ITEC 1011 Introduction to Information Technologies 2. Data Formats Chapt. 3.
Chapt 3 French
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Transcript of Chapt 3 French
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Lectures on
Oscillations and WavesBy
D. D. PantAssociate Professor
Department of Physics
BITS PilaniChamber No. 3258
Email Id [email protected]
Mobile No. 09950425605
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Text Book:
Vibrations and Wavesby
A. P. FrenchReference Book:
Waves and Oscillations
by
N. K. Bajaj
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Oscillations and Waves
Why s tudy o sc i llat ions and waves?
– Almost al l phy sical s i tuat ions involve periodic or
osc i l latory behavior
• Motion of the planets• Stable mechanical systems
• Electrical systems
• Fundamental forces
–Per iod ic mo t ion in con t inuous media
• Wave propagation
• Electromagnetic radiation (light/optics)
• Matter particles are “waves”.
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The world is full of oscillatory motions
-A child on a swing-A guitar string being played
-Swinging pendulum of wall clock
-Atoms in molecules or in solid lattice
-Air molecules as a sound wave passes by
-Radio waves, microwaves and visible light are
oscillating magnetic and electric field vectors
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Isaac Newton(1642-1727)
Early Studies of Oscillations
Robert Hooke(1635-1703)
Christian Huygens(1623-1697)
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Periodic motion:- Any motion that
repeats itself in equal intervals oftime.
Oscillatory motion:- If a particlemoves back and forth over the
same path.
Harmonic motion:- Oscillatory
motions which can be expressed in
terms of sine and cosine functions.
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Simple Harmonic Oscillators
A simple pendulum
A mass fixed to a wall via a spring
A frictionless U tube containing liquid
A hydrometer floating in a liquid
An inductor connected across a
capacitor carrying a charge q
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Floating objects
Example I : The up-down motion of a partiallyimmersed solid
x
Equilibrium Position Pushed down by x
F
xg A Force Buoyancy Additional F
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Equation of motion of the body is :
x g Adt xd m 2
2
Simple Harmonic motion withm
g A
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yy
2yg
L
M)y(U
M : Total mass of liquid
L : Total length of the water column
2yM
2
1KE
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Energy conservation :
0dt
dE
0yyL
Mg2yyM
0yL
g2y
SHM of frequency :
L
g2
22 ygL
MyM
2
1E
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Example III
Prob. 6.17 ( K & K): A rod of length l and mass m, pivoted at one end, isheld by a spring at its midpoint and a spring at its
far end, both pulling in opposite directions. The
springs have spring constant k, and at equilibrium
their pull is perpendicular to the rod. Find thefrequency of small oscillations about the
equilibrium position.
l
g
m
k
2
3
4
15
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tsin&tcos
0equationSHMof Solution 2..
x x
The two independent solutions are :
The most general solution of SHM equation is :
)sin()cos()( t Bt At x
(A & B are arbitrary constants)
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Any arbitrary initial condition on position and
velocity can be accommodated within the solution
of above kind, with appropriate values of A & B
Example :
Suppose the initial (t = 0) position and initial
velocity are respectively. Obtain thesolution.
00 v&x
Bv;Ax00
tsinv
tcosx)t(x 00
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The solution of a linear differential equation with
constant coefficient is an exponential function :
t jt j eC eC t x 21)(
t peC t x )(
Substituting this into the eq.
j p
So the most general complex solution is :
)tCos(A
)tCos(C2CC )()(
t jt j
ee
complexare& 21 C C
022
2
xdt
xd
We get
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The complex solution :
)( t je A
)(
t je A z
A
t
is thus, a rotating vector of fixed length ,
rotating counter-clockwise, with an angular
velocity
A
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)( t je A z
)tcos(Ax
The SHM is the projection of the vector on the x-
axis.
)sin( t A y
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Prob. 3.19
Mass m connected to two
springs on frictionless
horizontal table. Spring
constant k and unstretched
lengths of springs 0
x
y
(k) (k)
xk 2
dt
xdm
2
2
m
k 2
x
a) Eq. of motion along x
F = - 2kx
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y
2
22/1
2
2
2
y
1
y
1
)(2
y)( 0
2
00
yk 2F 0y
b) Eq. of motion along y : y
yk 2dt
yd
m 0
2
2
2/1
0y
m
k 2
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x
y
2/1
0y
x
T
T
t At x x
cos)( 0
t At y y cos)( 0
c) Ratio of periods along x & y
d) x & y as functions of time if and mass
starts from rest
0
00 A y x
00
A,A
at t = 0
x
y
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Damped Simple Harmonic Motion
1. Pendulum with air drag
In addition to the restoring force, there is adamping force, always opposing the motion
of the oscillator
2. U-tube with viscous liquid
The damping force is usually proportional tothe velocity of the oscillator :
dt
dx bv bF
damp
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Damped Simple Harmonic Motion
Equation of motion :
dt
dx bxk
dt
xdm
2
2
Or, 0xdt
dx
dt
xd 202
2
frequencynaturalorfrequencyundampedcalledisIt
absentisdampingwhenfrequencyangularisand 0m
k
frequencyof dimensionhaswhere
m
b
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Each value of this term describes particular
type of motion
negativeorzero positive, becan4
root termSquare 20
2
t
eC t x
20
2
42
)(
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Real exponential functions means
Non Oscillatory Motion
For example Pendulum inside thick syrup
Most general solution is
CCex(t) 212-
qt qt t
ee
Case 1: Heavily Damped or Over Damped
Motion
q2
0
2
4 writeusLet
2
0
2
4
or square root term is +ve
Or damping force > restoring force
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i) Initial conditions :Pendulum released from rest
0)0(;)0(i.e. 0 x x x
qt t qt
eqeqeq
xt x
224
)( 20
t
x(t)
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ii) With the initial conditions :
0v)0(x;0)0(x
t qt qt
eeeq
vt x
20
2)(
t
x(t)
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The most general solution is
t2e)tBA()t(x
Case 2: Critical Damping
2
0
2
4 i.e.
This is the limiting case of behaviour of case 1
as q changes from +ve to –ve value.
square root term is zero i.e. q = 0
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i) With the initial conditions :
0)0(x;x)0(x 0
t2
0 et21x)t(x
t
x(t)
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ii) With the initial conditions :
0v)0(x;0)0(x
t2
0
etv)t(x
t
x(t)
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Applications of Critical Damping Mechanism
In many systems, quick damping is desirable
to bring the system to a quick stop.
i) Needle in meters such as ammeter,
voltmeter etc.
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ii) Door closers :
Out of the two non-oscillatory damping – over
damping and critical damping – it is the latterthat brings the system back to equilibrium
quicker
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t
x(t) Critical Damping
Over Damping
0)t(x
)t(xim
od
cd
t
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motiondampedof frequencyangularis
4 Where
22
0
j p
p
2
i.e.
quantitycomplexaisSo
Case 3: Damped Simple Harmonic Motion
2
0
2
4
or square root term is -veOr damping force < restoring force
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The most general complex solution :
t jt j
t
e Be Aet z
2
)(
The most general real solution :
)tcos(eA)t(xt
20
This is a SHM with decaying amplitude
conditioninitialfromobtainedareand0
A
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Rotating vector representation of Damped SHO
A lit d f th ill t d ith ti
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Energy of the oscillator also decays with time as
t
e Amt Ak E
2
0
2
0
2
2
1
)(2
1
t
0 eE
4
),(2
2
0
2
0
t2
0
eA)t(A
Amplitude of the oscillator decays with time as
Frequency of damped oscillator is less than theundamped oscillator
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22
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1
22
2
2
2
2
0
2
m
E
x
m
E
x
This is an equation of an ellipse
P h i lli
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Path is an ellipse
x
x m
E2
2m
E2
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Q lit F t Q l
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Quality Factor or Q – valueIt describes the charactertics of Damped
Harmonic Motion
T2 timein thisradiansof Number
1 t
1
00
e E e E E t
Q
It is defined as the number of radians through
which damped oscillator oscillates as its energy
decays to e-1 of its initial value.
)t(x
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)t(x
t
)t(x
t
t
)t(x
1
2
3
Quality Factor
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4
Oscillator 4 is a better quality oscillator than
1 -3 , even though its amplitude decreases
faster
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The true quality of a damped SHO is not
measured by how long it lives (time in
which the amplitude drops substantially),but rather, by how many cycles of
oscillations it completes in this lifetime.
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Amplitude after n cycles :
Q
n
0n eAA
Energy after n cycles :
Q
n2
0n eEE
cycle perlostenergy
systeminstoredenergy2 Q
Q is also obtained from following energy relation
P b 3 16
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Prob. 3.16 According to classical electro-magnetic theory an accelerated electron
radiates energy at the rate :
3
22
c
aeK P
229106 C m N K
e = electron chargec = speed of light
a = instantaneous acceleration
) If l ill i l
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a) If an electron were oscillating along astraight line :
tsinAx
how much energy would it radiate in one cycle?
Ans : tsinAxa
2
tsin
c
AeK P
2
3
242
3
232T
0
cyclec
AeK dtPE
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b) What is the Q of this oscillator?
cycle perlostenergy
systeminstoredenergy2 Q
2
3
232
322
Ae2K
Am2
Ke
mcc
Q
c) After how many oscillations will the
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c) After how many oscillations, will theenergy be down to half the initial value?
2
EeEE 0Q
n2
0n
2n2Qn
d) Putting for the typical optical
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opt Ke
mcQ 2
3
115
opt
opt sec100.4c2
810
sec107.12nQn2
T 82/12/1
d) Putting for the typical opticalfrequency, find Q and the half life