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TRANSCRIPT
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2013-2015
Notes for School Exams
Physics XI
Simple Harmonic Motion
P. K. Bharti, B. Tech., IIT Kharagpur
2007 P. K. Bharti
All rights reserved.
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SSoommeeDDeeffiinniittiioonnss
Periodic motion: A motion which repeats itself after a
regular interval of timeis called periodic motion.
Oscillation/ Vibration: Those periodic motion which
repeats itself about equilibrium point are known as
oscillation or vibration.
Note: Equilibrium point is the point where net forceand/or net torque is zero.
e.g., uniform circular motion is a periodic motion, but it is
not oscillatory.
Every oscillatory motion is periodic, but every periodic
motion need not be oscillatory.
Difference between oscillation & vibration: When the
frequency is small, we call it oscillation. e.g., the
oscillation of a pendulum. When the frequency is high,
we call it vibration. e.g., the vibration of a string of a
guitar.
Time Period (T): The smallest interval of time after
which the periodic motion is repeated is called time
period.
S.I. unit: second (s)
Frequency (orf): The number of repetitions that occur
per unit time is called frequency of the periodic motion. It
is denoted by (Greek nu) or f. Frequency is the
reciprocal of time period T. Therefore,
1v f
T= = (relation between frequency and time
period)
S.I. unit: hertz (Hz).
1 Hz = 1 s-1
Physically, if a body repeats its motion faster, it will said
to have higher frequency.
Periodic, harmonic and non-harmonic functions
(Mathematically)
Any function that repeats itself at regular intervals of its
argument is called a periodic function. The periodic
functions which can be represented by a sine or cosine
curve are calledharmonic functions.
All harmonic functions are necessarily periodic but all
periodic functions are not harmonic. The periodic
functions which cannot be represented by single sine or
cosine function are called non-harmonic functions. The
following sine and cosine functions are periodic with
period T:
f(t) = sin t= sin2
tT
and g(t) = cos t= cos2 t
T
SSpprriinnggmmaassssssyysstteemmoonnaaffrriiccttiioonnlleessssssuurrffaaccee
Let us consider a mass attached to a spring which in turn
attached to a rigid wall. The spring-mass system lies on a
frictionless surface.
We know that if we stretch or compress a spring, the mass
will oscillate back and forth about its equilibrium
(mean) position. Equilibrium position is the point where
net force and net torque is zero.
The point at which the spring is fully compressed or fully
stretched is known as extreme position.
The maximum displacement of the body oscillation on
either side of the equilibrium position is called the
amplitude. In other language, we can say thatamplitude
is the distance between mean position and extremeposition. Amplitude is denoted by letterAand its SI uni
is m.
If we observe motion of the block carefully, we find that
speed i.e., magnitude of velocity is maximum at mean
position. Similarlyspeed is minimum i.e., zero at extreme
positionsas block stops momentarily at extreme positions
Since, equilibrium position is the point where net force
and net torque is zero. Therefore, acceleration of the mass
is zero at equilibrium point. Magnitude of acceleration is
maximum at extreme positions.
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SSiimmpplleeHHaarrmmoonniiccMMoottiioonn
Let us again consider the spring-mass system lies on a
frictionless surface. We know that if we stretch or
compress the spring, the mass will oscillate back and forth
about its equilibrium (mean) position.
Let us displace spring by a distance x towards right.
As we displace it towards right, spring force will try to
bring mass m towards left. Thus at a displacement x, a
spring forceFdevelops in the spring in the left direction.
We also say this force Fas restoring force as it tries to
bring back mass m towards the mean position.
As this restoring force F is opposite to that of
displacement, therefore, we can write from Hookes Law
F= kx (S.H.M.) (A)
(negative sign because Fis opposite tox)
F x (S.H.M.) (B)
(because kis a constant)
Thus, the resultant restoring force F acting on the body
is proportional to the displacement x from the equilibrium
position and is directed opposite to the displacement, i.e.,towards the equilibrium point. This kind of motion is
known as simple harmonic motion (S. H. M.).
Again, from Newtons 2nd
Law, we have
F = ma
Therefore, using (A),
kx ma =
ka x
m = (i)
Since k and m are constants, acceleration a of the
oscillating body is directly proportional to its
displacement from the equilibrium position and is
directed opposite to the displacement, i.e.,
a x (S.H.M.) (C)
Thus, acceleration a of the body is proportional to the
displacement x from the equilibrium position and is
directed opposite to the displacement, i.e., towards the
equilibrium point. This kind of motion is known as simple
harmonic motion (S. H. M.)
Definition of SHM
We can defineSHM as an oscillatory motion in which net
restoring force or acceleration of the oscillating body is
directly proportional to its displacement from the
equilibrium position and is directed towards the mean
position.
The body performing SHM is known as a simple
harmonic oscillator (SHO).
If we put 2k
m= in eqn. (i), we get
2a x= (S.H.M.) (D)
where,
is known asangular frequency of SHM.
Relation between angular frequency () with time
period (T) and frequency (f)
Loosely speaking, we can consider angular frequency to
be the angular velocity when a body moves in uniform
circular motion.
Clearly, the particle covers an angular displacement 2
rad in a time equal to its time period T. Therefore,
2
T
= (angular frequency in terms of time period)
SI unit of is s1
.
Since, frequency f is given by f = 1/T, therefore we can
write
22 f
T
= =
Time period of spring mass oscillator,
2T
=
2 m
Tk
= (Time period of spring-mass oscillator)
Clearly,
2 2kk m
m = =
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SSHHMM((QQuuiicckkRReevviissiioonn))
A motion is linear SHM if given conditions are satisfied:
1. Motion must be oscillatory and hence periodic.
2. Force or acceleration of the particle is directly
proportional to its displacement from the equilibrium
position.
3. Force or acceleration is always directed opposite to the
displacement i.e., towards the mean position. Expressions
of S.H.M. are
F x
F= kx
a x2
a x=
2 2with & 2k
fm T
= = =
LLiinneeaarrSSHHMM
AApppplliiccaattiioonnmmeetthhoodd
STEP I: Find out the equilibrium position: At
equilibrium position net force and net torque is zero. For
linear SHM net force should be zero at equilibrium
position.
STEP II: Assume x = 0 at the equilibrium position.
Displace particle at a distance x from the equilibrium
position.
STEP III: Draw FBD of the particle when the particle is
at a distancexfrom the equilibrium.
STEP IV:Write Newtons 2
nd
law. Write this equation inthe form of 2a x= and find out.
STEP V: Use T = 2/to find out time period.
EExxaammppllee:: A mass m is attached to a vertical spring of
spring constant k. Suppose the mass is displaced from the
equilibrium position vertically. Find the time period of
the resulting oscillation.
Solution: Let us use step by step method to find out the
time period of the oscillation.
STEP I: Find out the equilibrium position. Let the
elongation of the spring beyat the equilibrium position.
We draw FBD to find out equilibrium position. Clearly
forces on the mass are: weight mg downward and spring
force ky upward . Therefore, we get,
mg ky= 0
y = mg/k (i)
Therefore, equilibrium position is at a distance y = mg/k
below the natural length of the spring.
STEP II: Assume x = 0 at the equilibrium position
Displace particle at a distance x from the equilibrium
position.
STEP III: Draw FBD of the particle when the particle is
at a distance x from the equilibrium. Forces are:
Weight mg (downward)
Spring force k(x+y) upward.
(because net compression from natural length is (x+ y) in
this case)
STEP IV: Using Newtons 2nd
Law in the downward
direction, we have,
mg k (x + y) = ma
a = g k (x + y)/m (ii)
Putting y = mg/k from eqn. (i) in eqn. (ii) we get,
a = g k (x + y)/m = g k (x + mg/k)/m
a = (k/m)x (iii)
Clearly, equation (iii) is in the form of 2a x=
Therefore, motion is SHM.
Therefore, comparing 2a x= with equation (iii), we get
2 = k/m
= (k/m) (iv)
STEP V: Use T = 2/to find out time period.
Time period,
2T
= 2
mT
k= (Ans).
Equilibrium
position
Normal
Length
y + x
x
y
Displaced
position
m
mmg ky
mg k(y+x)
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SSIIMMPPLLEEPPEENNDDUULLUUMM
A simple pendulum is an idealized model consisting of a
point mass (which is known as bob) suspended by a
massless, unstretchable string.
When the point mass is pulled to one side of its
equilibrium position and released, it oscillates in a
circular arc about the equilibrium position. We shall show
that, provided the angle is small (less than about 10), the
motion is that of a simple harmonic oscillator.
Let us consider the bob of mass mis suspended by a light
string of lengthL that is fixed at the upper end.
Clearly the equilibrium position is the lowest position of
the bob. Let the bob is rotated by an angle from
equilibrium. We have to show net torque is directly
proportional to angular displacement and is directed
opposite to .
Forces acting on the particle are:
Weight mg downward andTension Talong the string.
Now net torque about suspension point is given by
= (mg sin) L (i)
As the amplitude is small (less than about 10),
sin
Hence, eqn. (i) becomes
= mgL (ii)
(negative sign because torque is in clockwise direction,
whereas angular displacement is in anticlockwise
direction)
Thus, from (ii),
Hence, motion is SHM.
Now, from (ii)
= mgL
I= mgL
mL2= mgL (I = mL
2)
g
L =
2 =
whereg
L=
Time period is given by:
2T
=
2 L
Tg
= (time period of a Simple Pendulum)
Linear SHM Kinematics
Displacement
We know that a motion is SHM if a = 2x.
From Kinematics we know that acceleration is given by
2
2
d xa
dt=
Thus, a = 2x
22
2
d xx
dt =
which is a differential eqn. of 2nd
order.
Solution of this differential eqn. is given by
x = A sin (t + )
(displacement of a particle executing Linear SHM )
where,
x= displacement of particle from mean position at time t
A= amplitude
= angular frequency
(t+ ) = phase
= phase constant or phase difference
Thus, any eqn., where displacement can be written in the
form of x = A sin (t + ),represents SHM.
Note & Remember:
If you study different books you will find different
expressions for SHM, i.e., you may get
x = A cos(t + )instead of x = A sin(t + ).
You can use either of eqns. Both are correct. Thus
displacement:
x = A sin(t + )
or x = A cos(t + )
mgcos mgsin
L
T
O
mg
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Velocity & Acceleration
Eqn. of displacement by a particle executing SHM is
given by
x = A sin(t + ) (1)
Differentiating this eqn. wrt. time twe get velocity:
v= dx/dt
v=Acos(t + ) (2) Differentiating, velocity vwrt. time t, we get acceleration.
a= dv/dt
v=A2cos(t + ) (3)
If you compare eqn. (1) and (3) you will get
a = 2x
which represents SHM
Using eqns. (1) & (2) and little Trigonometry, we can find
the relation between velocity and displacement of the
particle undergoing SHM. This eqn. is given by
2 2v A x=
(relation between velocity & displacement in SHM)
Energy of the simple harmonic oscillator
Let the displacement and velocity of the mass executing
SHM at a particular instant of time bex& vrespectively.
We can writex& vin SHM as:
x = A sin (t + )
and v=Acos(t + )
Hence, kinetic energy of mass :
K = mv2
= m A2
2
cos2
(t+ ) (1)
Similarly, potential energy of spring :
U = kx2
= k A2
sin2
(t + ) (2)
Using, ( )2 ... 3k
k mm
= =
Thus from (1), we have
K = mv2
= m A2
2
cos2
(t+ )
K = k A2
cos2
(t+ ) (4)
Hence, Mechanical Energy of the system at that instant
ME = K+U = kA2
cos2
(t + ) +kA2
sin2
(t + )
ME = k A2
(Using cos2
(t + ) + sin2
(t + ) = 1)
Hence, Mechanical energy of the Simple harmonic
oscillator is given by:
ME = k A2
= m A2
2
(Mechanical Energy of Simple Harmonic Oscillator)
That is, the total mechanical energy of a simple harmonic
oscillator is a constant of the motion and is proportiona
to the square of the amplitude.
Note that U is small when K is large, and vice versa
because the sum must be constant.
Since, K = mv2
= m A2
w2
cos2
(w t + ) and
U = kx2
= k A2
sin2
(w t + ), we can plot energy
diagram as shown below:
Effective Spring Constant
Let n ideal springs of spring constants k1, k2, k3, , kn.Let k
effbe effective spring constant. Then,
Series combination:
1 2 3
1 1 1 1 1...
eff nK k k k k
= + + + +
Parallel combination:
keff
= k1+ k
2+ k
3+ + k
n
Time period spring mass system is given by:
2eff
mT
k
=
If a spring of spring constant k is broken into different
pieces then,
k x = k1x
1= k
2x
2= k
3x
3= = k
nx
n
and x = x1+ x
2++ x
n
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Effective g
Case 1: If a simple pendulum is in a carriage which is
accelerating with acceleration , then
effg g a=
& 2
eff
lT
g=
e.g., if the acceleration a
is upward, then
and 2eff
lg g a T
g a= + =
+
If the acceleration a
is downwards, then (g > a)
and T=2eff
lg g a
g a=
If the acceleration a
is in horizontal direction, then
2 2
effg a g= +
In a freely falling lift geff= 0 and T = , i.e., the
pendulum will not oscillate .
Case 2: If in addition to gravity one additional constant
force ,F
(e.g., electrostatic forcee
F
) is also acting on the
bob, then in that case,
eff
Fg g
m= +
& 2
eff
lT
g=
Here, mis the mass of the bob.
Physical Pendulum
Any rigid body suspended from a fixed support
constitutes a physical pendulum . A circular ring
suspended on a nail in a wall, a heavy metallic rod
suspended through a hole in it etc. are example of
physical pendulum. for small oscillations, the motion is
nearly simple harmonic. The time period is
2 I
Tmgl
=
(time period of a Simple Pendulum)
whereI= moment of inertia about suspension point and
l = distance between point of suspension and centre of
gravity
Example: A uniform rod of length 1.00 m is suspended
through an end is set into oscillation with small amplitude
under gravity. Find the time period of oscillation.
Solution : For small amplitude the angular motion is
nearly simple harmonic and the time period is given by
2
2
32 2
1.00 =2 2 1.16 .
3 3 9.80
ml
IT
mgl mgl
l ms
mg s
= =
= =
Oscillations of a liquid column in a U-tube
Suppose the U-tube of cross-section A contains liquid o
density upto height h.
If the liquid in one arm is depressed by distance x, it rises by
the same amount in the other arm. If the left to itself, the liquid
begins to oscillate under the restoring force,
F= Weight of liquid column of height 2 xF = A2x g= 2A g x (i)
i.e., Fx
Thus the force on the liquid is proportional to displacement
and acts in its opposite direction. Hence the liquid in the U
tube executes SHM. Comparing equation (i) with F = k x
we have
k= 2A g
The time-period of oscillation is
22 2 2
2
m A h hT
k A g g
= = =
If lis the length of the liquid column, then
2 and 2 .2
ll h T
g= =
Equilibrium level x
h
2x
x
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Oscillations of a body dropped in a tunnel along the
diameter of the earth
Let us consider earth to be a sphere of radius Rand centre O.
A straight tunnel is dug along the diameter of the earth. Let g
be the value of acceleration due to gravity at the surface of the
earth.
Suppose a body of mass mis dropped into the tunnel and it is
at point P at a depth d below the surface of the earth at any
instant. If g is acceleration due to gravity at P, then
' 1 d R d
g g gR R
= =
If x is distance of the body from the centre of the earth
(displacement from mean position), then
'
R d x
yg g
R
=
=
Therefore, force acting on the body at pointPis
' ...(i)
. .,
mgF mg x
R
i e F x
= =
Thus the body will execute SHM with force constant,
Comparing equation (i) with F= k x, we have
mgk
R=
The period of oscillation of the body will be
2 2 2 ./
m m RT
k mg R g = = =
Oscillation of a floating cylinder
In equilibrium, weight of the cylinder is balanced by the
upthrust of the liquid.
Let the cylinder be slightly depressed through distancexfrom
the equilibrium position and left to itself. It begins to oscillate
under the restoring force,
F= Net upward force = Weight of liquid column of height x
or,F= Axlg= Algx (i)
i.e., Fx.
Negative sign shows that F and x are in opposite directionsHence the cork executes SHM with force constant, k=A lg
Also, mass of =Ah
Period of oscillation of the cork is
2 2 2l l
m A h hT
k A g g
= = =
Oscillation of a ball in the neck of an air chamber
Let us consider an air chamber of volume V, having a neck of
area of cross-section Aand a ball of mass m fitting smoothly
in the neck. If the ball be pressed down a little and released, i
starts oscillating up and down about the equilibrium position
If the ball be depressed by distance x, then the decrease in
volume of air in the chamber is V=Ax.
Volume strainV Ax
V V
= =
If pressure Pis applied to the ball,
then hydrostatic stress = P
Bulk modulus of elasticity of air,
or/ /
P P EAE P x
V V Ax V V = = =
Restoring force,2
...(i)EAx EA
F PA A xV V
= = =
Thus F is proportional to xand acts in its opposite direction
Comparing equation (i) with F= k x, we have,
2EA
kV
=
Period of oscillation of the ball is
2 22 2 2
/
m m mV T
k EA V EA = = =
(a)
If the P-Vvariations are isothermal, thenE= P,
22 .
mVT
PA =
(b) If the P-Vvariations are adiabatic, thenE= P
22 .
mVT
PA
=
R
A
O
Pd
x
l
Ph
Equilibrium
positionl
l
x
Air
x
V
m
A
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1. Free oscillations:If a body, capable of oscillation, is
slightly displaced from its position of equilibrium and left
to itself, it starts oscillating with a frequency of its own.
Such oscillations are called free oscillations. The
frequency with which a body oscillates freely is called
natural frequency and is given by
01
2kvm
=
Some important features of free oscillations are
(a) In the absence of dissipative forces, such a body vibrates
with a constant amplitude and fixed frequency, as shown
in figure. Such oscillations are also called undamped
oscillations.
(b) The amplitude of oscillation depends on the energy
supplied initially to the oscillator.
(c) The natural frequency of an oscillator depends on its
mass, dimensions and restoring force i.e., on its inertial
and elastic properties (mand k).
Examples.
(i) The vibrations of the prongs of tunning fork struck
against a rubber pad.
(ii) The vibrations of the string of a sitar when pulled aside
and released.
(iii)The oscillations of the bob of pendulum when displaced
from its mean position and released.
2. Damped oscillations: The oscillations in which the
amplitude decreases gradually with the passage of time
are called damped oscillations.
In actual practice, most of the oscillations occur in
viscous media, such as air, water, etc. A part of the energy
of the oscillating system is lost in the form of heat, in
overcoming these resistive forces. As a result, the
amplitude of such oscillations decreases exponentiallywith time. Eventually, these oscillations die out.
In an oscillatory motion, friction produces three effects:
(i) It changes the simple harmonic motion into
periodic motion.
(ii) It decreases the amplitude of oscillation.
(iii) It slightly reduces the frequency of oscillation.
Examples.
(i) The oscillations of a swing in air.
(ii) The oscillations of the bob of a pendulum in a fluid.
Differential equation for damped oscillators and its
solution
In a real oscillator, the damping force is proportional to the
velocity v of the oscillator.
Fd = bv
where b is damping constant which depends on thecharacteristics of the fluid and the body that oscillates in it
The negative sign indicates that the damping force opposes the
motion.
Total restoring force = kx bv
2
2
2
2
or
or 0
d x dx dxm kx b v
dt dt dt
d x dxm b kx
dtdt
= =
+ + =
This is the differential equation for damped S.H.M.
The solution of the equation is
x(t) =Ae
bt/2m
cos (dt+ )The amplitude of the damped S.H.M. is
A =Aebt/2m
where A is amplitude of undamped S.H.M. Clearly, A
decreases exponentially with time.
The angular frequency of the damped oscillator is
2
24d
k b
m m =
Time period,2
2
2 2
4
d
d
Tk b
m m
= =
The mechanical energy of the damped oscillator at any instantis given by
( ) 2 2 /1 1
'2 2
bt mE t ka ka e
= =
Obviously, the total energy decreases exponentially with time.
As damping constant, b= F/v
SI unit of2
1
1 1
N kg mskg s
ms msb
= = =
x(t) t
A
0
+AConstant amplitude
Gradually falling amplitude
x(t) t
A
0
+A
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Resonance
Figure shows the variation of the amplitude of forced
oscillations as the frequency of the driver varies from zero
to a large value. Clearly, the amplitude of force
oscillations is very small for v > v0. But
when0,v v the amplitude of the forced oscillations
becomes very large. In this condition, the oscillator
responds most favourably to the driving force and draws
maximum energy from it. The case v = v0 is called
resonance and the oscillations are called resonant
oscillations.
Resonant oscillations and resonance:It is a particular
case of forced oscillations in which the frequency of thedriving force is equal to the natural frequency of the
oscillator itself and the amplitude of oscillation is very
large. Such oscillations are called resonant oscillations
and phenomenon is called resonance.
Examples.(a) An aircraft passing near a building shatters its window
panes, if the natural frequency of the window matches the
frequency of the sound waves sent by the aircrafts
engine.
(b) The air-column in a reasonance tube produces a loud
sound when its frequency matches the frequency of the
tuning fork.(c) A glass tumbler or a piece of china-ware on shelf is set
into resonant vibrations when some note is sung or
played.
Principal of tuning of a radio receiver
Tuning of the radio receiver is based on the principal of
resonance. Waves from all stations are present around the
antenna. When we tune our radio to a particular station,
we produce a frequency of the radio circuit which
matches with the frequency of that station. When this
condition of resonance is achieved, the radio receives andresponds selectively to the incoming waves from that
station and thus gets tuned to that station.
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vv0
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8/12/2019 SHM feb 13sa
11/12
S.H.M. Author: Pranjal Sir (B. Tech., IIT Kharagpur) Concept,, Sec 4, JB-20, Bokaro Ph. 7488044834
11 www.vidyadrishti.org An education portal for future IITians by Ex-IITians
Physics Classes by Pranjal Sir(Admission Notice for XI & XII - 2014-15)
Batches for Std XIIth
Batch 1 (Board + JEE Main + Advanced): (Rs. 16000)
Batch 2(Board + JEE Main): (Rs. 13000)
Batch 3(Board): (Rs. 10000)
Batch 4(Doubt Clearing batch): Rs. 8000
About P. K. Bharti Sir (Pranjal Sir)
B. Tech., IIT Kharagpur (2009 Batch)
H.O.D. Physics, Concept Bokaro Centre
Visiting faculty at D. P. S. Bokaro
Produced AIR 113, AIR 475, AIR 1013 in JEE -
Advanced
Produced AIR 07 in AIEEE (JEE Main)
Address:Concept, JB 20, Near Jitendra Cinema, Sec 4
Bokaro Steel City
Ph: 9798007577, 7488044834Email:[email protected]
Website:www.vidyadrishti.org
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
Basics from XIth Vectors, FBD, Work, Energy, Rotation,SHM
3r Mar to 4t Apr 14
1. Electric Charges and
Fields
Coulombs Law 5th
& 6th
Apr 5th
& 6th
Apr 5th
& 6th
AprElectric Field 10
th& 12
thApr 10
th& 12
thApr 10
th& 12
thApr
Gausss Law 13t
& 15t
Apr 13t
& 15t
Apr 13t
& 15t
AprCompetition Level NA 17
th& 19
thApr 17
th& 19
thApr
2. Electrostatic Potential
and Capacitance
Electric Potential 20th
& 22nd
Apr 20th
& 22nd
Apr 20th
& 22nd
AprCapacitors 24
th& 26
thApr 24
th& 26
thApr 24
th& 26
thApr
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
May3. Current Electricity Basic Concepts, Drift speed, Ohms
Law, Cells, Kirchhoffs Laws,Wheatstone bridge, Ammeter,
Voltmeter, Meter Bridge, Potentiometer
etc.
6th
, 8th
, 10th
, 13th
May
6th
, 8th
, 10th
, 13th
May
6th
, 8th
, 10th
, 13th
May
Competition Level NA 15th
& 16th
May 15th
, 16th
, 17th
, 18th
&
19th
May
PART TEST 2 Unit 3 18th
May NA NA
NA 20
th
May 20
th
MaySUMMER BREAK 21stMay 2013 to 30
thMay 2013
4. Moving charges and
Magnetism
Force on a charged particle (Lorentz
force), Force on a current carrying
wire, Cyclotron, Torque on a currentcarrying loop in magnetic field,
magnetic moment
31stMay, 1st&
3rdJun
31stMay, 1st&
3rdJun
31stMay, 1st& 3r Jun
Biot Savart Law, Magnetic field due
to a circular wire, Ampere circuitallaw, Solenoid, Toroid
5th, 7th& 8thJun 5th, 7th& 8thJun 5th, 7th& 8thJun
Competition Level NA 10t & 12t Jun 10t , 12t , 14t & 15t
Jun
PART TEST 3 Unit 4 15th
Jun NA NA
NA 22nd
Jun 22nd
Jun
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8/12/2019 SHM feb 13sa
12/12
S.H.M. Author: Pranjal Sir (B. Tech., IIT Kharagpur) Concept,, Sec 4, JB-20, Bokaro Ph. 7488044834
12 id d i hti A d ti t l f f t IITi b E IITi
5. Magnetism and
Matter
17th, 19th& 21st
Jun
17th, 19th& 21st
Jun
Not in JEE Advanced
Syllabus6. Electromagnetic
Induction
Faradays Laws, Lenzs Laws, A.C.
Generator, Motional Emf, Induced Emf,Eddy Currents, Self Induction, Mutual
Induction
24t , 26t & 28t
Jun
24t , 26t & 28t
Jun
24t , 26t & 28t Jun
Competition Level NA 29t Jun & 1stJul 29t Jun, 1st, 3r & 5t
Jul
PART TEST 4 Unit 5 & 6 6t
Jul NA NA
NA 13th
Jul 13th
Jul
7.
Alternating current AC, AC circuit, Phasor, transformer,resonance, 8th
, 10th
& 12th
Jul
8th
, 10th
& 12th
Jul
8th
, 10th
& 12th
Jul
Competition Level NA 15th
July 15th
& 17th
July8. Electromagnetic
Waves
19th
& 20th
July 19th
& 20th
July Not in JEE Advanced
Syllabus
PART TEST 5 Unit 7 & 8 27th
Jul 27th
Jul 27th
Jul
Revision Week Upto unit 8 31stJul & 2
nd
Aug
31stJul & 2
nd
Aug
31stJul & 2
ndAug
Grand Test 1 Upto Unit 8 3r
Aug 3r
Aug 3r
Aug
9.
Ray Optics
Reflection 5t
& 7t
Aug 5t
& 7t
Aug 5t
& 7t
AugRefraction 9th& 12thAug 9th& 12thAug 9th& 12thAugPrism 14t Aug 14t Aug 14t AugOptical Instruments 16t Aug 16t Aug Not in JEE Adv
SyllabusCompetition Level NA 19th& 21stAug 19th, 21st, 23rd, 24thAu
10.
Wave Optics
Huygens Principle 26thAug 26thAug 26thAugInterference 28t & 30t Aug 28t & 30t Aug 28t & 30t AugDiffraction 31stAug 31stAug 31stAugPolarization 2ndSep 2ndSep 2ndSepCompetition Level NA 4t & 6t Sep 4t , 6t , 7t , 9t , 11t Se
PART TEST 6 Unit 9 & 10 14th
Sep 14th
Sep 14th
Sep
REVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13th
Sep to 27th
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
Radiation and Matter
Photoelectric effect etc 9t & 11t Oct 9t & 11t Oct 9t & 11t Oct
Grand Test 3 Upto Unit 10 12th
Oct 12th
Oct 12th
Oct
12. Atoms 14th& 16thOct 14th& 16thOct 14th& 16thOct
13. Nuclei 18t & 19t Oct 18t & 19t Oct 18t & 19t Oct
X-Rays NA 21stOct 21st& 25t Oct
PART TEST 7 Unit 11, 12 & 13 26thOct NA NA
14. Semiconductors Basic Concepts and Diodes, transistors,logic gates
26t , 28t , 30t
Oct & 1stNov
26t , 28t , 30t
Oct & 1stNov
Not in JEE Adv
Syllabus15. Communication
System
2nd
& 4th
Nov 2nd
& 4th
Nov Not in JEE Adv
Syllabus
PART TEST 8 Unit 14 & 15 9thNov 9thNov NA
Unit 11, 12 & 13 Competition Level NA 8t
, 9t
& 11t
Nov
8t
, 9t
, 11t
, 13t
& 15Nov
PART TEST 9 Unit 11, 12, 13, X-Rays NA 16th
Nov 16th
Nov
Revision Round 2
(Board Level)
Mind Maps & Back up classes for late
registered students18
thNov to
Board Exams
18th
Nov to
Board Exams
18th
Nov to Board
Exams
Revision Round 3
(XIth portion for JEE)
18th
Nov to JEE 18th
Nov to JEE 18th
Nov to JEE
30 Full Test Series Complete Syllabus Date will be published after Oct 2014
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