5.4)
f = 2p n
For destructive interference the phase difference
f = 2p ( n + 1 / 2 ) --------(2.5.5)
where
(n= 0,1,2,……)
23
f two waves,
phi=0).
The resultant wave of two waves,
which are out of phase (phi=180).
The intensity at a given point
depends on the relative phase
of the two waves at that point.
2.5.1 Interference of two sound sources in a room:
Consider the famous classroom demonstration of interference of two sound waves from two
speakers (S1 and S2) . The speakers were set at distance d (approximately 1 m) apart and produced
identical tones. The two sound waves traveled through the air in front of the speakers, spreading out
through the room in spherical fashion. A snapshot in time of the appearance of these waves is
shown in the diagram below.
In the diagram, a thick line represents the compressions
of a wavefronts and thin lines represent the rarefactions.
These two waves interfere in such a manner as to
produce locations of some loud sounds (Constructive
Interference, CI) and other locations of no sound
(Destructive Interference, DI). Of course the loud
sounds are heard at locations where compressions meet
compressions or rarefactions meet rarefactions and the
"no sound" locations appear wherever the compressions
of one of the waves meet the rarefactions of the other
wave. If you were to plug one ear and turn the other ear
towards the place of the speakers and then slowly walk across the room parallel to the plane of the
speakers, then you would encounter an amazing phenomenon. You would alternatively hear loud
sounds as you approached anti-nodal locations and virtually no sound as you approached nodal
locations. (In reality, however, the nodal locations were not true nodal locations due to reflections
of sound waves off the walls, which tended to fill the entire room with reflected sound. Even though
the sound waves which reached the nodal locations directly from the speakers destructively
interfered, other waves reflecting off the walls tended to reach that same location to produce a
pressure disturbance.)
Consider a general point P at a distance L from the speakers
as sown in the figure.
24
The path difference between two wave fronts reaching P at a given instant = S 2 P - S1 P = d sinq
A path difference of l is equivalent to a phase difference of 2p .
2p
Therefore, the phase difference between two wave fronts reaching P at a given instant =
d sin q
Therefore, from (2.5.4), for points at which CI occurs
2p
-------- (2.5.6)
d sin q = 2pn
l
l
2p
and for DI
l
d sin q = 2p ( n + 1 / 2 )
-------- (2.5.7)
n= 0,1,2,…….is the order of interference.
Therefore, if one moves perpendicular to the symmetry axis away from the axis, q increases from
zero to larger values. When q is zero on the symmetry, according to the (2.5.6) one should here a
maxima of intensity and consecutively minima and maxima should be heard.
2.6 Reflection of Waves from Boundaries
When an object, like a ball, is thrown against a rigid wall it bounces back. This "reflection" of the
object can be analyzed in terms of momentum and energy conservation. If the collision between ball
and wall is perfectly elastic, then all the incident energy and momentum is reflected, and the ball
bounces back with the same speed. If the collision is inelastic, then the wall (or ball) absorbs some
of the incident energy and momentum and the ball does not bounce back with the same speed.
Waves also carry energy and momentum, and whenever a wave encounters an obstacle, the obstacle
reflects them. This reflection of waves is responsible for echoes, radar detectors, and for allowing
standing waves, which are so important to sound production in musical instruments.
Wave pulse traveling on a string:
The speed of a wave pulse travelling on a string depends on the elastic properties, namely the
tension of the string T in this particular example and the inertia, the mass per unit length of the
T
string µ . The speed v is given by v =
. --------(2.6.1)
µ
Reflection from a HARD boundary:
Consider a wave pulse, which is traveling from left to right
on a string towards the end, which is rigidly clamped. As the
wave pulse approaches the fixed end, the internal restoring
forces, which allow the wave to propagate, exert an upward
force on the end of the string. But, since the end is clamped,
it cannot move. According to Newton's third law, the wall
must be exerting an equal downward force on the end of the
string. This new force creates a wave pulse that propagates from right to left, with the same speed
and amplitude as the incident wave, but with opposite polarity (upside down). (see the animation in
the class)
25
=> at a fixed (hard) boundary, the displacement remains zero and the reflected wave changes its
polarity (undergoes a 180o phase change)
Reflection from a SOFT boundary:
Consider a wave pulse on a string moving from left to right towards the end which is free to move
vertically (imagine the string tied to a massless ring which slides frictionlessly up and down a
vertical pole). The net vertical force at the free end must be
zero. This boundary condition is mathematically equivalent to
requiring that the slope of the string displacement be zero at
the free end (look closely at the movie (in the class) to verify
that this is true). The reflected wave pulse propagates from
right to left, with the same speed and amplitude as the incident
wave, and with the same polarity (right side up).
=> at a free (soft) boundary, the restoring force is zero and the reflected wave has the same polarity
(no phase change) as the incident wave
Reflection from an impedance discontinuity
When a wave encounters a boundary which is neither rigid (hard) nor free (soft) but instead
somewhere in between, part of the wave is reflected from the boundary and part of the wave is
transmitted across the boundary. The exact behavior of reflection and transmission depends on the
material properties on both sides of the boundary. One important property is the characteristic
impedance of the material. The characteristic impedance of a material is the product of mass density
and wave speed, z = µ v . If a wave with amplitude A1 in medium 1 encounters a boundary with
medium 2, the amplitudes of the reflected and transmitted waves are determined by
Ar 1 =
z1 / z 2 - 1
A1 --------(2.6.2)
z1 / z 2 + 1
At 2 =
2
A1 ---------(2.6.3)
z 2 / z1 + 1
In the examples below, two strings of different densities are connected so that they have the same
tension. The density of the thick string is four times greater than the thin string. Since the speed of
waves on a string is related to density and tension by equation (2.6.1) the wave speed of the wave
on the low-density string should be greater than that of high density.
Example 1: Wave propagating from low density (high speed) to high density (low speed):
Compare the amplitudes, polarity and the width of the
transmitted and the reflected waves, respectively, with those of
the incident wave (see the animation in the class).
Example 2: Wave propagating from high density (low speed) to low density (high speed):
Repeat the same comparison here as in Example 1.
26
2.7 Refraction of a sound wave
Sound waves can be altered by several ways in creating acoustic shadows. Acoustical shadows can
usually be traced to one or more of three causes: absorption, wind direction and wind shear, or
refraction.
Absorption - Sometimes the material between a sound source and an observer will render the sound
inaudible due to the absorption. The material can be soil, forest, snow or a variety of other
substances.
Wind direction and wind shear - In general, sounds are more likely to be heard downwind of a
sound source than upwind. Since winds aloft are usually faster than at ground level, the upper part
of a sound wave will travel faster than the lower part when traveling with the wind and more slowly
when against the wind. This will cause refraction towards the ground in the former case and away
from the ground in the latter case.
Refraction of sound waves –
The speed of a sound wave in air depends on the temperature (331.36+0.6067T) where T is the
temperature in oC. Often the change in the wave speed, and the resulting refraction, is due to a
change in the local temperature of the air. For example, during the day the air is warmest right next
to the ground and grows cooler above the ground. This is called a temperature lapse. Since the
temperature decreases with height, the speed of sound also decreases with height. This means that
for a sound wave traveling close to the ground, the part of the wave closest to the ground is
traveling the fastest, and the part of the wave farthest above the ground is traveling the slowest. As a
result, the wave changes direction and bends upwards. This can create a "shadow zone" region into
which the sound wave cannot penetrate. A person standing in the shadow zone will not hear the
sound even though he/she might be able to see the source. The sound waves are being refracted
upwards and will never reach the observer.
Sometimes upwardly refracted waves hit a warmer layer higher up and are refracted back down,
creating rings of audibility.
27
A temperature inversion is when the temperature is coolest right next to the ground and warmer as
you increase in height above the ground. Since the temperature increases with height, the speed of
sound also increases with height. This means that for a sound wave traveling close to the ground,
the part of the wave closest to the ground is traveling the slowest, and the part of the wave farthest
above the ground is traveling the fastest. As a result, the wave changes direction and bends
downwards. Temperature inversions most often happen on clear, cool nights (and in the morning
following them) after the sun goes down when the ground (or water in a lake) cools off quickly,
while the air above the ground remains warm. This downward refraction of sound is why you can
hear the conversations of people across the lake, when otherwise you should not be able to hear
them.
2.8 Resonance and Standing Waves
Barton’s pendulum is a simple device that can be used to demonstrate the principle of
resonance. When the driver pendulum is set to oscillate with its natural frequency, the energy
would be transferred to other pendulums. However, the maximum amplitude can be observed
in the pendulum with the same height, i.e. the same natural frequency, as the driver pendulum.
Therefore, it demonstrates that the energy transfer is highest when the frequency of the driving
force is equal to the natural frequency of the driven pendulum.
28
A simple pendulum has only one natural frequency. However, in general objects have more than
one natural frequency.
All objects have a frequency or set of frequencies with which they naturally vibrate when struck,
plucked, strummed or somehow disturbed. Each of the natural frequencies at which an object
vibrates is associated with a standing wave pattern. When an object is forced into resonance
vibrations at one of its natural frequencies, it vibrates in a manner such that a standing wave is
formed within the object. A standing wave pattern was described in section 2.4 as a resultant wave
of two identical waves travelling in opposite directions. Practically, this situation arises when a
wave is reflected from one end of a medium causing the interference between the incident wave and
the reflected wave in such a manner that specific points along the medium appear to be standing
still. Such patterns are only created within the medium at specific frequencies of vibration; these
frequencies are known as harmonic frequencies, or merely harmonics. At any frequency other than
a harmonic frequency, the interference of reflected and incident waves results in a resulting
disturbance of the medium, which is irregular and non-repeating.
So the natural frequencies of an object are merely the harmonic frequencies at which standing wave
patterns are established within the object. These standing wave patterns represent the lowest energy
vibrational modes of the object. While there are countless ways by which an object can vibrate
(each associated with a specific frequency), objects favor only a few specific modes or patterns of
vibrations. The favored modes (patterns) of vibration are those, which result in the
highest amplitude vibrations with the least input of energy. Objects favor these natural
modes of vibration because they are representative of the patterns, which require the least
amount of energy. Objects are most easily forced into resonance vibrations when
disturbed at frequencies associated with these natural frequencies.
The diagrams at the right show two of the most common standing wave patterns for the
Chladni (square metal plate fixed at the center and salt was sprinkled upon the plate in an
irregular pattern) plate. Two plates are at resonance with a violin at two different natural
frequencies of the plate. The white lines represent the concentration of salt locations
29
(nodal positions).
Consider the case when a system is driven by an external driving force.
Example: When pushing a child on a swing. Graph (a) shows what the
child’s x-t graph would look like as you gradually put more and more
energy into her vibrations. A graph of your force versus time would
probably look something like graph (b).
It turns out, however, that it is much simpler mathematically to consider a
vibration with energy being pumped into it by a driving force
that is itself a sine-wave. A good example of this is your eardrum being
driven by the force of a sound wave. In a realistic system, the system will
reach a steady state between the energy supplied and the energy loss due
to damping. The frequency of oscillations would be the frequency of the
driving force. However, if we change the frequency of the driving force, when it equals to the natural
frequency of the object, the object resonates with the highest amplitude. These two cases could be
summarized as
(1) The steady-state response to a sinusoidal driving force occurs at the frequency of the force, not at the system’s own natural frequency of vibration.
(2) A vibrating system resonates at its own natural frequency. That is, the amplitude of the steadystate response is greatest in proportion to the amount of driving force when the driving force matches
the natural frequency of vibration.
Example1: an opera singer breaking a wineglass
In order to break a wineglass by singing, an opera singer must first tap the glass to find its natural frequency of
vibration, and then sing the same note back.
See the following site: http://www.acoustics.salford.ac.uk/feschools/waves/wine1video.htm
Example2: Collapse of the Nimitz Freeway in an earthquake
The collapse of a section of the Nimitz Freeway in Oakland, CA.,
during a 1989 earthquake is however a simpler example to analyze. An
earthquake consists of many low-frequency vibrations that occur
simultaneously, which is why it sounds like a rumble of indeterminate
pitch rather than a low hum. The frequencies that we can hear are not
even the strongest ones; most of the energy is in the form of vibrations
in the range of frequencies from about 1 Hz to 10 Hz.
Now all the structures we build are resting on geological layers of dirt, mud, sand, or rock. When an
earthquake wave comes along, the topmost layer acts like a system with a certain natural frequency
of vibration, sort of like a cube of jello on a plate being shaken from side to side. The resonant
frequency of the layer depends on how stiff it is and also on how deep it is. The ill-fated section of
the Nimitz freeway was built on a layer of mud, and analysis by geologist Susan E. Hough of the
U.S. Geological Survey shows that the mud layer’s resonance was centered on about 2.5 Hz, and
had a width covering a range from about 1 Hz to 4 Hz. When the earthquake wave came along with
30
its mixture of frequencies, the mud responded strongly to those that were close to its own natural
2.5 Hz frequency. Unfortunately, an engineering analysis after the quake showed that the overpass
itself had a resonant frequency of 2.5 Hz as well! The mud responded strongly to the earthquake
waves with frequencies close to 2.5 Hz, and the bridge responded strongly to the 2.5 Hz vibrations
of the mud, causing sections of it to collapse.
Example 3: Collapse of the Tacoma Narrows Bridge
Mile-long bridge, constructed in July 1940 & collapsed
due to 42-mile-per-hour wind on November 7 with the
sides vibrating 8.5 meters (28 feet) up and down.
Let’s now examine the more conceptually difficult case of
the Tacoma Narrows Bridge. The surprise here is that the
wind was steady. If the wind was blowing at constant
velocity, why did it shake the bridge back and forth? The
answer is a little complicated. Based on film footage and
after-the-fact wind tunnel experiments, it appears that two different mechanisms were involved.
The first mechanism was the one responsible for the initial, relatively weak vibrations, and it
involved resonance. As the wind moved over the bridge, it began acting like a kite or an airplane
wing. As shown in the figure, it established swirling patterns of airflow around itself, of the kind
that you can see in a moving cloud of smoke. As one of these swirls moved off of the bridge, there
was an abrupt change in air pressure, which, resulted in an up or down force on the bridge. We see
something similar when a flag flaps in the wind, except that the
flag’s surface is usually vertical. This back-and-forth sequence of
forces is exactly the kind of periodic driving force that would excite a
resonance. The faster the wind, the more quickly the swirls would get
across the bridge, and the higher the frequency of the driving force
would be. At just the right velocity, the frequency would be the right
one to excite the resonance. The wind-tunnel models, however, show
that the pattern of vibration of the bridge excited by this mechanism
would have been a different one than the one that finally destroyed
the bridge.
The bridge was probably destroyed by a different mechanism, in which its vibrations its own natural
frequency of 0.2 Hz set up an alternating pattern of wind gusts in the air immediately around it,
which then increased the amplitude of the bridge’s vibrations. This vicious cycle fed upon itself,
increasing the amplitude of the vibrations until the bridge finally collapsed.
However, the modern analysis of the Tacoma Bridge incidence does not favour the forced
resonance, with the wind providing the external frequency that match with the natural frequency of
the bridge, as the cause of the vibrations of the bridge. It has been demonstrated that the ultimate
failure of the bridge was in fact related to an aerodynamically induced conditions of self-excitation
in a torsional degree of freedom, which is a fundamentally different phenomena from that of
resonance.
31
As long as we’re on the subject of collapsing bridges, it is worth bringing up the reports of bridges
falling down when soldiers marching over them happened to step in rhythm with the bridge’s
natural frequency of oscillation. This is supposed to have happened in 1831 in Manchester,
England, and again in 1849 in Anjou, France. Many modern engineers and scientists, however, are
suspicious of the analysis of these reports. It is possible that the collapses had more to do with poor
construction and overloading than with resonance.
Example 4: Emission and absorption of light waves by atoms
In a very thin gas, the atoms are sufficiently far apart that they can act as individual vibrating
systems. Although the vibrations are of a very strange and abstract type described by the theory of
quantum mechanics, they nevertheless obey the same basic rules as ordinary mechanical vibrations.
When a thin gas made of a certain element is heated, it emits light waves with certain specific
frequencies, which are like a fingerprint of that element. As with all other vibrations, these atomic
vibrations respond most strongly to a driving force that matches their own natural frequency. Thus
if we have a relatively cold gas with light waves of various frequencies passing through it, the gas
will absorb light at precisely those frequencies at which it would emit light if heated.
Example 5: Nuclear Magnetic Resonance
NMR is a technique used to deduce the molecular structure of unknown chemical substances, and it
is also used for making medical images of the inside of people’s bodies. If you ever have an NMR
scan, they will actually tell you are undergoing “magnetic resonance imaging” or “MRI”. In fact,
the nuclei being referred to are simply the non-radioactive nuclei of atoms found naturally in your
body. (for more information, http://en.wikipedia.org/wiki/Magnetic_resonance_imaging).
Here’s how NMR works. Your body contains large numbers of hydrogen atoms, each consisting of
a small, lightweight electron orbiting around a large, heavy proton. That is, the nucleus of a
hydrogen atom is just one proton. A proton is always spinning on its own axis, and the combination
of its spin and its electrical charge cause it to behave like a tiny magnet.
Now a proton in one of your body’s hydrogen atoms finds itself surrounded by many other whirling,
electrically charged particles: its own electron, plus the electrons and nuclei of the other nearby
atoms. These neighbors act like magnets, and exert magnetic forces on the proton. The k of the
vibrating proton is simply a measure of the total strength of these magnetic forces. Depending on
the structure of the molecule in which the hydrogen atom finds itself, there will be a particular set of
magnetic forces acting on the proton and a particular value of k. The NMR apparatus bombards the
sample with radio waves, and if the frequency of the radio waves matches the resonant frequency of
the proton, the proton will absorb radio-wave energy strongly and oscillate wildly. Its vibrations are
damped not by friction, because there is no friction inside an atom, but by the reemission of radio
waves. By working backward through this chain of reasoning, one can determine the geometric
arrangement of the hydrogen atom’s neighboring atoms. It is also possible to locate atoms in space,
allowing medical images to be made.
Finally, it should be noted that the behavior of the proton cannot be described entirely correctly by
Newtonian physics. Its vibrations are described by the laws of quantum mechanics. It is impressive,
however, that the few simple ideas we have learned about resonance can still be applied
successfully to describe many aspects of this exotic system.
32
Quality Factor: It is customary to describe the amount of damping with a quantity called the quality
factor, Q, defined as the number of cycles required for the energy to fall off by a factor of 535. (The
origin of this obscure numerical factor is e 2p , where e=2.71828... is the base of natural logarithms.)
The terminology arises from the fact that friction is often considered a bad thing, so a mechanical
device that can vibrate for many oscillations before it loses a significant fraction of its energy would be
considered a high-quality device.
Example: exponential decay in a trumpet
Question: The vibrations of the air column inside a trumpet have a Q of about 10. This means that
even after the trumpet player stops blowing, the note will keep sounding for a short time. If the
player suddenly stops blowing, how will the sound intensity 20 cycles later compare with the sound
intensity while she was still blowing?
Solution: The trumpet’s Q is 10, so after 10 cycles the energy will have fallen off by a factor of
535. After another 10 cycles we lose another factor of 535, so the sound intensity is reduced by a
factor of 535x535=2.9x10*5.
The decay of a musical sound is part of what gives it its character, and a good musical instrument
should have the right Q, but the Q that is considered desirable is different for different instruments. A
guitar is meant to keep on sounding for a long time after a string has been plucked, and might have a Q
of 1000 or 10000. One of the reasons why a cheap synthesizer sounds so bad is that the sound
suddenly cuts off after a key is released.
Example: Q of a stereo speaker
Stereo speakers are not supposed to reverberate or “ring” after an electrical signal that stops
suddenly. After all, the recorded music was made by musicians who knew how to shape the decays
of their notes correctly. Adding a longer “tail” on every note would make it sound wrong. We
therefore expect that stereo speaker will have a very low Q, and indeed, most speakers are designed
with a Q of about 1. (Low-quality speakers with larger Q values are referred to as “boomy.”)
(3) When a system is driven at resonance, the steady-state vibrations have amplitude that is
proportional to Q.
This is fairly intuitive. The steady-state behavior is an equilibrium between energy input from the
driving force and energy loss due to damping. A low-Q oscillator, i.e. one with strong damping,
dumps its energy faster, resulting in lower-amplitude steady-state motion.
Our fourth and final fact about resonance is perhaps the most surprising. It gives us a way to determine numerically how wide a range of driving
frequencies will produce a strong response. As shown in the graph, resonances do not suddenly fall off
to zero outside a certain frequency range. It is usual to describe the width of a resonance by its full
width at halfmaximum (FWHM) as illustrated on the graph.
(4) The FWHM of a resonance is related to its Q and its resonant
frequency fres by the equation
FWHM =
f res
Q
------------(2.8.1)
(This equation is only a good approximation when Q is large.) Why? It is not immediately obvious that
there should be any logical relationship between Q and the FWHM. Here’s the idea. As we have seen
already, the reason why the response of an oscillator is smaller away from resonance is that much of the
driving force is being used to make the system act as if it had a different k. Roughly speaking, the halfmaximum points on the graph correspond to the places where the amount of the driving force being wasted
in this way is the same as the amount of driving force being used productively to replace the energy being
33
dumped out by the damping force. If the damping force is strong, then a large amount of force is needed to
counteract it, and we can waste quite a bit of driving force on changing k before it becomes comparable to it.
If, on the other hand, the damping force is weak, then even a small amount of force being wasted on
changing k will become significant in proportion, and we cannot get very far from the resonant frequency
before the two are comparable.
Example: Changing the pitch of a wind instrument
Question: A saxophone player normally selects which note to play by choosing a certain fingering,
which gives the saxophone a certain resonant frequency. The musician can also, however, change
the pitch significantly by altering the tightness of her lips. This corresponds to driving the horn
slightly off of resonance. If the pitch can be altered by about 5% up or down (about one musical
half-step) without too much effort, roughly what is the Q of a saxophone?
Solution: Five percent is the width on one side of the resonance, so the full width is about 10%,
FWHM / f res=0.1. This implies a Q of about 10, i.e. once the musician stops blowing, the horn will
continue sounding for about 10 cycles before its energy falls off by a factor of 535. (Blues and jazz
saxophone players will typically choose a mouthpiece that has a low Q, so that they can produce the
bluesy pitch-slides typical of their style. “Legit,” i.e. classically oriented players, use a higher- Q
setup because their style only calls for enough pitch variation to produce a vibrato.)
Example: decay of a saxophone tone
Question: If a typical saxophone setup has a Q of about 10, how long will it take for a 100-Hz tone
played on a baritone saxophone to die down by a factor of 535 in energy, after the player suddenly
stops blowing?
Solution: A Q of 10 means that it takes 10 cycles for the vibrations to die down in energy by a
factor of 535. Ten cycles at a frequency of 100 Hz would correspond to a time of 0.1 seconds,
which is not very long. This is why a saxophone note doesn’t “ring” like a note played on a piano or
an electric guitar.
Example: Q of a radio receiver
Question: A radio receiver used in the FM band needs to be tuned in to within about 0.1 MHz for
signals at about 100 MHz. What is its Q?
Solution: Q = fres / FWHM =1000. This is an extremely high Q compared to most mechanical
systems.
Musical Instruments:
Guitar Strings -A guitar string has a number of frequencies at which it will naturally vibrate.
These natural frequencies are known as the harmonics of the guitar string. Each of these natural
frequencies or harmonics is
associated with a standing wave
pattern. The graphic below
depicts the standing wave
patterns for the lowest three
harmonics or frequencies of a
guitar string.
34
The relationship between L and l can be summarized as
n
2L
for n= 1,2,3…. ---------(2.8.2)
L = l or l =
2
n
v
v
The resonance frequencies are f = = n for n= 1,2,3 ---------(2.8.3)
l 2l
Equation (2.8.3) tells us that the resonance frequencies are integer multiples of the lowest resonance
frequency, f = v/2l, which corresponds to n=1. The oscillations with the lowest frequency is the
fundamental mode or the first harmonic. n=2, n=3 are called 2nd harmonic (or 1st overtone) , 3rd
harmonic (2nd overtone) etc. The collection of all possible oscillation modes is called the harmonic
series, and n is called the harmonic number.
Open-End Air Columns- Many woodwind instruments consist of an air column enclosed inside of
a hollow metal tube. If both ends of the tube are uncovered or open, the musical instrument is said
to contain an open-end air column. A variety of instruments operate on the basis of open-end air
columns; examples include the woodwinds such as the flute and brass instruments such as the
saxophone and oboe. Even wind chimes and some organ pipes serve as open-end air columns.
Standing Wave Patterns for the first three Harmonics.
The resonance frequencies correspond to the wavelengths l =
nv
, n=1,2,3….
l 2L
Closed-End Air Columns-A closed-end instrument is an instrument in which one of the ends of the
metal tube containing the air column is covered and not open to the surrounding air. Some pipe
organs and the air column within the bottle of a pop-bottle orchestra are examples of closed-end
instruments. First three harmonics are shown below.
resonance frequencies are
f =
v
2L
, n= 1,2,3…… and the
n
=
.
l = 4L
l = 4L / 3
l = 4L / 5
4L
v nv
, n= 1,3,5,…. and f = =
, n= 1,3,5….. Note that only odd
n
l 4L
harmonics can exist in a pipe with only one open end. For example, the second harmonic, with n=2,
cannot be set up in such a pipe.
Wavelengths l =
35
2.9 The Human Ear
Understanding how humans hear is a complex subject involving the fields of physiology,
psychology and acoustics. Here, we will focus on the acoustics (the branch of physics pertaining to
sound) of hearing. We will attempt to understand how the human ear serves as an astounding
transducer, converting sound energy to mechanical energy to a nerve impulse, which is transmitted
to the brain. The ear's ability to do this allows us to perceive the pitch of sounds by detection of the
wave's frequencies, the loudness of sound by detection of the wave's amplitude and the timbre of
the sound by the detection of the various frequencies, which make up a complex sound wave.
The ear consists of three basic parts - the outer ear, the middle ear, and the inner ear. Each part of
the ear serves a specific purpose in the task of detecting and interpreting sound. The outer ear serves
to collect and channel sound to the middle ear. The middle ear serves to transform the energy of a
sound wave into the internal vibrations of the bone structure of the middle ear and ultimately
transform these vibrations into a compressional wave in the inner ear. The inner ear serves to
transform the energy of a compressional wave within the inner ear fluid into nerve impulses, which
can be transmitted to the brain. The three parts of the ear are shown below.
The outer ear consists of an ear flap and an approximately 2-cm long ear canal. The ear flap
provides protection for the middle ear in order to prevent damage to the eardrum. The outer ear also
channels sound waves which reach the ear through the ear canal to the eardrum of the middle ear.
Because of the length of the ear canal, it is capable of amplifying sounds with frequencies of
approximately 3000 Hz. As sound travels through the outer ear, the sound is still in the form of a
pressure wave, with an alternating pattern of high and low pressure regions. It is not until the sound
reaches the eardrum at the interface of the outer and the middle ear that the energy of the
mechanical wave becomes converted into vibrations of the inner bone structure of the ear.
The middle ear is an air-filled cavity, which consists of an eardrum and three tiny, interconnected
bones - the hammer, anvil, and stirrup. The eardrum is a very durable and tightly stretched
membrane, which vibrates as the incoming pressure waves reach it. As shown at the right, a
compression forces the eardrum inward and a rarefaction forces the eardrum outward, thus vibrating
the eardrum at the same frequency of the sound wave. Being connected to the hammer, the
movements of the eardrum will set the hammer, anvil, and stirrup into motion at the same frequency
of the sound wave. The stirrup is connected to the inner ear; and thus the vibrations of the stirrup
36
are transmitted to the fluid of the middle ear and create a compression wave within the fluid. The
three tiny bones of the middle ear act as levers to amplify the vibrations of the sound wave. Due to a
mechanical advantage, the displacements of the stirrup are greater than that of the hammer.
Furthermore, since the pressure wave striking the large area of the eardrum is concentrated into the
smaller area of the stirrup, the force of the vibrating stirrup is nearly 15 times larger than that of the
eardrum. This feature enhances our ability of hear the faintest of sounds. The middle ear is an airfilled cavity, which is connected by the Eustachian tube to the mouth. This connection allows for
the equalization of pressure within the air-filled cavities of the ear. When this tube becomes clogged
during a cold, the ear cavity is unable to equalize its pressure; this will often lead to earaches and
other pains.
The inner ear consists of a cochlea, the semicircular canals, and the auditory nerve. The cochlea and
the semicircular canals are filled with a water-like fluid. The fluid and nerve cells of the
semicircular canals provide no roll in the task of hearing; they merely serve as accelerometers for
detecting accelerated movements and assisting in the task of maintaining balance. The cochlea is a
snail-shaped organ which would stretch to approximately 3 cm. In addition to being filled with
fluid, the inner surface of the cochlea is lined with over 20 000 hair-like nerve cells which perform
one of the most critical roles in our ability to hear. These nerve cells have a differences in length by
minuscule amounts; they also have different degrees of resiliency to the fluid which passes over
them. As a compressional wave moves from the interface between the hammer of the middle ear
and the oval window of the inner ear through the cochlea, the small hair-like nerve cells will be set
in motion. Each hair cell has a natural sensitivity to a particular frequency of vibration. When the
frequency of the compressional wave matches the natural frequency of the nerve cell, that nerve cell
will resonate with a larger amplitude of vibration. This increased vibrational amplitude induces the
cell to release an electrical impulse which passes along the auditory nerve towards the brain. In a
process which is not clearly understood, the brain is capable of interpreting the qualities of the
sound upon reception of these electric nerve impulses.
2.10 Energy and Power of a travelling wave
A wave, mechanical or electromagnetic, carries a momentum and energy. In a mechanical wave
energy includes both kinetic energy and potential energy. In an electromagnetic wave, it includes
both electric energy and magnetic energy.
Consider a wave propagating on a string:
Consider the length dx of the string at the
two locations (1) and (2) of the wave as
shown in the diagram. It is clear that the
(1)
stretch of the string is highest at the location
(2)
(1) and it is lowest at the location (2).
Therefore, the potential energy should be a
maximum at (1) and it is lowest at (2).
dx
Now, if we consider the transverse motion
dx
of the string, where on the string the kinetic
energy would be a maximum? The kinetic
energy is a maximum when the displacement is zero. Therefore, the kinetic energy is a maximum at
locations like (1) and it is minimum at locations like (2). Therefore, both the potential energy stored
37
on the string and the kinetic energy takes their maximum values at locations like (1) and they are
minimum at locations like (2).
Consider a small element of mass dm on the string. The kinetic energy of the element,
dK = 1 2 dm u 2 ,
where u is the transverse speed of the string element.
dy
u=
dt
dy
For
u=
= - y m w cos ( kx - wt )
y = y m sin ( kx - wt ) ,
dt
Let dm = µ dx , where µ is the mass density (mass/unit length).
Then dK = 1 µ dx ( wy m ) 2 cos 2 ( kx - wt )
2
The rate at which the kinetic energy is transported along the string is
dK 1
=
µ v ( wy m ) 2 cos 2 ( kx - wt )
2
dt
dx
where
v=
dt
The time average rate is (
dK
1T
)ave = ò 1 µ v ( wy m )2 (cos 2 ( kx - wt ))dt
dt
20 2
1
= 1 µ v ( wy m ) 2 = 1 µ v w 2 y m2 --------(2.10.1)
2
4
2
When we consider the time-averaged values, the average K.E. is same as the average P.E.
Therefore, the average rate of transport of P.E. is same as (2.10.1). Hence, the total average power
dK
transported Pav = 2(
)ave = 1 µ vw 2 y m2 . ----------(2.10.2)
2
dt
µ and v depend on the material and the tension of the string. w and y m depend on the process that
generates the wave.
2.11 Derivation of the wave speed
2.11.1 Wave speed on a stretched string
Consider a single symmetrical pulse
travelling to the left. Choose the
reference frame in which the pulse
remains stationary, that is the frame,
which moves with the pulse. In this
frame, while the pulse remains
stationary the string moves from left to
right with speed v.
Fig. 2.11.1
38
Consider the forces acting on the string element Dl . Horizontal forces cancels out and the net force
F is in the downward direction.
F = 2T sin q
= T 2q
for small q
and 2q =
Dl
R
\ F = TDl / R
Mass of the string element, Dm = µ Dl , where µ is the mass density. The string element Dl moves
along the arc of radius R as shown by the circle in the diagram.
v2
Therefore, the centrifugal force acting on the element = Dm
R
2
2
Dl
v
v
\ F =T
= Dm
= µ Dl
R
R
R
T
-------(2.11.1)
v=
µ
So, the speed depends on the elasticity properties (T) and inertia ( µ ) of the string as mentioned
earlier. Note: One could obtain the above expression by dimensional analysis.
2.11.2Wave speed of a sound wave in air
In a string, the potential energy is associated with the stretching of the string elements when the
wave passes through them. As a sound wave passes through air, potential energy is associated with
periodic compression and expansions (or rarefactions) of small volume elements of the air. The
property that determines the extent to which an element of the medium changes its volume as the
pressure applied to it depends on the Bulk Modulus B, which is defined by
Dp
--------(2.11.2)
B=DV / V
Note that the minus sign is included in (2.11.2) so that B is always positive, because the signs of Dp
and Dv are always opposite. Instead of the linear mass density, µ , for the string in (2.11.1), the
corresponding quantity for the volume element of the air would be the density r . By replacing µ
and T by r and B, respectively in (2.11.1), we get the speed of sound in air v.
v=
B
r
----------(2.11.3)
Proof:
Consider a single compressional pulse travelling with seed v, from right to left, in an air column
as shown in the following figure. Consider the reference frame, the frame travelling with the pulse,
in which the pulse appears stationary.
39
In this frame, the pulse is at rest, but the air is moving to the right. Consider a slice of air of
thickness Dx and area A moving
towards the pulse with speed v, Fig.
2.11.2 (a)
(a)
p- pressure of the undisturbed air.
p + Dp - pressure inside the pulse.
Compression zone
Dp is positive since it is inside a
(b)
compression.
As the slice of air enters the pulse,
its leading face encounters a region
of higher pressure, ( p + Dp ). As a
result the speed will reduced. Lets
Fig. 2.11.2
call it v + Dv , but Dv is negative.
This slowing of the speed will
Dx
continue until the rare surface enters the pulse, which requires a time interval Dt =
v
The average force on the slice of air during this period
F = pA - ( p + Dp ) A = - Dp A
Mass of the slice = rADx = rAvDt
Dv
Average acceleration during this period =
Dt
Dv
à - Dp = r v 2
By applying the Newton’s second law - Dp A = ( rAvDt )
Dv
Dt
v
The air that occupies a volume V ( = AvDt ) outside the pulse is compressed by DV ( = ADv Dt ) as
it enters the pulse.
DV Dv
- Dp
B
Therefore,
=
r v2 =
=Bà v=
DV
V
v
r
V
2.12 Travelling Sound Waves
The variation of the displacement and the pressure associated with a sound wave travelling
in air will be discussed here. Figure 2.12.1(a) displays such a wave travelling towards the
right through a long air-filled tube. Consider a thin slice of air of thickness Dx , at the
location of x along the tube. As the wave passes through x, this element of air oscillates left
and right in simple harmonic motion about its equilibrium position (Fig. 2.12.1 (b). Thus the
oscillations of this air element due to a sound wave are similar to the oscillations of a string
element of a string wave, except that the air element oscillates longitudinally where as the
string element oscillates transversely.
40
Compression
v
x
(a)
Rarefaction
(b)
Oscillating air element
sm
s
sm
Equilibrium Position
Fig. 2.12.1
The displacement s of the air element with respect to its equilibrium position is a simple harmonic
motion, which can be written as either sine or cosine form.
Let s = s m cos( kx - wt ), -----(2.12.1)
where s m is the displacement amplitude. The wavelength l is now the distance in which the pattern
of compression or rarefaction due to the wave begins to repeat itself.
As the wave moves, the air pressure at any position x in the fig. (a) rises and falls with time, the
variation being given by
Dp = Dpm sin( kx - wt ) ------(2.12.2) (see the proof below)
A negative value of Dp in the above equation corresponds to an expansion, and a positive value
corresponds to a compression.
Fig. 2.12.1(b) shows an oscillating element of air of area A and thickness Dx , with its center
displaced from its equilibrium position by s.
From equation (2.11.2), the pressure variation in the displacement element is
DV
------(2.12.3)
Dp = - B
V
The quantity V is the volume of the element, given by V = ADx -------(2.12.4)
The quantity DV is the change in volume that occurs when the element is displaced. This volume
change comes about because the displacements of the two faces of the elements are not quite the
same (due to pressure differences), differing by some amount Ds . Thus we can write the change in
volume as
DV = ADs ------(2.12.5)
Substituting (2.12.5) and (1.12.4) into (2.12.3) and passing to the differential limit
Ds
¶s
-----(2.12.6)
Dp = - B
= -B
Dx
¶x
41
From equation (2.12.1)
Substituting in (2.12.6)
¶s
= - ks m sin ( kx - wt )
¶x
Dp = B ks m sin ( kx - wt )
In comparing with (2.12.2) and using (2.11.3)
Dpm = ( Bk )sm = ( v 2 rk )sm -------(2.12.7)
!v = w / k
= ( rwv )s m = ( 2p frv )s m ------(2.12.8)
\ sm =
Dp m
------(2.12.9)
2pr v f
The maximum pressure variation that human ear can tolerate in loud sound is about 28Pa. The
density of air r = 1.21 kg/m3 and the speed of sound in air = 343 m/s. For a sound of frequency f =
1000Hz,
From (2.12.9)
s m = 11µm .
Note that the 28Pa is very small, in compared with that of the atmospheric pressure, 105Pa.
For the faintest detectable sound at 1000 Hz à Dpm = 2.8 x10 -5 Pa
à sm = 1.1 x10 -11 m= 11 pm
The ear is indeed a very sensitive detector of sound waves. It can detect a pulse of sound whose
total energy is as small as a few electron-volts, which is in the same order of magnitude as the
energy needed to remove an electron from a single atom.
2.13 The Intensity of a Sound Wave
The average rate of energy transmitted per unit cross-section area by the wave is defined as the
intensity of the wave.
1
I = r v s m2 w 2 ------(2.13.1)
2
Proof: Consider a thin slice of air of mass dm and length dx oscillating back and forth. The crosssection area of the slice is A.
The displacement of the element about the equilibrium position
can be written as,
s = s m cos ( kx - wt ) ------(2.13.2)
1
The kinetic energy of the slice dK = dm v s2 ,
2
where, v s is the speed of the oscillating element of air.
From (2.13.2),
¶s
v s = = - s m w sin( kx - wt )
¶t
The mass of the element
dm = r A dx
1
\ dK = r A dx s m2 w 2 sin 2 ( kx - wt )
2
dm
A
dx
42
Therefore, the rate at which the K.E. is transmitted along the wave:
dK
1
= r A v s m2 w 2 sin 2 ( kx - wt ) -----(2.13.3)
dt
2
dx
where, v =
, is the wave speed. Taking the time average of (2.13.3)
dt
1T 2
1
dK
1
à
sin ( kx - wt ) dt =
(
)ave = r A v s m2 w 2
ò
dt
4
T0
2
The time-averaged potential energy is same as that of the kinetic energy, i.e., ( K .E )ave = ( P .E )ave .
dK
(
)ave
DP 2
1
The intensity
I = 2 dt
= r v s m2 w 2 = ( or u sin g 2.12.9 ) m -----(2.13.4)
A
2
2 rv
For humans,.
for the faintest detectable sound , sm = 10 -11 m
for the loudest tolerable sound,
sm = 10 -5 m
the ratio is 106
The intensity I µ S m2 ,
The ratio at the above two limits ~ 1012
The humans can hear over an enormous range of intensities!!
In order to deal with such an enormous rang of values, the logarithms are used in general.
Suppose y = log x , then
y ¢ = log ( 10 x ) = 1 + log x = 1 + y
y ¢¢ = log ( 10 10 x ) = 10 + log x = 10 + y
Therefore, instead of speaking of the intensity, I, of a sound wave, the sound level b is used, where
b is defined as
I
b = ( 10 dB ) log ( ) -----(2.13.5)
I0
W
where, I 0 = 10 - 12 2 is the standard reference intensity, which is near the lower limit of the
m
human range of hearing. As seen in the following table, when the intensity increases by a factor of
10, the sound level is increased by 10.
I
I0
100
101
102
103
104
……
1012
b (dB)
0
10
20
30
40
……
120
43
This is just the way that human hearing system works. The response of the hearing system to sound
is not the same at every frequency.
Fig. 2.13.1
Fig. 2.13.1 shows the frequency dependence of the response of the human hearing system to sound.
As can be seen both thresholds intensities depend on the frequency of the sound wave. Especially,
the threshold of pain (or the Threshold of audibility) largely depends on the frequency of the sound.
Furthermore, it is clear that the ear is most sensitive to frequencies in the rang of 3000-4000 Hz.
44
3. The Doppler Effect
3.1 Introduction
The motion related frequency changes, as proposed by an Austrian Physicist J.C. Doppler in 1842
and tested in 1845, is known as the Doppler effect. The Doppler effect holds not only for sound
waves but also for electromagnetic waves, including microwaves, radio waves and visible light.
Police use the Doppler effect with microwaves to determine the speed of a car: a radar unit beams
microwaves of a certain frequency f toward the oncoming car. The microwaves that reflect from the
metal portion of the car back to the radar unit have a higher frequency f ¢ owing to the motion of
the car relative to the radar unit. The difference between f ¢ and f provides the information on the
speed of the car, the larger the difference is, the higher the speed of the car, and the radar detector
displays the speed of the car.
However, the displayed speed is equal to the actual speed of the car only if the car is moving
directly towards the radar detector. Any deviation will reduce f ¢ and hence the detector will record
a lower speed than the speed of the car. If the radar beam is perpendicular to the velocity of the car
then f ¢ is equal to f and the radar unit will display a speed of zero for the speed of the car.
Similarly, if a radar gun is used to measure the speeds of a ball, the gun should be aimed toward an
incoming ball for accurate measurements.
45
The astronomers have used the Doppler effect of visible light to determine the speed of distance
stars and galaxies relative to the Earth. Astronomers have analyzed the wavelengths of the light that
our telescopes intercept from distance galaxies. It has been discovered that the wavelengths of the
light received by the telescopes are shifted towards the higher values (red shifted) or the frequencies
are shifted towards the lower values in compared with that of what one would observe from a
stationary galaxy. The above observation indicates that the galaxies are moving away from the
earth. It turns out that the more distance galaxy are moving away faster than the near by galaxies.
This lead to the conclusion that the universe is expanding, which in turn lead to the Big Bang model
of the Universe.
In following analysis, we consider only the Doppler effect of sound waves propagating in air, which
is at rest (no effects due to wind). The body of air is considered as the reference frame, which is
equivalent to a reference frame fixed with respect to the earth. Furthermore, we assume that the
source of sound S and the detector D are moving along the line joining them with speeds relative to
the body of air are less than the speed of sound.
When there is a relative motion between the source of sound S and the detector (or observer) D, the
frequency as measured by the detector f ¢ is given by:
The velocity of sound with respect to the Detector
-----(3.1.1)
f¢=
The wavelength as measured by the Detetor
S
Medium
Air at rest
D
Case I. Consider the case when the source is stationary and the detector is moving.
The source S is at rest and the detector is moving with
speed v D as shown in the Fig. 3.1.1. The source emits a
sound wave of frequency f and wavelength l . The speed of
the sound wave is v. The wave fronts are drawn one
wavelength ( l ) apart. The frequency detected by D is the
rate at which the detector intercepts wave fronts (or
individual wavelengths).
If D were stationary, the rate would be f. Now the detector
would intercept the wave fronts at a higher rate. Therefore,
the detected frequency f ¢ would be greater than f.
First consider the situation in which D is stationary as shown in Fig.
3.1.2.
The distance moved to the right in time t = vt
Fig. 3.1.1
Fig.3.1.2
46
The number of wavelengths in that distance vt is the number of wavelengths intercepted by D in
vt
time t, and that number is .
l
vt
The rate at which D intercepts wavelengths=ferquency f =
\f =
v
l
t
l.
-----(3.1.2)
In this situation, with D stationary (S also stationary), there is no Doppler effect. Frequency
detected is same as the frequency emitted by S.
Now consider the case again in which D moves opposite the wave as
shown in Fig.3.1.3.
In time t,
The distance moved to the right by the wave front = vt
The distance moved to the left by the detector
= vDt
The distance moved by the wave front relative to D = v D t + vt
v t + vt
The number of waves intercepted by D = D
l
The rate at which D intercepts wavelengths in this situation is:
v D t + vt
l = v + v D -------(3.1.3)
f¢=
t
l
Using equation (3.1.2)
v + vD
v + vD
--------(3.1.4)
f¢=
= f
v
v
f
Note that f ¢ is always greater than f unless v D =0.
Similarly, we can find the frequency detected by D if D moves away from the source. In this
situation, the wave front moves a distance vt - v D t relative to D in time t. Then the frequency
measured by the detector f ¢ is given by,
v - vD
v - vD
-----(3.1.5)
f¢=
= f
v
v
f
f ¢ is always less than f unless v D =0.
We can summarize the two results (3.1.4) and (3.1.5) for the moving detector with stationary source
as
v ± vD
------(3.1.6)
f¢= f
v
You can remember the sign using the physical result: when the detector moves towards the source,
the frequency is greater, which requires a plus sign in the numerator. Otherwise a minus sign is
required.
47
Equation (3.1.6) is same as f ¢ =
v ± vD
l
, where l is the wavelength of the wave. This relationship
is in agreement with the expression (3.1.1), as can be seen, the numerator is the velocity of sound
relative to the detector and the wavelength determined by the detector is l , which is same as the
emitted wavelength.
Case II. Consider the case when the detector is
stationary and the source is moving.
Let the detector D be stationary and the source is moving
towards the detector with speed v s relative to the body of
air as shown in the figure 3.1.4. Again the wave fronts are
drawn one wavelength apart. As can be seen from the
figure, the wavelength measured by the detector would be
different from that of the emitted wave. For the wave
reaching the detector, l is shorter and, it is larger when
receding from the detector. Therefore, the frequency
would be higher when the source moving towards the
detector.
Let T = 1 / f be the time between the emission of any pair of successive wave fronts W1 ,W2 ....
During the time t:
The distance traveled by the wave front W1 = vT
The distance traveled by the source
= v sT
The wave front W 2 is emitted at the end of period T.
Therefore, the distance between the two wave fronts as they reach the detector = vT - v s T
v
v
v
=
=
l ¢ vT - v sT v / f - v s / f
v
------(3.1.7)
= f
v - vs
The frequency measured by the detector f ¢ =
Note that f ¢ is always greater than f unless v s =0.
If the source is moving away from the detector, the wavelength of the wave is
l ¢ = vT + v sT
The frequency of the sound as measured by the detector:
f¢= f
v
-----(3.1.8)
v + vs
Now f ¢ must be less than f unless v s =0.
For the stationary detector with a moving source, the equations (3.1.7) and (3.1.8) can be
summarized as f ¢ = f
v
--------(3.1.9)
v ! vs
48
When the source is moving towards the detector, physically, f ¢ must be greater than f, and hence
the minus sign in (3.1.9) should be used. Similarly, plus sign should be used when the source is
moving towards the detector.
Again the equation (3.1.9) is in agreement with the equation (3.1.1). Equation (3.1.9) can be written
as
f¢=
v
v
(3.1.10)
=
T( v ! vs ) l ¢
The numerator is the speed of sound with respect to the detector and the denominator is the
wavelength as measured by the detector.
Case III. Consider the case when both the detector and the source are moving.
This is the combination of the above two cases. We can combine the equations (3.1.6) and (3.1.9) to
produce the general expression for the Doppler effect. Replacing the f (the frequency of the source)
in equation (3.1.6) with the f ¢ (the frequency of the source associated with the motion) in equation
(3.1.9) leads to
f¢= f
v ± vD
--------(3.1.11)
v ! vs
Again, the + and – signs can be selected as described previously (motion towards each other means
greater frequency). Putting v D = 0 and v s = 0 in equation (3.1.11) leads to (3.1.9) and (3.1.6),
respectively.
The Doppler effect at low speeds:
The Doppler effects for a moving source (3.1.9) and for a moving detector (3.1.6) are different even
the source and the detector are moving at the same speed. However, at very low speeds ( v D << v
and v s << v ), the frequency changes produced by the above two cases are the same.
v
v( 1 ± D )
v
v
v
where D << 1 & s << 1
f¢= f
v
v
v
v( 1 ! s )
v
n
n( n - 1 ) 2
The binomial theorem can be written in the form, ( 1 + x ) n = 1 + x +
x + .......
1!
2!
For small x, ( 1 + x )n = ( 1 + nx ) neglecting higher order terms. Using this result
v
(1± D )
v = f ( 1 ± v D )( 1 ! v s )- 1
f¢= f
v
v
v
(1! s )
v
49
v
v
vD
v
)( 1 ± s ) = f ( 1 ± D ± s )
v
v
v
v
v ± vs
f ¢ = f (1± D
)
v
u
f ¢ » f ( 1 ± ) (at low speeds) ------(3.1.12)
v
where, u = v D ± v s is the relative speed of the source with respect to the detector. The rule for sign
remains the same. If the source and the detector were moving towards each other then the frequency
would be higher; this requires that we choose the plus sign in equation (3.1.12).
f ¢ = f (1±
3.2 Supersonic Speeds
In the previous section, we considered the case when
the speed of the source or the detector is smaller than
the speed of sound (i.e. v D and v s < v). Fig. 3.2.1
shows the propagation of wave fronts when the
source of sound is moving towards a stationary
detector with the speed equal to the speed of sound,
that is v s = v. As can be seen from the figure, all
wave fronts meet the source in the forward direction
keeping the pace of source with its own wave front, and the detected frequency f ¢ would be
infinitely large.
What happens when the speed of the source exceeds
the speed of sound? Fig.3.2.2 illustrates what happens
when the speed of the source v s is greater than the
speed of sound v. The equation 3.1.9 is not applicable
under this situation. The wave front W1 was generated
when the source is at S 1 and the wave front W 5 is
generated when it is at S 5 . The radius of any wave
front in the figure is vt, where t is the time that has
elapsed since the source emitted that wave front. All
the wave fronts expand at the speed of sound v and
bunch along the V-shaped envelope. In three
dimensions the envelope is a cone, which is called the
Mach cone. A shock wave is said to exist alone the
surface of this cone, because the bunching of wave
fronts causes an abrupt rise and fall of air pressure as the surface pass through any point. The half
angle of the cone, q , as marked on the figure, which is called the Mach cone angle, is given by
sin q =
vt
v
= ------(3.2.1)
vst vs
50
The ratio v s
is known as the Mach number. If an airplane was flown at Mach 2.5, it means that
v
its speed was 2.5 times the speed of sound in the air through which the plane was flying.
The shock wave generated by a supersonic aircraft or a projectile produces a burst of sound called
sonic boom. A similar effect occurs for visible light when electrons travel through water or any
transparent medium at speeds that are greater than the speed of light in that medium. The blue glow
that emanates from the water in which highly radioactive reactor fuel rods are stored is caused by
this effect. This could be considered as “optical sonic boom”. The emitted light (or radiation) is
called Cerenkov radiation.
First supersonic flight - 1947 Bell X-1 rocket plane.
Examples of shock waves generated by a bullet (two upper figures) and airplanes are shown below.
A bullet travelling at Mach 1.01
A bullet travelling at Mach 2.45
F-18 at the exact instant it goes
supersonic
The sudden decompression of the air causes water
droplets to condense, forming a cloud in the shape
of a cone.
When Mach number > 1 à Supersonic flights
< 1 à Subsonic flights
> 5 à Hypersonic flights
Shock waves increase the drag on the aircraft. A plane designed for supersonic flights has features
that reduce the drag. Some features are:
51
• Sharply pointed nose,
• Sharp and thin edge wings that can knife through the air,
• The wings may be angled backward.
• Need powerful jet or rocket engines.
After a supersonic airplane flies overhead, people on the ground may here a sharp “boom” or
“bang”, the sonic boom, which resembles that of an explosion. Two booms, a second or two apart,
may be heard from some airplanes due to shock waves from the front and back. However, if they
are too closed only one can be heard physically. A sonic boom may be strong enough to break
windows or damage buildings. Its strength depends on the Mach number, altitude and the shape of
the plane. In general, the planes flying faster at lower elevations could produce strong sonic booms.
3.3 The Doppler effect for light
Can we use the expressions derived for sound waves with the replacement of the speed of sound
with the speed of light? This is not possible because, the sound waves –like all other mechanical
waves- need a medium for their propagation, but light waves do not require a medium. The speed of
sound is always with respect to the medium which is not applicable to light waves since the medium
is not essential. The speed of light always has the same value c, in all directions and in all inertial
frames.
For sound waves, the expressions derived above for Doppler effect depend on whether the source is
moving or the detector is moving. For light waves, however, all that can possibly matter is the
relative motion of the source and the detector. There should not be any difference whether the
source is moving or the detector is moving. The expression for the Doppler effect of light has to be
derived on the basis of the special theory of relativity and the expression would be different from
that of for the sound.
However, the relativistic expressions should lead to the non-relativistic results approximately at low
enough speeds. Thus the equation 3.1.12 with v replaced by c should holds for light waves if
u << c , where u is the relative speed between the source and the detector.
u
That is
f ¢ » f ( 1 ± ) (light waves; u << c ) ------(3.3.1)
c
For the case when the source and the detector are approaching each other, the plus sign in the
expression should be selected. In astronomy, it is the Doppler effect of the wavelength of light from
distance galaxies measured directly, not the frequency. By replacing f ¢ and f in equation (3.3.1) by
c / l ¢ and c / l , respectively, we find
u
u
l ¢ = l ( 1 ± ) - 1 » l ( 1 ! ) ------(3.3.2)
c
c
l¢ - l
u
We can write this as
-------(3.3.3)
=!
l
c
or
u=
Dl
c , -------(3.3.4)
l
where, Dl is the magnitude of the Doppler wavelength shift. If the wavelength decreases its called
blue shifted, because the blue side of the visible spectrum has lower wavelengths or higher
frequencies. If this is the case, the distance galaxy should be moving towards the earth. However,
52
all observations indicate that the wavelength of distance galaxies shifted towards higher
wavelengths (or red shifted) due to Doppler effect, which indicates that the galaxies are moving
away from the earth.
3.3.1 The Expansion of the Universe
In the late 1920's, Hubble, building on results obtained earlier by Slipher, combined Doppler shift
measurements of radial velocities with distance measurements to conclude that almost all galaxies
were flying away from the Milky Way, and that the velocity of recession was proportional to the
distance from us: the further the galaxy from us, the faster it was receding.
The details of this expansion are dictated by the Speed
value of the Hubble Constant. The objects furthest
away from us appear to be receding at near the
Slope = Hubble
velocity of light. This expansion of the universe is
constant
a result of the original explosion that created the
universe-the big bang. The big bang did not
happen in space and in time; our modern
Distance
understanding is that space and time as we
presently experience them are themselves created in the big bang. Therefore, it makes no more
sense to ask what was before the big bang than to ask what is north of the North Pole.
4. Ultrasound
4.1 Introduction
53
Sound with frequencies higher than the highest audible frequency, 20000 Hz, are called ultrasound.
However, many animals can here ultrasound. Many animals use ultrasound to detect obstacles in
their way and to detect the movement of other animals who they are hunting for. Scientists and
engineers have invented ultrasound devices, especially for medical purposes and industrial use.
Example for animals: Bats and Dolphins use ultrasound to locate objects in the dark and to
determine the motion of prey. They use the technique called echolocation. They send out pulses of
ultrasound and hear the echoes of the reflected pulses from obstacles around them. Using this
technique they could locate the direction and the distance to obstacles. At the same time by noticing
the shift in the frequency due to Doppler effect they could determine the motion of the objects or
other animals around them.
4.1.1 Ultrasound devices:
There are some simple ultrasound devices like ultrasonic whistles, which produces ultrasound in air.
Since the dogs can here ultrasound, some people use these whistles to call their dogs. There are
several complex devices that use ultrasound. Ultrasonic transducers, for example, can produce and
send ultrasound through liquids and solids and receive the echoes of the ultrasound. These devices
convert electrical energy into ultrasound and vice versa. These devices are made of materials such
as quartz and some ceramics that are piezoelectric.
The piezoelectric materials vibrate when an electric voltage is applied to them. Inversely, they
produce a voltage when a sound wave causes them to oscillate. Ultrasonic transmitters and receivers
can be made using Electronic circuits.
4.1.2 Medical uses of ultrasonic machines:
•
•
•
To detect heartbeats of fetuses.
To produce images of normal and diseased tissues inside the body.
To make images of fetuses.
•
•
•
To detect and evaluate cancer.
To detect fetal abnormalities and other conditions.
The instruments that use the Doppler effect can measure the flow of blood in the heart and
blood vessels.
A lithotripter uses pulses of ultrasound to break up gallstones (chiefly of cholesterol crystals)
or kidney stones.
•
Other uses:
• Burglar alarms.
• Automatic door openers.
• To detect flaws in metal parts.
• In machines that weld plastics.
• Tools that cut metals.
54
•
Under water sonar devices, which operate much like radar, to measure distances to the ocean
floor.
To detect submarines and other vessels.
Ultrasonic cleaning instruments: use waves to loosen dust and other contaminants from small,
delicate products such as watches and electronic components.
•
•
Ultrasound was used to make this image of a
fetus. The resolution of the image is related to
the wavelength, since details smaller than about
one wavelength cannot be resolved. High
resolution therefore requires a short wavelength,
corresponding to a high frequency.
A Doppler ultrasound study may be part of an ultrasound examination.
Doppler ultrasound is a special ultrasound technique that evaluates blood as it flows through a blood
vessel, including the body's major arteries and veins in the abdomen, arms, legs and neck.
There are three types of Doppler ultrasound:
§
§
§
Color Doppler uses a computer to convert Doppler measurements into an array of
colors to visualize the speed and direction of blood flow through a blood vessel.
Power Doppler is a newer technique that is more sensitive than color Doppler and
capable of providing greater detail of blood flow, especially in vessels that are located
inside organs. Power Doppler, however, does not help the radiologist determine the
direction of flow, which may be important in some situations.
Spectral Doppler Instead of displaying Doppler measurements visually, Spectral
Doppler displays blood flow measurements graphically, in terms of the distance traveled
per unit of time.
55
5. Dispersion
5.1 Dispersion
A truly sinusoidal wave has no
beginning or end, either in space or
time, and all its intervals of length
l are identical (Fig. 5.1 (a). Such a
wave cannot carry any signal from
point to point because it is
unchanging. We could send a
v
(a)
t=0
v
(b)
Fig. 5.1
t=0
56
l
signal with a pulse (Fig. 5.1 (b)), which could represent the quantity integer 1 in a digitized signal.
The pulse in Fig. 5.1(b) is not a pure sinusoidal wave because it is finite both in space and time.
Fourier Analysis: The pulse in (b) can be build up by combining sinusoidal waves of chosen
frequencies and amplitudes. If the combining waves all travel through a medium with the same
speed, the shape of the pulse remains the same when traveling. Then the medium and the waves are
said to be dispersionless. If the speeds of waves depend on the frequency, then the different
combining waves move at different speeds and hence the pulse changes its shape as it moves. Then
the medium and the waves are said to exhibit dispersion.
The speed of sound in air is independent of the frequency of the sound wave. All sound wave
propagates at the same speed. If you shout towards a distance cliff the returning echo can be
recognize.
Light waves are dispersionless in vacuum, but exhibit dispersion in a transparent medium. The
function ω(k), which gives ω as a function of k, is known as the dispersion relation, where ω is the
wave's angular frequency and k is the wave number.
5.2 Group Velocity and Phase Velocity
The phase velocity of a wave is the rate at which the phase of the wave propagates in space. This is
the velocity at which the phase of any one frequency component of the wave will propagate. You
could pick one particular phase of the wave (for example the crest) and it would appear to travel at
the phase velocity. The phase velocity is given in terms of the wave's angular frequency ω and wave
vector k by
Note that the phase velocity is not necessarily the same as the group velocity of the wave, which is
the rate that changes in amplitude (known as the envelope of the wave) will propagate.
The phase velocity of electromagnetic radiation
may under certain circumstances (e.g. in the case of
anomalous dispersion) exceed the speed of light in a
vacuum, but this does not indicate any superluminal
information or energy transfer.
The group velocity of a wave is the velocity with
which the variations in the shape of the wave's
amplitude (known as the modulation or envelope
of the wave) propagate through space (see the
figure). The group velocity is defined by the equation:
57
where: vg is the group velocity, ω is the wave's angular frequency and k is the wave number.
The group velocity is often thought of as the velocity at which energy or information is conveyed
along a wave. In most cases this is accurate, and the group velocity can be thought of as the signal
velocity of the waveform. However, if the wave is travelling through an absorptive medium, this
does not always hold. For example, it is possible to design experiments where the group velocity of
laser light pulses sent through specially prepared materials significantly exceeds the speed of light
in vacuum. However, superluminal communication is not possible, since the signal velocity remains
less than the speed of light. It is also possible to reduce the group velocity to zero, stopping the
pulse, or have negative group velocity, making the pulse appear to propagate backwards.
If the dispersion relationship is such that, ω is directly proportional to k, then the group velocity is
exactly equal to the phase velocity. Otherwise, the envelope of the wave will become distorted as it
propagates. This "group velocity dispersion" is an important effect in the propagation of signals
through optical fibers and in the design of short pulse lasers.
Two sources of dispersion are material dispersion and waveguide dispersion. The dispersion caused
due to the frequency-dependent response of a material is called material dispersion. Waveguide
dispersion occurs when the speed of a wave in a waveguide depends on its frequency. Dispersion in
a waveguide used for telecommunication results in signal degradation, because the varying delay in
arrival time between different components of a signal "smears out" the signal in time.
58