Download Intro SHO Damped

Welcome to
Subject No. – PH11001
Subject Name – Physics 1
Credit - 4 (3-1-0-4) (Lecture-Tutorial-Practical-Total)
Dr. Shivakiran Bhaktha B.N.
Department of Physics
May 2, 2017
Introduction
1
You can contact me at:
o My office : First Floor of Physics Dept (C-220)
o My contact Phone Number : 283802
o My E-mail: kiranbhaktha@phy.iitkgp.ernet.in
o For any discussion I am available on Thursdays 5-6 PM
May 2, 2017
Introduction
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Course Breakup
o 3 Lecture Classes per week
Class Room: F 142 (Raman Auditorium)
(Sections 6 and 7)
Monday: 10:30 AM;
Tuesday: 08:30 AM;
Thursday:11:30 AM.
o 1 Tutorial Class per week (Check your groups)
Tutorials start from Tuesday(7th Jan, 2014)
More Details Displayed on 1st Year Notice Board
in Department of Physics
May 2, 2017
Introduction
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Marks Breakup
o Mid Semester Exam : 30%
o End Semester Exam : 50%
o Tutorial : 20%
Mid Sem. Exams: 17th Feb 2014 to 25th Feb 2014
End Sem. Exams: 21st Apr 2014 to 29th Apr 2014
May 2, 2017
Introduction
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Course Contents
1.
Overview of vibrations (mechanical, electrical, optical).
2.
Free, damped, forced oscillations.
3.
Coupled oscillations.
4.
Wave motion.
5.
Electromagnetic waves.
6.
Radiation.
7.
Optical phenomena (interference, diffraction).
8.
Wave mechanics – failure of classical oscillators , quantum
oscillators.
May 2, 2017
Introduction
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References - Books
1.
“Lecture Notes & Problems bank for Physics”
by R.S. Saraswat and G.P. Sastry
(Available at THACKERS Book shop at Tech. Market)
2.
“The Physics of Vibrations and Waves” by H.J. Pain ( Wiley)
3.
“Feynman Lectures on Physics: Vol- I”
4.
“Optics” by Eugene Hecht
5.
PPT slides: Will be made available at regular intervals.
http://cts.iitkgp.ernet.in/Phy_1st/
Please note: For completeness, you MUST consult books and take
Slides as reference.
Note: If any other source is used then it would be told in the class
May 2, 2017
Introduction
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Acknowledgements
©
©
©
©
©
©
©
©
©
©
May 2, 2017
A. K. Das
A. Dhar
P. Khastgir
S. Bharadwaj
A. Roy
P. Roy Chaudhuri
S. Kar
S. Majumder
A. Chandra
S.K. Varshney
SHO
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No use of mobile phones in
the class
May 2, 2017
Introduction
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Lets begin...
May 2, 2017
Introduction
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Chapter 1
Oscillations
May 2, 2017
Introduction
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Importance of Chapter 1: Oscillations
• Oscillations in nature are ubiquitous (found everywhere)
• Some oscillations are visible; some are subtle but it is difficult to find
something which never exhibits oscillations.
• Examples:
– A mosquito’s wings, for example, vibrate hundreds of times per
second and produce an audible note.
– The whole earth, after being jolted by an earthquake, may continue
to vibrate at the rate of about one oscillation per hour.
– The human body itself is a treasure-house of vibratory phenomena.
“After all, our hearts beat, our lungs oscillate, we can hear and
speak because our eardrums and larynx vibrate. The light waves
which permit us to see entail vibration. We move by oscillating our
legs, even the atoms of which we are constituted vibrate…”
May 2, 2017
Introduction
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As a result…
¤
They have an enormous impact on understanding how things work.
¤
In astrophysics, thermal physics, quantum mechanics, optics,
condensed-matter physics, mechanics, atmospheric and planetary
physics, etc.
“So it’s basic literacy in physics”
May 2, 2017
Introduction
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Examples
Obvious Oscillations
»
»
»
»
water waves
pendulums
earthquakes
car springs, shock absorbers
Less-Obvious Oscillations
»
»
»
»
»
»
May 2, 2017
musical instruments - guitar, piano, flute
suspension bridges
lasers
quartz-crystal electronic watches
radio antenna
fiber optics
Introduction
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More examples
Subtle Oscillations
» heat in a solid
» structure of an atom
» superconductivity
» heat radiation
May 2, 2017
Introduction
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Similarly…..
“Waves are everywhere”
• Mechanical waves (sound, water, phonons…)
• Electromagnetic waves (radiation, visible light…)
• Matter waves (atoms)
• Gravitational waves (neutron stars, black holes..)
May 2, 2017
Introduction
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Before starting...
Lets learn some important notations
May 2, 2017
Introduction
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Notations
May 2, 2017
Introduction
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Complex Numbers
 zˆ  x  iy 
• We will extensively use complex numbers throughout
this course.
• Their use is NOT mandatory.
• But the use gives tremendouse convenience in
classical wave physics.
May 2, 2017
Introduction
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Complex Numbers
x  1
2
x ?
2
Definition :
x  1  i
2
2
i  1
And also satisfied by -i
May 2, 2017
Introduction
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Complex Numbers
ẑ  x  iy
x & y : real numbers
i : unit imaginary number
iy : pure imaginary number
May 2, 2017
Introduction
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Imaginary Exponent
i
e  x  iy
eiϕ = cos ϕ + i sin ϕ
The combination of exponential series with the complex number
notation i is very convenient in physics.
Mathematically it is convenient to express sine or cosine oscillatory
behaviour in the form of eix.
May 2, 2017
Introduction
21
Argand/Phasor Diagram
Representation of a complex number in terms of real and imaginary
components
Im
Complex Plane
zˆ  r
^z
r sin 

r cos 
b
= tan-1(b/a)
Re
a
z=a+ib
z = r e i
May 2, 2017
: Cartesian representation
: Polar representation
Introduction
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In general
zˆ  r (cos   i sin  )
Complex conjugate
zˆ*  r (cos   i sin  )
May 2, 2017
Introduction
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i
e  x  iy
May 2, 2017
Introduction
An Argand diagram is a plot of
complex numbers as points
z=x+iy in the complex
plane using the x-axis as the real
axis and y-axis as the imaginary
axis. In the plot the circle
represents the complex modulus
|z| of z and the angle represents
its complex argument.
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Phasor
The complex argument is also called the phase.
Phasor =
Rotating Arrow + Associated Phase Angle
May 2, 2017
Introduction
25
SIMPLE
HARMONIC OSCILLATION
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OSCILLATION
Free
Forced
OSCILLATION
OSCILLATION
Under NO damping
(Undamped Oscilation)
Under damping
(Damped Oscillation)
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Free Oscillation SHO
A simple harmonic oscillator is an oscillating system which
satisfies the following properties.
1. Motion is about an equilibrium position at which point
no net force acts on the system.
2. The restoring force is proportional to and oppositely
directed to the displacement.
3. Motion is periodic.
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Elements of an Oscillator
o
need inertia, or its equivalent
mass, for linear motion
moment of inertia, for rotational motion
inductance, e.g., for electrical circuit
o
need a displacement, or its equivalent
amplitude (position, voltage, pressure, etc.)
o
need a negative feedback to counter
inertia
displacement-dependent restoring force:
spring, gravity, etc.
electrical potential restoring charges
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Model System
Fs  k x
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Hooke’s Law:
Equation of SHM
Angular frequency
Time period
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Example
l
Note: Small angle approximation is valid till ~ 0.4 radians (= 23)
32
Example
Torsional Oscillation
Where,
I = Moment of Inertia
θ = Angular displacement
 = Restoring couple
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Example
34
Free Oscillation SHO
Let‘s find the general solution...
The equation of motion is given by:
2
d x
k
2
2
 o x  0 where o 
2
dt
m
This is a second order linear homogeneous equation with
constant coefficients.
The general solution is given by:
iot
iot
x c e
c e
1
2
The constansts c1 and c2 can be determined by the initial conditions.
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i t
i t
x  c e o c e o
1
2
Special cases
• The mass is pulled to one side and released from rest at t=0
x  a , and dx  0 at t  0
dt
0
x a cos  t
0
0
• The mass is hit and is given a speed v0 at its equilibrium position
at t=0
v
dx
x  0, and
v
at t  0
x  sin  t
0
dt

0
0
0
• The mass is given a speed v at a displacement a at t=0
dx
x  a, and
 v at t  0
dt
x a cos t   
0
0
 v
where a  a  

2
0
0



2
 v 

and   arctan 
  a
0
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Finding Solution
Equation of
SHM
i t
i t
x  c e o c e o
1
2
A=Amplitude,
=Phase
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Complex Representation
x̂(t )
© SB
38
Meaning/Significance
The real part of the complex number
x̂(t )
Represents the oscillating quantity
© SB
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Complex Velocity
xˆ (t )
ˆv(t )
x̂(t )
Note: i acts as an operator
xˆ (t )
© SB
40
Velocity
A=2 units
41
© SB
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What does moving along a circular path have to do with moving back
& forth in a straight line (oscillation about equilibrium) ??
SHM Example
x=ASin0t
t
x=Asin(0t+p/3)
t
x=Asin(0t+p/2)
t
© Hecht
43
Simple harmonic motion along x <=> x component of uniform circular motion
since q =  t
x = A cos q = A cos (t),
x
x
1
1
2
A
3
A
8
q
2
3
y
7
4
6
5
0
-A
p
2
q
p
4
6
5
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Energy of SHO
Harmonic Oscillator Potential
The potential energy is found by summing all the small elements of
x
work: (kx.dx).
1
 kx.dx  2 kx
2
0
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Energy of SHO
Because,
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Potential
Kinetic
(ot   )
• Energy of SHO = Sum of potential and kinetic energy remains a constant.
• Assumption: Ideal case, total energy remains constant.
• All P.E. becomes K.E. and vice versa.
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48
Time Average Energy of SHO
=
HOW???
49
REFRESH
Time Average
50
Average of Oscillations
Q
1.0
0.5
0.0
0
-0.5
10
20
30
40
50
(0t+)
-1.0
51
Time Average
1 T /2
2
 Q  limT   [Q(t )] dt
T T / 2
2
1
sin (0t   ) 
2
2
1
(1  cos 2(0t   ))
2
Note: Here ‘T’ is the total time of observation.
52
Time Average Energy of SHO
=
Root Mean Square (RMS)
x
2
 xˆ.xˆ / 2
*
54
You are aware…
In case of a sinusoidal wave, the RMS value is easy to calculate. If we define Ip to be
the amplitude, then:
where t is time and ω is the angular frequency (ω = 2π/T, where T is the period of
the wave). Since Ip is a positive constant:
Using a trigonometric identity:
but since the interval is a whole number of complete cycles (per definition of
RMS), the sin terms will cancel out, leaving:
= 0.707 Ip
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SUMMARY:
Undamped Free OSCILLATION
x(t)=Real part of z(t); z(t) = x+iy
A=Complex amplitude
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Superposition of Two SHMs
57
Superposition of SHMs in 1D
Case I- oscillation frequencies are the same
0
Additions of two SHMs becomes convenient with imaginary exponents
With,
[
+
]
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Superposition of SHMs in 1D
Case II- oscillation frequencies are different
For simplicity we write the solutions as,
x1  a sin 1t 
and
x2  a sin 2t 
Assuming that 2>1
Resulting displacement:
x  x1  x2
 a[sin( 1t )  sin( 2t )]
 1  2 t   2  1 t 
 2a sin 
 cos

2
2

 

59
 1  2 t   2  1 t 
x  2a sin 
 cos

2
2

 

 1   2 t 
sin 

2


 2  1 t 
cos

2


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Superposition of two SHM in perpendicular directions
Case I:
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Eliminate ‘t’
Square and sum the above two equations to obtain:
General equation for an ellipse
In most general case the axes of the ellipse are inclined to the x and y axes.
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The equation simplifies when the phase difference:
(2  1 ) 
(2  1 ) 
p
2
2
2
x
y
 2 1
2
a
b
Circle
Straight line
Ellipse otherwise
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Case II:
Lissajous figures
65
Lissajous figures
SHO-2
66
Damped Free Oscillations
67
Damping
(free oscillation case)
In many real systems, non-conservative
forces are present.
• The system is no longer ideal.
• Friction/drag force are common nonconservative forces.
• In this case, the mechanical energy of
the system diminishes in time, the
motion is said to be damped.
* A conservative force is a force with the property that the work done in moving a
particle between two points is independent of the path taken.
68
Damping
(free oscillation case)
Damping occurs due to
coupling
of
energy
of
macroscopic oscillator to its
surroundings, even if weakly
Dissipation of energy takes
place.
We assume: viscous damping
force or drag force that acts in
opposite to the velocity
69
Examples of Damped Harmonic Oscillations
a)
The mass experiences a frictional force as it moves
through the air.
b)
When the mass oscillates horizontally attached to a
string, then there exists frictional forces between mass
and surface.
c)
There are resistive force acting on the charge in LC
circuit, due to wires and internal resistance of the
devices.
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Damped Free OSCILLATION
Resistive force is proportional to velocity F
F  mx   rv  kx
mx  rv  kx  0
r
k
x 
x 
x0
m
m
2
x  2  x  o x  0
drag
 rv
r
k
2
; 0 
Where,  
2m
m
Or sometimes given in the form...
..
2x  0
x x o
k
Where,  =r/m and  
m
2
0
71
Solution
Linear differential equation of order n=2
homogeneous
or
d 2 x(t )
dx(t )
a2
 a1
 a0 x(t )  0
dt
dt
2
d
inhomogeneous a2 x(t )  a1 dx(t )  a0 x(t )  f (t )
dt
dt
General solution = Complimentary + Particular solution
For Complementary solution :
1.
2.
3.
4.
Take trial solution : x=emt, m is constant
m1, m2,……….will be the roots. If all roots are real and distinct, then
solution x=c1em1t+c2em2t+…………….
If some roots are repeated, say m1 repeated k times, then solution will be
(c1 + c2t+ …..cktk-1)em1t
If some roots are complex, (if a+ib then a-ib are roots) solution will be
eat(c1 cos(bt) +c2 sin(bt)) +………
For Particular solution : Trial solution to be assumed depending
on the form of ƒ(t)
Hyperbolic functions
Just as the points (cos t, sin t) form a circle with a unit radius, the points
(cosh t, sinh t) form the right half of the equilateral hyperbola.
73
Ref: Wikipedia
74
75
Solution
x  2  x  o2 x  0
• The equation is a second order linear homogeneous equation
with constant coefficients.
• Solution can be found which has the form: x = Cept where C
has the dimensions of x, and p has the dimensions of T-1.
x  pCe pt ; x  p 2Ce pt
mx  rx  kx  0
Ce pt (mp2  rp  k )  0
x  Ce pt  0
Trivial solution
mp 2  rp  k  0
• Solving the quadratic equations gives us the two roots:
2
p1, 2  
r
k
 r 
 
 
2m
 2m  m
p1, 2      2  o2
• The general solution takes the form:
x  x1  x2  A1e p1t  A2e p2t
r

2m
k
2
0 
m
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r
2m
k
 02 
m

Case I: Overdamped
(Heavy damping)
The square root term is +ve: The damping resistance term dominates the
p1t
p2 t
x

A
e

A
e
1
2
stiffness term.
xe
2
Let:
k
 r 
q 
 
 2m  m
 t
( A1e  A2e
qt
 qt
)
Now, if:
 A qt  qt B qt  qt 
Then displacement is: x  e
 (e  e )  (e  e ) 
2
2

x  e  t  A(cosh( qt ))  B(sinh( qt )) 
 t
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xe
 t
 A(cosh( qt ))  B(sinh( qt )) 
• Non-oscillatory behavior can be observed.
• But, the actual displacement will depend upon the boundary
conditions
78