Unit-1 Simple Harmonic motion

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Unit-1 Simple Harmonic motionSimple harmonic motion can be defined as the simplest kind of oscillatory motion

in which displacement varies sinusoidally with time i.e. displacement is sine or

cosine function of time.

Let us consider a particle Pat a point P(x,y) rotating on a circumference of a circle

of radius A with uniform angular velocity ω. O is the origin (centre of circle).

Let a time t=0, P lies on x-axis i.e. at P0. At an arbitrary time t, the point will be at

P where angle POX = ωt.OP is the position vector of the particle. Let, N be the

foot of the perpendicular (projection on point P on x-axis).

i.e. ON=A cos ωt& OM =A sin ωtLet, ON = x: so x= a cosωt.

As particle (P) rotates on the circumference of the circle N moves to and fro about

the origin on the diameter. The position on the particle P and that of N at different

times:

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times:

1












times:

2

SerialNo.

Position ofparticle

(P)

Position of projectionon

X-axis (N)

Time as afraction of time

period (T)

Angledescribed byradius vector

(θ)1. P0 P0 (extreme position) 0 02. P1 O (mean position) T/4 π/23. P2 P2(Extreme position) T/2 π4. P3 O (mean position) 3T/4 3π/25. P4 P0 (extreme position) T 2 π

Thus, it is evident that as P moves on the circumference of the circle, N moves to

and fro about the centre O along the diameter POP2. Hence, when a particle rotates

on the circumference of a circle with an angular velocity the foot of the

perpendicular on its diameter executes simple harmonic motion.

Characteristics of simple harmonic motion

Amplitude: The maximum displacement of N on either side of the mean

position O is called amplitude of simple harmonic motion (S.H.M) i.e. OP0 = A,

OP2= −A.

Time period (t): Time required to complete one revolution i.e. ωt = 2π.

t = . This t is called time period of S.H.M. and denoted as t.

Frequency (υ): Inverse of time period i.e. υ = = . Hence, ω = 2πυ.

Phase (ϕ):The choice of time t= 0 is arbitrary and one could have chosen t=0 to

be instant when P was at P’ (not a P0).

Now if angle P’OX= θ and POX = (ωt+θ). Hence, ON= A cos (ωt+θ). Thequantity (ωt+θ) is known as the phase of the motion. Θ represents initial phase.The value of θ is arbitrary and depends on the instants from which one startmeasuring time.

1.1 Equation of motion of a particle executing simpleharmonic motion (SHM):

The displacement of a particle executing SHM:

3

2 2

22

2

2

cos

cos( )

x A t

dxv

dt

A x

d xa A t

dt

a x

Now, we know that, F=ma or F=-mω2xAgain, F=-kx[if we write k=mω2=force constant]

2

2

2

2

22 2

2

0

0

d xm kx

dt

d x kx

dt m

d xx k m

dt

The solution of this equation:

cos sin cos sin

cos sin

where c= +cos sin

- are const.

i t i tx e e

t i t t i t

t i t

C t D tD i

1.2.Graphical representation of SHM

2

cos

sin

cos

x A t

v A t

a A t

t ωt x v a0 0 A 0 -Aω2

T/4 π/2 0 -Aω 0T/2 π -A 0 Aω2

3T/4 3π/2 0 Aω 0T 2π A 0 -Aω2

Again, let

C=Acos

D=-Asin

cosx A t

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Time vs. displacement plot:

Energy in SHM:

When a particle executes SHM, its total energy consists of potential energy andkinetic energy.

Let us consider, a particle of mass m executing SHM along X axis OX. If theparticle is displaced through a distance OL= S from the mean position O, therestoring forced acting on the particle is= − ; = force constant

Let the particle be further displaced by distance dS; then work done on the particleagainst force F is: = . = cos 180°= − = −(− ) =

o L NM

=

===

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Let at any instant t the displacement of the particle from the mean position be x.

Total work done on the particle for displacement x is given by:= = = 12 = 12This work done stored as potential energy, which is given by:= 12 = 12 ( + )And kinetic energy, is:= 12 = 12 ( + ) = 12 ( − )So, the total energy is given by:= + = 12 + 12 ( − ) = Hence, the total energy is independent of time.

Sl. No. Physicalquantity

Mean positionx=0

Extremeposition x=A

At = √1. 0 (min) (max) 142. (max) 0 (min) 143. 12 12 14 Variation of , and as function of position:

1.3.Damped Harmonic OscillationDamping force acts in direction opposite to the motion or displacement for smallvelocity, proportional to velocity.1. Frestoring=-k0x2. Fdamping=-rdr/dt; r=damping constant.

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3. Equation of motion of the damped oscillatory particle

2

2

2

2

2

2

22

2

o

0

0

2 0.................................. 1

2 damping coefficient.

natural frequency of undamped

net restoring damping

o

o

o

o

F F F

d x dxm k x r

dt dt

d x dxm r k x

dt dt

kd x r dxx

dt m dt m

d x dxb x

dt dtr

bm

oscillation

Solution of this equation:

2 2

2 2

2 22

2 2

22 2

02

( )

.

2

From equation 1 ;

2 2

bt

bt bt

bt

bt bt

bt

bt bt

x t t e

dx de be

dt dtd

b edt

d x d d db e b b e

dt dt dt dt

d x d db b e

dt dt dt

d x d db b e b b e

dt dt dt

22 2

02

2 20

0

0

Now, can be positive, negative or zero.

btt e

db

dt

b

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Case 1:

2 2

2 2

2 2

;damping is small (Underdamping)

t cos t+θ

cos t+θ

o

o

bto

b

A b

x t Ae b

Logarithmic decrement:- The rate at which the amplitude dies away:

2 2

2 2

1 2 3

cos t+θ

Let θ=0;

cos t+θ

At t=0; x=A

Suppose A ,A ,A ,.................are the amplitudes at t=T,2T,3T,..................

bto

bto

x t Ae b

x t Ae b

1

22 1

33

Then, A

A A

A A

bt bT

bT

bT

Ae Ae

e

e

Hence, the decrement in amplitudes;32

1 3 4

............. btAAAd e

A A A

2 2o

2 bNatural log of decrement; =bT=

b

Mean life time (τm):-Time interval in which the amplitudes of the under-damped oscillator falls to 1/e ofits initial value.

1

1

11;

m

m

b

b

m m

A Aee

ee

bb

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Quality Factor(Q):-Half of the mean life time is called relaxation time (τ). Now, quality factor(Q)=τωo. (ωo- natural frequency)

Case 2:

2 2 2 2o o

2 2 2 2o o

2 2o

22 2

o2

1 2

1 2

(Overdamping)

d0

( )b t b t

b b t b b t

b

bdt

t A e A e

x t A e A e

These two power are negative, the displacement decreases exponentially to zerowhen time increases without performing any oscillation.

Case 3:

2 20

2

2

critical damping

d0

;C and D are const. of Integrationbt

b

dtt ct D

x t ct D e

displacement approaches zero asymptotically

1.3.Forced OscillationThe Phenomenon of setting body into vibrations with the help of periodic forcehaving frequency different from the natural frequency of the body is called forcedvibrations.Example: vibration of bridge under the influence of marching soldiers. Let usconsider external periodic force Fcosωt, where ω= frequency of applied force.The equation of motion

Fnet = Frestor+ Fdamp+Fext.

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and

2

2

2

2

22 2

0 02

dcos

dcos

d F2 cos ;where G= .;2 ; .............. 1

m

x dxm kx r F t

dt dt

x r dx k Fx t

dt m dt m m

x dx r kb x G t const b

dt dt m m

Equation (1) is an inhomogeneous linear second order differntial equation.Solution is given by:x(t)=xc(t)+xp(t); xc(t):complimentary solution; xp(t)- Particular integralThe complimentary solution;

Assuming 2 20b ; (weak damping or under damping).

2 20cosbt

cx t Ae b t

The particular integral;Let's try

22

2

2 20

2 20

cos

sin

cos

From equation(1):

cos 2 sin cos cos

cos 2 sin cos cos cos sin sin ........ 2

p

p

p

x t a t

dxa t

dt

d xa t

dt

a t ba t a t G t

a t ba t a t G t G t

Equating coefficients of cos(ωt-φ) and sin(ωt-φ)

2 20 cos

.................... 32 sin

a G

ab G

using equation (3);

10

22 2 2 2 2 2 20

22 2 2 2 2 20

22 2 2 20

22 2 2 20

4

4

................... 44

cos

cos .................. 54

p

a a b G

a b G

Ga

b

x t a t

Gt

b

Thus general solution x(t)=xc(t)+xp(t);

=

2 20

2 20 22 2 2 2

0

cos cos

cos cos ............ 64

c p

bt

bt

x t x t

Ae b t a t

GAe b t t

b

First term of RHS:Transient solutions (corresponding to natural vibrations which dies awayexponentially with time).Second term of RHS:Steady state solutions (corresponds to forced vibrations by external force. Nowfrom equation (3):

2 20

12 20

2tan ,

phase difference between oscillator and external driving force

2= tan ................... 7

b

b

Amplitude(a):

22 2 2 20

............... 74

Ga a

b

'a' will be maximum when 22 2 2 20 4b is minimum.

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Resonance: frequency of driving force = natural frequency of body.

So, the condition of resonance; 22 2 2 20 4b is minimum

22 2 2 20

2 2 20

2 2 20

2 2 20

22 20 0 2

0

.

4 0

2 2 8 0

2 0

2

22 1 ................. 8

i e

db

d

b

b

b

bb

When damping is extremely small (b tends to zero), ω=ω0 [from equation 8].

(I) Resonant frequency: 2

0 20

21 ............ 8R

ba

(II) Resonant Amplitude: From equation 7(a) taking ω=ωR

22 2 2 20

22 2 2 20

2 2 2 4 2 2 4 2 20 0 0

4

2 4 2

F, as G= .............. 9

m4 4 8 4 4 2

R

Ga

b

G

b b b

G G F

b b b b b bm b

For very weak damping b2<<ω02;

0 0

r, as 2b= ................ 10

2 m

1, .............

R

R

F Fa

bm r

Thus ar

(III) Sharpness of Resonance(SOR): Indicates how fast the amplitude falls oneither side of resonant frequency. SOR depends upon 'b' [equation (8)a]

12

0

1 0

2 0

1 2

0

1 2

2 width of resonance curve.

If 'b' is very small

If 'b' is small : small and resonance is sharp. i.e

1SOR ................... 12

b

R

R

b

b

b

b

(IV) Quality Factor(Q): 0average energy stored per unit cycle2 .

average energy dissipated per unit cycle 2Q

b

Unit-1 Waves and Wave function

1.4 Waves and wave functionA wave function is described by the periodic variation of an associated physical quantity inspace and time. This is known as wave function (r,t) .Wave is a pattern generated in medium when a disturbance (energy ) travel from one toanother point with no transport of particles. Mainly waves are of two types – mechanicaland non-mechanical waves.

The waves which can propagate only in a material ,medium are called mechanical waves:e.g. sound waves through solid, liquid and gaseous medium . The waves which can moveeven without a medium are called non- mechanical waves or electromagnetic waves; e.g.light waves, radio waves, etc. in mechanical waves displacement of particles of the mediumform equilibrium position varies periodically.

In electromagnetic waves electric and magnetic fields vector vary periodically in space andtime. The associated physical quantity the periodic variation of which in space and timedescribes the wave motion is called wave function.

Longitudinal waves Transverse Wavesυ= √(E/ρ)

E=mod of elasticity; ρ=density of medium

υsolid=√(η/ρ) ;η=rigidity mod

υsolid=√(Y/ρ) ; Y= young’sMod

υstreched string=√(T/μ); T= tensionstring;μ=Mass per unit length

υLiquid= √(K/ρ) ; K= BulkMod

υe.m. in free space=1/√(μ0 ε0);μo=magnetic permeability of freespace; εo=electric permeability infree space

υGas= √(γp/ρ) ; p= pressureof gas medium; γ=cp/cv

























Reflection and transmission of a wave

Reflection at denser medium (Z2>Z1) ; Z= impedance;; =mass per unit length; =wave velocityB1 = -A1

A2 < A1

Reflection at rarer medium(Z2<Z1)

B1 = A1

A2 = 2A1 A1= amplitude of incidentB1= amplitude of reflected waveA2= amplitude of transmitted wavePhase velocity, Group velocity, wave packet

A harmonic wave moving along positive X-direction is represented by the wave function.( , )= A cos ( − );Where, ( − )= phase of the wave.

The locus of the points at which the wave function ( , ) has the same phase is calledsurface of constant phase or the wave front. The velocity at which the wave front is thephase velocity.

During the propagation of the wavefront in space the quantity ( − ) remains constanton its surface; i.e. ( − )=constant

( − )=0

k – =0

= = p= phase velocity

Or, p= = =

Thus phase velocity is equal to the wave velocity of the harmonic wave.

It is usually observed the a wave packet is formed when a larger no. of harmonic waveswith slightly different amplitudes, frequency and wavelengths superpose such that theresultant wave function is non-zero only over a finite region of space and zero outside theregion . Thus a wave packet is localizes wave train, the amplitude of which is zero except in

a small region of space. Example pulse travelling along a string. The velocity ofpropagation of the envelop or the amplitude of the wave packet is called group velocity.

# Group velocity, g=

Phase velocity p= ; or = k p; -----------------------------Eqn 1

Group velocity g= = (k p)= p+k ;-----------------Eqn 2

Now, we know , k=2 ; or = 2Thus = -2 ;---------------Eqn 3

From eqn 2

g= p+k. .

= p+k . .(-2 )

= p+ .(-2 )

= p - ;-----------------------------Eqn 4

Case-1:

>0; g< p

This phenomenon is called normal dispersion and the medium causing such dispersion iscalled normal dispersive medium.E.g.; formation of rainbow due to passing of sunlight through rain drops.

1.5 Principal of superstition of waves:It states that when two or more waves move through a point simultaneously , the wavefunction resultant wave is equal to the sum of the wavelength of the individual waves.

If 1, 2, 3,…… N are the wave function of the individual waves then the wavefunction of the resultant wave will be:

= 1+ 2+ 3+…..+ N

Case-2

<0; g> p

This phenomenon is called anomalous dispersion and the medium causing it calledanomalous dispersive medium. E.g.; propagation of em wave through ionized media.

Case-3 =0; g= p

e.g.; propagation of em wave through vacuum.In such cases, since all the components of wave move with same velocity , they are notseparated out. Hence, dispersion does not take place. This medium is called non-dispersivemedium.

#Superposition of two similar waves in opposite direction(stationary waves):When two waves of same frequencies , wavelengths and amplitude travel with same speed

in opposite directions, they superpose on each other to produce a new type of wave calledstanding or stationary wave. They are called stationary because flow of energy along thewave is zero.

Let, 1= A cos ( − )So, 2=A cos (- − )

= A cos ( + )Thus, = 1 + 2 Formula:⇒ = 2A cos kx cos cos A+ cos B= 2 cos cos

Thus = R cos ; where, R=2A cos kx

Equation 1 represents SHM with angular frequency and amplitude R . R is not constantrather varies as x. the position of point in the region where the stationary wave is formed.

Antinodes: Nodes:

The amplitude r is max. at positions; The positions at which R vanishi.e..

Cos kx= ± 1 cos kx=0

Thus kx= n ; n=0,1,2,3,4…… Thus kx= (2n+1) ;

n=0,1,2,3…..⇒ =n ⇒ = (2n+1)

X= x= (2n+1)

So, at x= 0, , , ,……… Thus at x= ,

, ,………..Max. amplitude. This point are called the resultant wave vanishes.Antinodes; and R=±2A

Superposition of two waves of same frequency moving in same direction

# consecutive nodes are antinodes are separated by .

# at antinodes R=2A

Superposition of two waves of same frequency moving in same direction :-When 1= A1 cos ( − + 1 ) and

2= A2 cos ( − + 2 ) two harmonic waves having same frequency, butdifferent amplitudes and different phases superpose then the resultant wave function:

= 1+ 2

When two harmonic waves superpose, the wavefunction of resultant wave is equal to sum

of the wave-function of resultant wave is equal to sum of the wave functions of individual

waves. But , intensity of the resultant waves is not equal to the sum of intensities of

individual waves . Rather , the resultant intensity depends on the intensities of component

waves as well as on the phase difference between the two waves.

Let us consider, two harmonic waves same frequency but different amplitudes and differentphase.

1= A1 cos ( − + 1 );--------------------------------eqn 1

2= A2 cos ( − + 2 );--------------------------------eqn 2

Therefore = 1+ 2= A1 cos ( − + 1 ) + A2 cos ( − + 2)= A1[cos ( − ) cos 1 – sin ( − ) sin 1] + A2[cos ( − ) cos

2

– sin ( − )sin 2]

= cos ( − ) [A1 cos 1+ A2 cos 2]- sin ( − )[ A1 sin 1+ A2 sin

2]

Let , A1 cos 1+ A2 cos 2 = R cos ;------------------------eqn 3and; A1 sin 1+ A2 sin 2 = R sin ;-------------------------eqn 4. . = R cos cos ( − ) - R sin sin ( − )

= R cos ( + − )= R cos ( − + ); -----------------------------eqn 5

This shows that the resultant wave is harmonic wave with amplitude R and phase .# Determination of R and : squaring and adding eqn 3 and 4

R2= A12cos2

1 + A22 cos2

2+2A1 A2cos 1cos 2+ A12 sin 2

1+ A22 sin2

2 + 2A1

A2sin

1sin 2

= A12 + A2

2 + 2A1 A2cos ( 2- 1) ……………..eqn 6 [ . cos ( 2- 1)= cos( 1- 2)]

Also eqn 4 ÷ eqn 3

Tan = ………………………eqn 7

1.5 Determination of intensity (I):Intensity (I) {amplitude (R)}2∴ I=kR2 [k= proportionality constant]∴ I= k[A1

2 + A22 + 2A1 A2cos ( 2- 1)]∴ I= I1 + I1 + 2 I I cos ( 2- 1);----------------------eqn 8

Thus , I depends upon I1 and I2 of individual waves and on the interference termcos ( 2- 1).

# For Coherent superposition of Coherent superposition of no. ofwaves

two waves:Phase difference ( 2- 1)= const.

a. If ( 2- 1)=± 2n ; a. Imax =∑ ( )Where , n=0,1,2,3… If all the waves have same amplitude

Then cos( 2- 1)=1 then, Ii = I1

From eqn 6 ∴ Imax= N2IR2=(A1 + A2)

2 or, Imax N2

From eqn 8

Imax=( + )2

b. If 2- 1 = ±(2n+1) b. Imin=∑ − ∑ ; ,Cos ( 2- 1)= -1From eqn 6R2= (A1 - A2)

2

From eqn 8

I=( − )2

#For Incoherent superposition:-The value of cos( 2- 1) will vary from -1 to +1.So, time average of cos( 2- 1) at a point is zero..

i.e. <cos ( 2- 1)> = 0from eqn 6: R2=(A1 + A2)

2 For two Wavesfrom eqn 8: I= I1 + I1 * for no. of waves

I=∑ =NI1 [ifIi=I1]

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Unit-1 Interference of light waves

1.1 IntroductionWave interference is the phenomenon that occurs when two waves meet

while traveling along the same medium. The interference of waves causes themedium to take on a shape that results from the net effect of the two individualwaves upon the particles of the medium. The phenomenon of the interference oflight has proved the validity of the wave theory of light. In physics, interference isa phenomenon in which two waves superpose to form a resultant wave of greater,lower, or the same amplitude. According to it, when two light waves of the samefrequency and having a constant phase difference traverse simultaneously in thesame region of a medium and cross each other, then there is a modification in theintensity of light in the region of the superposition, which is in general differentfrom the sum of intensities due to individual waves at that point. This modificationin the intensity of light resulting from the superposition of two (or more) waves oflight is called interference. At certain points the waves superimpose in such a waythat the resultant intensity is greater than the sum of the intensities due toindividual waves. The interference produced at these points is called constructiveinterference or reinforcement, while at certain other points the resultant intensity isless than the sum of the intensities due to individual waves. The interferenceproduced at these points is called destructive interference. Beyond the region ofsuperposition the waves come through completely uninfluenced by each other.

1.2.Types of interferenceThe phenomenon of interference may be grouped into two categories namely: (a)Division of Wavefront, (b) Division of Amplitude.(a) Division of Wavefront:Under this category, the coherent sources are obtained by dividing the wavefront,originating from a common source, by employing mirrors, biprisms or lenses. This

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1





2

class of interference requires essentially a point source or a narrow slit source. Theinstruments used to obtain interference by division of wavefront are the Fresnelbiprism, Fresnel mirrors, Lloyd's mirror, lasers, Youngs double slit experiment,etc.

(b) Division of Amplitude: In this method, the amplitude of the incident beam isdivided into two or more parts either by partial reflection or refraction. Thus wehave coherent beams produced by division of amplitude. These beams traveldifferent paths and are finally brought together to produce interference. The effectsresulting from the superposition of two beams are referred to as two beaminterference and those resulting from superposition of more than two beams arereferred to as multiple beam interference. The interference in thin films, Newton'srings, and Michelson's interferometer are examples of two beam interference andFabry-Perot's interferometer is an example of multiple beam interference.

1.3.Young's double slit experiment

Figure 1.1 Interference pattern from two slitsIn 1801, Thomas Young performed an experiment that lead to the establishment ofthe wave theory of light. This arrangement consists of two small, closely spacedholes or slits, S1 and S2, in an opaque surface with a light source S placed behind it,as shown in Figure 1.2. Here S1 and S2 serve as two coherent sources of light of thesame intensity. Young used sunlight as a light source and observed an interference

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pattern on a screen placed parallel to the two sources S1 and S2. A series of a fewcolored bright and dark vertical lines called fringes appear on the screen. As amodification of the original experimental arrangement, sunlight is replaced bymonochromatic light and then we get alternatively dark and bright bands.

Figure 1.2 Young's double slit experimentLet S be a narrow slit illuminated by monochromatic light of wavelength ?, and S1

and S2 be the two narrow slits separated by a small distance 2d. Let a screen XY beplaced at a distance D from the coherent sources S1 and S2. Let O be the foot of theperpendicular drawn from C, the mid-point of S1 and S2 on the screen. Nowconsider a point P on the screen at a distance x from O at which the condition forbright and dark fringes are to be determined. Draw S1M and S2N perpendicularsfrom S1 and S2 on the screen and join S1P and S2P.

The path difference between the waves reaching at P from S1 and S2 is given by:

2 1Δ=S P-S PFurther, in the right angled triangle S2NP, we have:

2 2 2 222 2

12 21

2 222

S P S N NP D + x+d

x+dS P D + x+d 1+

DD

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Using the binomial theorem and neglecting higher terms, since D >> x or d, weget:

Similarly in a right angled triangle S1MP, we have:

Thus, the path difference will be:

2 2

2 2

x+d x-dΔ= +

2 21

x+d x-d2

4xd 2xdΔ=2D D

D DD D

D

For simplicity of calculation, if the distance between the slits will be taken as'd' instead of '2d' then the above path difference can be written as:

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xdΔp=D

Now, the phase difference will be :

2π 2π xdΔφ= .Δp= .λ λ D

(i) Condition for Bright FringesFor bright fringes and for maximum intensity at P.

Δp=2n. ; 0,1,2,3,.....2

xd

D n

n

n Dn x x

d

So, bright fringes will be obtained at :

2 40, , , ,..........................

D D D

d d d

(ii)Condition for Dark FringesFor dark fringes and for minimum intensity at P.

Δp= 2n+1 . ; 0,1,2,3,.....2

xd2n+1 . 2n+1 .

D 2 2dn

n

Dx x

So, dark fringes will be obtained at :

3 5, , ,..........................

2 2 2

D D D

d d d

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(iii)Calculation of Fringe Width (β)Fringe width is the distance between consecutive dark and bright fringes. It isdenoted by 'β'. In case of constructive interference fringe width remains constantthroughout. It is also known as linear fringe width.

For Bright fringe:

1

1n n

n D n D Dx x

d d d

For dark fringe:

1

2 3 2 1' '

2 2n n

n D n D Dx x

D d d

Thus, the distance between any two consecutive maxima and minima:

2 1'

2 2n n

n Dn D Dx x

d d d

i.e No dependence on 'n'; which means equally spaced.

(iv)Shape of fringes:

2 1

2 1

Δp=S P-S P=n maxima

Δp=S P-S P=(2n+1) minima2

So, for a given 'n', the locus of points of maximum and minimum intensity is givenbyS2P-S1P=const.

This is the equation of hyperbola with foci S1 and S2.

1 2

1 2

S S AB circular

S S AB straight

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(iv) Intensity distribution:

1 2 1 2I=I I 2 I I cos φ=1

If φ=0, 2 , 4 ,..............cos φ=1 ; so intensity will be maximum.

If φ= , 3 , 5 ..............cos φ=-1 ; so intensity will be minimum.

If I1=I2=Io ; Imax=4Io

& Imin=0

1.6 Fresnel's BiprismA Fresnel Biprism is a thin double prism placed base to base and have very

small refracting angle ( 0.5º). This is equivalent to a single prism with one of itsangle nearly 179º and other two of 0.5º each. Monochromatic light through anarrow slit S falls on biprism , which divides it into two components.

Figure 1.3 Interference through Fresnel Biprism

Fresnel produced the interference fringes in the laboratory by deriving twocoherent sources S1 and S2 from a single monochromatic source S by using the

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deviation produced by a biprism. The experimental arrangement is shownschematically. The biprism consists of two acute angled prisms with their bases incontact. It is actually a single prism of an obtuse angle of about 179º . The acuteangle on both sides is about 30' (or 1/2). The prism is placed with its refractingedge parallel to the slits such that S is normal to the face BC of the prism. Lightfrom a narrow vertical slit S, illuminated with monochromatic light of wavelengthλ, is allowed to fall symmetrically on the biprism. When light falls from S on thelower half of the prism, it bends upwards and appears to come from S2. Similarlythe light from S, which falls on the upper half of the prism, is bent downwards andappears to come from S1. The virtual images S1 and S2, being the images of S, actas two coherent sources. The source S1 illuminates the portion EY of the screenwhile the source S2 illuminates the portion XF. In the common shaded region thecoherent beams overlap and therefore interfere. In the region XY the interferencefringes have equal width.

1.7 Newton's Ring ExperimentNewton's rings, in optics, a series of concentric light- and dark-coloured

bands observed between two pieces of glass when one is convex and rests on itsconvex side on another piece having a flat surface. Thus, a layer of air existsbetween them.

Figure 1.4 Newton's RingsBasically, it is a phenomenon in where an interference pattern is created by thereflection of light between two surfaces: a spherical surface and an adjacenttouching flat surface. The phenomenon is named after Isaac Newton, whoinvestigated the effect in his 1704 treatise Opticks. When viewed withmonochromatic light, Newton's rings appear as a series of concentric, alternating

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bright and dark rings centered at the point of contact between the two surfaces.However, when the same phenomena was viewed with white light, it forms aconcentric ring pattern of rainbow colors, because the different wavelengths oflight interfere at different thicknesses of the air layer between the surfaces.

Figure 1.5 Setup for Newton's ring experiment

If S be the source of monochromatic light. When light from the source (S) isallowed to fall on a convex lens, then it render’s a parallel beam of light .Thisparallel beam of light is allowed to fall on plane glass plate. That is placed at anangle to the direction of incident beam of light. Then the glass plate reflects theincident beam of light normally towards the air film enclosed between the Planoconvex lens and the glass plate. First of all this light is allowed to fall on a planesurface of Plano convex lens. Then a part of this light is reflected and a part oflight is transmitted, then this transmitted light is allowed to fall on a curve surfaceof Plano convex lens .Then a part of this transmitted light is reflected and comesout in the form of ray no. 1 and a part of light is transmitted, after that thistransmitted light is allowed to fall on a plane glass late then a part of light isreflected and comes out in the form of ray no. 2 and a part of light is transmitted

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and comes out in the form of ray no. 3. Thus at a particular constant thicknessinterference take place due to the reflected ray no. 1 and 2. Due to convexity of thePlano convex lens and at the particular constant thickness the radii or foci areconstant so that the interference pattern is take place in the form of concentric ring.Thus interference pattern is either dark or bright depend upon path differencebetween the two reflected rays. Thus the path difference between two reflected ray

will be:- 2 tcos(r+ )+2

.However, in the experiment rays are incident normally

thus for normal incidence. Angle of refraction is zero (r=0). For air film µ=1 andfor small wedge angle 'α' is very small. Thus, the value of (r+α) tends to zero, inturn cos (r+α)=1.Hence, the path difference between both of the reflected ray will be: (2t+λ/2). Butat the point of contact thickness of air film is zero so that when t=0 then pathdifference between the two reflected ray is which is condition of the minima so’that at the point of contact in case of reflected light interference pattern will bedark.

Figure 1.6 Interference in Newton's ring experiment

(a) Condition for Bright fringe

2t+ 2 . ; 0,1,2,3,.......2 2

2t (2 1)2

n n n

n

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(b) Condition for Dark fringe

2t+ (2 1). ; 0,1, 2,3,.......2 2

2t

n n n

n

NB: Here, it is seen that for t=0; n=0. That is the central fringe is dark.

(c)Thickness of the filmR= radius of curvature of lensrN= radius of Newton's ring for film thickness t.

From ∆ CPQ:2 2 2

2 2 2

2 2 2 2

2 2

2

2

(CQ) (CP) (PQ)

( )

2

2

2

2

N

N

N

N

N

R R t r

R R Rt t r

r Rt t

r Rt

rt

R

(Neglecting 't' as t<<R)

(d) Diameter of Bright fringe:

2

2

2

2

2 (2 1)2

2. (2 1)2 2

2 1.

2

2 1.

4 22(2 1).

2 (2 1) 2 1 2 .

2 1

N

N

N

N

N

N

t n

rn

Rn

r R

D nR

D n R

D R n K n K R const

D n

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(e)Diameter of dark fringe:

2

2

2

'2.

2'

N

N

t n

rn

R

r n R

2

2

'

4' 4

' 4 ' ' 4 .

'

N

N

N

N

Dn R

D n R

D R n K n K R const

D n

1.8 Determination of wavelength of lightWe know,

If the contact of lens and the glass plate is not perfect the order of nth ring cannotbe known precisely, due to uncertainty in the thickness of the film. To overcomethis; diameter of 'n' and (n+p)of dark ring is measured.

2' 1.4

ND slope

n R R

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2

2

2 2

2 2

' 4

' 4( )

' ' 4

' '

4

n

n p

n p n

n p n

D n R

D n p R

D D p R

D D

PR

1.9 Determination of refractive index of liquidLet us place liquid instead of air in between glass plate and Plano-convex lens:Now condition for dark ring:

2

2 ' 2

' 2

2

22

4

4

N

N N

N

t n

rn

Rn R n R

r D

n RD

Again, in case air film; we can write;

' 2 ' 2 4 ...................... 1n p nD D p R

and in case of liquid;

' 2 ' 2 4...................... 2n p n

p Rd d

From equation 1 &2;

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' 2 ' 2

' 2 ' 2

' 2 ' 2

' 2 ' 2

44

. ;

n p n

n p n

n p n

n p n

D D p Rp Rd d

D Di e

d d

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Assignment-Unit I

Self Study

1. Derive an expression of time period for a compound pendulum.

2. Give a brief account of sonometer with diagram and its use to verify the laws of

vibration.

3. Derive an expression for the path difference between two reflected light rays from

upper and lower surface of a wedge-shaped film.

4. Explain the working of Michelson interferometer with diagram.

5. What do you mean by a coupled oscillator? Derive an equation of motion for normal

modes of vibration.

6. What are normal mode frequency? Two identical simple pendulums each of mass m

and length l are coupled through a massless spring of force constant k. Calculate

their normal mode frequencies? Objective type questions carrying 2 marks

1. Define Simple Harmonic Motion (SHM) and write down the differential equation of motion for SHM.

2. What is resonance and sharpness of resonance? How does the sharpness of resonance depend on damping.

3. What do you mean by a wave? What are transverse wave and longitudinal wave, give

one example for each wave.

4. What is interference? Mention the conditions of sustained interference.

5. Show that y = 3sin2t + 4cos2t represents SHM. Find the time period and amplitude.

6. The differential equation of motion of a freely oscillating body is given as

2 Calculate the natural frequency of the body.

7. Find the position at which the potential energy is equal to kinetic energy in case of SHM.

8. If two pendulum each of mass 0.6 kg and length 1 meter are connected by spring of force constant 0.3 N/m. Calculate the normal modes of frequencies for coupled

system.

9. Find the speed of a longitudinal wave propagating in a medium of density 1.5 × 103

g/cm3 and bulk modulus 6.0 × 107 dynes/cm2.

10. Two harmonic waves given by ψ1 = 10sin(2πt − 3x) and ψ2 = 5sin(3πt − 4x) are

superposed to produce a single wave. Find the group velocity of the resultant wave, where t and x are in second and cm respectively.

11. 10 waves of equal amplitude superpose coherently to produce resultant intensity of

10 watt/m2. What would be their intensity if they superpose incoherently?

12. Find the ratio of maximum intensity and minimum intensity for two waves having

amplitude 2 units and 3 units respectively to be coherent.

13. In a Newton’s rings experiment the diameter of the 5th ring is reduced to three fourth of its value after introducing a liquid below the convex surface. Calculate the

refractive index of the liquid.

14. In a Newton’s rings experiment, the air film is replaced by a liquid film of refractive

index 4/3. In what proportion would the diameters of the rings change?

Subjective type questions carrying 5 marks

15. Write the expression for displacement, velocity, acceleration and find the positions of maxima and minima for displacement, velocity, acceleration.

16. What is meant by damped harmonic oscillations. Set up differential equation of a damped harmonic oscillator.

17. Starting from the differential equation for a damped harmonic oscillator discuss

underdamped, critically damped, overdamped oscillations. Represent these

oscillations on displacement time graph.

18. Explain the terms logarithmic decrement, relaxation time, quality factor with reference to under damped oscillator and co-relate them.

19. What is forced vibration? Set up a differential equation for a forced oscillator subjected to an external periodic force.

20. State and derive the differential form of wave equation.

21. Define phase velocity and group velocity and find a relation between them.

22. What is stationary wave? Find the positions of node and antinode for a stationary wave.

23. Distinguish between stationary wave and progressive wave.

24. Discuss coherent superposition of N-waves having same frequency and a constant successive phase difference between the successive waves.

25. Obtain an expression for fringe width in the interference pattern due to Young’s double slit.

26. With a neat labelled diagram show how Newton’s rings are formed. Derive an

expression for the radius of nth dark ring.

27. Explain how the wavelength of monochromatic light is determined using Newton’s rings experiment.

28. Describe how the refractive index of an unknown liquid can be obtained by using Newton’s ring experiment.

29. The equation of a particle executing SHM is, where x is in

meter and t is in seconds. Find the amplitude, time period, maximum speed and maximum acceleration. Also find the velocity and acceleration at t = 0.5s.

30. The quality factor of a tuning fork is 9.42 × 103 having frequency 300 Hz. Find the

time after which its energy reduces to 10 % of its initial value.

31. The amplitude of an under damped oscillator falls to th of its initial value after 10 number of oscillations. If time period is 2 s, calculate (a) damping coefficient, (b)

logarithmic decrement, (c) time during which energy falls to of its original value, (d) relaxation time, (e) quality factor.

32. A body of mass 10 gm is acted upon by a restoring force per unit displacement of 107

dyne cm−1, a frictional force per unit velocity of 4 × 103 (dyne/cm)s and a driving force of 105 cosωt dyn. Find the value of maximum amplitude.

33. A wave is represented by ) where x and y are given in

centimeter and t in second. Calculate wavelength, amplitude, frequency, velocity and

initial phase constant.

34. In a Young’s double slit experiment green light of wavelength 5000A0 was used. The

separation between two consecutive dark fringe was found to 5 mm when a screen was placed at distance of 1 m from the slits. Find the separation between the slits.

35. In Newton’s rings with reflected light are observed between a planoconvex lens of radius of curvature of 120 cm and a plane glass plate. The diameter of the 15th bright

ring is 0.58 cm. Calculate the diameter of the 25th bright ring and the wavelength of the light used.

36. Green light of wavelength 5100 A0 from a narrow slit is incident on a double slit. If

the overall separation of 10 fringes on a screen 200 cm away is 2 cm, find the slit

separation.

37. In young double slit experiment, while using a source of light of wavelength 5000 A0,

the fringe width obtained is 0.6 cm. If the distance between the screen and the slit is reduced to half, what should be the wavelength of the source to get fringes 0.003m

wide?

38. Newton’s rings are observed normally in transmitted light of wavelength 5.9×10−5

cm. The diameter of the 10th bright ring is 0.5 cm. Find the radius of curvature of the lens.

39. In a Newton’s arrangement a source of light having two wavelengths 6000 A0 and

4500 A0 is used. It is found that nth dark ring due to 6000 A0 coincides with (n+1)th

dark ring due to 4500 A0. Calculate the radii of nth dark rings 6000 A0 and 4500 A0, if radius of curvature of planoconvex lens is 100 cm.

40. In a Newton’s arrangement, the diameter of a bright ring is 0.5 cm. What would be

the diameter of the ring if the lens placed on the plane glass plate is replaced by

another one having double the radius of curvature?

Unit-1 Simple Harmonic motion

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