10 Harmonic Oscillator
description
Transcript of 10 Harmonic Oscillator
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PHY F111
Mechanics, Oscillations and Waves
InstructorsDr. V. Satya Narayana MurthyDr. Asrarul HaqueBITS Pilani Hyderabad Campus
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Part - I
Oscillations and Waves
V. Satya Narayana MurthyA217BITS Pilani Hyderabad CampusHyderabad
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Topics to be covered
Kleppner & KolenkowCh 10 Harmonic oscillator
A P FrenchCh 3 Free vibrations of a physical system
Oscillations involving massive springs
Ch 4 Forced vibrations and resonanceThe power absorbed by a driven oscillator
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Ch 2 Super position of periodic motions
Ch 5 Coupled oscillators & normal modes
Ch 6 Normal modes of continuous systems
Free vibrations of a stretched stringSuperposition of modes on a stringForced harmonic vibrations of a
stretched string
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Ch 7 Progressive wavesNormal modes and travelling wavesProgressive wavesDispersion, phase and group velocityThe energy in a mechanical wave
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The Harmonic Oscillator
Kleppner & Kolenkow (CH 10)A P French (CH - 3 & 4)
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Topics to be covered
Periodic motions
Simple harmonic motion
Damped harmonic oscillator
Forced harmonic oscillator
The power absorbed by a driven oscillator
Oscillations involving massive springs
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In every day life we come across various
things that move
The motion of physical systems can be
classified into 2 broad categories
1 Translational motion
2 Vibrational motion
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Periodic motions
Vibratory / Oscillatory motions
SHM
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Periodic motions
A movement that repeats with periodicity
Ex:
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The pattern that repeats may be simple or complicated
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Vibratory / oscillatory motion
What is the difference between oscillatory and vibratory motion?
A body in periodic motion moves back and forth over the same path
In oscillation time taken to complete one cycle is constant, in vibration it may not beOscillations occur in physical or biological systemsVibrations occur in mechanical systems
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Every oscillatory motion is periodic
but every periodic motion need not be oscillatory
EX. the oscillations of a pendulumthe vibrations of a string of a guitar
Uniform circular motion is a periodic motion, but it is not oscillatory
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Simple Harmonic Motion / Sinusoidal Motion
Simple periodic motion
In many systems a small displacement (x) from the equilibrium position sets up SHM
Restoring force = -kx
Where k is a constant (stiffness or spring constant)
SHM No friction force
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ldl
x
In equilibrium net force acting on the mass, F = -k dl = Mg
Now M is displaced from equilibrium position by a distance x
MgMg
Total net force acting on M is F net = -k (dl + x) Mg = -kx
Vertical spring mass system
Fnet
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The equation for the motion of SHM is:
)1( 0xmkx
kxxm
How to solve?
kxF
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The solution for this equation is of the form:tCcosx o
Another possible solution is : tBsinx o
Therefore the most general solution will be
)2( tCcostBsinx oo
)3( )tAcos(x 0
Equ. (2) can be written in convenient form as
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Rotating vector representation
SHM can be represented as geometric projection of uniform circular motion
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Acosx If we take counter clock wise direction as +vethen = 0t +
)3( )tAcos(x 0
The value of is determined from the value of x at t=0
0
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tCcostBsinx oo )2(
)tAcos(x 0 )3(
AcosC
AsinB
or
where
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)tcos(Ax 02
0
)4( 0xx 20
Comparing eq. (1) & (4)
)5(mk
0
)tAcos(x 0 )3(
0xmkx )1(
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Examples of SHMSimple pendulum
l
mmgcos
s
mgsin
mg
For small angular displacementssin
lS
l S
Eq. Of motion is
mgsinSm
(6) 0 gsinS
and
0 SlgS )6(
lg
0
kxxm
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Oscillations of floating bodies
When a body is in equilibrium,the weight is balanced by thebuoyant force
xgxm A mAg
0
mDisplace the body from its equilibrium position by an amount xthe extra buoyancy force is given by:
mg
Fb
x
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Harmonic oscillations of an LC circuit
LC
Ksource
voltage across capacitor
2
2
dtqdL
dtdiLv
voltage across inductor Cqv
qLCq
0qLC1q
LC1
0
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Nomenclature
)tAcos(x 0 )3(
x = instantaneous displacement of theparticle at time tA = amplitude (maximum displacement)o = angular frequency = 2/T = phase factor or phase angle
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A and are fixed by initial conditions
Let at t = 0, the position of the mass is x(0)and its velocity is v(0)
cos)0( Ax sin)0( 0Av
)0()0(tan
0
1
xv
2
0
2 )0()0(
vxA
How to calculate A and ?
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Energy of an oscillatorThe total energy (potential +kinetic) is a constant for an undamped oscillator
22 mv21kx
21KUE
tsinAm21
tcoskA21E 0
22200
22
2kA21E
The individual values of P.E and K.E will vary with time
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Time average values
f(t)
t1 t2
fWhat is time average?
dttf2
1
t
t
What is area under the curve between t 1 and t2 ?
dttfttf2
1
t
t12 dttftt
1f2
1
t
t12or
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Examples
0 1 2 3 4 5 6-1.0
-0.5
0.0
0.5
1.0
Sin
()
(radian)
0sin t
0 1 2 3 4 5 60.0
0.2
0.4
0.6
0.8
1.0
(radian)Si
n2(
)
21sin2 t
21)(sin
2sin
2
0
22
dtttMathematically
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What is the time average values of P.E. or K.E. over one period?
22241sin
21.. kAtkAEP
22241cos
21.. kAtkAEK
.... EPEK
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Time average value
When friction is present, this is no longer true
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Calculating 0 or T from ESpring mass system
m
22 xm21kx
21KUE
Since E is constant 0dtdE
0xmkx
mk
0
km2T
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Simple pendulum
ms
l-y
y
m
l
22
222
y2lysylsl
For small
sy
2lsy
2
mgymv21E 2
22
dtdsv
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22 sl
mg21sm
21E
2lsy
dtdsv
2
22
Since E is constant 0dtdE
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0dtdE
0slgs
lg
0
gl2T
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mgymv21E 2
Simple pendulum
E interms of
l cos
ms
my
ldtdl
dtdsv
22
2 mgl21
dtdml
21E
0dtdE
0lg
lg
0
For small
2lcos1ly
2
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Recap
Periodic motionVibratory / Oscillatory motionEqu. of motion for different SHOHow to guess a solution for second order differential equ. having constant coefficientsTime average values of KE & PE
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Today's topics
Complex numbersDamped harmonic oscillator
Equ. of motionLightly dampedHeavily dampedCritically damped
EnergyQuality factor
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Complex numbers
)tAcos(x 0 Sol. of a SHM
)t( sinAx 00
)tcos(Ax 02
0
To simplify the calculations we use complex numbers
What is the use of complex numbers in harmonic oscillator?
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Complex numbers are represented by z = x + iy
x is the real part and y is the imaginary part
Graphical representation of complex numbers
y
x
A
A cos
A sin
Imaginary axis
Real axis
z = x + iy = A (cos + i sin )
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z = x + iy = A (cos + i sin )
z = A ei
y
x
A
xy - complex plane
vector of length A makes an angle with the real axis
Geometrically what is the meaning?
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O X
Y
) ( i) ( i21 112 e eAAz t
2 1 1
Add vector of length A2at angle (2- 1) to A1 1
Turn it by an angle (t+ 1)
O XA1
A212
1 t
A1
12
A2
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)tAcos(x 0
How to represent in complex form ?
)t sin(Ay 0
Consider the imaginary component
)t sin(A i )t( cos A Z 00
)t( i 0e A Z Calculation becomes simpler
Real part represents the equ. of SHM
Real part represents the equ. of SHM
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)t( i 0e A Z
)t( i 0
0ei A Z
)t( sinAx 00 Real part
)t( i 20
0e A Z
)tcos(Ax 02
0 Real part
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Damped Harmonic OscillatorSHM No friction forceWhat is the effect of friction on the harmonic oscillator?
Assume a special form of friction force viscous force velocity f = - bv
Condition: Viscous force arises when an object moves through a fluid at speeds which are not so large to cause turbulence
b = coefficient of damping force
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Total force acting on m is F = Fspring + f
bvkxF
xbkxxm
0xxx
0xmkx
mbx
20
Equ. of motion is
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In complex form
020 xxx
How to solve ?
To convert into complex form use the companion equation
020 yyy
020 zzz
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The solution will be of the form, tezz 0
Substituting the solution back into the original equation gives us:
0202
0 )(ezt
020 zzz
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20
2
4
2
The most general solution will be:
tB
tA
21 ezezz
Here zA and zB are constants and 1 and 2are the two roots
0202
0 )(ezt
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20
2
21 42
,
tB
tA ezezz 21
2o
2
4 2o
2
4 2o
2
4
Case (i) Case (ii) Case (iii)
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2o
2
4 Case (i) Light Damping
orUnder Damping
22
4 o
is imaginary
1
22o i2
4
i2
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ti2ti1t/2 11 ezezez The solution to the differential equation is:
tCsintBcosex 11t/2 Real part of x is
tA(t)costcosAex 112t
or
The solution is oscillatory, but with a reduced frequency and time varying (exponentially decaying) amplitude
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4
2
2o1
2o
2
4
01
tA(t)tAext
112 coscos
-
2o
2
4 Case (ii)
Heavy Dampingor
Over Damping
2o
2
4 is real
2
2o
412
2
Both roots are negative
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This represents non-oscillatory behaviorThe actual displacement will depend upon the initial conditions
tt 21 BeAex
Real part of the solution is
tt ezezz 21 21 Solution is
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01
2o
2
4
-
2o
2
4 Case (iii) Critical Damping
t/2Cex
The sol. to a 2nd order differential equ.should have two independent constantswhich are to be fixed by the initial conditions
2
Sol. is
The solution is incomplete Why?
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solution will be of the form teBtAx )2/(
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teBtAx )2/(tt 21 BeAex tAext
12 cos
Air Thickoil Water
Light Heavy Critical
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Energy of a Damped Harmonic Oscillator
frictionW0EtE From work energy theorem
Wfriction = work done by the friction forcefrom time 0 to t
f = -bvopposes the motion
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x(t)
x(0)f fdxW
t
0
2dtbv 0 Friction force dissipates energy
E(t) decreases with time
22
21
21 xmkxK (t)U (t)E (t)
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For the lightly damped oscillator)tt/2)cos(Aexp(x 1
)t(t)sinexp(2mAmv
21K(t) 1
221
22
can be neglected
)tcos(2
)tsin(Aev 11
1
t2
1
12
1
2o
2
4 01
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)t(cosekA21kx
21tU 1
2t22
)t(kcos)t(sinmeA21
tE
12
122
1t2
For light dampingmk
202
1
t2ekA21tE
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At t=0 20 kA2
1E
In general t0eEtE
0 1 2 3 4 5 6 7
0
1
2
3
4
5E
time(s)
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The decay is characterized by a time , damping time, during which the energy falls to e-1 of its initial value
t0eEtE 00 0.368E
eEtE
When 1
Time constant
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Recap
Complex numbersDamped harmonic oscillatorEqu. of motion xbkxxm
20
2
21 42
,
tB
tA ezezz 21
-
2o
2
4 2o
2
4 2
o
2
4
Case (i) Case (ii) Case (iii)
teBtAx )2/(tt 21 BeAex tAext
12 cos
Light Heavy Critical
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t0eEtE
0 1 2 3 4 5 6 7
0
1
2
3
4
5
E
time(s)
00 0.368EeEtE
1
Energy of a lightly damped harmonic oscillator
Time constant ()
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Todays topics
Q factor of DHO
Forced Harmonic oscillator
Undamped FHOEqu. Of motionSolutionResonance+ve and ve aspects
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Quality factorThe damping can be specified by a dimensionless parameter Q
radianperdissipatedenergyoscillatortheinstoredenergyQ
Rate of change of energy EeEdtdE t
0
Energy dissipated in a time T is ETTdtdE
E(t)
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T = 2 / 1 oscillates through 2 radians
Energy dissipated per radian is 1
E
EEQ 01
1
Light damping Q>>1Heavy damping Q is lowUndamped oscillator Q is infinite
ETTdtdE
Energy dissipated in 2 radians
01 Q
1
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In an experiment, a paperweight suspendedfrom a hefty rubber band had a period of1.2 s and the amplitude of oscillationdecreased by a factor of 2 after threeperiods. What is the estimated Q of thesystem?
2t
AeA(t)
Solution
-
1.8
(0)
AeAe2
ln21.8
10.39s
131.2*0.39
2
Q 1
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10.3 In an undamped free harmonic oscillator the motion is given by x = A sin0t. The displacement is maximum exactly midway between the zero crossings. In a damped oscillator the motion is no longer sinusoidal and the maximum is advanced before the midpoint of zero crossings. Show that the maximum is advanced by a phase angle given approximately by = 1/2Q
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Forced Harmonic OscillatorUndamped Forced Oscillator
Equ. of motion of a SHO
tcosFkxxm 0
kxxm
Driving force
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tcosFkxxm 0
How to solve?
Try the solution
tcos A x
RHS of equ. has cos t
LHS of equ. must also have cos t
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tcosFkxxm 0
tcos A x
Equ. of motion
solution
tFtAkm o coscos2
2mkFA o
22
1
o
o
mFA
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The solution is tmFx
cos1
220
0
Incomplete solution ???
No arbitrary constants
Must able to specify x0 and v0
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Complete solution is
)tBcos(cos
1mFx 022
0
0
t
Steady state solution
General solution ofundamped oscillator
020 xx
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Resonance22
1
o
o
mFA
0A
= 0 A is finite
A 0Resonance
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0 20 40 60 80 100
-0.010
-0.005
0.000
0.005
0.010
0
A
221
o
o
mFA
00;A
00;A
The displacement is opposite to the directionof the force!There is a phase difference of between thedisplacement and the applied force
0;A
-ve A ?
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The phenomenon of resonance has both +veand ve aspects+ve aspectsSmall driving force gives large amplitudeTuning radios to the desired frequency
-wave ovenfood with no water content cannot be heated
Applied -wave frequency is equal to the H2O molecules (non zero dipole moment) natural frequency
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-ve aspects
To reduce response at resonance dissipative friction force is needed - Forced Damped Harmonic Oscillator
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Recap
Q 01
tcosFkxxm 0
solution tmFx
cos1
220
0
Forced Undamped oscillator
Q factor of DHO
-
Complete solution is
)tBcos(cos
1mFx 022
0
0
t
Steady state solution
General solution ofundamped oscillator
020 xx
-
221
o
o
mFA
0 20 40 60 80 100
-0.010
-0.005
0.000
0.005
0.010
0
A
-ve A ?
displacement is opposite to the direction ofthe force!There is a phase difference of between thedisplacement and the applied force
tcosFkxxm 0
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Todays topics
Forced damped harmonic oscillator
Equ. Of motionSolutionResonanceEnergyQuality factor
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Forced Damped Harmonic Oscillator
Undamped FHO drivingspring FFF
tcosFkxxm 0
SHM springFF kxxm
Actual motion is the superposition of oscillations at two frequencies and 0
Transient behaviorTransient behavior
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drivingviscousspring FFFF Damped FHO
tcosFbv -kxxm 0
In the initial stage transient state exists
After a sufficiently long time the natural oscillations dies out because of the damping force Now the oscillator oscillates at the frequency of the driving force Steady state
-
tmFxxx oo cos
2
Will x =A cos t satisfy this differential equ.?No!
The velocity term gives sin t
tcosFbv -kxxm 0
tcosmFx
mkx
mb x 0
-
tmFxxx oo cos
2
How to find the solution?
Write the above equation in complex form
tiemFzzz 020
Solution will be of the form z = zo eit
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Real part of z = zo eit gives the solution to Forced damped harmonic oscillator
Substituting z = zo eit in complex equation
imFz
emFiez titi
220
00
020
20
1
)(
-
i1
mFz
220
00
In polar form
220
1
21
22220
0*00
0
tan
)()(1
Re
mFzzR
z i
22220
2200
)()(i(
mF
)
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Real part
The complete solution is z = zo eit
titi-i ReeRez
)cos( tRx
1/22222oo
1mFRA
22
1tan
o
Phase difference betweenthe driving force and the displacement
-
1/22222oo
1mFRA
A is constant for a given frequency
0dtdA
21
20m 2Q11
At = max
-
For light damping, A is maximum for = oand the amplitude at resonance is:
o
oo m
FA )(
The behavior of A and as functions of ,depends on the ratio / o
1/22222oo
1mFRA
-
10
1
Q 01 1/22222o
o
1mFA
21
20m 2Q11
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As increases, the maximum amplitude occurs at a frequency less than the resonant frequency
10
21
20m 2Q11
1/22222o
o
1mFA
-
22
1tan
o
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0 20 40 60 80 100
-0.010
-0.005
0.000
0.005
0.010
0
A
Undamped FHO Damped FHO
1/22222oo
1mFA
22
1tan
o
221
o
o
mFA
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EnergyFor steady state motion amplitude is constant in time
tAx cos
tAv sin
)sin(21
21)( 222 tAmmvtK
)(cos21
21)( 222 tkAkxtU
22
41 AmK
22
41 AkU
-
2222
222 )(41
o
oo
mFE
)(mA41E 20
22
22
41 AmK
22
41 AkU
Steady state
- Light Damping
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Resonance curve or lorentzian
-
222
2/1
81
o
o
mFE maximum
height 24
Falls to one half maximum
2
20 2
20
FWHM = 2(-0) = 2
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Forced damped harmonic oscillator
tcosFbv -kxxm 0
Recap
Equ. Of motion
Steady state solution
1/22222oo
1mFRA
22
1tan
o
tAx cos
-
gA is constant for a given frequency
21
20m 2Q11
1/22222oo
1mFRA
-
22
1tan
o
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Steady state 22
2
2/1
81
o
o
mFE
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Todays topics
Power absorbed by an oscillatorSimilarity between the power and energy curvesQ factor calculation from resonance curvesOscillations involving massive springs
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Power absorbed by an oscillator
How to maintain the amplitude of a forced harmonic oscillator constantly?
Rate at which energy is supplied to a driven oscillator to maintain its amplitude constantly is
FvdtdxF
dtdwP
-
tcos A x 221
o
o
mFAwhere
t sinA x v
t2 sin2AF - t cos t sinAF P 00 Fv
t cos F F 0 Driving force
&
Undamped FHO
-
t2 sin2AF P 0 Fv
t2 sin P
0 P
Energy is fed into the system in one half cycle and is taken out again during next half cycle
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Damped FHO
2/12222 ])()[(1
o
o
mFA
)(t tsincosAFFv P 0
tAx cos
tAv sin
-
)sin(cos P 0 ttAFFv
tcos)sinAF(
tcostsin)cosAF(2
0
0
Average value is zero
sinAF21P 0
-
sin21 0 AFP
2222022
0
1m
F21 P
For light damping 0 =
2
20
20
2
1mF
81 P
-
220
20
/21
mF
81 P
222
2/1
81
o
o
mFEAverage energy is
-
Resonance curve or lorentzian
-
maximum height 2
4
Falls to one half maximum
2
20 2
20
220
20
/21
mF
81 P
222
2/1
81
o
o
mFE
-
E or P
20
0- +
= Full width at half maximum / resonance width
half maximum2E
Width of the curve
2P or
-
Quality factor
curve resonance ofwidth frequency frequency resonance
oQ
Gives the frequency selective property of an oscillator
Sharpness of resonance curve means the system will not respond unlessdriven very near its resonance frequency
Q = 10 is more selective
-
Response in time vs response in frequency
50 100 150
0.00
0.02
0.04
0.06
0.08
0.10
0.12
E
FWHM
P
-
Oscillators which are very frequencyselective also have weak damping.
So such an oscillator does not recover from adisturbance or does not respond quickly.
The damping time and the resonance curvewidth obey 1
1
-
Oscillations involving massive springs
mk M
Total energy is a constant
2
21 kxU
E = K + U = Constant
K = K spring + K mass
l0
dtdE
How to calculate the KE of the spring ?
What is the frequency of oscillation?
-
The spring oscillations are not so large that they cause the spring coils to bump into each otherStretching force is same at all points along the springAll the points in the spring undergo displacements proportional to their distances from fixed end Static extensionVelocity is the same for all the elements of the spring
Assumptions
-
l/3 l/3 l/3m
mx
32x
3x
l
ms ds
dsl
MdM
2
21 (dM)dvdK
dtdx
lsdv
xls
Displacement of ds
-
22
21
21
dtdx
lsds
lM(dM)dvdK
dssdtdx
lMdK 2
2
32
l
spring dssdtdx
lMK
0
22
32
2
6
dtdxMK spring
-
E = PE spring + KE spring + KE mass
222
21
621
dtdxm
dtdxMkxE
0dtdE
3Mm
k
-
3Mm
k
Suppose m = 0
Mk3
The above calculation is not exact Why?Because of the assumptions (i) Extension of the spring is proportional
to the distance from the fixed end(ii) Velocity (dx/dt) is the same for all the
elements of the spring
-
Is only an approximation
It will hold if M
-
10.7 Find the driving frequency for whichthe velocity of a forced damped oscillator isexactly in phase with the driving force.
-
10.10 A small cuckoo clock has a pendulum25 cm long with a mass of 10 g and aperiod of 1s. The clock is powered by a 200g weight which falls 2 m between dailywindings. The amplitude of the swings is0.2 rad. What is the Q of the clock? Howlong would the clock run if it were poweredby a battery with 1 J capacity?