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# WAVES Author: Pranjal K. Bharti (B. Tech., IIT Kharagpur) www.vidyadrishti.

org

## Notes for School Exams

Physics XI
Waves
Author: P. K. Bharti (B. Tech., IIT Kharagpur)

## H. O. D. Physics, Concept Bokaro Centre

Mb: 7488044834

2013-2015

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## Definition of wave Travelling wave equation (Waveform)

• A type of motion of energy where transportation of energy • Waveform means shape of the wave or simply equation of
the wave.
occurs without any bulk motion of material together with it
• Travelling wave has the form:
is called wave motion.
Classification of wave motion
f ( kx − ωt ) ;
y= along + ve x - axis
y f ( kx + ωt ) ;

• Wave motion can be broadly categorized in two categories = along -ve x - axis
depending on medium: •Here f denotes any arbitrary function. Some examples of
1. Mechanical waves: These waves require material medium solution of wave equation are:
(solid, liquid or gas) for propagation. Examples include = y A sin ( kx − ωt )
sound waves, water waves, Rayleigh waves etc.
y = Ae(
kx + ω t )
2. Non-mechanical waves: These waves do not require ; A : constant
y = ( kx − ωt ) + ( kx − ωt )
2 3
material medium for propagation. Example include light
wave. • The wave of the first example is called harmonic travelling
• Wave motion can be broadly categorized in two categories wave.
depending on propagation:
1. Longitudinal wave: In a longitudinal wave the particle The sinusoidal wave:
displacement is parallel to the direction of wave • When the left end of the string is continuously disturbed in
propagation. Example: sound wave. simple harmonic motion (SHM), the wave generated is
2. Transverse wave: In a transverse wave the particle called sinusoidal wave or sine wave. The shape of the string
displacement is perpendicular to the direction of wave at some time is shown in figure.
propagation. Example: water wave, wave on a string.
y v
Waves on a string: Travelling Wave
x
• Waves can be produced in a string by disturbing the string at
one or more points. Characteristics of the wave produced on
the string depend on elastic and inertia properties of the Sine wave
string. • The displacement y of any point with coordinate x at time t
• When a string is disturbed at some point, bump is produced can be written as
that propagates through the string.
• If the source of disturbance is active for a very short time, =y A sin ( kx − ωt + φ )
(sinusoidal waveform)
the bump produced will be localized only to a small part of or=y A cos ( kx − ωt + φ )
space at a time. This is called a wave pulse.
• Some terms associated with sine wave are described below:

Crest
Wave pulse λ
• If the source is active for some extended time repeating its A
motion several times, a collection of bumps are generated
which is called wave train or wave packet.
λ
Trough

## Wave train • Amplitude: The amplitude A of a wave is the magnitude of

• These wave pulse and wave train are called travelling the maximum displacement of the elements from their
waves. equilibrium positions as the wave passes through them.
• Speed of transverse wave on a string is given by • Crest: The highest points reached by the wave.
T • Trough: The lowest points reached by the wave.
v= (speed of wave on a string)
µ • Time period: The time period of oscillation T of a wave is
defined as the time any string element takes to move
where T = tension in the string through one complete oscillation.
and μ = mass per unit length of the spring
• Wavelength (λ): The distance between any two adjacent
crests and troughs.
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Note that wavelength is the distance travelled by the wave Total mechanical energy: From the expression of kinetic and
in the interval of one time period of disturbation SHM. So, potential energy, total mechanical energy transmitted per unit
if T is the time period of the disturbation SHM, we can length is equal to:
write λ = vT. 1 1
K .E. unit length + P.E. unit length =µω 2 A2 + µω 2 A2
M .E. unit length =
This is a very important relationship for sine wave. Also 4 4
note that two particles in the medium separated by distance 1
equal to integer multiple of wavelength, have identical ⇒ M .E. unit length = µω 2 A2
4
motion.
Average power: To calculate the power transmitted, we have to
• Frequency (f or ν ): The frequency of the disturbation
calculate the energy transmitted through the string per second.
SHM is called the frequency of the wave. This is commonly
As the wave speed is v m/s, the power transmitted will be equal
λ
denoted as ν or f. So we have, =
v = λf to:
T
1
• Angular wave number or propagation constant (k): The
Power = µ vω 2 A2
2
quantity 2π/λ is called the wave number. Physically wave
number denotes number of repeating units of the sine wave
per unit of space. We can write,
Interference (superposition)

k= •
λ Interference is the phenomenon that occurs when two or
more waves superpose (meet) while traveling along the
• Important relation: same medium.
λ ω • The principle of superposition states that interfering waves
=
v = λ f=
T k algebraically add to produce a resultant wave (or a net
wave). The principle implies that the interfering waves do
• Some alternative form of sine wave equation using relation not, in any way, alter the travel of each other.
λ ω
=
v = λ f= can be written as follows: • If we have two or more waves moving in the medium the
T k resultant waveform y is the sum of wave functions of
=y A sin(ωt − kx + φ ) individual waves y1 , y2 ,..., yn . Thus,
⇒ y A sin {ω (t − x / v) + φ }
= y = y1 + y2 + ... + yn
• Please note that during interference (destructive or
⇒ y A sin {k (vt − x) + φ }
=
constructive) the total energy of two waves remains
  t x 
⇒ y A sin 2π  −  + φ 
= conserved.
 T λ  

## Energy transmission along a sine wave Interference of two waves

The harmonic oscillator that is used for producing the Let a wave travelling along a stretched string is given by
disturbance at the left end imparts energy to the string at the left =y1 A1 sin(ωt − kx)
end, which is transmitted along the string by the wave. The wave and another wave, shifted from the first by a phase φ ,
carries the energy imparted in the form of kinetic and potential
=y2 A2 sin(ωt − kx + φ )
energy of the string.
Kinetic energy: As the particles of the string are moving, there By the superposition principle, the resultant displacement at
must be some kinetic energy associated with per unit length of point is
the string. Kinetic energy transmitted per unit length of the string ⇒ y = y1 + y2
is given by: ⇒ y A1 sin ( kx − ωt ) + A2 sin ( kx − ωt + φ )
=
1
K .E. unit length = µω 2 A2 ⇒ y A sin ( kx − ωt + δ )
=
4
A2 A
Potential energy: As the length of the string extends when wave where,
passes through the string, there must be some potential energy
A= A12 + A22 + 2 A1 A2 cos φ φ δ
associated with per unit length of the string. Potential energy
A2 sin φ
transmitted per unit length of the string is given by: tan δ = A1
1 A1 + A2 cos φ
P.E. unit length = µω 2 A2
4
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Constructive interference: The resultant amplitude A will • From the principle of superposition, the resultant
be maximum when displacement of the particles in the string after interference
cos φ = 1 can be written as:
y= y1 + y2
⇒ φ 2nπ , n ∈ I
= (Constructive interference)
⇒ y A {sin ( kx − ωt ) + sin ( kx + ωt )}
=
∴ A= A + A + 2 A1 A2 cos φ =
2 2
A + A + 2 A1 A2 (1)
2 2

2 A sin kx cos ωt
⇒y=
1 2 1 2
... (i)
⇒ A= ( A1 + A2 )
2
(standing wave)

⇒ A= ( A1 + A2 ) (Constructive interference) • It can be observed form this equation that each particle
executes SHM: y = Ax cos ωt with amplitude Ax depending
Destructive interference: The resultant amplitude A will be
minimum when on the position of the particle as: Ax = 2 A sin kx
cos φ = −1 • Now we will discuss two boundary conditions of the string:

## ⇒ φ= ( 2n + 1) π , n∈I (Destructive interference) a) String fixed at both ends:

∴ A= A12 + A22 + 2 A1 A2 cos φ = A12 + A22 + 2 A1 A2 (−1) • In this case nodes are formed at the fixed ends all the time.
Therefore, the boundary conditions are:
⇒ A= ( A1 − A2 )
2

## y = 0 (node) at x = 0 for all t

⇒ A = A1 − A2 (Destructive interference)
y = 0 (node) at x = L for all t
• Using these boundary conditions on Eq. (i), we arrive:

= L , n∈I
Standing waves 2
• Standing waves are produced when two waves with v
• Using, f = we get:
identical frequency interfere with one another while λ
travelling in opposite direction in the same medium. nv
= f , n∈I ... (ii)
• Due to constructive and destructive interferences at different 2L
points of the string, all particles of the string execute simple n F
harmonic motion with different amplitudes. =
⇒ f , n∈I
2L µ
• At the points of maximum destructive interferences, the
(String fixed at both ends)
particles remain stationary. These points are called nodes.
• At the points of maximum constructive interferences, the • So there are infinite numbers of frequencies at which a
particles execute SHM with maximum amplitude. These string can vibrate. These frequencies are called natural
points are called antinodes. frequencies or resonant frequencies.
• The lowest of all these frequencies is called fundamental
frequency f1 or 1st harmonic. Putting n = 1 in (ii) we get,
v
f1 = (fundamental frequency or 1st harmonic)
2L
• As the nodes are formed at some fixed points of the string,
the entire pattern gives an appearance of standing still. Thus
it is called standing or stationary wave.
• In standing waves, there is no net transport of energy in
either direction, although there is energy in the system. • Putting n = 2 in (ii) we get,
v
Standing waves: Analytically f2 = (2nd harmonic or 1st overtone)
L
•To analyze the standing string wave, let us consider two
waves of same amplitude and frequency travelling along a
same string but in opposite direction. Equation of these
waves can be written as follows:
• Putting n = 3 in (ii) we get,
= y1 A sin ( kx − ωt )
3v
=y1 A sin ( kx + ωt ) f3 = (3rd harmonic or 2nd overtone)
2L

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SOUND WAVES

## Basic Concepts of Sound Waves

• The collection of all possible modes is called the harmonic • The most general definition of sound is a longitudinal wave
series and n is called the harmonic number. in a medium.
• Sound waves may be described in terms of variations of
b) One end fixed and other end free pressure at various points. In a sinusoidal sound wave in air,
the pressure fluctuates above and below atmospheric
• In this case nodes are formed at the fixed ends all the time
pressure P0 in a sinusoidal variation with the same
and antinodes at the free end all the time . Therefore, the
frequency as the motions of the air particles. The human ear
boundary conditions are: Here the boundary conditions are:
operates by sensing such pressure variations.
i) y = 0 (node) at x = 0 for all t. • The disturbance (compression and rarefaction) produced
ii) As y = L is a free end, antinodes will form there. travels in the medium at a speed which depends on the
• Using these boundary conditions on Eqn. (i), we arrive at: elastic and inertia properties of the medium.
λ • When this propagating disturbances or waves strike the ear,
L= ( 2n − 1) , n∈I
4 sound is perceived. Human ear can perceive sound whose
v frequency lies between 20 Hz and 20 KHz.
• Using, f = we get:
λ
Displacement and pressure wave
( 2n − 1) v
=f , n∈I ... (iii) • In case of string wave, the wave was described by the
4L
displacement of the string particles.
2n − 1 F
=
⇒ f , n∈I • In case of sound waves, the wave can be described by either
4L µ
(i) displacement of the medium particles as:
(String fixed at one end) =s s0 sin ( kx − ω t )
• Putting n = 1 in (iii) we get, or
v (ii) the excess pressure generated as:
f1 = (fundamental frequency or 1st harmonic)
4L =p p0 sin ( kx − ω t )
• A unique relation between these displacement and pressure
wave exists which is given as
p0 = Bks0
• Putting n = 2 in (iii) we get, • Thus pressure amplitude is Bk times the displacement
3v amplitude and pressure amplitude differs in phase by π/2
f2 = (3rd harmonic or 1st overtone) from the displacement amplitude.
4L

## Speed of sound in solid

• For one dimensional solid structures (like beam), only
longitudinal sound waves are generally observed. The speed
of sound wave in this case is given by:
• Putting n = 3 in (iii) we get,
Y
f3 =
5v
(5th harmonic or 2nd overtone) v=
3L ρ

material.

## Speed of sound in liquid

Resonance: If the driving frequency is equal to any natural
• Speed of sound in liquid medium is given by
frequency of the string, the string will vibrate with a larger
amplitude. This phenomenon is known as resonance. B
Because the string has a large number of natural v=
ρ
frequencies, resonance can occur at many different
frequencies. where B = bulk modulus which is an elastic property and ρ
= density which is an inertia property of the fluid.

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## Newton’s formula: • In this case pressure node (or displacement antinode) is

formed in the open end and pressure antinode (or
• The first theoretical expression for speed of sound wave in displacement node) is formed in the closed end.
gases was given by Isaac Newton. He suggested that it is an • So the boundary conditions are:
isothermal process. This implies Boyle’s law is applicable p = 0 (node) at x = 0 for all t.
As x = L is a free end, antinodes will form there.
to the layers through which sound is propagating.
• Using these boundary conditions on Eqn. (i), we arrive at:
Thus:
λ
L= ( 2n − 1) , n∈I
PV = constant 4
v
∆P • Using, f = we get:
⇒ P ∆V + V ∆P = 0 ⇒ − = P λ
∆V / V
⇒B=
P ( 2n − 1) v
=f , n∈I ... (ii)
4L
So, (Closed organ pipe)
B P
=v = • Putting n = 1 in (ii) we get,
ρ ρ
v
• But results obtained from this formula don’t resemble the f1 = (fundamental frequency or 1st harmonic)
4L
experimental results. For example in STP, the calculated
value of speed of sound is 280 m/s whereas the measured
speed is 332 m/s.

## Laplace’s correction: • Putting n = 2 in (ii) we get,

3v
• Laplace suggested that this is an adiabatic process. Thus: f2 = (3rd harmonic or 1st overtone)
4L
PV γ = constant
⇒ ln P + γ ln V =
constant
∆P ∆V ∆P
⇒ +γ = 0⇒− = γP
P V ∆V / V
• Putting n = 3 in (ii) we get,
⇒B= γP
5v
f3 = (5th harmonic or 2nd overtone)
B γP 3L
=
So, v =
ρ ρ

## • At STP putting γ = 1.4 (for air), we get v = 332 m/s which

matches with the experimental result.
• Sound speed is generally largest in solids, then in liquids
and lowest in gases.
Open organ pipe

## • In this case pressure node (or displacement antinode) is

Vibration of air columns formed at both ends. So the boundary conditions are:
p = 0 (node) at x = 0 for all t.
•Vibration of air columns in closed and open pipe is very
p = 0 (node) at x = L for all t.
similar to the standing waves formed in string. Consider two
sound waves having same amplitude and frequency with • Using these boundary conditions on Eq. (i), we arrive:
phase difference δ between them travel in the opposite nλ
= L , n∈I
direction in a medium. The equation for the pressure waves 2
can be written as: v
= p1 p0 sin(kx − ω t ) • Using, f = we get:
λ
= p2 p0 sin(kx + ω t )
nv
• The resultant pressure at a point x at a point t can be written = f , n∈I ... (iii)
2L
as:
=
p p1 + p2 n F
=
⇒ f , n∈I
2L µ
2 p0 sin kx cos ω t
⇒ p= ... (i)
(String fixed at both ends)

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## • Putting n = 1 in (iii) we get, Beats

v
f1 = (fundamental frequency or 1st harmonic) • The phenomenon of periodic variations in intensity of sound
2L when two waves of slightly different frequencies and
amplitudes travelling in the same direction, are
superimposed on each other is called beats.
• One cycle of maximum and minimum intensity is called one
beat. The numbers of beats heard in one second is called
beat frequency and this is equal to the difference in
• Putting n = 2 in (iii) we get, frequency of the interfering waves. Thus,
v
f2 = (2nd harmonic or 1st overtone) Beat frequency = no. of beats per sec = f1 − f 2
L

Consider two sound waves having same amplitude and
slightly different frequencies travel in the same direction of
some medium. Equations of the corresponding pressure
waves can be written as:
= p1 p0 sin ω1 ( t − x / v )
• Putting n = 3 in (iii) we get,
=p2 p0 sin ω2 ( t − x / v )
3v
f3 = (3rd harmonic or 2nd overtone) • The resultant change in pressure due to interference can be
2L
written as:
=p p1 + p2
 ω − ω2 ω +ω
=⇒ p 2 p0 cos  1 ( t − x / v ) sin  1 2 ( t − x / v )
 2   2 
 ∆ω
=⇒ p 2 p0 cos  ( t − x / v ) sin ω ( t − x / v )
 2 
Sound intensity
 ω1 − ω2 ω + ω2 
 Assuming = ∆ω and 1 =ω
• The amount of energy crossing a unit cross-sectional area  2 2 
perpendicular to the direction of propagation of the wave in • The above equation can be interpreted as the equation of a
unit time is known as the intensity of the sound wave. travelling wave of angular frequency ω, whose amplitude
• Therefore average intensity, which is power transmitted per
 ∆ω
unit area, is given by 2 p0 cos  ( t − x / v ) oscillates between 0 to 2p0. The
 2 
Power Bω 2 s02 p02 v p02 scenario is shown in Fig a and b.
=I av = = =
Area 2v 2B 2ρ v
• It can be seen that sound intensity is directly proportional to
the square of the pressure amplitude and inversely
proportional to the density of the medium.
Loudness
• The perceived loudness of sound is the ear’s response to
intensity. Higher the intensity of the sound wave, higher
will be the loudness the sound and vice versa.
• Loudness is correlated with the intensity by the using the
parameter known as sound level which is expressed in
decibels (dB). This is expressed as follows:

I
β (dB ) = 10 log10 • So the frequency of variation of amplitude is
I0
( ∆ω / 2 ) / 2π . Thus the frequency of amplitude
• Here I the intensity of the sound of interest and I0 is a
reference intensity with value 10e– 12 W/m2. variation = 2 × {( ∆ω / 2 ) / 2π } =∆ω / 2π =f1 − f 2
• As the intensity is proportional to the pressure
amplitude, the intensity varies periodically with
frequency f1 − f 2 .

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Doppler effect
2) Source stationary and listener moving:
• The apparent change in the frequency of sound when the
• Suppose the source is stationary with respect to the medium
source, the observer and the medium are in relative motion
and the listener is moving towards the source at a speed vo.
is called Doppler effect.
• When the listener receives a compression pulse, the next
• We consider the following three cases for Doppler effect:
compression pulse must be at a distance vot away from the
1) Listener stationary and source moving: listener. The relative velocity of the listener and the next
compression pulse is vo + v. Therefore time interval between
• Suppose the listener is stationary with respect to the
the two consecutive compression pulses received by the
medium and the source is moving towards the listener at a
listener is:
speed vs.
v
x T' = T
v + vo

S vs S vs O • v0
Compression Compression Compression
t=0 t=t
S O
vs t x– vs t
• Suppose at some point of time the source emits a vT
compression pulse towards the listener. Let’s assume that at
this moment distance between the source and the listener is v0
x. This compression pulse will reach the listener at time t1 = S O
x/v where v is the speed of sound in medium.
• The source will emit the next compression pulse after time T
where T is the time period. The source has travelled a vT’ v0T’
distance of vsT at this time, so current distance between • Therefore the apparent frequency experienced by the
source and listener will be x–vsT. This compression wave listener will be:
will reach the listener at a time t2 = T + (x– vsT)/v.
v + vo
• Therefore time interval between the two consecutive f '= f
v
compression pulses received by the listener is:
• It can be proved similarly that instead of approaching, if the
x − vs T x v − vs
T ' = t 2 − t1 = T + − = T listener recedes from the source at a speed of u, the apparent
v v v
frequency will be:
• Therefore the apparent frequency experienced by the
v − vo
listener will be: f '= f
v
1 v
f=' = f • Thus if the listener is moving and the source is stationary,
T ' v − vs apparent frequency increases as the listener approaches the
• It can be proved similarly that instead of approaching, if the source and decreases as the listener recedes from the source.
source recedes from the listener at a speed of u, the apparent 3) Both listener and source are moving
frequency will be:
• Suppose the source and listener are moving towards each
v
f '= f other with speeds vs and vo with respect to the medium. Let
v + vs
at distance between them be x at t = 0.
• Thus if the source is moving and the listener is stationary,
x
apparent frequency increases as the source approaches the
listener and decreases as the source recedes from the S vs v0 O
listener.
t=0

S vs v0 O

vs t vo t
t=T
• The speed of the sound wave relative to the observer is v+v0.
Therefore, the observer will receive the first compression
pulse at time,
x
t1 =
v+v0
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## • After one time period t = T, both the source and observer

have moved towards each other covering distances vsT and
v0T respectively. The new distance between the source and
the observer is x – (vs+v0)T.
• The 2nd compression pulse will reach the observer at time,
x − ( v s +v 0 ) T
t2= T +
v+v0
• The time interval between two successive compression
pulses or the period of the waves as recorded by the
observer is
x − ( v s +v 0 ) T x
T ' = t2 − t1 = T + −
v+v0 v+v0
 v−vs 
⇒T'=  T
 v+v0 
• Therefore the apparent frequency experienced by the
listener will be:
 v+vo 
f '= f
 v−vs 

## • Let vo be the speed of the listener w.r.t. the medium,

considered positive when it moves towards the source and
negative when it moves away from the source.
• Let vs be the speed of the source w.r.t. the medium,
considered positive when it moves towards the listener and
negative when it moves away from the listener.
• Then combining results of case 1 and 2, the expression for
apparent frequency can be written as:
v + vo
f '= f
v − vs

## • Thus apparent frequency increases as separation between

source and listener decreases and decreases as separation
between source and listener increases.
• Following two points are to be kept in mind:
1. It has been assumed that the motion of source or
listener is along the line joining the two. If the motions
are in other directions, components of the velocity
should be taken along the joining line before using the
formulas.
2. It has been assumed that the medium is in rest. If the
medium itself is moving appropriate corrections are to
be made in determining the speeds of the source and the
listener.

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## About P. K. Bharti Sir (Pranjal Sir)

Physics Classes by Pranjal Sir • B. Tech., IIT Kharagpur (2009 Batch)
• H.O.D. Physics, Concept Bokaro Centre
(Admission Notice for XI & XII - 2014-15) • Visiting faculty at D. P. S. Bokaro
• Produced AIR 113, AIR 475, AIR 1013 in JEE -
Batch 1 (Board + JEE Main + Advanced): (Rs. 16000)
• Produced AIR 07 in AIEEE (JEE Main)
Batch 2 (Board + JEE Main): (Rs. 13000)
Batch 3 (Board): (Rs. 10000) Address: Concept, JB 20, Near Jitendra Cinema,
Batch 4 (Doubt Clearing batch): Rs. 8000 Sec 4, Bokaro Steel City
Ph: 9798007577, 7488044834
Email: pkbharti.iit@gmail.com

Physics Class Schedule for Std XIIth (Session 2014-15) by Pranjal Sir
Sl. No. Main Chapter Topics Board level JEE Main Level JEE Adv Level
Vectors, FBD, Work, Energy, Rotation, rd th
Basics from XIth 3 Mar to 4 Apr 14
SHM

1. Electric Charges and Coulomb’s Law 5th & 6th Apr 5th & 6th Apr 5th & 6th Apr
Fields Electric Field 10th & 12th Apr 10th & 12th Apr 10th & 12th Apr
Gauss’s Law 13th & 15th Apr 13th & 15th Apr 13th & 15th Apr
Competition Level NA 17th & 19th Apr 17th & 19th Apr
2. Electrostatic Potential Electric Potential 20th & 22nd Apr 20th & 22nd Apr 20th & 22nd Apr
and Capacitance Capacitors 24th & 26th Apr 24th & 26th Apr 24th & 26th Apr
Competition Level NA 27th & 29th Apr 27th & 29th Apr, 1st,
3rd & 4th May
PART TEST 1 Unit 1 & 2 4th May NA NA
NA 11th May 11th May
3. Current Electricity Basic Concepts, Drift speed, Ohm’s 6th, 8th, 10th, 13th 6th, 8th, 10th, 13th 6th, 8th, 10th, 13th May
Law, Cells, Kirchhoff’s Laws,
Wheatstone bridge, Ammeter,
May May
Voltmeter, Meter Bridge, Potentiometer
etc.
Competition Level NA 15th & 16th May 15th, 16th, 17th, 18th &
19th May
PART TEST 2 Unit 3 18th May NA NA
NA 20th May 20th May
SUMMER BREAK 21st May 2013 to 30th May 2013
4. Moving charges and Force on a charged particle (Lorentz 31st May, 1st & 31st May, 1st & 31st May, 1st & 3rd Jun
force), Force on a current carrying 3rd Jun 3rd Jun
Magnetism wire, Cyclotron, Torque on a current
carrying loop in magnetic field,
magnetic moment
Biot Savart Law, Magnetic field due 5th, 7th & 8th Jun 5th, 7th & 8th Jun 5th, 7th & 8th Jun
to a circular wire, Ampere circuital
law, Solenoid, Toroid
Competition Level NA 10th & 12th Jun 10th, 12th, 14th & 15th
Jun
PART TEST 3 Unit 4 15th Jun NA NA
NA 22nd Jun 22nd Jun

P a g e | 10 CONCEPT: JB-20, Near Jitendra Cinema, City Centre, Bokaro Mb: 7488044834
WAVES Author: Pranjal K. Bharti (B. Tech., IIT Kharagpur) www.vidyadrishti.org
5. Magnetism and 17 , 19 & 21st
th th
17th, 19th & 21st Not in JEE Advanced
Matter Jun Jun Syllabus
6. Electromagnetic Faraday’s Laws, Lenz’s Laws, A.C. 24th, 26th & 28th 24th, 26th & 28th 24th, 26th & 28th Jun
Generator, Motional Emf, Induced Emf,
Induction Jun Jun
Eddy Currents, Self Induction, Mutual
Induction
Competition Level NA 29th Jun & 1st Jul 29th Jun, 1st, 3rd & 5th
Jul
PART TEST 4 Unit 5 & 6 6th Jul NA NA
NA 13th Jul 13th Jul
7. Alternating current AC, AC circuit, Phasor, transformer, 8th, 10th & 12th 8th, 10th & 12th 8th, 10th & 12th Jul
resonance,
Jul Jul
Competition Level NA 15th July 15th & 17th July
8. Electromagnetic 19th & 20th July 19th & 20th July Not in JEE Advanced
Waves Syllabus
PART TEST 5 Unit 7 & 8 27th Jul 27th Jul 27th Jul
Revision Week Upto unit 8 31st Jul & 2nd 31st Jul & 2nd 31st Jul & 2nd Aug
Aug Aug
Grand Test 1 Upto Unit 8 3rd Aug 3rd Aug 3rd Aug
9. Reflection 5th & 7th Aug 5th & 7th Aug
5th & 7th Aug
Refraction 9th & 12th Aug 9th & 12th Aug
9th & 12th Aug
Ray Optics Prism 14th Aug 14th Aug14th Aug
Optical Instruments 16th Aug 16th Aug
Syllabus
Competition Level NA 19th & 21st Aug 19th, 21st, 23rd, 24th Aug
th th
10. Huygens Principle 26 Aug 26 Aug 26th Aug
th th th th
Interference 28 & 30 Aug 28 & 30 Aug 28th & 30th Aug
st st
Wave Optics Diffraction 31 Aug 31 Aug 31st Aug
nd nd
Polarization 2 Sep 2 Sep 2nd Sep
th th
Competition Level NA 4 & 6 Sep 4th, 6th, 7th, 9th, 11th Sep
th th
PART TEST 6 Unit 9 & 10 14 Sep 14 Sep 14th Sep
th th
REVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13 Sep to 27 Sep
Grand Test 2 Upto Unit 10 28th Sep 28th Sep 28th Sep
DUSSEHRA & d-ul-Zuha Holidays: 29th Sep to 8th Oct
11. Dual Nature of Photoelectric effect etc 9th & 11th Oct 9th & 11th Oct 9th & 11th Oct
Grand Test 3 Upto Unit 10 12th Oct 12th Oct 12th Oct
12. Atoms 14th & 16th Oct 14th & 16th Oct 14th & 16th Oct
13. Nuclei 18th & 19th Oct 18th & 19th Oct 18th & 19th Oct
X-Rays NA 21st Oct 21st & 25th Oct
PART TEST 7 Unit 11, 12 & 13 26th Oct NA NA
14. Semiconductors Basic Concepts and Diodes, transistors, 26th, 28th, 30th 26th, 28th, 30th Not in JEE Adv
logic gates
Oct & 1st Nov Oct & 1st Nov Syllabus
15. Communication 2nd & 4th Nov 2nd & 4th Nov Not in JEE Adv
System Syllabus
PART TEST 8 Unit 14 & 15 9th Nov 9th Nov NA
Unit 11, 12 & 13 Competition Level NA 8th, 9th & 11th 8th, 9th, 11th, 13th & 15th
Nov Nov
PART TEST 9 Unit 11, 12, 13, X-Rays NA 16th Nov 16th Nov
Revision Round 2 Mind Maps & Back up classes for late 18th Nov to 18th Nov to 18th Nov to Board
registered students
(Board Level) Board Exams Board Exams Exams
Revision Round 3 18th Nov to JEE 18th Nov to JEE 18th Nov to JEE
(XIth portion for JEE)
30 Full Test Series Complete Syllabus Date will be published after Oct 2014

P a g e | 11 CONCEPT: JB-20, Near Jitendra Cinema, City Centre, Bokaro Mb: 7488044834
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