Chapter 15 Waves
CHAPTER NO.15 WAVES
18.1 INTRODUCTION
In the previcus Chapter, we studied the motion of
objects oscillating in isolation. What happens in a system, which is a
collection of such objects? A material medium provides such an exampke. Here,
clastic forces bind the constituents to each other and, therefore, the motion
of one affects that of the other. If you drop a little pebble in a pond of
still water,the water surface gets disturbed. The disturbance does not remain
confined to one place, but propagates outward along
a circle. If you continue dropping pebbles in the
pond, you see circles rapidly moving outward from the point where the water
surface is disturbed. It gives a feeling as if the water is
moving cutward from the point of disturbance. If you
put some cork pieces on the disturbed surface, it is seen that the cork pieces
move up and down but do not move away from the centre of disturbance. This
shows that the water
mass does not flow outward with the circles, but
rather a moving disturbance is created. Similarly, when we speak,
the sound moves outward from us, without any flow of
air from one part of the medium to another. The disturbances
produced in air are much less obvious and only our
ears or a microphone can detect them. These patterns, which move
without the actual physical transfer or flow of
matter as a whole, are called waves. In this Chapter, we will study such waves.
Waves transport enemy and the pattern of disturbance
has information that propagate from one point to another. All our
communications essentially depend on transmission of sig-nals through waves.
Speech means production of sound
waves in air and hearing amounts to their detection.
Often,communication involves different kinds of waves. For exam-ple, sound
waves may be first converted into an electric cur-rent signal which in tim may
generate an electromagnetic wave that may be transmitted by an optical cable or
via a
satellite. Detection of the original signal will
usu-ally involve these steps in reverse order.
Not all waves require a medium for their
propagation. We know that light waves can travel
through vacuum. The light emitted by stars, which are hundreds of light years
away,Teaches us through inter-stellar space, which is practically a vacuum,
The most familiar type of waves such as waves ona
string, water waves, sound waves, seismic waves, etc. ia the so-called mechanical
waves.These waves require a medium for propagation,
they cannot propagate through vacuum. They involve
oscillations of constituent particles and depend on the elastic properties of
the medium.The electromagnetic waves that you will learn in Class XII are a
different type of wave.Electromagnetic waves do not necessarily require a
medium - they can travel through vacuiun.Light,radiowaves, X-rays, are all
electromagnetic waves. In vacuum, all electromagnetic waves
have the same speed c, whose value ts :
¢= 299, 792, 458 ms". (15.1)
A third kind of wave is the so-called Matter waves.
They are associated with constituents of matter: electrons, protons, neutrons,
atoms and molecules. They arise in quantum mechanical
description of nature thet you will learn in your
ater studies. Though conceptually mare abstract than mechanical or
electro-magnetic waves, they have already found applications tn several devices
basic to modern technology; matter waves associated with electrons are employed
in electron microscopes.
In this chapter we will study mechanical
waves, which require a material medium for
their propagation.
The aesthetic influence of waves on art and
Hterature is seen from very early times; yet the first scientific analysis of
wave motion dates back te the seventeenth century. Some of the famous
scientists associated with the physics of wave
motion are Christiaan Huygens (1629-1695),Robert Hooke and Isaac Newton. The
understanding of physics of waves followed the physics of oscillations of
masses tied to springs
and physics of the simple penduhumn. Waves in
elastic media are intimately connected with harmonic oscillations. (Stretched
strings, cofled springs, air, etc., are examples of elastic media).
We shall illustrate this connection throngh simple
examples.
Consider a collection of springs connected to one
another as shown tn Fig. 15.1. Ifthe apring at one end is pulled suddenly and
released, the
disturbance travels to the other end. What has A
:Pig. 15.1 A collection of springs connected to each other. The end A ts pulled
suddenly generating a disturbance, which then propagates fo the other
end.happened? The first spring 1s disturbed from its equilfprium length. Since
the second spring ts
connected to the first, it is also stretched or
compressed, and so on. The disturbance moves from one end to the other; but
each spring only executes small oscillations about its equilibrium
posttion. As a practical example of this
situation,consider a stationary train at a railway station. Different bogies of
the train are coupled to cach
other through a spring coupling. When an
engine is attached at one end, ft gives a push to
the bogie next. to it; this push is transmitted from one bogic to another
without the entire train
being bodily displaced.
Now let us consider the propagation of sound waves
in air. As the wave passes through atr, it compresses or expands a small region
of air. This causes a change in the density of that region,
say dp, this change induces a change in pressure,dp,
in that region. Pressure js force per unit area,so there is a restoring force
proportional to
the disturbance, fust like in a spring. In this
case, the quantity similar to extension or compression of the spring is the
change in density. Ifa region is compressed, the molecules fn that region are
packed together, and they tend
to move out to the adjoining region, thereby
increasing the density or creating compression in the adjoining region.
Consequently, the air in the first region undergoes rarefaction. If a
region is comparatively rarefied the surrounding air
will rush in making the rarefaction move to the adjoining region. Thus, the
compression or
rarefaction moves from one region to another,making
the propagation of a disturbance possible in air.
In solids, similar arguments can be made. In a
crystalline solid, atoms or group of atoms are arranged in a periodic lattice.
In these, each
atom or group of atoms is in equilibrium, due to
forces from the surrounding atoms. Displacing
one atom, keeping the othera fixed, leada to
restoring forces, exactly as in a spring. So we can think of atoms in a lattice
as end points,with springs between pairs of them.
In the anbsequent sections of this chapter we are
going to discuss various characteristic properties of waves.
15.2 TRANSVERSE AND LONGITUDINAL
WAVES
We have seen that motion of mechanical waves
involves oscillations of constituents of the medium. If the constituents of the
medium oscillate perpendicular to the direction of wave
propagation, we call the wave a transverse wave.If
they oscillate along the direction of wave propagation, we call the wave a
longitudinal wave.
Fig.15.2 shows the propagation of a single pulse along a string, resulting from a single up and down jerk. If the string is very long compared
to the size of the pulse, the pulse will damp out before it reaches the other end and reflection from that end may be ignored. Fig. 15.3 ahows a similar situation, but this time the external agent gives a continuous periodic sinusoidal 1p
and down jerk to one end of the string. The
resulting disturbance on the string ‘s then a sinusoidal wave. In either case
the elements of the string oscillate about their equilibrium mean
position as the pulse or wave passes through them.
The oscillations are normal to the direction of wave motion along the string,
so this is an example of transverse wave.
We can look at a wave in two ways. We can fix an
instant of tine and picture the wave in space.This will give us the shape of
the wave as a whole in space at a given instant. Another way is to fix a
location ie. fix our attention on a particular element of string and see its
oscillatory motion in time.
Fig. 15.4 describes the situation for
longitudinal waves in the most familiar example of
the propagation of sound waves. A long pipe filled with air has a piston at one
end. A single sudden push forward and pull back of the piston
will generate a pulse of condensations (higher
density) and rarefactions {lower density) in the medium (air). Ifthe push-pull
of the piston is continuous and periodic (sinusoidal), a sinusoidal wave will
be generated propagating
in air along the length of the pipe. This is clearly
an example of longitudinal waves.
The waves considered above, transverse or
longitudinal, are travelling or progressive waves since they travel from one
part of the medium to another. The material medium as a whole does not move, as
already noted. A stream, for
example, constitutes motion of water as a whole.Ina
water wave, it is the distirbance that moves,not water as a whole. Likewise a
wind (motion of air as a whole) should not be confused with a sound wave which
is a propagation of disturbance (in pressure density) in air, without the
motion of air medium as a whole.
Mechanical waves are related to the elastic
properties of the medium. In transverse waves,the constituents of the medium
oacillate perpendicular to wave motion causing change is shape. That is, each
element of the medium in subject to shearing stress. Solida and atrings
have shear modulus, that is they sustain
shearing stress. Fluids have no shape of their own -
they yield to shearing streaa. This ts why transverse waves are possible in
solids and strings (under tension) but not in fluids.However, solids as well as
fluida have bulk
modulus, that is, they can sustain compressive strain.
Since longitudinal waves involve compressive stress (pressure), they can be
propagated through solids and fluids. Thus a
steel bar possessing both bulk and sheer elastic
moduli can propagate longitudinal as well as transverse waves. But air can propagate
only
longitudinal pressure waves (sound). When a medium
such as a steel bar propagates both longitudinal and transverse waves, their
apeeda
can be different since they arise from different
elastic moduli.
Example 15.1 Gtven below are some
examples of wave motion. State in each case ifthe
wave motion ts transverse, longttudinal or a combination of both:
(a) Motionofa kink na longitudinal spring produced
by displacing one end of the spring sideways.
(b) Waves produced in a cylinder
containing a liquid by moving its piston
back and forth.
(c) Waves produced by a motorboat sailing in water.
(d) Ultrasonic waves in air produced by a vibrating
quartz crystal.
Answer
(a) Transverse and longthudinal
(b) Longitudinal
(c) Transverse and longitudinal
(d) Longitudinal
18.3 DISPLACEMENT RELATION IN A
PROGRESSIVE WAVE
For mathematical deacription of a travelling ‘wave,
we need a function of both position x and time t. Such a function at every
instant should
give the shape of the wave at that instant. Also at
every given location, it should describe the motion of the constituent of the
medium at that
location. If we wish to deacribe a sinusoidal
travelling wave (such as the one shown in Fig.15.3) the corresponding function
must also be sinusoidal. For convenience, we shall take the
wave to be transverse ao that if the position of the
constituents of the medium is denoted by x,the displacement from the
equilibrium position
may be denoted by y. A sinusoidal travelling wave is
then described by :
y(x,t) = asin(kx — wt + ) (15.2)
The term ¢ in the argument of sine function means
equivalently that we are considering a linear combination of sine and cosine
functions:y(«.t) = Asin(kx - wt) + Beos(kx -— at) (15.3)From Equations (15.2)
and (15.3),
: { B a=VA2+ BP and $= tan (5)
To understand why Equation (15.2)
represents a sinusoidal travelling wave, take a
fixed instant, say t = ¢,. Then the argument of the sine function in Equation
(15.2) is simply kx + constant. Thus the shape of the wave (at any fixed
instant) as a function of x is a sine wave. Similarly, take a fixed location,
say x= x,.Then the argument of the sine function in Equation (15.2) is constant
-wt. The displacement y, at a fixed location, thus varies sinusoidally with
time. That is, the constituents
of the medium at different positions execute simple
harmonic motion. Finally, as tincreases,xmust increase in the positive
direction to keep kx-ot+9 constant. Thus Eq. (15.2) represents
a sinusiodal (harmonic) wave travelling along the
positive direction of the x-axis. On the other hand, a function
y(x,t) = asin(kx + wt + 0) (15.4)
represents a wave travelling in the negative
direction of x-axis. Fig. (15.5) gives the names of the various physical
quantities appearing in Eq.
(15,2) that we now interpret.
Fig. 15.6 shows the plots of Eq. (15.2) for
different values of time differing by cqual intervals of time. In a wave, the
crest is the point of maximum positive displacement, the trough
is the point of maximum negative displacement.To see how a wave travels, we can fix attention ona crest and see how it progresses with time.In the figure, this is shown by a cross (x) on the crest.
In the aame
manner, we can see the motion of a particular constituent of the mediwun at a
fixed location, say at the origin of the x-axis. This is shown by a solid dot
(*) The plots of Fig. 15.6 show that with time, the solid dot (*) at the origin
moves periodically Le. the particle at the origin
oscfilates about its mean position as the wave progresses. This is true for any
other location also. We also see that during the time the solid
dot (*) has completed one full oscillation, the
crest has moved further by a certain distance.Using the plots of Fig. 15.6, we
now define the various quantities of Eq. (15.2).
15.3.1 Amplitude and Phase
In Eq. (15.2), since the sine function varies
between 1 and -], the displacement y (x.§ varies between aand —a. We can take
ato be a positive
constant, without any loss of generality. Then a represents the maximum
displacement of the constituents of the medium from their equilfbrium poattion.
Note that the displacement
y may be positive or negative, but a is positive.It
is called the amplitade of the wave.
The quantity (kx - wt + ¢ appearing as the argument
of the sine function in Eq. (15.2) 1s called the phase of the wave. Given the
amplitude a, the phase determines the displacement of the wave at any position
and at any instant. Clearly ¢ is the phase at x= 0 and
t=0. Hence ¢ is called the initial phase angle.By
suttable choice of origin on the x-axis and the intia] time, it 1s possible to
have ¢=0. Thus there is no loss of generality in dropping 4, 1.c..
in taking Eq. (15.2) with ¢=0.
15.3.2 Wavelength and Angular Wave Number The
minimum distance between two points having the same phase is called the wave
length of the wave, usually denoted by 4. For simplicity,we can choose points
of the same phase to be crests or troughs. The wavelength is then the distance
between two consecutive crests or troughs in a wave. Taking ¢ = 0 in Eq.
(15.2),the displacement at t= 0 is given by y(x,0) = asin kx (15.5)Since the
sine function repeats its value after every 2x change in angle, siniex =
sin(lex + 2nz) = sink | xX+ at
That is the displacements at points x and at x +
2nnr k are the same, where n=1,2,3,... The least distance between points with
the same displacement (at any given instant of time) is obtained by taking n=
1. ), is then given by
, Qa Qa Azs ka .k A 015.6)k is the angular wave
number or propagation
constant; its SI unit is radian per metre or rad m=*
15.3.3 Period, Angular Frequency and Frequency Fig. 15.7 shows again a
sinusoidal plot. It describes not the shape of the wave at a certain
instant but the displacement of an element (at any
fixed location) of the medium as a function of time. We may for, simplicity,
take Eq. (15.2)
with ¢=0 and monitor the motion of the element say
at x=(. We then get
‘y(0.t) = asin(-at)‘=-asinot
'@ 1s called the angular frequency of the wave.Its
SI untts is rad s“. The frequency v is the number of oscillations per second.
Therefore,
y nn 15.9)T 22 (15.8)'y is usually measured in
hertz.
In the discussion above, reference has always been
made to a wave travelling along a string or a transverse wave. In a
longitudinal wave, the displacement of an element of the medium is parallel to
the direction of propagation of the
wave. In Eq. (15.2), the displacement function for a
longitudinal wave is written as.s{x, 9 = asin (loc— owt + ¢) (15.9)where s(x,
i) is the displacement of an element of the medium in the direction of
propagation of the wave at position xand time ¢. In Eq. (15.9),
ais the displacement amplitude; other
quantities have the same meaning as in case of a
transverse wave except that the displacement function y (x, t) is to be
replaced by the function s (x, é).
Example 15.2 A wave travelling along a
string is described by,yt, t) = 0.005 sin (80.0
x-3.0 4,in which the numerical constants are in SI units (0.005 m, 80.0 rad
mm”, and 3.0 rad s°). Calculate (a) the amplitude,(b) the wavelength, and (c)
the period and frequency of the wave. Also, calculate the displacement y of the
wave at a distance x= 30.0 cm and time f= 208?
Answer On comparing this displacement
equation with Eq. (15.2),y (x. }= asin (kx ot),we
find
{a) the amplitude of the wave is 0.005 m = 5 mm.
{b) the angular wave number k and angular frequency
@ are k=80.0 mand #=3.0s"We then relate the wavelength A to k through Eq.
(15.6),A=2n/k
oe 80.0 m™ = 7.85 cm
(c) Now we relate Tto w by the relation
T=2n/a om 3.0 s7]=2.09s and frequency, v = 1/T=0.48
Hz The displacement y at x = 30.0 cm and time t = 20 9 is given by
y =(0.005 m) sin (80.0 x 0.3 - 3.0 x 20)
= (0.005 m) sin (-36 + 12a)
= (0.005 m} sin (1.699)
= (0.005 m) sin (979 = 5 mm
16.4 THE 8PEED OF A TRAVELLING WAVE
To determine the speed of propagation of a
travelling wave, we can fix our attention on any particular point on the wave
(characterized by some value of the phase) and see how that point moves in
time. It is convenient to look at the
motion of the crest of the wave. Fig. 15.8 gives the
shape of the wave at two instants of time which differ by a small time internal
At. The entire wave pattern is seen to shift to the right
(positive direction of x-axis) by a distance Ax In particular the crest shown by a cross () moves
a distance Ax in time Aft, The
speed of the wave
is then Ax/At. We can put the cross (x) on a point
with any other phase. It will move with the same speed v {otherwise the wave
pattern will not remain fixed). The motion of a fixed phase point
on the wave is given by kx—- ot = constant
(15.10)Thus, as time t changes, the position x of the fixed phase point must
change so that the phase remains constant. Thus kx- ot = ket AX — aft+Ab
ork Ax- @At=0 Taking Ax, Ait vanishingly small, this
gives wk 2, (15.11)dt ik
Relating » to Tand k to A, we get
b= OR ava Om YT (15.12)Eq. (15.12), a general
relation for all progressive waves, shows that in the time required for one
full oscillation by any
constituent of the medium, the wave pattern travels
a distance equal to the wavelength of the wave. It should be noted that the
speed of a mechanical wave is determined by the inertial
(linear mass density for strings, mass density in
general) and elastic properties (Young's modulus for near media/ shear moduhus,
bulk modulus) of the medium. The medium determines the speed; Eq. (15.12) then
relates
wavelength to frequency for the given speed. Of
course, as remarked earlier, the medium can support both transverse and
longitudinal waves,which will have different speeds in the same medium. Later
in this chapter, we shall obtain
specific expressions for the speed of mechanical
waves in some media.
15.4.1 Speed of a Transverse Wave on Stretched
String
The speed of a mechanical wave is determined by the
restoring force setup in the medium when it is disturbed and the inertial properties
(mass density) of the medium. The speed is expected to
be directly related to the former and inversely to
the latter. For waves on a string, the restoring force is provided by the
tension Tin the string.The inertial property will in this case be Imear mass
density jz, Which is mass m of the string divided by its length L. Using
Newton's Laws of Motion, an exact formula for the wave speed on a string can be
derived, but this derivation is
outside the scope of this book. We shall,therefore,
use dimensional analysis. We already know that dimensional analysis alone can
never
yield the exact formula. The overall
dimensionless constant is always left
undetermined by dimensional analysis.
The dimension of x is [ME"] and that of Tis
like force, namely [MLT?]. We need to combine these dimensions to get the
dimension of speed v [LT"]. Simple inspection shows that the quantity T/p
has the relevant dimension [mur]
bod Lv T?
[ ML! | [ ]Thus if T and p are assumed to be the
only relevant physical quantities,v=C fe (15.13)
where C is the undetermined constant of
dimensional analysis. In the exact formula, it turms
out, C=1. The speed of transverse waves on a stretched string is given by v= iE
(15.14)Note the important point that the apeed uv
depends only on the properties of the medium T and y
(T ia a property of the stretched string arising due to an external force). It
does not depend on wave length or frequency of the wave
ftself. In higher studies, you will come across
waves whose speed is not independent of frequency of the wave. Of the two
parameters 4 and y the source of disturbance determines the frequency of the
wave generated. Given the speed of the wave in the medium and the frequency Eq.
(15.12) then fixea the wavelength
Aaa (15.15)Example 15.3 Astee! wire 0.72 m long has
amass of 5.0 x10 kg. If the wire is under a tension of 60 N, what is the speed
of transverse waves on the wire ?
You can casily sce the motion of a pulsc on a rope.
You can also sce ite reflection from a rigid boundary and measure its velocity
of travel.You will uced a rope of diameter 1 to 3 cm, two hooks and some
weights. You can perform this experiment in your classroom or
laboratory.
Take a long rope or thick string of diameter 1 to 3
cm, and tc it to
hooks on opposite walls in a hall or laboratory. Let
one and pass on
a hook and hang some weight (about 1 to 5 k@ to itt.
The walls may be about 3 to 5 m apart.Take a stick or a rod and strike the rope
hard at a point ncar once end. This creates a pulse on the rope which now
travels on it. You
can sec it reaching the end and reficcting back from
it. You can
check the phase relation between the incident pulec
and refiectod Pulse. You can casily watch two or threc reflections before the pulse dica out. You
can take a stopwatch and find the time for the pulsc to travel the diatance
between the walls, and thus measure its velocity. Compare it with that obtained
from Eq. (15.14).
This js also what happens with a thin metallic
string of a musical instrument. The major difference is that the velocity on a
string fa fairly high because of low mass per unit length, as compared to that
on a thick rope. The low velocity on a rope allows us to watch the motion and
make mcasurcaments beautifully.
Arsaver Maas per unit length of the wire,_5.0x10 kg
#=—O 72m = 6.9 x10° kg m7"Tension, T= 60 N The speed of wave on the wire
fs given by = [T = J GON = 5!
. i 6.9x10Skg nv! Ooms
16.4.2 Speed of a Longitudinal Wave (Speed of Sound)
In a longitudinal wave the constituents of the
medium oscillate forward and backward in the direction of propagation of the
wave. We have already seen that the sound waves travel in the form of
compressions and rarefactions of small
volume elements ofair. The elastic property that
determines the stress under compressional strain is the bulk moduhus of the
medium defined
by (see chapter 9), AP B=-F (15.16) Here the change
in pressure AP produces a
AV volumetric strain Vv" Bhas the same
dimenaton @s pressure and given in SJ units in terms of pascal (Pa). The
inertial property relevant for the
propagation of wave in the mass density p, with
dimensions [ML*]. Simple inspection reveals that quantity 3/p has the relevant
dimension:
[Mu 1? .[mui T"] =[V T?] (15.17)
[ML]Thus if Band p are considered to be the only
relevant physical quantities,
v=c & (15.18)p where, as before, C is the
undetermined constant from dimensional analysis. The exact derivation shows
that C=1. Thus the general formula for Jongitudinal waves in a medium is: v= E
(15.19)p
Fora linear medium lfke a solid bar, the lateral expansion
of the bar is negligible and we may consider it to be only under longitudinal
strain.In that case, the relevant modulus of elasticity
in Young's modulus, which has the same
dimension as the Bulk modulus. Dimensional analysis
for this case is the same as before and yields a relation like Eq. (15.18),
with an undetermined C which the exact derivation shows to be untty. Thus the
speed of longitudinal
‘waves in a solid bar ie given by
Y v= je 15.20)& (15.20)
where Y is the Young's
modulus of the material
of the bar. Table 15.1 gives the speed of sound in some media.Table 16.1 Speed of Sound in some Media
Liquids and solids generally
have higher speeds of sound than in gases. [Note for solids,
the speed being referred to is the speed of
longitudinal waves in the solid]. This happens because they are much more
difficult to compress than gases and so have much higher values of bulk
modulus. This factor more than compensates for their higher densities than
gases.We can estimate the speed of sound in a gas in
the ideal gas approximation. For an ideal gas,the
pressure P, volume V and temperature T are related by (see Chapter 11).PV= Nk,T
(15.21)where Nis the number of molecules in volume
V, k, is the Boltzmann constant and T the
temperature of the gas (in Kelvin). Therefore, for an isothermal change it
follows from Eq.(15.21)that VAP + PAV =0
or . AP P AV/V Hence, substituting in Eq. (15.16),
we have B=P Therefore, from Eq. (15.19) the speed of a longitudinal wave in an
ideal gas is given by,v= & (15.22)This relation was first given by Newton
and is known as Newton's formula.
Example 15.4 Estimate the speed of
sound in air at standard temperature and
pressure. The mass of 1 mole of air is
29.0 x10 kg.
Answer We know that 1 mole of any gas
occupies 22.4 litres at STP. Therefore, density of
air at STP is :
Pp, = (nass of one mole of air)/ (volume of one mole
of air at STP)_ 29.0x10™ kg
22.4x107 m° = 1.29 kg m*
According to Newton’s formula for the speed of sound
in a medium, we get for the speed of sound in air at STP,
pq | Odo N m™ Pf 1 “| 1.29kg m™* =280ms? (15.25)
The result shown in Eq.(15.23) is about 15% smaller
as compared to the experimental value of 331 m s* as given in Table 15.1. Where
did we go wrong ? If we examine the basic assumption made by Newton that the
pressure variations in a medium during propagation of
sound are isothermal, we find that this is not
correct. It was pointed out by Laplace that the pressure variations in the
propagation of sound wavea are so fast that there is little time for the
heat flow to maintain constant temperature.These
variations, therefore, are adiabatic and not isothermal. For adiabatic
processes the ideal gas satisfies the relation,PV’ = conatant
Le. A(PV") =0 or PyVt" AV+ VTAP=0
Thus, for an ideal gas the adiabatic bulk modulus is
given by,= AP
Baé 077 = yP where yis the ratio of two specific
heats,C,/C,, The speed of sound is, therefore, given by,v= [re 15.
3 (15.24)This modification of Newton's formula is
referred to as the Laplace correction. For air 7=7/5. Now using Eq. (15.24) to
estimate the speed of sound in air at STP, we get a value 331.3 m a",
which agrees with the measured speed.
16.6 THE PRINCIPLE OF SUPERPOSITION OF
WAVES
What happens when two wave pulses travelling in
opposite directions cross each other? It turns out that wave pulses continue to
retain their
identities after they have crossed. However,during
the time they overlap, the wave pattern is different from either of the pulses.
Figure 15.9
shows the situation when two pulaes of equal and
oppostie shapes move towards each other.When the pulses overlap, the resultant
displacement is the algebraic sum of the displacement due to each pulse. This
is known as the principle of superposition of waves.According to this
principle, each pulse moves as if others are not present. The constituents of
the medium therefore suffer displacments due to both and since displacements can be poattive and negative, the net displacement is an algebraic sum of the two. Fig. 15.9 gives graphs of the wave shape at different times. Note the dramatic effect in the graph (c); the displacements due to the two pulses have exactly
cancelled each other and there is zero displacement throughout.To put the
principle of superposition
mathematically, let y, (x9 and y, (x,9 be the displacements
due to two wave disturbances in the medium. If the waves arrive in a region
simultaneously and therefore, overlap, the net displacement. y (x, is given by
yi O= y,&O+ yl. 8 (15.25)If we have two or more
waves moving in the medium the resultant waveform is the sum of
wave functions of individual waves. That is, if the
wave functions of the moving waves are Y, =fv0,¥, = £008,UY, = J, O08 then the
wave function describing the
disturbance in the medium is y= flx— 00+ £(x- v9+
...+ flx— ud -E Jp Gory (15.26)
The principle of superposition is basic to the
phenomenon of interference.For simplicity, consider two harmonic
travelling waves on a stretched string, both with
the same (angular frequens) and k (wave number), and, therefore, the same
wavelength A. Their wave speeds will be identical. Let us further assume that
their amplitudes are equal
and they are both travelling in the positive
direction of x-axis. The waves only differ in their initial phase. According to
Eq. (15.2), the two
waves are described by the functions:
yx = aain (kx- of (15.27)and ylx, = asin (kx- mt +
9} (15.28)The net displacement is then, by the principle of superposition,
given by y (& t) = aain (kx- mf + asin (kx- et + 9)(15.29)
nel ai] ee) *#) ot 2 2 (15.30)
where we have used the familiar trignometric
identity for sin A+ sin B. We then have y(x,t)=2a cos sa(te—ar + Jason Eq.
(15.31) is alao a harmonic travelling wave in the positive direction of x-axis,
with the same
frequency and wave length. However, ita initial
phase angle is a the algnificant thing is that its amplitude is a function of
the phase difference ¢ between the constituent two waves:
A(@ = 2acoa 4% (15.32)For » = 0, when the waves are in phase,y(x,t}=2a sin(kr—- ar) (15.33)fe. the resultant wave has amplitude 2a, the largest possible value for A. For ¢=#, the
waves are completely, out of phase and the reaultant wave has zero diaplacement
everywhere at all times y (&f) =0 (15.34) Eq.
(15.33) refers to the so-called constructive interference of the two waves
where the amplitudes add up in
the resultant wave. Eq. (15.34)
is the case of destructive intereference where the
amplitudes subtract out in the
resultant wave. Fig. 15.10 shows these two caaes of
interference of wavea arising from the principle of superposition.
15.6 REFLECTION or WAVES
So far we considered waves propagating in an
unbounded medium. What happens if a pulse or a wave meets a boundary? If the
boundary is rigid, the pulse or wave gets reflected. The phenomenon of echo is
an example of reflection by a rigid boundary. If the boundary is not completely
rigid or is an interface between two different elastic media, the situation is
some what complicated. A part of the inckient wave is reflected and a part is
transmitted into the
second medium. Ifa wave is incident obliquely on the
boundary between two different media the transmitted wave is called the
refracted wave. The incident and refracted waves obey Snell's law of
refraction, and the incident and
reflected waves obey the usual laws of
reflection.
Fig. 15.11 shows a pulse travelling along a
stretched string and being reflected by the boundary. Assuming there is no
absorption of energy by the boundary, the reflected wave has
the same shape as the incident pulse but it suffers
a phase change of x or 180° on reflection.This is because the boundary is rigid
and the disturbance must have zero displacement at all
times at the boundary. By the principle of
superposition, this is possible only if the reflected and incident waves differ
by a phase of x, so that the resultant displacement is zero. This
reasoning is based on boundary condition on a rigid
wall. We can arrive at the same conclusion dynamically also. As the pulse
arrives at the wall,it exerts a force on the wall. By Newton's Third
Law, the wall exerts an equal and opposite force on
the string generating a reflected pulse that differs by a phase of x.
If on the other hend, the boundary point is not
rigid but completely free to move (such as in the case of a string tied to a
freely moving ring ona rod), the reflected pulse has the same phase
and amplitude (assuming no energy dissipation)as the
incident pulse. The net maximum displacement at the boundary is then twice the
amplitude of each pulse. An example of non- rigid
boundary is the open end of an organ pipe.
To summarize, a travelling wave or pulse
suffers a phase change of r on reflection at a Tigid
boundary and no phase change on Teflection at an open boundary. To put this
mathematically, let the Incident travelling wave
be y2 (x,t) = asin (kx a)At a rigid boundary, the
reflected wave is given
by y,0< 0 = asin (kx- ost + x).=- asin (kx- ot)
(15.35)At an open boundary, the reflected wave is given by y,0c 0 = asin (xc—
oot + 0).= asin (kx- at) (15.36)
Clearly, at the rigid boundary, y= y, + y, =0 at all
times.
16.6.1 Standing Waves and Normal Modes
We considered above reflection at one boundary.But
there are familiar situations (a string fixed at either end or an air column in
a pipe with
either end closed) in which reflection takes place
at two or more boundaries. In a string, for example, a wave going to the right
will get Teflected at one end, which in turn will travel and get reflected from
the other end. This will
go on unt] there is a steady wave pattern set up on
the string. Such wave patterns are called standing waves or stationary waves.
To see this mathematically, consider a wave travelling along the positive
direction of x-axis and a Teflected wave of the same amplitude and wavelength
in the negative direction of x-axis.From Fg. (15.2) and (15.4), with ¢= 0, we
get:
yx, t) = asin (kx at)y,0< t) = asin (kx + at)
The resultant wave on the string is, according to
the principle of superposition:yi d= ule 8 tye 8
=a [sin (kx- mt) + sin (kx + at)]
Using the familiar trignometric identity
Sin (A+B) + Sin (A-B) = 2 sin A cosB we get,yl, ft]
= 2asin kx cos at (15.37)
Note the tmportant difference in the wave pattern
described by Eq. (15.37) from that described by Eq. (15.2) or Eq. (15.4). The
terms kx and ot appear separately, not in the combination kx- wt, The amplitude
of this wave is 2a sin kx. Thus in this wave pattern, the amplitude varies from
potnt to point, but each element of the string oscillates with the same angular
frequency or time period. There is no
phase difference between oscillations of different
elements of the wave. The string as a whole vibrates in phase with differing
amplitudes at different points. The wave pattern is neither
moving to the right nor to the left. Hence they are
called standing or stationary waves. The amplitude is fixed at a given location
but, as remarked earlier, it is different at different
locations. The points at which the amplitude is zero
{l.c., where there is no motion at all) are nodes; the points at which the
amplitude fs the largest are called antinedes. Fig. 15.12 shows
a stationary wave pattern resulting from
superposition of two travelling waves in opposite
directions.
The most significant feature of stationary ‘waves ie
that the boundary conditions constrain the possible wavelengths or frequencies
of
vibration of the system. The system cannot oscillate
with any arbitrary frequency (contrast this with a harmonic travelling wave),
but is characterized by a set of natural frequencies or
normal modes of oscillation. Let ue determine these
normal modes for a stretched string fixed at both ends.
First, from Eq. (15.37), the positions of nodes
(where the amplitude ts zero) are given by sinkx=0. which implies
kx=nx n=0,1,2,3,...Stoce k =22// , we get nA x=
in=0,1,2,3,... (15.38)
Clearly, the distance between any two
successive nodes is < in the same way, the
positions of antinodes (where the amplitude is the largest) are given by the
largest value of sin kx:lsin kxl = 1
which implies kx= (n+) xz;n=0,1,2,3,...
With k = 22/4, we get x=(n+ vy 4 ;n=0,1,2,3,...
(15.39)Again the distance between any two consecutive antinodes ts 4. Eq.
(15.38) can be applied to
the case of a stretched string of length L fixed at
both ends. Taking one end to be at x= 0, the boundary conditions are that x = 0
and x= L are positions of nodes. The x = 0 condition is already satisfied. The
x = L node condition
requires that the length L is related to 4 by L=n4;
n=1,2,3,.. (15.40)Thus, the possible wavelengths of stationary waves are
constrained by the relation
as ze; nH1,2,3,.. (15.41)n with corresponding
frequencies ve 2 for n=1,2,3, (15.42)2L We have thus obtained the natural
frequencies - the normal modes of oscillation of the system.
The loweat possible natural frequency of a system is
called its fandamental mode or the first harmonic. For the stretched string
fixed “v at efther end it is given by v= op’ responding to n= 1 of Eq. (15.42).
Here v is the speed of
wave determined by the properties of the
medium. The n = 2 frequency is called the second
harmonic; n = 3 is the third harmonic and so on. We can label the various
harmonics by the symbol v, (n= 1, 2, sods Fig. 15.13 shows the first six
harmonics of a stretched string
fixed at either end. A string need not vibrate in one of these modes
only. Generally, the vibration of a string will be a superposition of
different modes; some modes may be more strongly
excited and some less. Musical instruments Ife sitar or violin are based on
this principle. Where the string is plucked or bowed, determines
which modes are more prominent han others.Let us
next consider normal modes
of oscillation of an air column with one end closed and the other open.
A glass tube partially filled with water fllustrates thie system. The
end in contact with water is a node,while the open end is an antinode.
At the node the pressure changes are the largest, while the displacement is minimum (zero). At the open end - the antinode, it is just the other way - least pressure change and maximum
amplitude of displacement. Taking the end in contact
with water to be x= 0, the node condition (Eq. 15.38) is
already satisfied. If the other end x
= Lis an antinode, Eq. (15.39) gives
lla L= (n+ >) qforn=0, 1,2,3,...
The possible wavelengths are then restricted by the
relation :
A= 4 _, forn=0, 1.2.3... (15.43)
(n + 1/2)The normal modes — the natural frequencies
—of the system are
+) nn 801,23 18 vent a)op ine @ Be hay wre ove ( AA)The fundamental frequency corresponds to n=
0, and ia given by —— . The
higher frequencies are odd harmonics, i.c., odd multiples of the
fundamental frequency : 3—, 5 ete.
Fig. 15.14 shows the first six odd harmonics of air
column with one end closed and the other open. For a pipe open at both ends,
each end is an antinode. It is then easily seen that an open air cohimn at both
ends generates all harmonics (See Fig, 15.15).
The syatems above, strings and air columns,can also
undergo forced oscillations (Chapter 14). If the external frequency is close to
one of
the natural frequencies, the system shows resonance.
Normal modes of a circular membrane rigidly clamped
to the circumference as in a tabla are determined by the boundary condition
that no point on the circumference of the membrane vibrates. Estimation of the
frequencies of normal modes of this system is more complex. This problem involves
wave propagation in two dimensions. However, the underlying physics is the
same.
Example 15.5 A pipe, 30.0 cm long, is
open at both ends. Which harmonic mode
of the pipe resonates a 1.1 kHz source? Will
resonance with the same source be
observed if one end of the pipe is closed 7 Take the
speed of sound in air as 330ms".
Answer The first harmonic frequency is given by Y=
Z=77 (open pipe)where Lis the length of the pipe. The frequency
of its rth harmonic ts:v,= or for n=1, 2, 3, ...
(open pipe)First few modes of an open pipe are shown in Fig. 15.14.
For L= 30.0 cm, v=330ms",‘nx 330 (ms™)
%.= o6m = 550
ns Clearly, a source of frequency 1.1 kHz will resonate at v,, i.e. the second
harmonic.
Now ifane end of the pipe is closed (Fig. 15.15),it
follows from Eq. (14.50) that the fundamental frequency is
v,= Gra (pipe closed at one end)and only the odd
numbered harmonics are present ;
“3p 7 v, = ~, v,= Fp + and 80 on.
For L = 30 cm and uv = 330 m a", the
fundamental frequency of the pipe closed at one end
is 275 Hz and the source frequency corresponds to its fourth harmonic. Since
this harmonic is not a posafble mode, no resonance will be observed with the
source, the moment
one end is closed.
15.7 BEATS
‘Beats’ is an interesting phenomenon arising from interference of waves. When two harmonic sound waves of close (but not equal) frequencies
are heard at the same time, we hear a sound of similar frequency
(the average of two close
frequencies), but we hear something else also.We
hear audibly distinct waxing and waning of the intensity of the sound, with a
frequency equal] to the difference in the two close frequencies. Artists use
this phenomenon often
while tuning their instruments with each other.They
go on tuning until their sensitive ears do not detect any beats.
To see this mathematically, let us consider two
harmonic sound waves of nearly equal angular frequency @, and m, and fix the
location to be x = O for convenience. Eq. (15.2) with a suitable choice of
phase ($ = */2 for each) and,
assuming equal amplitudes, gives
&, =acosa@t and s,=acos wt (15.45)
Here we have replaced the symbol y by s.
since we are referring to longitudinal not
tranaverse displacement. Let m, be the (slightly)greater of the two
frequencies. The resultant displacement is, by the principle of superposition,
$= 5,+5,=a(cos@,t+cos 9 Using the familiar
trignometric identity for
cas A+ cosB, we get =2a cos Ait cos ant (15.46)
which may be written as :
s=[2a cos wt] cos mt (15.47)If |e, -—a,1 <<a,
w,, w, >>, th where a, = (@-%) and @, = (4+%)2 2 Now if we assume | @,
—m,! <<o,, which means Musical Pillars
Temples often have some pillars portraying human
figures playing musical instru- ments, but sekiom do these pillars themse}ves produce music. At the
Nellaiappar temple in Tamil Nadu,gentle taps on a cluster of pillars carved out
ofa single piece of rock produce the basic notes of Indian classical music,
viz. Sa, Re. Ga, Ma, Pa, Dha.Ni, Sa. Vibrations of these pillars depend on
elasticity of the stone used, tts density and shape.
Musical pillars are categorised into three types:
The first is called the Shrati Pillar,as it can produce the basic notes — the
“swaras”, The second type is the Gana Thoongal, which gencrates the basic tuncs
that make up the “ragas”. The third variety ja the Laya Thoongal pillars that
produce “taal” (beats) when tapped. The pillars at the Nellaiappar temple are a
combination of the Shruti and Laya types.
Archacologists date the Nelliappar
temple to the 7th century and claim ft was built by
successive mulers of the Pandyan dynasty.
The musical pillars of Nelliappar and
several other temples tn southern India like those
at Hampi (picture), Kanyalommari, and Thinivananthapuram are unique to the
country and have no parallel in any other part of the world.
@, >> @,, we can interpret Eq. (15.47) aa
follows.The resultant wave is oscillating with the average angular frequency ,;
however ita amplitude is not constant in time, unlike a pure harmonic wave. The
amplitude is the largest when the term cos @,¢ takes ite limit +1 or —1. In
other words, the intensity of the resultant wave waxes
and wanes with a frequency which is 2m, = @, - @,.
Since m= 2av, the beat frequency v,_. is given by Voeat = V1 Ve (15.48)Fig.
15.16 illustrates the phenomenon of beats for two harmonic waves of frequencies
11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency
of 2 Hz.
Example 15.6 Two sitar strings Aand B
playing the note ‘Dha’ are slightly out of tune and
produce beats of frequency 5 Hz. The tension of the string B is slightly
increased and the beat frequency is found to decrease to 3 Hz. What is the
original frequency of B if the frequency of A 1s 427 Hz?
Answer Increase in the tension of a string increases
its frequency. If the original frequency of B (v,) were greater than that of A
(v,), further
increase in v, should have resulted in an increase
in the heat frequency. But the beat frequency is found to decrease. This shows
that v, < v,. Since v,—- v, = 5 Hz, and v, = 427 Hz, weget v, = 422 Hz.
15.8 DOPPLER EFFECT
It is an everyday experience that the pitch (or
frequency) of the whistle of a fast moving train Reflection of sound in an open
pipe When ae high pressure pulse of air travelling down an open pipe
reaches the other end, its momentum
drags the air out into the open, where
pressure falls rapidly to the
atmospheric pressure. AS a result the air following
after it in the tube is
pushed out. The low pressure at the end of the tube
draws air from further up the tube.The air gets drawn towards the open end forcing
the low pressure region to move upwards. As a result a pulse of high pressure
air travelling down the tube turns into a pulse of fow pressure air travelling
up the tube. We say a pressure wave has been reflected at the open end with a
change in phase of 180°. Standing waves in an open pipe organ like the fhite is
a result of this
phenomenon.Compare this with what happens when a
pulse of high pressure air arrives at a closed end: it collides and as a result
pushes the air back in the opposite direction. Here,we say that the pressure
wave Js reflected,
with no change in phase.decreases as it recedes
away. When we approach a stationary source of sound with high
speed, the pitch of the sound heard appears to be
higher than that of the source. As the observer recedes away from the source,
the observed pitch (or frequency) becomes lower than that of the source. This
motion-related
frequency change is called Doppler effect. The
Austrian physicist Johann Christian Doppler first proposed the effect in 1842.
Buys Ballot in
Holland tested it experimentally in 1845.Doppler
effect is a wave phenomenon, it holds not only for sound waves but also for
electromagnetic waves. However, here we shall consider only sound waves.We
shall analyse changes in frequency under three different situations: (1)
observer is
stationary but the source is moving, (2) observer is
moving but the source is stationary, and (3)both the observer and the source
are moving.The situations (1) and (2) differ from each other
Decause of the absence or presence of relative
motion between the observer and the medium.Most waves require a medium for
their propagation; however, electromagnetic waves do not require any medium for
propagation. If there
is no medium preaent, the Doppler shifts are same
irrespective of whether the source moves or the observer moves, since there is
no way of distinction between the two situations.
16.8.1 Source Moving ; Observer Stationary Let us
choose the convention to take the direction from the observer to the source as
the positive direction of velocity. Considera source S moving with velocity v,
and an observer
who is stationary in a frame in which the medium is
also at rest. Let the speed of a wave of angular frequency @ and period T, both
measured by an observer at rest with respect to
the medtum, be v. We assume that the observer has a
detector that counts every time a wave crest reaches it. As shown in Fig.
15.17, at time t = 0 the source is at point S,located at a distance L from the
observer, and
emits a crest. This reaches the observer at
time t, = L/v.
At time t = T, the source has moved a distance u,T, and is at point S., located at a distance (L + v,T,) from the observer. At S.,, the
source emits a second crest. This reaches the
observer at ty = T, +G+ at)
At time n T,, the source emits fta {n+1)® crest and
this reaches the observer at time tor = NT {et nut) net Hence, in a time
interval | + (+ net) -4 v v the observer's detector counts n cresta and the
observer recorda the period of the wave as T given by T = et + (L+ nv.Ty) -E
iin v v pg ele = vs = 1, (: + 2) (15.49)
Equation (15.49) may be rewritten in terms of the
frequency v, that would be measured if the source and observer were stationary,
and the frequency v observed when the source is moving, a5 v, Y'
veX% (: + =) (15.50)If v, is small compared with the
wave speed u,taking binomial expansion to terms in first order in v,/v and
neglecting higher power, Eq. (15.50)may be approximated, giving vey (: - 2)
(15.51)For a source approaching the observer, we replace u, by -v, to get ve
vfs *) (15.52)The observer thus measures a lower frequency
when the source recedes from him than he does when
it ia at rest. He measures a higher frequency when the source approaches him.
18.8.2 Observer Moving;
Source Stationary Now to derive the Doppler shift when the observer is moving
with velocity v, towards the source and the source is at reat, we have to
proceed in a different manner. We work in the
reference frame of the moving observer. In this
reference frame the source and medium are approaching at speed uv, and the
speed with which the wave approaches ia u, + v. Following a similar procedure
as in the previous case, we find that the time interval between the arrival
of the first and the (n+1) th crests is
boa ~ t, Bn Ty ~ nwo Uy tv The observer thus,
measures the period of the
wave to be vo yf Y% yel T, jr “ |
giving vy [1+] (15.53)Ls i is small, the Doppler
shift is almost same whether it is the observer or the source moving since Eq.
(15.53) and the approximate relation Eq. (15.51 ) are the same.
18.8.3 Both Source and Observer Moving
We will now derive a general expression for Doppler
shift when both the source and theobserver are moving. As before, let us take
the direction from the observer to the source as the positive direction. Let
the source and the
observer be moving with velocities v, and
uv,reapectively as shown in Fig.15.18. Suppose at time t = 0, the observer is
at O, and the source
is at S,, O, being to the left of S,. The source
emits a wave of velocity v, of frequency v and period 7, all measured by an
observer at rest
with respect to the medium. Let L be the
distance between O, and S, at £ = 0, when the source
emits the first crest. Now, since the observer is moving, the velocity of the
wave relative to the observer is v+up. Therefore, the
firat creat reaches the observer at time t, =
L/{v-+u,). At time t = T,, both the observer and the source have moved to their
new poattions O, and S, respectively. The new distance between the observer and
the source, O, S,, would be L+(u,— Vg) Tol. At S,, the source emits a second
crest.
Application of Doppler effect The change in
frequency caused by a moving objec due to Doppler effect ia used to measure the
velocities in diverse areas such as military medical acience, astrophysica,
etc, It is also uses by police te check over-speeding of vehicles.A sound wave
or electromagnetic wave o known frequency is sent towards a moving object Some
part of the wave is reflected from the objec and ite frequency is detected by
the monitorin;
station, This change in frequency is called Dopple:
ohift.It 19 used at airports to guide aircraft, and 11 the military to detect
enemy aircraft
Astrophysicists use it to measure the velocitie of
atara.
Doctors use it te study heart beata and bloa:flew in
different parts of the body. Here they us:ulltrasonic waves, and in common
practice, it 1 called sonography. Ultrasonic waves enter th
body of the person, some of them are reflecte back,
and give information about motion of bloo and pulsation of heart valvea, as
well as pulsation
of the heart of the foetus. In the case of heart the
picture generated is called echocardiogram This reaches the observer at time.=
T, + IL+ (¥,-v)T I /e +2)
At time nT, the source emits its {(n+1) th cre and
this reaches the observer at time ta = NT, + (b+ nv, —vJT,)] /(v + v0)Hence, in
a time interval £,,,, -t, i.e., ewnflt 0 EP a os fee ean VRS YE shen noe FEF
shee ¢:the observer counts n crests and the observer records the period ofthe
wave as equal to Tgiven by
T= fT, ( p Uso 7B fr exe Us D+ VEU,
(15.54)The frequency v observed by the observer is
given by vay, fore “| (15.55)U+D,
Consider a passenger sitting in a train moving on a
straight track. Suppose she hears a whistle sounded by the driver of the train.
What frequency will she measure or hear? Here both the observer and the source
are moving with
the same velocity, so there will be no shift in
frequency and the passenger will note the natural frequency. But an observer
outside who is stationary with respect to the track will note
a higher frequency if the train is approaching hfm
and a lower frequency when it recedes from him.
Note that we have defined the direction from the
observer to the source as the positive direction. Therefore, if the observer is
moving towards the source, vu) has a positive (mumerical)value whereas if O is
moving away from 5, vu,
has a negative value. On the other hand, if S is
moving away from O, u, has a positive value whereas if it is moving towards O,
v, has a negative value. The sound emitted by the source travels in all
directions. It is that part of sound
coming towards the observer which the observer
recefvea and detects. Therefore, the relative velocity of sound with respect to
the observer is
o+u, in all cases.
Exampte 15.7 A rocket is moving at a
speed of 200 m s° towards a stationary
target. While moving, it emits a wave of
frequency 1000 Hz. Some of the sound
reaching the target gets reflected back to the
rocket as an echo. Calculate (1) the frequency of the sound as detected by the
target and (2) the frequency of the echo as detected by the rocket.
Answer (1) The observer is at rest and the source is
moving with a speed of 200 m s“'. Since this is comparable with the velocity of
sound,330 m s“, we must use Eq. (15.50) and not the approximate Eq. (15.51).
Since the source is approaching a stationary target, v, = 0, and v,
must be replaced by -v,. Thus, we have
yl v =»)v= 1000 Hzx [1-200 ms'/330ms'P"
= 2540 Hz
{2) The target is now the source (because it is the
source of echo) and the rocket’s detector is now the detector or observer
(because it detects
echo). Thus, v, = 0 and v, has a posttive value.The
frequency of the sound emitted by the source {the target) 1s v, the frequency
intercepted by
the target and not v,. Therefore, the frequency as
registered by the rocket is
U+0 v= ‘on o ] . v =2540 Hex 200ms” +330m Ss”330 ms
= 4080 Hz
SUMMARY
1. Mechanical waves can exist in material media and
are governed by Newton's Laws.
2 Transverse waves are waves in which the particles
of the medium oscillate perpendicular to the direction of wave propagation.
3. Longitudinal waves are waves in which the
particles of the medium oscillate along the direction of wave propagation.
4. Progressive wave is a wave that moves from one
poimt of medium to another.
5. The displacementin a sinusoidal wave propagating
in the positive x direction is given.by ylx $a sin(kx— wt + 9}
where a is the amplitude of the wave, kis the
angular wave number, is the angular frequency, (kx —- wt + ¢) ts the phase, and
¢ ts the phase constant or phase angle.
6. Wavelength J of a progressive wave is the
distance between two consecutive points of the same phase at a given time. In a
stationary wave, it is twice the distance between two consecutive nodes or
antinodes.
7. Pertod T
of oscillation of a wave ie defined as the time any element of the medium takes
to move through one complete oscillation. It is related to the angular
frequency ca through the relation pTalt a
8 Frequency v of a wave 1s defined as 1/T and 1s
related to angular frequency by ar)v=— 20
9. Speed of a progressive wave ia given by y= 2 = *
- iy
10. The speed of a transverse wave on a stretched
string is eet by the properties of the string. The speed on a string with
tenaion T and linear mase denaity jis v= f—
11. Sound waves are longitudinal mechanical waves
that can travel through solids, liquids,or gases. The apeed v of sound wave in
a fluid having bulk modulus Band denaity pis . E vcs [—
pf The speed of longitudinal waves in a metallic bar
is t= j— Pp For gases, since B = yP, the speed of sound is
?on fe Pe
12. When two or more waves traverse the same medium,
the displacement of any element of the medium ie the algebraic sum of the
displacements due to cach wave. This is known as the principle of superposttion
of waves yay Sx - vt)
ial
13. Two sinusoidal waves on the same string exhibit
interference, adding or cancelling according to the principle of superposition.
If the two are travelling in the same direction and have the same amplitude a
and frequency but differ in phase by a phase constant ¢, the result is a aingle
wave with the same frequency a:= 14] sinfex—ot +20 y lx, § [200s 59| sin( dex
wt 59 If $=0 or an integral multiple of 22, the waves are cxactly in phase and
the interference is constructive; if ¢= #, they are exactly out of phase and
the interference is destructive.
14. Atravelling weve, at a rigid boundary or a
closed end, is reflected with a phase reversal but the reflection at an open
boundary takes place without any phase change.For an incident wave
yx. 0 = aain (hx — ot)the reflected wave at a rigid
boundary is y.& § =— aain (lec + wt) For reflection at an open boundary
u,bx.t) = @ ain (hoc + anf
15. The interference of two identical waves moving
in oppoalte directions produces standing waves, For a string with fixed ends,
the standing wave is given by y (x. § = Baain kx] cos at
Standing waves are characterised by fixed locations
of zero displacement called nodes and fixed locations of maximum displacanents
called antinodes. The separation between two consecutive nodes or antinodes is
4/2.Aatretched string of length L fixed at both the ends vibrates with
frequencies given by
v-- n=12,3,..2 2L The set of frequencies given by
the above relation are called the normal modes of oacillation of the system.
The oacillation mode with lowest frequency is called the fundamental mode or
the first harmonic. The second harmonic is the oacillation mode
with n= 2 and ao on.A pipe of length L with one end
closed and other end open (such as air columns)vibrates with frequencies given
by va(n+%) =, n=#0,1,2,3,..The set of frequencies represented by the above
relation are the normal modes of oscillation of such a system. The lowest
frequency given by v/4L is the fundamental mode or the first harmonic.
16. Astring of length L fixed at both ends or an air
column closed at one end and open at the other end, vibratea with frequencies
called ita normal modes. Each of these frequencies is a resonant frequency of
the system.
17. Beats arise when two waves having slightly
different frequencies, y, and v, and comparable amplitudes, are auperposed. The
beat frequency is
Veeat = V; ~ Va
18. The Doppler effect ia a change in the obaerved
frequency of a wave when the source and the observer O moves relative to the
medium. For sound the observed frequency vis given in terms of the source
frequency v, by v+ Yo
v=yv, | ——_ v tv,here via the speed of sound through
the medium, v, is the velocity of obecrver relative to the medium, and v, ie
the source velocity relative to the medium. In using this
formula, velocities in the direction OS should be
treated as positive and those opposite to it should be taken to be negative.
POINTS TO PONDER
1. A-wave ia not motion of matter as a whole in a
medtum. A wind is different from the sound wave in air. The former involves
motion of afr from one place to the other. The latter involves compressions and
rarefactions of layers of air.
2. Ina wave, energy and not the matter is
transferred from one paint to the other.
3. Energy tranafer takes place becauae of the
coupling through clastic forces between neighbouring oscillating parts af the
medinm.
4 Tranaverse waves can propagate only nm medtum with
shear modulus of elastictty,
Longitudinal wavea need bulk modulus of elasticity
and are therefore, posaibie in all media, sclids, liquids and gases.
5. In a harmonic progressive wave of a given
frequency all particles have the same amplitude but different phases at a given
instant of time. In a stationary ware, all particles between two nodes have the
same phase at a given instant but have different amplitudes.
6. Relative to an observer at rest in a medium the
speed of a mechanical wave in that medhim (tj depends anly on elastic and other
properties (auch aa masa denaity) of the medium. It does not depend on the
veloctty af the source.
7. For an observer moving with velocity v, relative
to the medium, the speed of a wave is obviously different from v and is given
by ut v,.
EXERCISES
18.1 Astring of mass 2.50 kg is under a tension of
200 N. The length of the stretched string is 20.0 m. If the traneveree jerk is
struck at one end of the string, how long does the disturbance take to reach
the other end?
15.2 Astone dropped from the top of a tower of
height 300 m high splashes into the water ofa pond near the base of the tower.
When is the splash heard at the top Siven that the speed of sound in air ie 340
m s'7 (g = 9.8 m 84}
15.3 A steel wire haa a length of 12.0 m and a mass
of 2.10 kg. What should be the tension in the wire so that speed of a
transverse wave on the wire equals the speed of sound in dry air at 20°C = 343
ms".[xe
15.4 Use the formula t= to explain why the speed of
sound in air P
(a) = ia independent of pressure,
(b) increases with temperature,
(c) mereages with humidity.
15.6 You have Icarnt that a travelling wave in one
dimension is represented by a function y = f ix, } where x and ¢ must appear in
the combination x -— vt or x+v Le.y =f (x + v @. Ie the converse truc? Examine
if the following functions for y can possibly represent a travelling wave :
(a) (x—vt}* th) log [fx + v4/x](cc) 1/0e+ 04
15.6 A bat emits ultrasonic sound of frequency 1000
kHz in air. If the sound meets a water eurface, what is the wavelength of (a)
the reflected aound, (b) the tranamftted sound? Speed of sound in air is 340 m
s— and in water 14866 m s*.
15.7 Ahospital usea an ultrasonic scanner to locate
tumours in a tlasue. What is the wavelength of sound in the tissue in which the
speed of sound is 1.7 kam 5"? The operating frequency of the scanner is
4.2 MHz.
15.8 A traneveree harmonic wave on a string is
described by yx, 0 =3.0 sin (96 t + 0.018 x + 5/4)where x and y are in cm and
fin s. The positive direction of x is from left to right.
(a) Ie this a travelling wave or a stationary wave
?If it ts travelling, what are the speed and direction of its propagation ?
(b) What are ite amplitude and frequency ?
(c) ‘What is the initial phase at the origin ?
(d) What ie the least distance between two
successive crests in the wave 7?
15.9 For the wave described in Exercise 15.8, plot
the displacement (yj versus (4 graphs for x =0, 2 and 4 cm. What are the ahapea
of these graphs? In which aspects does the oscillatory motion in travelling
wave differ from one paint to another: amplitude,frequency or phase ?
15.10 For the travellmg harmonic wave
ylx, § = 2.0 cos 2x (10t- 0.0080 c+ 0.35)where xand
y are in cin and tin s. Calculate the phase difference between oscillatory
motion of two pointe separated by a distance of fa) 4m,
fb) 0.5m,(c)
A/2,(aq) ava
15.11 ‘The transverse displacement of a string
(clamped at its both ends) ts given by ‘(aR yx, § = 0.06 ain (=. con (120 79
where «and y are in m and tin s. The length of the string is 1.5 m and its mass
is 3.0 x10%kg,
Answer the following :a) Does the function represent
a travelling wave or a stationary wave?
(b) Interpret the wave as a uperposition of two waves travelling in
opposite directions. What is the wavelength, frequency, and speed of each wave
?
(c) Determine the tension in the string.
16.12 @ For the wave on a string described in
Exercise 15.11, do all the pointe on the string oscillate with the same (a)
frequency. (b) phase, (c) amplitude? Explain your answers. (i) What is the
amplitude of a point 0.375 m away from one end?
16,13 Given below are eome functione of x and {to
represent the displacement (transverse or longitudinal) of an elastic wave.
State which of these represent (i) a travelling wave, Gi) a stationary wave or
(il) none at all:
(a) y=2 cos (3% ain (108
(b) y =2fx — vt
(c) y=3S ain (6x— 0.56 + 4 cos (6x — 0.60
d) yecos xsin t+ cos 2xain 2t
15.14 Awire stretched between two rigid supports
vibrates in ite fundamental mode with a frequency of 45 Hz. The maas of the
wire ia 3.5 x 10 kg and ite linear masa density ie 4.0x 107 kg m“. What is (a)
the speed of a transverse wave on the string. and
(b) the tension in the string?
15.15 Ametre-long tube open at one end, with a
movable piston at the other end, shows resonance with a fixed frequency source
{a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm.
Estimate the speed of sound im afr at the temperature of the experiment. The
edge effecte may be neglected.
15,16 A steel rod 100 cm long is clamped at ite
middle. The fundamental frequency of longitudinal vibrations of the rod are
given to be 2.53 kHz. What is the speed of sound in steel?
13.17 A pipe 20 cm long is closed at one end. Which
harmonic mode of the pipe is
reaonantly excited by a 430 Hz source ? Will the
same source be in resonance with the pipe if both ends are open? (speed of
sound in afr is 340 m s*}.
15.18 Two sitar strings A and B playing the note
‘Ge’ are alightly out of tune and produce beats of frequency 6 Hz. The tension
in the string A is slightly reduced and the heat frequency is found to reduce
to 3 Hz. If the original frequency of A ia 824 Hz,what is the frequency of B?
13.19 Explain why (or how):
(a) ma sound wave, a displacement node ia a pressure
antinode and vice versa,
(b) «= bats can ascertain distances, directions,
nature, and aizes of the obstaclea without any “eyes”,
(c) a violin note and sitar note may have the same
frequency, yet we can
distinguish between the two notes,
(d) solide can support both longitudinal] and
transverse waves, but only longitudinal wavea can propagate in gasea, and
e) = the shape of a pulse gete distorted during
propagation in a dispersive medium.
18.20 Atrain, standing at the outer signal of a
railway station blows a whistle of frequency 400 Hz in etill air. @) What is
the frequency of the whistle for a platform observer
when the train {a) approaches the platform with a
speed of 10 m s", (b) recedes from the platform with a speed of 10 m 9‘?
(i) What ia the speed of sound in cach case ? The speed of sound im atill air
can be taken as 340 ms".
15.21 A train, standing in a station-yard, blows a
whistle of frequency 400 Hz in atill air. The wind starte blowing in the
direction from the yard to the station with a speed of 10 m s?. What are the
frequency, wavelength, and speed of sound for an observer standing on the
station's platform? Is the situation exactly identical to the case when the air
is still and the observer runs towarda the yard at a speed of 10 m 8"? The
speed of sound in still air can be taken as 340 m s*
Additional Exercises
15.22 A travelling harmonic wave on a string is
described by yx. § = 7.5 sin (0.0050x +12 + 2/4)
(a)what are the displacement and velocity of
oscillation of a point at x= 10cm, and t= 15? Is this velocity equal to the
velocity of wave propagation?
(b) Locate the points of the string which have the
same traneverse displacements and velocity as the x = 1 an point at (= 2 e, 53
and 11s.
15.23 A narrow sound pulee ffor example, a short pip
by a whistle) is sent acroae a medium. (a) Does the pulse have a definite (i)
frequency, i) wavelength, (ii) speed of propagation? (b) If the pulse rate is 1
after every 20 a, (that is the whistle is blown for a split of second after
every 20 s}, is the frequency of the note produced by the whistle equal to 1/20
or 0.05 Hz ?
15.24 One end of a long string of linear mass
denaity 8.0 x 10% kg mis connected to an.electrically driven tuning fork of
frequency 256 Hz. The other end pasees over a pulley and is tied to a pan
containing a mass of 90 kg. The pulley end absorhe all the incoming energy so
that reflected waves at this end have negligible amplitude.At ¢= 0, the left
end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and
is moving along posttive y-direction. The amplitude of the wave ia 5.0
om. Write down the transverse displacement y as function of x and tthat
deacribea the wave on the string.
15.26 ASONAR system fixed in a submarine operates at
a frequency 40.0 kHz. An encmy submarine moves towards the SONAR with a speed
of 360 km h™. What ia the
frequency of sound reflected by the submarine ? Take
the speed of sound in water to be 1450 m a7.
15.26 Earthquakes generate sound waves inside the
earth. Unlike a gas, the earth can experience both transverse (5) and
longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km 3’,
and that of P wave is 6.0 km s'. A selsmograph records Pand S- waves from an
earthquake. The first P wave arrives 4 min before the first S wave. Assuming
the waves travel in straight line, at what distance does the
earthquake occur ?
15.27 A bat is flitting about in a cave, navigating
via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40
kHz. During one fast swoop directly toward a flat wall surface, the bat is
moving at 0.03 times the speed of sound in air.What frequency does the bat hear
reflected off the wall 2?
