1
Wave Motion in One
Dimension
Wave is one of primary subjects of Physics. This topic is quite familiar to you, and revision of
concepts that you have already studied in XII will help you to understand things better. Importance
of study of waves is highlighted in brief below.
Waves ari
se in a wide range of physical phenomena.
In our common day experience light radiation
from sun, radio transmission from broadcasting station, sound/speech reaches us as waves.
Wave
occur
s
as ripples on a pond and as seismic waves following an earthquake.
Furthermore, we
communicate with each other through a variety of different waves.
(Radio waves, sound)
At the
microscopic level, the particles of matter have a wave nature as expressed by quantum wave
mechanics.
T
herefore
it may be said
that waves are at th
e heart of many branches of the physical
sciences including optics, electromagnetism, quantum mechanics and acoustics.
Most of us experienced waves as kid when we dropped a pebble into a pond. At the point where the
stone/pebble dropped in water, waves ar
e created. The waves move outward towards shore from the
point of creation. What
really
moves?
Have you played, Hello

hello using two
ice

cream
cups
in
your
childhood
? Is there any science behind it
?
Remark concerning
wave phenomena
(
A. Einstein &
L.
Infeld,
The Evolution of Physics,
1961)
A bit of gossip starting in Washington reaches New York [by word of mouth]
very quickly, even
though not a single individual who takes part in spreading it
travels between these two cities. There
are two quite differ
ent motions involved,
that of the rumor, Washington to New York, and that of the
persons
who spread the rumor. The wind, passing over a field of grain, sets up a wave
which spreads
out across the whole field. Here again we must distinguish between
the moti
on of the wave and the
motion of the separate plants, which undergo
only small oscillations... The particles constituting the
medium perform
only small vibrations, but the whole motion is that of a progressive wave. The
essentially new thing here is that f
or the first time we consider the motion of
something which is not
matter, but energy propagated through matter.
Interesting questions from Halliday and Resnik*
1)
In a long line of people waiting to buy tickets, the first person leaves and a pulse of
m
otion
occurs as people step forward to fill the gap. As each person steps forward, the
gap moves
through the line. Is the propagation of this gap transverse or longitudinal?
2)
Consider the “wave” at a baseball game: people stand up and shout as the wave arrives
at
their location, and the resultant pulse moves around the stadium. Is this wave transverse
or
longitudinal?
Important
B
ooks
*
Fundamentals of Physics
by Halliday, Resnik and Walker,
Fundamentals of vibration and waves by S.P. Puri
Dr. Devidas Gulwade,
Department of Physics, VES College
Feel free to get back to me for any queries/suggestions
2
A wave is generated by a particle of a body, which is executing periodic vibrations
that
due to
elasticity of medium gets transmitted
to other adjoining (neighbour
i
ng) particles. As a consequence,
each successive
particle
acquires kinetic energy due to its displacement
. Therefore,
the wave motion
is accompanied by energy through the medium.
Have a look at the equation that you have studied and revise the meaning
,
definitions
of each
parameter and significance. [Refer your school level books
or any other std. book
]

(E
quation I
)
Few points about the above equation
Let the wave be represented by equation at time t=0.
y(x)=A.sin(
(Why ?)
Let the velocity of wave is v.
After time t what will be the distance travelled by the wave?

What is the displacemen
t of particle at x after time t(or at any instant of time t)
y(x,t)=

(Equation II
)
*
*We are discussing the special case for sinusoidal wave and we will use this analogy for general
equation of wave.
Rearrange and do manipulations so as t
o convert Equation II to Equation I
.
[
Hint:
a)
Y
ou will have to
substitute some terms in equation
I.
[Use relation between λ & k;
What is relationship between λ, T
and velocity
of wave
?
; relation between T and ω
]
What is expression for angular velocity (ω)
and wave vector (k)
. Understand its significance.
Wave at time t
Wave at t=0
3
What is the velocity of particle and velocity of wave?
[
Understand the difference between these two
quantities
, v=d(?)/d(?). Let u be velocity of particle
(also called as transverse velocity)
then
u=
d(?)/d(?)
Write expressions for
u
. Comment on Phase difference between
x (
particle displacement
)
and
u
(
particle velocity
)
, What is maximum value of u and acceleration
(du/dt)
]
Differential equation of wave motion (Equation of wave)
a)
Start with
sinusoidal wave equation (Equation I)
b)
Calculate dy/dt, dy
2
/dt
2
and dy/dx, d
2
y/dx
2
. Establish relationship between second derivative
w.r.t. t and x)
c)
Construct
expression
[This is differential equation of wave motion and
sinusoidal wave is one of the solutions of the equations. We will study general solution of
this equation in the next section]
Problems
1.
The string is driven at a frequency of 5.00 Hz. The amplitude
of the motion is 12.0 cm, and
the
wave speed is 20.0 m/s. Determine the angular frequency and angular wave number
k
for
this wave, and write an expression for the wave function.
Calculate the maximum values for
the transverse speed and transverse accelerat
ion of any point on the string.
[Ans: ω=31.4
rad/s,
k=1.57 rad/m
,
y=(0.120 m) sin(1.57
x

31.4
t
),
3.77 m/s; 118 m/s
2
.
]
2.
A wave travelling in a string is given by a) y=10cos(0.1x

4.0t) b) 0.00327sin(72.1x

2.72t)
where x and y are in cm and t is in secon
ds. Calculate following parameters
Amplitude, T, frequency, velocity and wavelength
Max. Transverse speed and acceleration of the particle in the string
O
ne dimensional wave motion
The differential equation of wave equation is,
(Eqn. 1)
A function ψ is function of space and time and c=
√
. It will be shown later that c is velocity with
which the wave travels without change of the form. The wave variable ψ
may represent
transverse
displacement
in case of wave on strin
g, pressure in sound waves or the electric field in a radio wave.
It is function of x for plane waves but will become function of two coordinates (say x and y) for
water waves i.e. two dimensional waves. Similarly it will be function of three coordinates f
or three
dimensional waves, ex. sound waves in gas.
4
Rearranging the terms is Eqn 1.
(
)
(
)
(
)

(Eqn 2)
c velocity of wave is not a function of x or t there
fore the differential operator has been put in two
factors.
Let,
Now,
and
(
)
(Eqn 3)
Similarly it can be shown that
(
)
(Eqn 4)
Putting values of
and
in Eqn 2
Therefore wave equation (Eqn 1) becomes,
I
ntegrating this equation with respect to ξ, we get
Where
is an arbitrary
function.
Integrating once more
w.r.t.
we get
,
Where f1 and f2 are arbitrary functions. Inserting the values of
We have,
.................................................(Eqn 5)
5
This is called as D’Almberts’ solution of the wave equation representing on

dimensional wave.
Assume for the sake of simplicity let f
2
=0, therefore,
This
solution implies that in each plane x=constant, the field amplitude changes with time and at a
particular momen
t
, the field is different for different values of x. Thus, the field has the same value
for all those values of x and t, which give same
v
alue of
the argument x

ct ; that is x

ct=constant or
x=constant+ct
Calling two points (x
1
,
t
1
) and (x
2
,
t
2
)
where
has the same values, we get,
x
1

ct
1
=
x
2

c
t
2
c
=
(x
1

x
2
)/(t
1

t
2
)


(Equation
6
)
Therefore
, point moves with velocity c.
accordingly
f
1
(x

ct) represents a wave moving in the +ve x
direction with velocity c. Analogously, the solution f
2
(x+ct) represents plane wave moving with
velocity in
–
x direction.
1.
Verify whether following equations represen
ts one dimensional wave
a)
y
=Ax+Bt
b)
y
=
x
2
+c
2
t
2
c
)
y
=
10sinkx.sinct
d)
y
=
e
kx

ct
Classification of waves
The solution of wave discussed in previous section represents a wave which progresses with time.
These types of waves are called as travelling waves. The position of particle depends on time and
velocity of wave is given by Equation
6
.
The solution of wave
can be expressed in the form (by method of variable separation)
Ψ(x,t)=F
1
(x).F
2
(t)
Where F
1
and F
2
are new functions of x and t, respectively; in such a case x and t are not related and
the wave has no velocity. These types of waves are called as standing
or stationary waves. Examples
of stationary waves are musical drums, violin etc.
Waves can be further (of both kinds) can be classified in to two types depending upon direction of
vibration of particles and propagation of the wave.
Transverse waves
If t
he particles in the medium vibrate perpendicular to the direction of
propagation of wave then it is called as transverse wave.
Examples
...
Longitudinal waves
If the particles in the medium vibrate parallel to the direction of propagation of
wave then it
is called as transverse wave.
Examples
...
6
1
C
HARACTERISTICS OF WA
VE MOTION
1.
Wave motion is the disturbance produced in the medium by the repeated periodic motion in
the medium by the repeated periodic motion of particle or body
2.
Only wave travel
forward, whereas particles of the medium vibrate about their mean position
3.
There is a regular phase change between the various particles of medium. The
particles ahead
start
vibrating little later than a particle preceding it.
4.
The velocity of wave
(v=λ/T;
is different from the velocity with which
particles of medium. The
wave travel
with a uniform velocity while the velocity of the
particles is different at different position. It is maximum at the mean position and zero at
extreme points o
f the particle.
5.
The wave motion is of two types longitudinal and transverse. (However, waves may be
combination of two!!)
Exercise (
Regarding point 4 abov
e)
Let wave be represented by y=Asin(kx

wt). As per definition dy/dt
is the particle velocity and also
called as transverse velocity.
Draw waveform at t=0 for y, dy/dt and d
2
y/dt
2
and comment on it:
Step
1. y
t=0
=
Step
2
Calculate dy/dt
t=0
and d
2
y/dt
2

t=0
Step 3
Draw y
t=0
, dy/dt
t=0
and d
2
y/dt
2

t=0
one below the othe
r (keep same scale for x, don’t forget to
show labels on Y

axis)
S
tep
4
. Comment on
waveforms
(relative phase difference between x,dy/dt and x,d
2
y/dt
2
. Understand
when dy/dt is maximum relative to y and similarly d
2
y/dt
2
Step 5
.
Get back to me, if you could not complete step
4
!!!
1.1
T
RANSVERSE WAVES ON A
STRING
We will derive an expression for the velocity of a transverse wave travelling along a stretched string.
Let the string is uniform and perfectly flexible. Also, let us suppo
se that the displacement of the
string is small and the refl
ections from the ends are
ignored.
Consider that a transverse wave is
travelling along a string with velocity
‘
v
’
. Further, imagine that the whole string is being moved from
right to left with the
same velocity so that the wave remains fixed and the particles (of the string) go
around the curve successively.
Consider small portion (shaded in the
figure), having length Δs
. Let m be
the mass per unit length.
Therefore, the mass o
f the small
element of
length Δs
is m.
Δs. As this segment is moving along an arc of circle of
radius R, centripetal
force
is m.Δs.v
2
/R. This centripetal force is
provided by tension in the string.
The force T acts on each side of the segment. The horizontal
compon
ents
of the tension T cancel
each other. However,
tension T on each side
contributes
Tsinθ along the radial
direction (
along radius of circle, shown
as F
r
in the
figure). Therefore, total force along radial direction is 2Tsinθ.
Therefore, equation of moti
on becomes,
2Tsinθ =
m.Δs.v
2
/R
7
2T.θ = m.Δs.v
2
/R
by
definition of angle, 2θ = Δs./R (from the geometry shown in the figure )
Therefore,
2T.Δs./R = m.Δs.v
2
/R
i.e.
v =
√
Thus velocity of a wave along a string depends on the tension and the mass per
unit length.
Discuss how velocity changes with change in T and m
What happens when tension increases?
What if mass per unit length increases/
decreases
?
Problem
A uniform cord has a mass of 0.300 kg and a length
of
6.00
m. The cord passes over a
pulley and supports a 2.00
kg object. Find the speed of a pulse
travelling
along this cord.
1.2
L
ONGITUDINAL
W
AVES ON
R
OD
Consider an elastic rod of density ρ.
A point of cross

section is displaced through displacement ξ,
which i
s the same for all the points in cross

section.
We consider the portion of the rod of length δx lying between x and x
+ δx when the rod is at rest.
Under the influence of disturbance these coordinates become x+
ξ and x+
ξ+
δx+ δξ where δx+ δ
ξ i
s
strained
length.
F
orce per unit area =
Y
oung’s modulus X strain
and
strain = change in length / Original length
Strain
= (δξ+
δx

δx
)/
δx
=
δξ/
δx
Therefo
re,
To write the
equation of motion for the portion of rod
under consideration, we know force per unit
area is
at x and
therefore force
at x+
δx
at
(
)
(
)
Let the area of cross

section is S. Therefore the net force is given by
8
Applying Newton
’
s law of motion,
This is the equation of the wave motion on the rod. The velocity of rod is
√
. This is clear from
the e
xpression that velocity depends only on Young’s modulus and density
,
not on applied force or
the cross
sectional area.
1.3
P
RESSURE WAVES IN GAS
–
A
COUSTICS WAVES
The longitudinal waves may propagate through all phases of matter, solid, liquid, gases,
plasma in
the form of condensation and
rarefaction
. As a result there is a continuous variation of pressure all
along the direction of propagation.
Let us consider a fixed mass of a gas of volume V
o
,
pressure P
o
a
nd density
ρ
o
under
condition of
equilibri
um. When a longitudinal wave passes through it, these values are disturbed from the
equilibrium and(let it) becomes P
o
+p, V
o
+
v, and ρ
o
+
ρ
d
The
fractional change in volume
=
v/V
o
Fractional change in density
=
ρ
d/
ρ
o
Consider and infinitesimal thickness δx of gas containing in tube of unit cross

section
,
extending
between
the planes at x and x+δx. The planes get displaced to new positions x+ξ and x+δx+ξ+δ
ξ.
The volume of the gases gets disturbed because the pressure
on both sides does not balance. The net
force acting on the gas is
P
x

P
x+dx
= [P
x

(P
x
+
)]
=
dx
dx
dx

(Eqn 1)
The equation of motion for
the thin slice of gas, according to Newton’s law is
(equated it with EQN 1)
The bulk modulus of the gas,
The negative sign indicates that as volume increases, pressure decreases.
9
Now,
[
]
=
(Since Area=1 sq.units)
Substitutin
g this value of p in equation 2, we get,
The value of bulk modulus will depend on whether the changes in gas on the passage of wave are
isothermal or adiabatic.
[
Note: Isotherm
al consideration
(PV=constant, differentiating it
⁄
)
i.e. negligible temperature variation in layers of rarefaction and compression
. This
lead
to B =
p
and hence v=
√
this resulted in lesser velocity than the experimentally observed value
s
.
The Laplace suggested that the compression and rarefaction are rapid and hence there is no heat
exchange with the surrounding. This is also called as
adiabatic
correction this is us
ed in the
following derivation. This is for your information and understanding]
Let us assume that the changes are adiabatic and the gas obeys
PV
γ
= constant
Where
,
γ is the ratio of the specific heat of the gas at constant pressure to that at constant
volume.
Taking logarithms and differentiating, we get
Therefore,
The equation 4
is the differential equation of wave motion and velocity is
√
√
Speed of sound depends on bulk modulus and
density
.
1.3.1
V
ELOCITIES IN SOLIDS
AND LIQUIDS
b lk m d l s s l m ss b l y B β m ss b l y s l w s
solid i.e. bulk modulus is highest
; therefore
, velocity of sound will be highest in solids.
With
this logic it may
be argued that velocity in gas is lowest, higher in liquids and highest in solids.
(Gas: ~343 m/s
, Water: 1400
m/s
,
Aluminum and steel: ~
6
000
m/s
)
.
10
1.3.2
E
FFECT OF TEMPERATURE
AND HUMIDITY
For ideal gas for a given mass of gas
PV
/
T
is constant. As the
de
nsity of a given mass is
inversely proportional to its volume, the equ
ation may
b x ss d s P ρ s
P ρ s
Therefore, v=
√
√
. This means that velocity is proportional to
square
root of
temperature.
If
pressure is changed maintaining t he temperature constant
then density will change so that
P ρ s s d w ll s l g v l y
Alternately, PV=RT for a mole of gas. i.e.
where M is molar wt. Therefore,
and
hence, v
=
√
How velocity will change with change in humidity? (Density of moist air is less than the density
of dry air at same pressure)
1.4
P
ROBLEMS
1.
On a winter day sound travels with velocity 336 m/s. find the atmospheric temperature.
Speed of sound at 0
o
C is 332m/s.
2.
Suppose you create a pulse by
moving the free end of a
tring
up and down once with your
hand. The string is attached at its other end to a distant wall. The pulse reaches the wall in a
time
t
. Which of the following actions, taken by itself, decreases the time it takes the pulse to
reach the wall? More than one cho
ice may be correct. (a) Moving your hand more quickly,
but still only up and down once by the same amount. (b) Moving your hand more slowly, but
still only up and down once by the same amount. (c) Moving your hand a greater distance up
and down in the same
amount of time. (d) Moving your hand a lesser distance up and down
in the same amount of time. (e) Using a heavier string of the same length and under the same
tension. (f) Using a lighter string of the same length and under the same tension. (g) Using a
string of the same linear mass density but under decreased tension. (h) Using a string of the
same linear mass density but under increased tension.
3.
When all the strings on a guitar are stretched to the same
tension, will the speed of a wave
along the more
massive
bass strings be faster or slower than the speed of a wave
on the
lighter strings?
4.
If a long rope is hung from a ceiling and waves are sent up the rope from its lower end, they
do not ascend with constant speed. Explain. (Hint : 1. v will increase,
2. Try after solving
next problem)
5.
A uniform rope of length 12 m and mass 6 kg. hangs vertically from a rigid support. A block
of mass 2 kg is attached at the free end of the rope. A transverse wave of wavelength 0.06 m
is produced at lower end of the rope
. What is the wavelength of the pulse when it reaches the
top of the
rope?
(Ans: 0.12 m. hint: rope is heavy and tension is not same at different point)
6.
Find temperature at which the sound travels in H
2
, with the same speed as in O
2
at 1000
o
C.
Assume γ is same for both the gases. M
H2
/M
O2
=1/16 (Ans:

193
o
C. Hint: start with
expression of velocity and use gas equation for one mole)
7.
Show that velocity of sound in H
2
is 4 times the velocity of sound in O
2
(use data provided in
previous problems)
8.
T
he speed of sound in oxygen is 470 m/s., what will be the speed in hydrogen at same
temperature and pressure?
9.
Two blocks each of mass 3.2 kg are connected by a wire CD and the system
is suspended from the ceiling by another wir
e AB.(refer figure below). The
D
A
B
C
11
linear mass density of wire AB is 10gm/m. and that of CD is 8 gm/m. Find the speed of
transverse pulse produced in AB and in CD (77.9 m/s and 62.6 m/s)
10.
Choose correct options (may have multiple correct options). The density
of water is greater
than air. With this can you say that velocity in air a)is larger than in water b) smaller than
water c)is equal to that in water d) Can conclude anything from the given information
(insufficient information )
11.
Choose correct options. A
tuning fork sends sound waves in air. If the temperature of the air
increases, which of the parameter will change a) Displacement amplitude b)
Frequency c)
wavelength
d) time period e) velocity
12.
The speed of sound depends on a) elastic property but not in
ertial property b) inertial
property but not elastic property c) both elastic and inertial d) neither of them
13.
Velocity of sound in air at 14
o
C is 340 m/s, what will be its value at 100
o
C (γ=1.4 for air)
14.
Discuss, what will be effect on velocity if pressu
re is doubled (but temperature is constant) to
the velocity of the sound?
15.
Starting with general expression for the velocity of sound in air, show that Calculate the
increase in the velocity of sound in air per degree rise in temperature.
16.
State the wave equ
ation for one

dimensional motion and obtain its solution
17.
Verify that y=A.sin(kx

wt) represents solution of one

dimensional wave equation and hence
what is velocity of the wave. [The terms in the equation has usual meaning]
18.
Verify that y
=
f
1
(x

ct)
+
f
2
(x+
ct) represents general solution of one

dimensional
differential
wave equation
19.
Define particle velocity and acceleration
. Derive relation between particle velocity and wave
velocity [Assume y
=
A
.
sin(kx

wt)]
20.
A longitudinal disturbance generat
ed by an earthq
uake travels 100
km in 3 minutes. If
(average) density of rock is 2700 kg/m
3
. Calculate bulk modulus. [Ans: 8.3 x 10
10
N/m
2
]
21.
A wave is travelling along a string is
given by, y(x,t)=0.000327sin(72.1x

2.72t) a) what is
the amplitude b) What are T, freq,
velocity and wavelength of the wave c) what
is the
displacem
ent y at x=22.5 m and t=18.9 s
) what are transverse speed and acceleration of the
particle at t=8.9 s and x=0.225m
22.
A long string has a linear density of 0.01 kg/m and is subjected to a tension of
64 N along x

axis. Find the velocity of the transverse waves on the string.
23.
A string is stretched by suspending 5kg. The mass per unit length is 5gm/m. The travelling
waves are sent through the string by oscillating one of the e
nds with freq of 250 Hz and
amplitude
5mm. Calculate the velocity and length of the wave. Also write eqn of travelling
wave g=10m/s
2
(
A
ns: 100m/s.,
λ=0.4m)
24.
In metallic rod of density 7.5 gm/cc and Young
’
s modulus 7.5x10
11
dynes/cm
2
, the frequency
of wave is 300Hz, Find wavelength. (
Ans: 10.54 m)
25.
The freq. Of longitudinal wave produced in a solid rod is 1000Hz. Calculate the wavelength
of the waves if the density of the material is 9000kg/m
3
Y=9x10
11
N/m
2
. (Ans:
10m)
26.
Derive expression for velocity of longitudinal wave on rod
27.
Derive exp
ression for velocity of longitudinal wave in gas
28.
Derive expression for velocity of transverse wave on string
29.
What happens to the speed of a wave on a taut string when the frequency is doubled? Assume
that the tension in the string remains the same
12
2
U
LTRASONIC AND
A
COUSTICS OF
B
UILDINGS
We
are well aware of the fact that we can hear sound waves in the frequency range from 20 to
20,000 Hz and
this is called as audible range; it is not universal range also this range may change
with age.
Sound waves bey
ond frequ
ency more than 20k
Hz are called as ultrasonics. Obviously,
these waves travel
in air with the velocity of
~330 m/s. There frequency is higher therefore
wavelength is small.
2.1
I
NTRODUCTION
Sounds in the range 20

100
kHz are commonly used for commu
nication and navigation by
bats
,
dolphins
,
and some other species.
They use these waves for navigation and hunting bats can see with
the resolution of
~
mm; how is answered latter.
As we know,
wide variety of medical diagnostic applications use both the echo time and the Doppler
shift of the reflected sounds to measure th
e distance to internal organs and structures and
also the
speed of movement of the
se structures. Typical is the
2d

echocardiogram, in which a moving image
of the heart's action is produced in video form
that gives information about
the speed and direction
of
blood flow and heart valve movements. Ultrasound imaging near the surface of the body is capable
of resolutions less than a millimeter. The use of longer wave
lengths implies better resolution
and
hence high frequency (lower wa
velength) sound waves are u
sed.
2.2
P
ROPERTIES OF ULTRASO
NIC WAVES
1.
They have a high energy (higher frequency)
2.
Just like
sound waves, ultrasonic waves get reflected, refracted and absorbed.
3.
They show negligible diffraction because of their small wavelength. Hence they can be
transmitted
over long distances.
4.
They produce intense heating when passed through a substance
.
Two
important methods for generation of ultrasonic waves namely,
piezoelectric oscillator and
magnetostriction are discussed below
.
2.3
P
IEZOELECTRIC EFFECT
If one pair of opp
osite faces of a quartz crystal is subjected to pressure, the other pair of opposite
faces develops equal and opposite charges on them. The sign of the charge is reversed when faces are
subjected to tension rather than pressure /compression. The electric c
harge developed is proportional
to the amount of pressure or tension applied. This phenomenon is called as Piezoelectric effects.
The effect is reversible; if the electric field is applied to a pair of opposite faces then it result in
contraction or expa
nsion of the other pair of opposite faces.
When
the two opposite faces of a quartz crystal,
cut appropriately (perpendicular to optic axis!, axis
of hexagon, shape of quartz single crystal) subjected to alternating voltage, the other pair of faces
experien
ce stresses and strain. i.e. quartz crystal will undergo continuous contraction and expansion;
elastic vibrations are set up in the crystal.
13
When the frequency of alternating voltage is equal to the natural frequency of vibration of the crystal
it result
in resonance and exhibit large amplitude. These vibr
at
ions are longitudinal in nature. The
fundamental frequency of crystal of thickness t, exhibit a fundamental frequency of vibration n=
√
where Y is
Young’s modulus and is density.
The
inverse p
iezoelectric effect, when quartz crystal is subjected to an alternating potential it result
in elastic vibration in perpendicular direction. (This property is direction dependent and effect is
enhanced in certain direction therefore the appropriately cut c
rystal is used. Th
ere
is
hell lot of
incomplete and u
nnecessary information in text books
!
!)
This p
henomenon
/principle
is therefore
used to generate
ultrasonic waves
.
The
thickness of the crystal
is determined by the desired
frequency. A thin wafer element
vibrates with a wavelength that is twice its thickness. Therefore,
piezoelectric crystals are cut to a thickness that is 1/2 the desired radiated wavelength. The higher the
frequency of the transducer,
the thinner the active element but there will be limi
t to it as very thin
will be
too fragile.
The frequency of vibration is given by,
√
Where
Y
is
Y
oungs modulus and
ρ is density and t is thickness
Problem:
1. A quartz crystal
of thickness 0.001 m is vibrating at resonance. Calculate fundamental frequency
of vibration. Young’s modulus is 7.9 x10
10
N/m
2
and density is 2650 kg/m
3
. (Ans: 2.73x10
6
Hz)
2.
A piezoelectric crystal has a thickness of 1.5 mm. If the velocity of
propagation of longitudinal
sound
wave is 5760 m/s. Calculate fundamental frequency of vibration
(
Ans:
1.92x10
6
Hz)
3. Calculate thickness of quartz crystal for generating 500 MHz ultrasonic waves. Young’s modulus
is 7.9 x10
10
N/
m
2
and
density is 2650 kg/
m
3
.
2.3.1
A
DVANTAGES
1.
It can generate high frequencies up to 500MHz
2.
The output power is high. It is not affected by temperature or humidity
3.
Efficient than magnetostriction oscillator (discussed in next section)
4.
Produces constant and stable frequency of ultr
asonic wave
2.3.2
D
ISADVANTAGES
1.
The cost of quartz crystal is high
2.
Appropriate cutting and shaping is very complex
14
2.4
M
AGNETOSTRICTION
Ultrasonic waves may be produced using the principal of
magnetostriction
. When a rod of
ferromagnetic material like nickel i
s magnetized
,
it undergoes a change in length, this is called as
magnetostriction.
A nickel rod is placed in a rapidly varying magnetic field
,
alternately expands and contracts
with
twice the frequency of the applied magnetic field. By adjusting the freq
uency of the frequency
the
alternating field the resonance is achieved and as a result vibrations of large amplitude are produced.
The frequency of vibration of rod is given by,
√
Where l is length of rod,
and Y is
Young’s modulus and ρ density
2.4.1
A
DVANTAGES
1.
The construction cost is low
2.
They are capable of producing large acoustical power with fairly good efficiency (60%)
2.4.2
D
ISADVANTAGES
1.
It can produce frequencies up
to 3 MHz
2.
The frequency of oscillation depends on temperature
3.
I
t is diff
icult to get a
single frequency
; breadth of resonance curves us large.
2.5
D
ETECTION OF ULTRASON
IC WAVES
1.
Quartz crystal
method:
This method is based on the principal of piezoelectric effect. A
pair of
opposite faces of a quartz crystal
is
exposed to the ultrasonic
waves;
th
e opposite pair of faces
develops the charges. These charge produced is amplified and detected using an el
ectronic
circuit.
2.
Sensitive flame method: A narrow sensitive flame is moved along the medium. At the
position
of antinode
the flame is steady (because
antinode corresponds
to constant pressure).
The flame flickers at the positi
on of node because there is a change in pressure (because of
compression and rarefaction alternately). In this way position of nodes and antinodes can be
found out in the medium
and accordingly the wavelength can be found out. (The average
distance between two antinodes is equal to half the wavelength)
3.
Thermal detectors: A platinum wire
is placed in the region to be rested for ultrasonic waves.
At nodes, due to alternate compressi
ons and
rarefactions alternate
cooling and heating is
produced. Change in temperature leads to change in resistance and that is detected using
appropriate electronic circuit and hence t
he ultrasonic waves are detected
.
4.
Kundt
’
s tube method: If the waveleng
th of the ultrasonic
wave is
greater than a few
millimeters, Kundt’s tube can be used to form a stationary pattern with a well defined nodes
and antinodes. In liquid medi
um, powdered will fly from apart due to violent pressure
variation at antinodes wherea
s they remain undisturbed because particles there do not vibrate.
15
2.6
A
PPLICATIONS OF ULTRA
SONIC WAVES
1.
SONAR (
So
und
Na
vigation and
R
anging)
this
is used in submarines, it is equipped with
production, transmission and detection of ultrasonic waves. When
a p
ulse of ultrasonic wave
is sent through water it is reflected by the obstacles or other submarines present in the water
and the reflected waves are detected by the sender. By knowing time interval between the
interval between transmission and reception, th
e distance from the other objects/obstacles
may be calculated.
(Why radio waves are not used in this application?)
2.
Depth of Sea: (Using the same principle mentioned above). The waves are directed to sea

bed
from transmitter. They are reflected back from th
e sea

bed to the receiver. The time interval
between the transmission and reception (say t) gives the depth of the sea
(vt/2)
3.
Ultrasonic soldering: High

frequency sound waves are transmitted through molten solder to
remove undesirable surface oxide films f
rom a metal, thereby promoting wetting of the base
metal with solder.
High frequency waves
(i.e. violent pressure vibrations) helps to remove the
thin layer of oxide over the me
tal.
Using
ultrasonic waves,
metals such as A
luminium
,
susceptible to formation of oxide layer may be soldered and also this may b accomplished
without the use of flux.
4.
Investing of structure of Matter
We can determine the velocity of ultrasonics in
materials and its variation with
temperature
.
As the velocit
y depends on the elastic and inertial properties (bulk/young’s modulus and
density) of material that are in turn depend on atoms and their interactions with each other.
Therefore, studying velocity gives information about properties such as specific heat,
arrangements of atoms that depends on atoms and their interaction.
5.
Ultra
s
onic cleaning: This is used if
one need quality
cleaning of delicate objects such as
ornaments (which
are difficult to clean with
convention
al ways
as all parts are not
accessible
)
a
nd any materials/parts used in research
. The standing wave is produced in the water or some
other fluid and objects to be cleaned is immersed in it. Due to high frequency vibrations
during
rarefaction
leads to
formation of bubbles, (cavitation)
. These bubb
les formed bursts
during compression
on the
surface of
objects and
helps to
clean the object. The distance
between these bub
bl
es will depend on the
wavelength (that is ~mm in ultrasonics)
and hence
will affect the quality of cleaning
. This process is widel
y us
ed in research
labs where
cleanliness of objects is related with quality of results.
6.
Crack in metals: It is used to detect cracks in metal structures.
Ultrasonics are made to fall on
the metal sheet in which any fault/crack is
to be detected and
reflected si
g
n
al is detected by
detector. The reflected intensity should be same everywhere, any change in intensity
due to
crack may be detected and hence the
fault/crack in metal
sheet may be located
.
7.
Medical
Applications:
In general ultrasonics can be
used as a diagnostics tool to detect
tumors and some defect in the body such as stage of cancer, brain tumor, cardiograph, growth
of foetus. In 2d

echocardiograph
, it is used to measure the distance to internal organs and
structures and also the speed of
movement of these structures. A moving image (video) of the
heart's action (expansion and contraction) is investigated. Also, this gives information about
the speed and direction of blood flow and heart valve movements. Another example is gastric
sonograph
y.
16
Surgery:
Kidney stone
and brain tumors can be removed with the use of ultrasonic waves and
serve as painless surgery.
Sterilization: Ultrasonics waves can kill bacteria. Therefore they are used for sterilizing milk
or food products.
3
A
COUSTICS OF BUIL
DING
3.1
R
EVERBERATION
When a source produces sound wave inside a closed building, the waves are reflected by walls,
ceilings and other objects in the room
. The intensity of the sound wave decreases at every reflection
and finally sound becomes
inaudible
.
Therefore, listener receives a) direct waves 2) reflected waves
due to multiple reflections. Also, there is some gap between direct and successive/multiple reflected
waves received by the listener. Due to this effect, the sound persists for sometime even a
fter source
has stopped. This persistence of sound is called as reverberation. The time for which the sound
persists is called as reverberation time. The reverberation time is defined as the time taken by sound
(due to multiple reflections) to fall below
to one millionth of the original value. This time is a
measure of minimum audibility after the source of the sound has stopped.
The reverberation time will depend on the size of the room or the auditorium, the nature of reflecting
material on wall and ceil
ing and area of reflecting surfaces.
3.2
S
ABINE
’
S
F
ORMULA
Assumptions
1.
The average energy per unit volume is uniform
2.
The energy is not lost in the auditorium. The energy lost is only due to absorption of
materials of walls and ceiling. Also, due to the escape
through windows and ventilators.
Starting with these postulates, Sabine derived a formula for reverberation time given by,
∑
Where V is volume of t
he auditorium, A is area of the ceiling
and
wall etc. and α is
absorption.
The reverberation t
ime is proportional to
1.
Volume of the auditorium
2.
Inversely proportional to area of ceiling and walls
3.
Inversely proportional to total absorption plus transmission through open window
If the reverberation time is too small, the
sound dies away almost instantane
ously and gives the
hall a dead effect. If the reverberation time is too long, each syllable continues to be heard even
after the next syllable has been uttered. (
This
is mixing of direct sound with
the multiple reflected
previous sound and hence will be difficult to make any sense out of it this is called as
unintelligible
sound in the field of acoustics.)
This makes the sound (speech) unintelligible.
Therefore, the value of reverberation time ne
eds to be maintained at opt
i
mum value. The rooms
17
of different sizes have been studied the significance of Sabine’s formula and it is established that
the reverberation time is 1 to 2 seconds for speech. For music, it is between 2 to 2.5 seconds.
3.3
A
BSORPTION
If sound propagating in a
medium is incident on the surface
, a part of it is absorbed by the surface
and partly it is reflected. The amount of sound reflected and absorbed depends on nature of surface
on which sound is incident.
The absorption coefficient (
α
) is defined as the sou
nd energy absorbed to
the total energy incident on it.
Sabine chose one square meter of an open window
as a standard unit of absorption. All sound waves
falling on it (open window) pa
ss through it (transmitted completely) and can be said to be completely
absorbed
(
not reflected).
This unit is called as
S
abine and open window unit.
(O.W.U.)
The absorption coefficient of a material is defined as the ratio of sound energy absorbed by it
to that
absorbed by an equal area of an open window.
3.3.1
D
ETERMINATION OF ABSO
RPTION COEFFICIENT
Sabine’s method to calculate reverberation time: An organ pipe (of 512 Hz frequency) is sounded.
The reverberation time is determined using a chronograph (crudel
y we can say that a
stop watch is
started
when an organ pipe is vibrated and t
he reverberation time i.e. the time for the sound to
become inaudible is measured.
)
1.
The reverberation time (T
1
) without the absorbing materials in the hall
According to Sabine’s
formula
∑
Where
∑
is the total absorbing coefficient
,
in the hall without material and V
is volume of the hall
2.
The absorbing material is put inside the room. Let
α
1
is the coefficient of the absorbing
material an
d
s is surface area
∑
Therefore,
(
)
Knowing values of T
1,
T
2
, s, and V, value of absorption coefficient of material may be calculated.
3.4
A
COUSTICS OF
B
UILDING
The branch of Physics that deals with the design and constr
uction of buildings with good acoustic
properties is called as Acoustics. It deals with
manipulation of
1) reverberation time 2) noise
insulation and
reduction 3
) sound distribution and absorption
.
18
The few requirements of good auditorium are provided belo
w.
1.
The quality of sound must be unaltered i.e. relative intensity of the components of the sound
must be preserved
.
2.
The sound shou
l
d be sufficiently loud and intelligible in every part of hall
.
3.
Sound of each syllable should soon decay so that succeeding
syllable may be heard distinctly
(there should not be overlap between the direct sound with multiple reflected of earlier
syllable/word/sound). In brief, reverberation time should be optimum.
4.
No echoes should be present
.
5.
There should not be undesirable foc
usi
ng of sound in any part of hall. There should not be
any silence zone or regions of poor audibility in the hall.
6.
All extraneous noises must be shut out as
much
as possible.
3.4.1
F
ACTORS AFFECTING THE
ACOUSTICS OF
B
UILDING
The factors affecting acoustics of b
uilding are provided below
1.
Reverberation time
2.
Presence of echoes
3.
Adequate loudness
4.
Focusing and interference effects
5.
Extraneous and inside noise
3.4.1.1
REVERBERATION TIME
This is the important factor that affects the acoustics of sound. As discussed earlier it sh
ould be
optimum, smaller lead to dead effect and larger leads to unintelligible sound. After various trials and
experiments it is observed that reverberation time should be 1

2 s for speech and
2 to 2.3 for music
.
To adjust the reverberation time to the best optimum value, the
absorption of sound
may be increased
in following ways
1.
Provide windows and openings
2.
Cover
ceilings and walls with sound absorbing materials (ex. Asbestos)
3.
Cover floor with carpets
3.4.1.2
ECHOES
Ec
hoes can be avoided by covering the long distance walls
(responsible for echo)
with sound
absorbing materials to prevent reflection of sound.
3.4.1.3
A
DEQUATE LOUDNESS
Sufficient loudness in every portion of room is important factor for satisfactory hearing. If t
he
loudness of the sound is inadequate then it may be increased by
, low ceiling for the reflection of
sound towards the audience and/or by providing loud speakers at adequate distances.
3.4.1.4
F
OCUS
SING
(
SOUND DISTRIBUTION
)
If there is any concave surface in the
hall, sound is concentrated at its focus region. (
See
figure
below)
.
On the other hand, there might be dead places where intensity of sound is very low.
19
Therefore, these kinds of surfaces should be avoided or it should be covered with
absorbent
materials
.
Consider the following figure.
Listener L
,
will receive direct sound
as well as reflected from the
curved surface.
However there will be certain position where the intensity of sound will be less than
at L. The area where sound intensity is low is called a
s silent zone. In order to avoid this either such
shape should be avoided or good absorbing materials should be quoted on the surface.
3.4.1.5
E
XTRANEOUS NOISES
The noise which reaches the hall from outside the auditorium/hall through open doors/win
dows,
ventilato
rs.
To avoid this auditorium need to be insulated from the outside sound i.e. making the hall
air conditioned where it is completely closed,
3.4.1.6
I
NSIDE
N
OISE
The noise that is produced inside the hall is called as inside noise. It may be produced by m
achines
or air conditioners in the hall. The noise may be minimized by placing machines and air conditioner
on sound absorbent pads.
3.4.1.7
E
CHELON EFFECT
If regular structure
similar to stairs is present in the hall. The sound produced in front of such a
structu
re may produce a musical note due to successive echoes/reflection from the periodic structure.
Hence the structure should be avoided or remedy is to cover such a surface with absorbing materials
like carpet.
3.5
Q
UESTIONS
AND PROBLEMS
1.
What is reverberation
time? What mea
sures may be taken to achieve an
optimum value?
2.
Discuss reverberation.
3.
Describe the factors affecting acoustics in the hall.
4.
Discuss
piezoelectric and inverse piezoelectric effect.
5.
Calculate the coefficient of absorption of an auditorium of
90 m long, 15m wide and 8 m
high. The reverberation time is 1.2 s. Assume that the absorption of sound waves takes place
from the absorbing area of 3500 m
2
(can you solve this problem if the area is not specified!)
6.
An auditorium has a volume of 1200 m
3
,
h
as
a floor area 200 m
2
and ceiling area 200 m
2
and
wall area 180 m
2
. The absorption coefficient of wall surface, ceiling and floor surface is
0.032, 0.78 and 0.058 respectively. Calculate the average sound absorption coefficient and
reverberation time.
7.
A a
uditorium of 12 X10X5m dimensions has average absorption coefficient of 0.1. Calculate
reverberation time.
8.
An auditorium of volume 70000 cubic feet has a reverberation time of 0.9 s when empty.
Find the reverberation time when 250 persons are present in t
he auditorium. Absorption
coefficient of a person is 0.4. Average absorbing area of a person is 6 sq. feet.
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