lecture 2
Post on 02-Dec-2015
216 Views
Preview:
DESCRIPTION
TRANSCRIPT
RECAP
• Small oscillations in Bound System
• Periodic motion • Simple Harmonic Motion (Description, Examples of SHM) • Example Problem 4.15
• Example Problem 10.1
Total energy (potential +kinetic) is a constant for an undamped oscillator
22 mv2
1kx
2
1KUE
Energy of an oscillator
Total Energy
tkAtAmE 22222 sin2
1cos
2
1
tAmtAmE 222222 sin2
1cos
2
1
222
2
1
2
1kAAmE Expected
Work with harmonic oscillator : Time average values of sin(wt) and sin2(wt) over one cycle of oscillation
0 1 2 3 4 5 6
-1.0
-0.5
0.0
0.5
1.0
Sin
()
(radian)
0sin t
Area above the axis equals the area below the axis
0 1 2 3 4 5 60.0
0.2
0.4
0.6
0.8
1.0
(radian)
Sin
2(
)
2
1sin2 t
Mathematically
2
1)(sin
2sin
2
0
22
dttt
By symmetry
0cos t 2
1cos2 t
Question : What is the time averaged value of P.E. or K.E. over one period?
222
4
1sin
2
1. kAtkAEK
222
4
1cos
2
1. kAtkAEP
Similarly
.... EPEK
Next Topic
• Complex Numbers • Damped harmonic oscillator Equation of motion and solution Lightly damped Heavily damped Critically damped • Energy of the Damped Oscillator • Quality factor
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 )
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?
φ)tAcos(ωx 0
What is the use of complex numbers in harmonic oscillator?
Sol. of a SHM
φ)t(ω sinAωx 00
φ)tcos(ωAωx 0
2
0
To simplify the calculations
φ)tAcos(ωx 0
How to represent in complex form ?
φ)t sin(ωAy 0
Consider the imaginary component
φ)t sin(ωA i φ)t(ω cos A Z 00
φ)t(ω i 0e A Z
Calculation becomes simpler
Real part represents the equ. of SHM
φ)t(ω i 0e A Z
φ)t(ω i
00eiω A Z
φ)t(ω sinAωx 00 Real part
φ)t(ω i 2
00eω A Z
φ)tcos(ωAωx 0
2
0 Real part
Harmonic Oscillator : System is displaced from Equilibrium position (Experiences a restoring force)
Damping : Any effect that tends to reduce the amplitude of oscillations
Mechanics : Friction is one such damping effect
Damped harmonic oscillator
Consider the effect of friction on the harmonic oscillator Special form of friction force : Viscous force Forces arise : Object moves through a fluid
bvf
b depends on Shape of the mass Medium through which it is moving
Assuming a frictional force
Total force on the mass = Fspring+ f
m
bvkxF
0
0
2
0
xxx
xm
kx
m
bx
xbkxxm
m
b
m
k2
0
In complex form
02
0 xωxγx
How to solve ?
Companion equation
02
0 yωyγy
02
0 zωzγz
Solution will be of the form αtezz 0
Substituting the solution into the original equation:
02
0
2
0 )ωαγ(αez αt
02
0 zωzγz
2
0
2
ω4
γ
2
γα
Most general solution will be
tα
B
tα
A21 ezezz
zA and zB are constants 1 and 2 are the two roots
02
0
2
0 )ωαγ(αez αt
2
0
2
2142
ωγγ
α ,
tα
B
tα
A ezezz 21
Three possibilities
2
o
2
ω4
γ
2
o
2
ω4
γ
2
o
2
ω4
γ
Case (i)
Case (ii)
Case (iii)
Lightly damped and Oscillatory damped SHM
Heavily damped system
Critically damped system
2
o
2
ω4
γ Case (i)
Light Damping or Under Damping
22
4o
is imaginary
1
22
o iω2
γ
4
γωi
2
γα
2
0
2
2142
ωγγ
α ,
General solution :
tα
2
tα
121 ezezz
1
22
o iω2
γ
4
γωi
2
γα
tiω
2
tiω
1
γt/2 11 ezezez
tiω
2
tiω
1
γt/2 11 ezezez
Solution to the differential equation :
tCsinωtBcosωex 11
γt/2
Real part of the solution is
φtωA(t)cosφtωcosAex 112
γt
or
Solution is oscillatory, but with a reduced frequency and time varying (exponentially decaying) amplitude
4
γωω
22
o1
2
o
2
ω4
γ
01 ωω
φtωA(t)φtωAexγt
112 coscos
2
o
2
ω4
γ Case (ii)
Heavy Damping or Over Damping
2
o
2
ω4
γ is real
2
2
o
γ
4ω1
2
γ
2
γα
Both roots are negative
2
0
2
ω4
γ
2
γα
Represents non-oscillatory behavior
Actual displacement : Initial conditions
tαtα 21 BeAex
Real part of the solution is
tαtαezezz 21
21
Solution is
2
o
2
ω4
γ Case (iii) Critical Damping
tγ/2Cex
2
γα
Sol. is
2
0
2
ω4
γ
2
γα
tγ/2Cex
Sol. for 2nd order differential eqn. should have two independent constants which are to be fixed by the initial conditions
Sol. is
Solution is incomplete. Why?
Solution will be of the form
teBtAx )2/(
Energy considerations
Why the amplitude must decrease with time?
From work-energy theorem
tUtKtE
fW Work done by friction
frictionW0EtE
)(
)0(
tx
x
f fdxW
Work done by friction
t
fvdt0
t
dtbv0
2
0Frictional force dissipates energy
E(t) decreases with time
We can find how E(t) versus time
2
2
1..
dt
dxmEK
tAext
12 cos
)cos(2
)sin( 1
1
1
2
1
ttAevdt
dx t
)cos(2
)sin( 1
1
1
2
1
ttAevdt
dx t
)sin( 1
2
1
tAevdt
dx t
can be neglected
12ω
γ
1
2
o
2
ω4
γ 01 ωω
Therefore
)(sin2
1
2
11
222
1
2 teAmmvtK t
)(cos2
1
2
11
222 tekAkxtU t
)(cos)(sin2
11
2
1
22
1
2
tktmeA
tUtKtE
t
Light damping condition Again using
mk /2
0
2
1
tekAtE 2
2
1
)(cos)(sin2
11
2
1
22
1
2
tktmeA
tUtKtE
t
At t = 0 2
02
1kAE
In general teEtE 0
0 1 2 3 4 5 6 7
0
1
2
3
4
5E
time(s)
tekAtE 2
2
1
teEtE 0
00 368.0 E
e
EE
When 1
Max Energy
Characteristic time
b
m
1
Time constant (τ) : The decay is characterized by the time required for the energy to drop to 1/e times its initial value :
Quality factor
The damping can be specified by a dimensionless parameter Q
radianperdissipatedenergy
oscillatortheinstoredenergyQ
Rate of change of energy γEeγEdt
dE γt
0
Energy dissipated in a time T is γETTdt
dE
E(t)
Time to oscillate thru 2 radians 1
2
Time to oscillate thru one radian 1
1
Energy dissipated per radian is
1ω
γE
γ
ω
γ
ω
ω
γE
EQ 01
1
Q≫1 : Lightly damped
Q≪1 : Heavily damped
A3 220.00 157.
A#3/Bb
3 233.08 148.
B3 246.94 140.
C4 261.63 132.
C#4/Db
4 277.18 124.
D4 293.66 117.
D#4/Eb
4 311.13 111.
E4 329.63 105.
F4 349.23 98.8
F#4/Gb
4 369.99 93.2
G4 392.00 88.0
G#4/Ab
4 415.30 83.1
A4 440.00
Middle C
Note Freq(Hz) Musical note
A
A musician’s tuning fork rings at A above middle C, 440 Hz. A sound level meter indicates that the sound intensity decreases by a factor of 5 in 4 s. What is the Q of the tuning fork?
Example 10.2
54
0
0
0 eE
eE54 e 14.0
4
5ln
5ln4
s
700
4.0
44021
Q
Sound Intensity α Energy of oscillation
γt
0eEtE
top related