The motion which repeats after a regular interval of time is called periodic motion. In such a motion the particle always come to rest is called mean or equilibrium position. If the particle is displaced from mean position there exist certain kind of force which tries the particle back to its mean position, such a force is called restoring force. Restoring force is the function of displacement. Periodic motion can be expressed in terms of trigonometric function. So, it is called harmonic motion.
SIMPLE HARMONIC MOTION
Simple harmonic motion is the special kind of oscillatory motion in which the body moves again and again over the same path about the fixed point (generally called equilibrium position) in such a way that it is acted upon by a restoring force ( or torque) propotional to it’s displacement ( or angular displacement) and directed towards the mean position.
SHM is a special type of periodic motion in which particle oscillates in straight line in such a way that acceleration is directly proportional to the displacement from mean position & it is directed towards mean position.
SOME TERMS ASSOCIATED WITH S.H.M.
The maximum displacement of the particle from mean position is called amplitude of the oscillation.
As the particle displacement is, x=Asin(ωt+δ) and sin(ωt+δ) varies from -1 to +1
∴ Maximum displacement =±A
A is called amplitude of oscillation.
2. Time period:-
The time required to complete one oscillation is called time period. It is denoted by T.
We have, displacement of particle in SHM as, , x=Asin(ωt+δ)…(i)
The position of particle will be same after time T i.e. x=Asin[ω(T+t)+δ]….(ii)
∴ sin(ωt+δ) repeats its value every 2π or even integral of π.
∴ ω(T+t)+δ = ωt+δ+2π
This is the time period of S.H.M.
The number of oscillation in one second is called frequency. It is denoted by 'f' and given as,
4. Angular frequency (ω):-
5. Phase & phase constant:-
Phase is the status of the particle which executes S.H.M.
We have, x=Asin(ωt+δ) and v=Aωcos(ωt+δ)
(ωt+δ) is responsible for the status of particle therefore it is phase of the particle. It is denoted by ϕ.
The phase increases with time.
δ is called phase constant which depends upon the choice of instant time (t=0). If we chose the instant time as origin then, ωt+δ=0 ∴ δ=0
This means phase constant will be zero.
OSCILLATION OF SPRING MASS SYSTEM
Consider a spring of length 'l' which is supported by a fixed rigid support at one end and another end is free. Let 'm' be the mass attached to free end. Due to mass 'm' let 'l' be the elongation produced then, according to Hook's law, the force applied is directly proportional to the extension produced,
i.e. F1∝ l
-ve sign indicates that the force is restoring.
Now, further pull the mass through a distance 'x' then, the force will be,
∴ the net force applied so that the spring mass system sets into oscillation is,
According to Newton;s law,
from (iii) & (iv)
which is equation of S.H.M i.e. the spring mass system executes S.H.M.
This is time period of spring mass system.
ANGULAR HARMONIC MOTION
Angular harmonic motion is the periodic motion in which angular acceleration is directly proportional to angular displacement & it is always acting towards mean position.
If θ be the displacement & α be the angular acceleration, then,
α ∝ θ
or, α =-ω²θ
Limitation of simple pendulum
a. The point mass heavy bob is not possible.
b. The mass less inextensible spring is not possible.
c. In simple pendulum the centre of oscillation & centre of gravity lies at the same point which is practically not possible
d. Since the string has certain mass, it has certain moment of inertia which is not considered in it.
e. It is an ideal pendulum.
Compound pendulum is a rigid body of any shape capable of oscillating in horizontal axis in vertical plane not passing through centre of gravity.
Let 'mg' be the weight acting vertically downward through c.g. Let 'l' be the length of compound pendulum i.e. distance from centre of suspension to centre of gravity(G). Now, displace the pendulum through angle 'θ' & G' be the new position of c.g. , then the force at G'and its reaction at c.s constitutes couple.
Then, torque is given as,
Torque (τ)=Force × perpendicular distance from axis of rotation
or τ = mg × lsinθ
If the displacement is small then,
sinθ ≃ θ
which provides restoring force i.e. which tries the pendulum back to its mean position.
∴ Restoring torque (τ)= -mglθ………(i)
If I be the moment of inertia and α be the angular acceleration, then,
Torque (τ) = Iα ….(ii)
From (i) & (ii)
Iα = -mglθ
i.e. angular acceleration is directly proportional to angular displacement or it executes angular harmonic motion.
comparing with α =-ω²θ , we get,
PARALLEL AXIS THEOREM
If I0 be the moment of inertia through then, I0=mK²
where, K is radius of gyration. It is perpendicular distance from the centre of suspension to centre of gravity.
Then moment of inertia through C.S., I=I0+mass × distance²
If L is the length of simple pendulum then, the time period of compound pendulum & simple pendulum will be same. This length is called length of equivalent simple pendulum.
Since, k² is always +ve is always greater than length of compound pendulum l.
This implies that the centre of oscillation always lies beyond the centre of gravity.
*Interchangeability of centre of suspension & centre of oscillation.
We have, time period of compound pendulum,
Now, inverting the centre of suspension & centre of oscillation we get,
Then, time period of compound pendulum will be,
Time period is same when it is inverted i.e. the centre of oscillation & centre of suspension can be interchanged.
* Maximum & minimum time period of compound pendulum.
We have, time period of compound pendulum
∴ Tmax=∝ at l→0
Now squaring eqn. (i) we get,
Differentiating eqn. (ii) w.r.t. l,
∴ Tmin=0 at K=±l
Time period of compound pendulum is minimum when radius of gyration 'K' is equal to the length of compound pendulum.
* Energy conservation is S.H.M.
E= KE + PE (constant)
The equation of motion for particle executing S.H.M. is written as F=-kx….(i)
Where, F=restoring force, K=positive constant & x= displacement of particle.
The amount of work done to displace a particle through distance dx will be=> dW=F× dx….(ii)
from (i) &(ii) , dW=-kx × dx
The total work done to displace x distance is,
∫ dW = ∫-Kxdx
This amount of work done is changed into potential energy.
If 'm' be the mass & 'v' be the velocity of particle executing S.H.M. then,
but, x = A.sin(ωt+δ)
, which is independent of time i.e. total energy remains conserved.
Q1. 2 Kg. mass hang from a spring. A 300g body hang below the mass stretches the spring 2 cm further. If the 300g body is removed and the mass set onto oscillation. Find the period of motion.
or, F=Kx (for magnitude only)
Q2. A harmonic oscillator has a period of 0.5 sec. & amplitude of 10 cm. Find the speed of object when it passes through the equilibrium position.
Vmax = A.ω
= A× 2πf
Q3. A body of mass 0.3 Kg executes S.H.M. with a period of 25 sec. and amplitude of 4 cm. Calculate the max. velocity, acceleration & K.E.
m= 0.3 Kg, T=25 sec., A=4cm .
Q4. A small body of mass 0.1 Kg is undergoing S.H.M. of amplitude 1.0 m and period 0.2 sec (a) what is the maximum value of force acting on it?, (b) If the oscillations are produced by a spring, what is the constant of spring?
Q5. Two springs having force constants k1 & k2 respectively are attached to a mass and two fixed supports as shown. If the surfaces are frictionless, find the frequency of oscillation.
Total restoring force,
F = -k1x-k2x
F = -k1x-k2x
ma = -(k1+k2)x
Q6. Show that if a uniform stick of length 'l' is mounted so as to rotate about a horizontal axis perpendicular to the stick and at a distance'd' from the centre of mass. The period has a minimum value when d=0.289 l.
General, M.I. = mK², where K= radius of gyration.
M.I. of different surfaces about c.g.
Thin ring/hollow cylinder
∴ d = 0.289l proved.
Q7. A body executing S.H.M of amplitude 10cm & frequency 10 vibrations/sec. Calculate its acceleration after 0.015 sec.
Q8. At a time when the displacement is half the amplitude what fraction of the total energy is kinetic & what fraction is potential energy. in S.H.M ?At what displacement is the energy half K.E. & half P.E.
Q9. Uniform circular disc of radius R oscillates in a vertical pane about horizontal axis. Find the distance of the axis of rotation from the centre of which the period is minimum. What is the value of the period?
Q10. A particle executing S.H.M. of amplitude 'A' along the X-axis. At t=0, the position of the particle is and it moves along the positive X-direction. Find the phase constant δ. If the equation is written as x=sin(ωt+δ)
The disturbance which travels onward through the medium due to the periodic motion of particle about their mean position is called wave motion. Thus wave transfers energy from one place to another place without any bulk transformation of intervening particles of the medium, so wave is the mode of transfer of energy.
According to the medium required or not required wave is categorized into two types.
The wave which require material medium to travel or propagate onward is called mechanical wave for e.g. sound wave.
The wave which do not require material medium to travel or propagate onward is called electromagnetic wave for e.g. radio wave, light wave etc.
According to the vibrations of particles wave motion is of two types
Transverse wave motion:-
The wave motion in which particles of medium oscillates in perpendicularly to the propagation of wave is called transverse wave motion. Therefore it is in the form of crest and trough.
Longitudinal wave motion:
The wave motion in which particles of the medium vibrates parallel to the direction of propagation of wave is called longitudinal wave motion. Therefore it is in the form of compression & rarefraction. It is also called as Pressure wave.
The wave which travels onward through the medium in a given direction without alternation or with constant amplitude is called progressive wave. It may be transverse or longitudinal wave. In such wave all particles vibrate with same amplitude, the only difference is that they have different instant of time.
Consider a particle at 'O' executes periodic motion then, if 'y' be the displacement of particle then equation for displacement,
Let 'P' be the point at a distance 'x' from origin,
In one complete oscillation the phase is 2π or wave covers the distance λ.
when, path difference is λ then phase difference is 2π
when path difference is 1 then phase difference is 2π/λ
when path difference is x then phase difference is
∴ path difference between O & P will be
If the wave travels in –ve X direction then,
Wave velocity & particle velocity
The rate of change of wave displacement w.r.t. time is called wave velocity (phase velocity). It is denoted by 'u' and given by,
We have wave equation for progressive wave travelling in +ve direction,
Now, differentiating w.r.t. time we get,
For given wave ωt-kx=constant
Now, differentiating w.r.t. time we get,
∴ wave velocity = wavelength × frequency
The rate of change of displacement of the particle with time is called particle velocity. It is denoted by 'v' & given by,
but, y=Asin(ωt-k) ….(iv)
Differentiating w.r.t. x
From eqn. (v) & (vi) we get,
Particle velocity = wave velocity × (-slope displacement curve)
Intensity of wave:-
The flow of energy per unit area per unit time is called intensity of wave.
where, m=mass of particle, A=amplitude of oscillation & ω=angular frequency
Consider a table of cross section area's' and length 'l' when the wave energy is transmitted. Let 'n' be the number of particles per unit volume then the total number of particle in volume 'sl' will be 'nsl'.
Hence, Intensity ∝ ρ
Hence, Intensity ∝ u
Hence, Intensity ∝ f²
Hence, Intensity ∝ A²
VELOCITY OF TRANSVERSE WAVE ALONG STRETCHED STRING.
The velocity of transverse wave along stretched string can be expressed in terms of tension along the string & mass per unit length of the string.
When a jerk is given to the stretched string fixed at one end transverse wave can be produced.
Consider an element length Δl of string which forms an arc 'PQ' of radius 'R'. The tension produced on the string can be resolved into two perpendicular components. Tcosθ along the horizontal & Tsinθ along radial direction. The horizontal components cancel each other and the radial components provide the tension.
The resultant tension along string = Tsinθ +Tsinθ =2 Tsinθ…(i)
Energy Transmission along
If μ be the mass per unit length (m/Δl) of string then, m=μΔ L
STANDING OR STATIONARY WAVE
When two travelling waves of same amplitude & same wavelength travel in opposite direction with same velocity superimpose with each other then the resultant wave formed is called standing wave or stationary wave.
There are certain position like 'N' where the particle vibration is minimum i.e. particle doesn't vibrate is called node. There are certain positions like 'AN' where the particle vibration is maximum is called Anti-Node.
The standing wave confines in certain region & cannot propagate forward.
By the principle of superposition the resultant displacement of the wave is given as, y=
This is the resultant wave equation of standing wave with amplitude
For maximum amplitude, coskx=±1
or, coskx=cosnπ [n=0,1,2,3,….]
or, kx = nπ
For minimum amplitude, coskx=0
RESONANCE OF STANDING WAVE
In standing waves either end points are nodes or anti-nodes, particle vibrates with maximum amplitude is called resonance.
The minimum frequency at which resonance occurs is called fundamental frequency. It is denoted by 'f0'.
For fundamental frequency, if the ends are closed i.e. if there are two nodes there is only one anti-node.
when there are two segments in the wave,
Q1. A wave of frequency 500 cycles per sec has a phase velocity of 350 m/s. How apart are two points 60° out of phase ? (Ans. 0.12 m)
Q2. A wave of frequency 500 cycles per sec has a phase velocity of 350 m/s. What is the phase difference between two displacements of a certain point at time 0.001 sec apart? (Ans. π ͑)
Q3. The equation of a transverse wave travelling in a rope is given by y=-10sin(0.01x-2t). Find the amplitude, frequency, velocity and wave length.
Q4. A string of mass 2g for unit length carries progressive wave of amplitude 15 cm, frequency 60 sec-1 & speed 20m/s. Calculate a. Energy per meter length of wire, b. the rate of energy propagation in the wave.
(Ans. 6.39 J/s, 0.031 J)
Q5. Two waves are simultaneously passing through a string. The equation of waves are y1=A1sink(x-vt), y2= A2sink(x-vt+x0), where the wave no k=6.28 cm-1 & x0=1.50 cm, the amplitudes are A1=50mm & A2=40mm.
WAVE OPTICS (PHYSICAL OPTICS)
GEOMETRICAL PATH & OPTICAL PATH
The distance by the light through a certain distance for a certain time in a certain medium is called geometrical path.
The distance travelled by the light in vacuum through a certain distance for the same time in which it travel in medium is called optical path.
Suppose light travels through the medium of refractive index 'μ' with velocity 'v' for time't' then the geometrical path will be,
If 'c' be the velocity of light in vacuum for time 't' then distance travelled by light,
∴ Optical path = refractive index of medium times the geometrical path.
Interference of light waves
The source of light distribute its energy uniformly in all directions if there are two sources which distributes energy of same wavelength & frequency the energy distribution will not be uniform. The non uniform distribution of energy due to the superposition of two light waves of same amplitude and frequency is called interference of light waves.
At some point energy will be maximum called constructive interference & at some places energy will be minimum called destructive interference.
The sources are said to be coherent if they produce the light wave of same amplitude & frequency.
For the sustain interference following conditions are to be satisfied:-
1) Two sources should be coherent
2) Two sources emit light wave continuously.
3) They should be narrow sources
4) Sources should be close to each other
5) The distance between the sources and screen should be far
6) Sources should be monochromatic
Condition for interference maxima & minima
Consider two point sources S1 & S2 which produces the interference fringes. Let 'P' be the point where we have to find the interference fringe, which is either maxima or minima.
Let y1 = asinωt be the wave produced by S1 & y2= asin(ωt+δ) be the wave produced by S2.
The resultant displacement of waves by superposition principle can be written as,
y = y1+y2
= asinωt+ asin(ωt+δ)
= a[sinωt+ sinωt.cosδ+cosωt.sinδ]
Let a(1+cosδ) = Acosθ …..(ii)
asinδ = Asinθ …….(iii)
then, y = A[sinωt.cosθ+cosωt.sinθ]
, which is the resultant wave equation of the resultant wave.
Squaring & adding (ii) & (iv)
a²sin²δ+a²+a²cosδ+ a²cos² δ=A²
or, 2a²+2 a²cosδ=A²
since, intensity of the wave is directly proportional to the square of amplitude, we can write,
The intensity will be maximum if =1
for 2π phase difference he path difference will be λ
The point 'P' will correspond to maximum or bright fringe if the path diff = nλ
for 2π phase difference path difference = λ
Thin films is an optical medium where thickness is order of wavelength of light in visible region (400 nm t 760 nm). It may be the soap bubble, glass or air enclosed between two transparent glass plate whose thickness is 0.5μ m to 10μm.
*Interference due to thin films
*Interference due to reflection on thin film
Let AB be the incident light incident in upper surface, XY of the thin film of thickness't' and refractive index 'μ '. At point B, most of the part transmitted along BC and a point of it is reflected along BF.
Also, at point 'C', most part is transmitted along CN and a point of it is reflected along CD. Similarly at point 'D' most part transmitted along DC and a part of it is reflected along 'DE'.
Only the light wave reflected from point 'B' and 'C' are of appreciable strength.
Now, the path difference between BF and DG = BC +CD-BN
Optical path difference = μ(BC+CD)-BN…(i)
Now, from fig. in Δ BCM,
from, (iv) & (v)
BN = 2tanr.sini….(vi)
At 'B' reflection takes place from the surface of denser medium, therefore the additional path difference λ/2 should be added [At reflection the wave losses half of the wave]
for minima or dark fringe,
path difference = (2n+1) λ/2 [n=0,1,2,….]
Let 'S' be the source of monochromatic light. Light rays form 'S' are made parallel by lens L. These parallel rays are incident on glass plate 'G', kept 45° with horizontal. A part of these rays fall normally on plane convex lens 'P' kept on glass plate 'AB'. The Plano-convex lens and glass plate AB create air film of variable thickness.
The light ray reflected form upper part of thin air film and lower part of thin film interfere to produce interference fringes as from circle(ring). Newton's rings are the fringes of equal thickness.
Let 'R' be the radius of curvature of Plano-convex lens, 't' be the thickness of air film for which radius of nth ring will be 'rn'.
For the interfere due to reflection on thin film, the path diff=2μtr+λ /2
For air film, μ = 1
for normal incidence, r=0
∴ path diff=2t+λ /2
for bright fringe path difference = nλ
∴ 2t+λ /2 = nλ [n=0,1,2,….]
2t=nλ -λ /2
2t = λ/2 (2n-1)….(i)
for dark fringe, path diff = ((2n+1) λ /2
∴ 2t+λ/2 = (2n+1) λ /2
2t = nλ
from fig, in Δ PNM
PM² =PN² +NM²
R²=rn² + R²-2Rt+t²
but, the radius of curvature is very greater then thickness of air film.
→2t= rn²/R=Dn² /R
Where, Dn= diameter of an fringe.
for bright fringe, eqn. (i) & (iii) must be equal
If dm be the diameter of mth ring then
Now, subtracting (v) fro (iv)
If we know the diameter of nth and nth ring, we can determine the wavelength of monochromatic light.
Diffraction:- It is the phenomenon of bending of light round the sharp corner and spreading the light into the geometrical shadow region. As a result of diffraction, maxima and minima of light intensities are found which has unequal intensities. Diffraction is the result of superposing of waves, from infinite of coherent sources on the same wave front, after the obstacle has distorted the wave front.
There are two types of diffraction,
a. Fresnel diffraction :-
b. Fraunhoffer diffraction :-
TO observe the diffraction phenomenon the size of the obstacle must be of the order of the wavelength of the waves.
Fresnel diffraction :- It involves spherical wave fronts. If either source or screen or both are at a finite distance from the diffracting device , the diffraction is called Fresnel diffraction and the pattern is shadow of the diffracting device modified by diffraction effects. Diffraction at a straight edge, narrow wire or small opaque disc is familiar examples of this type. It explain diffraction in terms of “half period zones” , the area of which is independent of the order of zone.
Fraunhoffer diffraction :-
It is the diffraction when both source and screen are effectively at infinite distance from the diffracting device and pattern is the image of source modified by diffraction effects. Diffraction at single slit, double slit and diffraction grating can be cited as examples of this type. It follows that fraunhoffer diffraction is an important special case of Fresnel diffraction.