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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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S.H.M. Author: Pranjal Sir (B. Tech., IIT Kharagpur) Concept,, Sec 4, JB-20, Bokaro Ph. 7488044834

2 www.vidyadrishti.org An education portal for future IITians by Ex-IITians

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.

a

vv0

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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


& 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
http://www.vidyadrishti.org/http://www.vidyadrishti.org/mailto:[email protected]:[email protected]:[email protected]://www.vidyadrishti.org/http://www.vidyadrishti.org/http://www.vidyadrishti.org/http://www.vidyadrishti.org/mailto:[email protected]://www.vidyadrishti.org/

8/12/2019 SHM feb 13sa

12/12


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


July 15th

& 17th

July8. Electromagnetic

Waves

19th

& 20th

July 19th

& 20th

July Not in JEE Advanced

Syllabus


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


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

Documents

SHM feb 13sa