Mechanical Oscillation
PERIODIC MOTION:
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.
1. Amplitude:
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
or, ωT=2π
This is the time
period of S.H.M.
3.
Frequency:
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.
Phase (ϕ)=ωt+δ
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. F_{1}∝
l
or, F_{1}=kl…(i)
ve sign indicates
that the force is restoring.
Now, further pull
the mass through a distance 'x' then, the force will be,
F_{2}=k(l+x)…..(ii)
∴ the net force
applied so that the spring mass system sets into oscillation is,
F=F_{2}F_{1}
or, F=k(l+x)(kl)
or, F=kx….(iii)
According to
Newton;s law,
F=ma
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
COMPOUND PENDULUM
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
∴
τ=mglθ
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 I_{0} be the moment
of inertia through then, I_{0}=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=I_{0}+mass × distance²
or, I=mK²+ml²
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
∴
T_{max}=∝ at l→0
Now squaring eqn.
(i) we get,
Differentiating
eqn. (ii) w.r.t. l,
∴
K=±l
∴
T_{min}=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)
y=A.ω.cos(ωt+δ)
, which is
independent of time i.e. total energy remains conserved.
NUMERICAL
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.
F=Kx
or, F=Kx (for
magnitude only)
or, mg=Kx
=150 N/m
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.
V_{max }= 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.
Soln.
m= 0.3 Kg, T=25
sec., A=4cm .
V_{max}=A.ω
a_{max}=Aω²
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?
F_{max}=m._{Qmax}
Fmax=m.Aω²
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.
Soln.
Total restoring force,
F = k_{1}xk_{2}x
F = k_{1}xk_{2}x
ma = (k_{1}+k_{2})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.
Soln.
General, M.I. = mK², where K= radius of gyration.


M.I. of different surfaces about c.g.


Surface

M.I.

Rectangle


Thin ring/hollow cylinder

mR²

Hollow sphere


Solid sphere

∴
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.
a=Aω²sin(ωt)
=
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?
Soln.
Q10. A particle
executing S.H.M. of amplitude 'A' along the Xaxis. At t=0, the position of the
particle is
and it moves
along the positive Xdirection. Find the phase constant δ. If the equation is
written as x=sin(ωt+δ)
WAVE MOTION
Introduction
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.
Mechanical wave
The wave which require material medium to travel or propagate onward is
called mechanical wave for e.g. sound wave.
Electromagnetic 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.
Progressive 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,
y=Asinωt…..(i)
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,
y=Asin (ωt+kx)
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,
y=Asin(ωtkx)……(ii)
Now,
differentiating w.r.t. time we get,
For given wave
ωtkx=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(ωtk)
….(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
stretched string
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 AntiNode.
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 antinodes, particle vibrates with maximum amplitude is
called resonance.
The minimum frequency at which
resonance occurs is called fundamental frequency. It is denoted by 'f_{0}'.
For fundamental frequency, if the ends are closed i.e. if
there are two nodes there is only one antinode.
when there are two segments in
the wave,
NUMERICALS
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(0.01x2t). 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 y_{1}=A_{1}sink(xvt), y_{2}=
A_{2}sink(xvt+x_{0}), where the wave no
k=6.28 cm^{1} & x_{0}=1.50 cm, the amplitudes are A_{1}=50mm
& A_{2}=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,
x=ut
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.
Coherent source
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
y_{1} = asinωt be the wave produced by S1 & y_{2}= asin(ωt+δ)
be the wave produced by S2.
The resultant displacement of waves
by superposition principle can be written as,
y = y_{1}+y_{2}
= asinωt+
asin(ωt+δ)
= a[sinωt+
sinωt.cosδ+cosωt.sinδ]
= A[sinωt(1+cosδ)+cosωt.sinδ]….(i)
Let a(1+cosδ) = Acosθ
…..(ii)
asinδ = Asinθ
…….(iii)
then,
y = A[sinωt.cosθ+cosωt.sinθ]
= Asin(ωt+θ)….(iv)
,
which is the resultant wave equation of the resultant wave.
Squaring
& adding (ii) & (iv)
a²sin²δ+a²(1+cosδ)²=A²(sin²θ+cos²θ)
a²sin²δ+a²+a²cosδ+ a²cos² δ=A²
or, 2a²+2 a²cosδ=A²
or, 2a²(1+cosδ)=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:
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 +CDBN
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,….]
Newton's Ring:
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 Planoconvex 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 Planoconvex lens, 't' be the thickness of air film for
which radius of n^{th} ring will be 'r_{n}'.
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
(2n1)….(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=r_{n}²
+(Rt)²
R²=r_{n}² +
R²2Rt+t²
r_{n}²=2Rtt²
but, the radius of curvature is
very greater then thickness of air film.
→2RT≫t²
→r_{n}²
=2RT
→2t= r_{n}²/R=D_{n}²
/R
Where, D_{n}=
diameter of an fringe.
for bright fringe,
eqn. (i) & (iii) must be equal
If d_{m} be
the diameter of m^{th}_{ }ring then
Now, subtracting
(v) fro (iv)
If we know the diameter of n^{th}_{
}and n^{th }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.
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