SHM feb 13sa

Notes for School Exams

Physics XI Simple Harmonic Motion P. K. Bharti, B. Tech., IIT Kharagpur

© 2007 P. K. Bharti All rights reserved.

www.vidyadrishti.com/concept

2013-2015

S.H.M.

Author: Pranjal Sir (B. Tech., IIT Kharagpur)

Concept,, Sec 4, JB-20, Bokaro 2π

Some Definitions • •

• • •

f (t) = sin ωt = sin

Periodic motion: A motion which repeats itself after a regular interval of time is called periodic motion.

and

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 force and/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 (ν or f): The number of repetitions that occur per unit time is called frequency of the periodic motion. It is denoted by ν (Greek nu) or f. Frequency is the reciprocal of time period T. Therefore,

v= f=

1 T

T

Ph. 7488044834

t

g (t) = cos ωt = cos

2π t T

Spring mass system on a frictionless surface •

•

• •

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 that amplitude is the distance between mean position and extreme position. Amplitude is denoted by letter A and its SI unit is m.

(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 called harmonic 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:

2

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•

If we observe motion of the block carefully, we find that speed i.e., magnitude of velocity is maximum at mean position. Similarly speed is minimum i.e., zero at extreme positions as 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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S.H.M.


Concept,, Sec 4, JB-20, Bokaro

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Simple Harmonic Motion •

•

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 force F develops in the spring in the left direction. We also say this force F as restoring force as it tries to bring back mass m towards the mean position.

•

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 define SHM 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). k If we put = ω 2 in eqn. (i), we get m a = −ω 2 x

•

(S.H.M.)

…(A)

(negative sign because F is opposite to x) •

 F∝–x •

(S.H.M.)

…(B)

(because k is 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.). nd

•

Again, from Newton’s 2 Law, we have

•

•

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

3

(S.H.M.)


…(C)

2π T

(angular frequency in terms of time period)

SI unit of ω is s–1 . Since, frequency f is given by f = 1/T, therefore we can write 2π ω = 2π f = T

•

Time period of spring mass oscillator, 2π T= ω

− kx = ma

…(i)

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 2π rad in a time equal to its time period T. Therefore,

• •

Therefore, using (A),

k x m

Relation between angular frequency (𝛚𝛚) with time period (T) and frequency (f)

ω=

F = ma

⇒a= −

…(D)

where, 𝛚𝛚 is known as angular frequency of SHM.

As this restoring force F is opposite to that of displacement, therefore, we can write from Hooke’s Law F = – kx

(S.H.M.)

 •

T = 2π

m k

(Time period of spring-mass oscillator)

Clearly, k = ω 2 ⇒ k = mω 2 m


S.H.M.

Author: Pranjal Sir (B. Tech., IIT Kharagpur) SHM (Quick Revision)

• 1. 2.

3.

Concept,, Sec 4, JB-20, Bokaro •

A motion is linear SHM if given conditions are satisfied: Motion must be oscillatory and hence periodic. Force or acceleration of the particle is directly proportional to its displacement from the equilibrium position. 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 ∝ –x a = −ω 2 x 2 with ω =

k 2π ω = 2π f &= m T

•

•

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 distance x from the equilibrium.

• •

• • •

STEP IV: Write Newton’s 2 law. Write this equation in the form of a = −ω 2 x and find out𝜔𝜔. STEP V: Use T = 2π/𝜔𝜔 to find out time period.

Example: 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 be y at 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

4

…(i)


Displaced position

Equilibrium position y

m y+x x

ky

mg

m mg

k(y+x)

•

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) nd

•

STEP IV: Using Newton’s 2 direction, we have,

Law in the downward

mg – k (x + y) = ma

nd

•

Therefore, equilibrium position is at a distance y = mg/k below the natural length of the spring. Normal Length

Linear SHM Application method •

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 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 a = −ω 2 x Therefore, motion is SHM. Therefore, comparing a = −ω 2 x with equation (iii), we get

𝜔𝜔 2 = k/m

 𝜔𝜔 = √(k/m) …(iv)

STEP V: Use T = 2π/𝜔𝜔 to find out time period. •

Time period, T=

2π

ω



T = 2π

m k

(Ans).


S.H.M.

Author: Pranjal Sir (B. Tech., IIT Kharagpur) SIMPLE PENDULUM

•

•

• •

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 m is suspended by a light string of length L 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 θ from equilibrium. We have to show net torque τ is directly proportional to angular displacement θ and is directed opposite to θ.

Concept,, Sec 4, JB-20, Bokaro  

g θ L 2 α = –ω θ

α = –

g L Time period is given by: 2π T= ω where ω =

•

⇔T = 2π

L g

(time period of a Simple Pendulum)

Linear SHM Kinematics Displacement • •

We know that a motion is SHM if a = – ω2x . From Kinematics we know that acceleration is given by

d2x dt 2 Thus, a = – ω2x a=

θ •

T

L

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⇔

d2x = −ω 2 x 2 dt

nd

O mg sin θ •

θ

which is a differential eqn. of 2 order. mg cos θ •

x = A sin (ω t + ø)

mg Forces acting on the particle are:

(displacement of a particle executing Linear SHM )

Weight mg downward and

where,

Tension T along the string. •

As the amplitude is small (less than about 10°),

•

sin θ ≈ θ Hence, eqn. (i) becomes

τ = – mgL θ

• •

…(ii)

(negative sign because torque is in clockwise direction, whereas angular displacement is in anticlockwise direction) Thus, from (ii), τ ∝ –θ Hence, motion is SHM. Now, from (ii) τ = – mgL θ  Iα = – mgL θ  mL2 α= – mgL θ (∵I = mL2 )

5

A = amplitude

…(i)

•

•

x = displacement of particle from mean position at time t

Now net torque about suspension point is given by τ = (mg sinθ) L


Solution of this differential eqn. is given by

ω = angular frequency (ω t + ø) = phase ø = phase constant or phase difference •

Thus, any eqn., where displacement can be written in the form of x = A sin (ω t + ø), represents SHM.

• •

Note & Remember: If you study different books you will find different expressions for SHM, i.e., you may get x = A cos (ω t + ø) instead of x = A sin (ω t + ø). You can use either of eqns. Both are correct. Thus, displacement:

•

x = A sin (ω t + ø) or x = A cos (ω t + ø)


S.H.M.


Concept,, Sec 4, JB-20, Bokaro 2

Velocity & Acceleration •

•

2

ME = ½ k A = ½ m A ω

Eqn. of displacement by a particle executing SHM is given by x = A sin (ω t + ø) …(1) Differentiating this eqn. wrt. time t we get velocity: v = dx/dt

 v =A ω cos (ω t + ø) …(2) • Differentiating, velocity v wrt. time t, we get acceleration.

(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 proportional to the square of the amplitude. Note that U is small when K is large, and vice versa, because the sum must be constant. 2

•

a = dv/dt

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2

2

2

2

Since, K = ½ mv = ½ m A w cos (w t + ø) and 2

2

2

U = ½ kx = ½ k A sin (w t + ø), we can plot energy diagram as shown below:

…(3)  v = – A ω2 cos (ω t + ø) • If you compare eqn. (1) and (3) you will get a = – ω 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 v= ±ω A2 − x 2 (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 be x & v respectively. We can write x & v in SHM as: x = A sin (ω t + ø) and v =A ω cos (ω t + ø) Hence, kinetic energy of mass :

•

K = ½ mv = ½ m A ω cos (ω t + ø) Similarly, potential energy of spring :

•

2

2

2

2

2

2

•

2

2

... ( 3)

•

Series combination: 1 1 1 1 1 = + + + ... + K eff k1 k2 k3 kn

•

Parallel combination: k = k +k +k +…+k eff

•

2

 K = ½ k A cos (ω t + ø) •

2

2

2

ME = K+U = ½ kA cos (ωt + ø) + ½ kA sin (ωt + ø)  ME = ½ k A 2

1

2 2

(Using cos (ω t + ø) + sin (ω t + ø) = 1)

2

T = 2π

3

n

•

m keff

If a spring of spring constant k is broken into different pieces then, kx=k x =k x = k x =…=k x 1

1

2

2

3

and x = x + x +…+ x 1

•

2

3

n

n

n

Hence, Mechanical energy of the Simple harmonic oscillator is given by:

6

3

Time period spring mass system is given by:

…(4)

Hence, Mechanical Energy of the system at that instant 2

2

2

K = ½ mv = ½ m A ω cos (ω t + ø) 2

1

Let k be effective spring constant. Then,

…(2)

k ⇔ k= mω 2 m Thus from (1), we have 2

Let n ideal springs of spring constants k , k , k , …, kn. eff

Using, ω=

2

•

…(1)

U = ½ kx = ½ k A sin (ω t + ø) •

Effective Spring Constant



S.H.M.

Author: Pranjal Sir (B. Tech., IIT Kharagpur) Effective g

•

Case 1: If a simple pendulum is in a carriage which is accelerating with acceleration , then    g eff= g − a & T = 2π

l  g eff

 e.g., if the acceleration a is upward, then  l g eff = g+a and T = 2π g+a  If the acceleration a is downwards, then (g > a)

 g eff = g − a

and T=2π

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Solution : For small amplitude the angular motion is nearly simple harmonic and the time period is given by  ml 2    3  I T 2= = π 2π  mgl mgl l 1.00m = π =2π 2= 1.16 s. m 3g 3 × 9.80 2 s

Oscillations of a liquid column in a U-tube Suppose the U-tube of cross-section A contains liquid of density ρ upto height h.

l g −a

 If the acceleration a is in horizontal direction, then  g= a2 + g 2 eff

•


x

Equilibrium level

2x

x

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 force Fe ) is also acting on the bob, then in that case,    F g eff= g + & T = 2π m

l  g eff

Here, m is 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 T = 2π

I mgl

(time period of a Simple Pendulum)

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 x F = – A × 2x × ρ × g = – 2 A ρ g x …(i) i.e., F ∝ – x Thus the force on the liquid is proportional to displacement and acts in its opposite direction. Hence the liquid in the Utube executes SHM. Comparing equation (i) with F = – k x, we have k=2Aρg The time-period of oscillation is m T 2= = π 2π k

A × 2h × ρ h = 2π g 2 Aρ g

If l is the length of the liquid column, then where I = moment of inertia about suspension point and l = distance between point of suspension and centre of

l 2= h and T 2π =

l . 2g

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.

7



S.H.M.


Oscillations of a body dropped in a tunnel along the diameter of the earth Let us consider earth to be a sphere of radius R and 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. A d

P

R x


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Let the cylinder be slightly depressed through distance x from 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 = – A x ρl g = – A ρl g x …(i) i.e., F ∝ – x. Negative sign shows that F and x are in opposite directions. Hence the cork executes SHM with force constant, k = A ρl g Also, mass of = A ρ h ∴ Period of oscillation of the cork is

m Aρ h ρh T 2= π = 2π = 2π k Aρ l g ρl g

O

Oscillation of a ball in the neck of an air chamber Suppose a body of mass m is 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  d  R−d  g ' = g 1 −  = g    R  R  If x is distance of the body from the centre of the earth (displacement from mean position), then R−d = x y ∴g' = g R Therefore, force acting on the body at point P is mg F= x ...(i) −mg ' = − R i.e., F ∝ − x Thus the body will execute SHM with force constant, Comparing equation (i) with F = – k x, we have mg k= R The period of oscillation of the body will be

Oscillation of a floating cylinder In equilibrium, weight of the cylinder is balanced by the upthrust of the liquid. ρ

h

ρρl l

Equilibrium position

8


EA2 V Period of oscillation of the ball is k=

m m R T 2= 2π = 2π . = π k mg / R g

P

Let us consider an air chamber of volume V, having a neck of area of cross-section A and a ball of mass m fitting smoothly in the neck. If the ball be pressed down a little and released, it 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 ΔV = Ax. A ∆V Ax ∴ Volume strain = = V V m If pressure P is applied to the ball, then hydrostatic stress = P x ∴ Bulk modulus of elasticity of air, P P EA E= − = − − x or P = ∆V / V Ax / V V V Restoring force, Air EAx EA2 F= PA = − A= − x ...(i) V V Thus F is proportional to x and acts in its opposite direction. Comparing equation (i) with F = – k x, we have,

x ρl

ρ

m m mV T 2= 2π 2π = π = 2 k EA / V EA2 (a) If the P-V variations are isothermal, then E = P, mV . PA2 (b) If the P-V variations are adiabatic, then E = γ P ∴T = 2π

∴T = 2π

mV

γ PA2

.


S.H.M.



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+A 1.

Gradually falling amplitude

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

1 k 2π m • 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 (m and k).

0 x(t)

v0 =

Constant amplitude +A

0

x(t)

t

–A

t

–A 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 the characteristics 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 or m

d2x dx = −kx − b dt dt 2

dx   v =dt   

d2x dx + b + kx = 0 2 dt dt This is the differential equation for damped S.H.M. The solution of the equation is x(t) = A e–bt/2m cos (ωd t + ϕ) The amplitude of the damped S.H.M. is A’ = Ae–bt/2m where A is amplitude of undamped S.H.M. Clearly, A’ decreases exponentially with time. The angular frequency of the damped oscillator is or m

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 exponentially with 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.

9


= ωd

k b2 − m 4m 2

Td Time period, =

2π =

2π

ωd

k b2 − m 4m 2 The mechanical energy of the damped oscillator at any instant t is given by 1 2 1 2 − bt / m = E (t ) = ka ' ka e 2 2 Obviously, the total energy decreases exponentially with time. As damping constant, b = F/v ∴ SI unit of b =

N kg ms −2 = = kg s −1 ms −1 ms −1


S.H.M.



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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 and v >> v0. But when v ≈ v0 , 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.

a

v0

v

•

Resonant oscillations and resonance: It is a particular case of forced oscillations in which the frequency of the driving 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 aircraft’s 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 and responds selectively to the incoming waves from that station and thus gets tuned to that station.

10



S.H.M.



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

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

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, 7488044834 Email: [email protected] Website: www.vidyadrishti.org

Physics Class Schedule for Std XIIth (Session 2014-15) by Pranjal Sir Sl. No.

Main Chapter Basics from XIth

1.

Electric Charges and Fields

2.

Electrostatic Potential and Capacitance

PART TEST 1 3.

Current Electricity

PART TEST 2 SUMMER BREAK 4. Moving charges and Magnetism

PART TEST 3

11

Topics

Board level

JEE Main Level rd

Vectors, FBD, Work, Energy, Rotation, SHM

3 Mar to 4 Apr 14

Coulomb’s Law Electric Field Gauss’s Law Competition Level Electric Potential Capacitors Competition Level

5th & 6th Apr 10th & 12th Apr 13th & 15th Apr NA 20th & 22nd Apr 24th & 26th Apr NA

5th & 6th Apr 10th & 12th Apr 13th & 15th Apr 17th & 19th Apr 20th & 22nd Apr 24th & 26th Apr 27th & 29th Apr

Unit 1 & 2

4th May NA 6th, 8th, 10th, 13th May

NA 11th May 6th, 8th, 10th, 13th May

NA

15th & 16th May

Basic Concepts, Drift speed, Ohm’s Law, Cells, Kirchhoff’s Laws, Wheatstone bridge, Ammeter, Voltmeter, Meter Bridge, Potentiometer etc. Competition Level

18th May NA 21st May 2013 to 30th May 2013 Force on a charged particle (Lorentz 31st May, 1st & force), Force on a current carrying 3rd Jun Unit 3

wire, Cyclotron, Torque on a current carrying loop in magnetic field, magnetic moment Biot Savart Law, Magnetic field due to a circular wire, Ampere circuital law, Solenoid, Toroid Competition Level

Unit 4


JEE Adv Level

th

NA 20th May

5th & 6th Apr 10th & 12th Apr 13th & 15th Apr 17th & 19th Apr 20th & 22nd Apr 24th & 26th Apr 27th & 29th Apr, 1st, 3rd & 4th May NA 11th May 6th, 8th, 10th, 13th May

15th, 16th, 17th, 18th & 19th May NA 20th May

31st May, 1st & 3rd Jun

31st May, 1st & 3rd Jun

5th, 7th & 8th Jun

5th, 7th & 8th Jun

5th, 7th & 8th Jun

NA

10th & 12th Jun

15th Jun NA

NA 22nd Jun

10th, 12th, 14th & 15th Jun NA 22nd Jun


S.H.M.


5.

Magnetism and Matter Electromagnetic Induction

6.

PART TEST 4 7.

th

st

th

th

st

Ph. 7488044834

17 , 19 & 21 Jun 24th, 26th & 28th Jun

Not in JEE Advanced Syllabus 24th, 26th & 28th Jun

NA

29th Jun & 1st Jul

6th Jul NA 8th, 10th & 12th Jul NA 19th & 20th July

NA 13th Jul 8th, 10th & 12th Jul 15th July 19th & 20th July

29th Jun, 1st, 3rd & 5th Jul NA 13th Jul 8th, 10th & 12th Jul

27th Jul 31st Jul & 2nd Aug 3rd Aug 5th & 7th Aug 9th & 12th Aug 14th Aug 16th Aug

27th Jul 31st Jul & 2nd Aug 3rd Aug 5th & 7th Aug 9th & 12th Aug 14th Aug 16th Aug

Photoelectric effect etc

9th & 11th Oct

9th & 11th Oct

9th & 11th Oct

Upto Unit 10

12th Oct

12th Oct

12th Oct

14th & 16th Oct

14th & 16th Oct

14th & 16th Oct

18th & 19th Oct NA 26th Oct 26th, 28th, 30th Oct & 1st Nov 2nd & 4th Nov

18th & 19th Oct 21st Oct NA 26th, 28th, 30th Oct & 1st Nov 2nd & 4th Nov

9th Nov NA

9th Nov 8th, 9th & 11th Nov 16th Nov 18th Nov to Board Exams 18th Nov to JEE

18th & 19th Oct 21st & 25th Oct NA Not in JEE Adv Syllabus Not in JEE Adv Syllabus NA 8th, 9th, 11th, 13th & 15th Nov 16th Nov 18th Nov to Board Exams 18th Nov to JEE

Faraday’s Laws, Lenz’s Laws, A.C. Generator, Motional Emf, Induced Emf, Eddy Currents, Self Induction, Mutual Induction Competition Level

AC, AC circuit, Phasor, transformer, resonance, Competition Level

8.

Electromagnetic Waves PART TEST 5 Revision Week

th

17 , 19 & 21 Jun 24th, 26th & 28th Jun

Unit 5 & 6

Alternating current


Unit 7 & 8 Upto unit 8

15th & 17th July Not in JEE Advanced Syllabus 27th Jul 31st Jul & 2nd Aug

3rd Aug 9. Reflection 5th & 7th Aug Refraction 9th & 12th Aug Prism Ray Optics 14th Aug Optical Instruments Not in JEE Adv 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 Diffraction Wave Optics 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 Unit 9 & 10 PART TEST 6 14th Sep 14th Sep 14th Sep th th REVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13 Sep to 27 Sep Upto Unit 10 Grand Test 2 28th Sep 28th Sep 28th Sep Upto Unit 8

Grand Test 1

DUSSEHRA & d-ul-Zuha Holidays: 29th Sep to 8th Oct 11.

Dual Nature of Radiation and Matter Grand Test 3 12.

Atoms

13.

Nuclei X-Rays PART TEST 7 14. Semiconductors 15.

Communication System PART TEST 8 Unit 11, 12 & 13

PART TEST 9 Revision Round 2 (Board Level) Revision Round 3 (XIth portion for JEE) 30 Full Test Series

12

Unit 11, 12 & 13 Basic Concepts and Diodes, transistors, logic gates

Unit 14 & 15 Competition Level Unit 11, 12, 13, X-Rays Mind Maps & Back up classes for late registered students

NA 18th Nov to Board Exams 18th Nov to JEE

Complete Syllabus

Date will be published after Oct 2014



SHM feb 13sa

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