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(3)
Repeating the procedure to find the second derivative,
(4)
Similarly the Time-Derivatives is given by
(5) First derivative
(6)Second Derivative
Combining the results in the equations (4) and (6)
We can write the One-Dimensional differential wave equation as
(7) One-D Wave Equation
This is the general form of the wave equation in One-Dimension.
Solution of the wave equation in one-dimension:
1
2
2
22
2
t
y
vx
y
x
x!
x
x
The solution to this second order partial differential equation can be expressed in terms of a sine or
cosine function of position
The displacement of successive particles of the medium is given by sine or cosine function of
position.
The displacement y at t = 0 is given by
kxsinAy !.(i)
Where A and k are constants.
Disturbance is propagating along positive x-direction then
vtxksinAy ! (ii)The waveform in (i) is a sine function, it repeats itself at regular distances.
The first repetition would take place when
T! 2kx or kx
T!2
This distance after which repetition takes place is called the wavelength .P
Hence k
T!P2
or P
T!
2k
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Constant k is called propagation constant or wave vector.
Equation (ii) turns into
vtxsinAY P
T!
2
(iii)
AtxsinAy,t
P
T!!
20
The equation which leads the wave represented by (iii) by half the wavelength is given by
y =
xvtsinAvtxsinA P
T!
P
P
T 2
2
2
.(iv)
Relation between wavelength and velocity of propagation :
Time taken for one complete of wave to pass any point is the time period (T).
If in the equation (iv) we put Tt the term xvt
P
T2
undergoes one complete cycle i.e,
increased by T2 .
Hence _ a T!
P
T
P
T2
2xvtxTtv
orT!
P
T2
2vT
orP!vT
..(v)Relation shows that the time taken by a wave in propagating a distance P is equal to period T.
NowP!
P! f
Tv
(vi)
Different forms of simple harmonic waves
P
PT!
xvtsinAy 2
=
PTx
ftsinA 2..(vii)
=
P
Tx
T
tsinA 2
..(viii)
=
PTTx
T
tsinA 22
= kxtsinA [ .(ix)
= J[tsinA (x)
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Time Period ( T ):
The time period for a point on the string is the time taken to complete one cycle of its periodic
motion. It is exactly the same time that it takes for one wavelength to pass the point.
Frequency ( f ) :
The number of complete vibrations of a point on the string that occur in one second or, the number of
wavelengths that pass a given point in one second.
Wave Velocity ( v) :
Since in one period T the wave advances by one wavelength P , therefore, the wave velocity is
kf
Tv
[!P!
P!
Amplitude ( A ) :
The maximum displacement of a particle on the medium from the equilibrium position.
Wave fronts and Rays:
A wave front is defined as a surface joining the points of same phase.
The speed with which the wave front moves onwards from the source is called the phase velocity or wave
velocity.
The energy of the wave moves in a direction perpendicular to the wave front.
Figure shows light waves emitting out from a point source forming a spherical wave front in 3-D space.
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The energy travels outwards along straight lines emerging from the source, t
That is the radii of the spherical wave front. These lines are called the rays
Rays are perpendicular to wave fronts.
HUYGENS PRINCIPLE:-
1) Every point on the given wave front acts as a fresh source of new disturbance, called secondary wavelets
which travel in all directions with velocity of light.
2) A surface touching these secondary wavelets tangentially in the forward direction at any instant gives thenew wave front at that instant. It is called secondary wave front.
3) Wave front:-A wave front is defined as the continuous locus of all the particles of a medium, which are
vibrating in the same phase.
Depending upon the shape of the source of light, wave front can be of different types.
i)Spherical wave front:-source of light is a point source
eg:-
ii)Cylindrical wave front:- source of light is linear.
iii) Plane wave front:-When the point source or linear source is at very large distance, a small portion ofwave front appears to be plane which is called as planewavefront.
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PROBLEMS
Problem : 01
The equation for the displacement of a stretched string is given by
-
T!
100
x
02.0
t2sin4y where y and x
are in cm and t in sec. Determine the (a) direction in which wave is propagating (b) amplitude (c) time
period (d) frequency (e) angular frequency (f) wavelength (g) propagation constant (h) velocity of wave
(i) phase constant
SOLUTION :
Comparing the given equation with the general wave equation J[! kxtsinAy
, i.e.,
-
J
PT!x
T
tsinAy 2
we find that
(a) As there is negative sign between t and x terms, the wave is propagating
along positive x-axis.
(b) The amplitude of the wave A = 4cm = 0.04m.
(c) The time period of the wave T = 0.02s =
s
50
1
(d) The frequency of the wave f =
HzT
501
!
(e) Angular frequency of the wave T!T![ 1002 f rad/s.
(f) The wavelength of the wavemcm 1100 !!P
(g) The propagation constant. (= wave vector) =
m/radT!
P
T2
2
(h) The velocity of wave s/mfv 50150 !v!P! .
(i) The phase constant., i.e., initial phase 0!J .
Problem : 02 A progressive wave of frequency 500Hz is travelling with a velocity of 360m/s. How far
apart are two points060 out of phase.
SOLUTION : We know that for a wave P!Y f .
So, f
Y
!P
=m.720
500
360!
Now as in a wave path difference is related to phase difference by the relation,
Phase difference P
T!J(2
( path difference x( )
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Here, phase difference
T!v
T!!J(
360
180600
rad.
So path difference m.
.x 120
32
720
2!
Tv
T!J(
T
P!(
Problem : 03 The amplitude of a wave disturbance propagating in the positive x direction is given by
? A21
1
xy ! at t = 0 and ? A211
1
! xy at t = 2s where x and y are in m. The shape of the wave
disturbance does not change during the propagation. What is the velocity of the wave ?
SOLUTION :
In terms of power- function a pulse is given by 2
txb
ay
Y!
O
A comparison of given equations with the above reveals
22
1!Yx
t
xO
for st 2!
i.e., 1!Yt for t = 2s
s/m.502
1!
!Y
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Chapter 2
3-D Wave Equation and Principle of Superposition
ENERGYOF A PLANE PROGRESSIVE WAVE :
Wave motion is accompanied by the transmission of energy; It is communicated by the source to the
medium. Energy distribution of the wave along the direction of propagation is not uniform. However energy
density of the medium is independent of the position as well as time.
Consider an elastic medium and a wave xvt/2sinAy PT!
Consider an elementary oscillating layer of the medium of thickness dx.
IfV be the average density of the medium.
The mass of the particles contained in the layer is m = dxV , for unit area of cross-section.
Now kinetic energy of the layer
2
2
2
1
2
1
V!!
dt
dydxmu
=
dxxvtcosvA
P
TV
P
T 22
2
1 22
..(i)
Also the force acting on the layer
2
2
dt
yddxmaF V!!
The work done for a displacement dy of layer against this force will be
FdydW !
then potential energyFdydW !
Hence V
P
T!!
yy
dxydyv
dy.F.E.P0
2
0
2
=
dxyv 2
22
2
P
TV
=
dxxvtsinAv
P
T
P
VT 22 22
222
..(ii)
Hence total energy of the layer E = K.E. + P.E.
=
dxxvtsinAv
dxxvtcosvA
P
T
P
VT
P
TV
P
T 2222
2
1 22
222
2
2
=dx
Av2
2222
P
VT
The total energy per unit length = 2222
222
22
fAAv
VTP
VT!
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INTENSITYOF THE WAVE :
It is the energy crossing per unit cross sectional area per unit time. So the total energy per unit time
will be within a distance dx = v
Intensity vfAI VT2222!
or I w A2
3-Dimensional wave equation
The time taken for light to travel from one wave front to another is the same along any ring.
n vectorpropagatiothecallediskwhere
repeats.wavethespaceallfillingwaveinfiniteanFor
)rki(Ae)r(or)rkAcos()r(or)rkAsin()r(
:arewavesplaneharmonicsoandconst.rk
:isklar toperpendicuplanehegenerallytmoresoand
0zzk0
yyk0xxkconst.zzkyykxxk
:requiresBut this
0)0
z(zzk)0y(yyk)0
x(xxk
:or0k0rr
:klar toperpendicuisplanethedefinesand
:zeroiskand0rrofproductdotthekzkjykixkkSince
k)0
z(zj)0
y(yi)0
x(x0rr
:isplanein thevectorasoand
)0
z,0
y,0
(xk0
zj0
yi0
x0randz)y,(x,kzjyixr
:isO)-SystemandO'-(Systemplanein thepointsany twoforectorposition vThe
.kvectoralar toperpendicuplaneaforequationThe
wave.planeaisWaveD-3aforformSimplest
T
TTTTTTTTT
TT
T
TTT
T
TTTT
TT
TT
T
y!y!y!
!y
!!
!
!y
!
!
!!
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The Partial differential equation which satisfies such a three dimensional wave equation is a
generalization of Eqn.(7)In 3-Dimensions.If=is the 3-D wave function(Likef for1-D), then
=
=
=
=
(1)
=
=
(2)
(3)
The entire operator in the brackets is calledLaplacian Operator
The equation simply becomes
= =
(4)
Where the solution of this wave equation is given by
)rki(Ae)r(or)rkAcos()r(or)rkAsin()r(
TTTTTTTTT y!y!y! (5)
Superposition of Two Waves Acting Along The Same Line
When two sets of waves intersect each other then it is called superposition.
A very simple law can be used to explain the superposition of waves.
The resultant displacement of any point is merely the sum of the displacements due to each wave
separately.
This is known as the principle of superposition and was first stated by Thomas Young in 1802.
Mathematical representation of Superposition Principle
Let two simple harmonic vibrations be represented by the equations;
y1 = a1 sin ([t E1) (1)
y2 = a2 sin ([t E2) (2)
where
y1 and y2 are the displacements of a particle due to the two vibrations,
a1& a2 are the amplitudes of the two vibrations and
E1 & E2 are the epoch angles.
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The two vibrations are assumed to be of the same frequency and hence [ is the same for both the waves.
The resultant displacement y of the particles is given by;
y = y1 + y2 = a1 sin ([t E1) + a2 sin ([t-E2)
= a1 (sin [t cos E1- cos [t sin E1 ) +
a2 (sin [t cos E2- cos [t sin E2 )
= sin [t (a1cos E1+ a2cos E2 )
=cos [t (a1sin E1+ a2 sin E2 ) (3)
Since the amplitudes a1 & a2andE1&E2 are constants, the coefficients of (sin [t) and (cos [t) in equation
(3) can be substituted by;
A cos N = a1cos E1+ a2cos E2 (4)
A sin N = a1sin E1+ a2 sin E2 (5)
Squaring and adding equations (4) and (5);
A2
(sin2N + cos
2N = a1
2(sin
2E1 + cos
2E1) +
a22 (sin2E2 + cos
2E2) +
2a1a2 ( cos E1 cosE2 + sin E1 sinE2 )
@A2
= a12
+ a22
+ 2a1a2 cos (E1 -E2 ) (6)
from equations (4) and (5);
tan N = (a1sin E1+ a2 sin E2 )/(a1cos E1+ a2cos E2) (7)
Equations (6) and (7) give the value ofA and N in terms ofa1, a2, E1 and E2;
@ y = A sin [t cos N A cos [t sin N
y = A sin ([t N
Equation (8) is similar to the original equations (1) & (2). The amplitude of the resultant vibration is A and
having epoch angle N. The time period of the resultant vibration is the same as the original vibrations. The
values of A and N are given by equations (6)& (7).
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Thus the resultant of two simple harmonic vibrations of the same period and acting in the same line is also
a simple harmonic vibration with a resultant amplitude A and epoch angle N
The addition of three or more simple harmonic motions of the same frequency will give a resultant
motion of the same type. It may be noted that, the amplitude A depends, according to equation (6),
upon the amplitude a1 and a2 of the components motions and upon their difference of phase;
H = E1 E2
The intensity of light at any point will be proportional to the square of the resultant amplitude. In
case when (a1 = a2 = a), from equation (6) we get;
IEA2 = 2a2(1+cosH ) = 2a2 (2cos2H/2)
@I = 4a2
cos2H/2 (9)
when the phase difference H = 0, 2TTnTor
the path difference ; x = 0, PPnP
where n = 0, 1, 2, 3,
@I = 4a2
, as shown in figure (4a)
when the phase difference H = TTT2n+1)Tor
the path difference; x = PPP 2n+1)P
where n = 0, 1, 2, 3,
@I = zero, as shown in figure (4b)
It is found from equation (9), that the intensity at bright points is 4a2
and at dark points is zero.
According to the law of conservation of energy, the energy cannot be destroyed. Here also the energy is
not destroyed but only transferred from the points of minimum intensity to the points of maximum
intensity. The intensity varies from 0 to 4a2
as shown in figure (4c).
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1) Constructive interference: the crust and crust, trough and trough of two waves are
superimposed; as a result maximum intensity wave is formed
2) Destructive interference: when two waves are super impose crust of one wave combinedwith the trough of another waveand the intensity of resulting wave decreases