Download Harmonic Oscillator

Harmonic Motion (HM)
Oscillation with Laminar Damping
If you don’t know the units of a quantity you probably don’t understand its physical significance.
Simple HM
Hooke' s Law :
r
r
F = −k x
(2B/T)2 = k/m
x = A sin ( No + Tt )
T2 = k/m
v = A T cos ( No + Tt )
a = !A T2sin ( No + Tt )
Simple pendulum: k = mg/L
or
T / 2Bf
fT/1
definitions:
T2 = g/L
T2 = Dwater g/DcorkL
Cork on water: k = DwaterAg
kinetic energy = ½mv2
elastic potential energy = ½kx2
conservation of mechanical energy: KEf + PEf = KEi +PEi
if friction can be neglected ( Wnon-conservative = 0 )
Damped HM
To2 / k/m
damping force Fdamp = & cv
energy:
E = Eo e
quality:
Q≡
laminar damping force proportional to the velocity v
c
− t
m
E /T
Pave / 2π
1
=
τ
or
c
for laminar damping
m
∆E
∆t
=−
E
Q/ω
Q
ω
≡ τ decay constant
a) low damping ( under damped ): c2 < 4mk
position :
Amplitude:
x = Ao e
−
c
t
2m
sin( ωt + ϕo )
A = Ao e − β t
β=
k
c
ω =   − 
m  2 m 
2
c
for laminar damping
2m
Damping decreases the frequency of oscillation, and also the amplitude as time increases.
b) medium damping ( critical damping ): c2 = 4mk
The system returns to equilibrium in the shortest time without oscillating.
c) large damping ( over damped ): c2 > 4mk
The system takes a longer time to reach equilibrium without oscillating.
C. Deurzen 2005
Forced HM and Resonance
Under the influence of an external driving force F = Fo sin Tt how will the system respond?
Position:
x = A sin ( Tt + N )
Amplitude A =
a)
displacement always lags (N < 0) the driving force
Fo / m
(ω
2
o
−ω
2
)
2
ω
+  
τ
tan ϕ = −
2
At low driving frequency T << To
Response controlled by spring (k)
b)
At resonance driving frequency T = To
Response controlled by damping (c)
c)
At high driving frequency T >> To
Response controlled by inertial mass (m)
ω /τ
ωo 2 − ω 2
A=
Fo / m
A=
Fo / m Fo / ωo
=
ωo / τ
c
A=
Fo / m
ω0
2
ω2
=
Fo
k
ϕ→0
Fo / ω 2
=
m
ϕ→−
π
2
ϕ → −π
A(ω = ωo ) τ / ωo
response at resonance
=
= ωo τ = Q
=
response at zero frequency
A(ω = 0) 1 / ωo2
Power Absorption:
Power is the time average of the force times the velocity (see exercise 2):
2
ω / τ)
(
Power =< F ⋅ v >= Pres
2
2
2
2
ω
ω
ω
/
τ
−
+
(
)
(o )
2
where
Near resonance (T.To) the half-power points occur when:
or
1 F
Pres = m  o  τ
2 m 
To2 & T2 = T/J
2 To )T . T/J
Thus the full width (2 )T) of the resonance at half maximum power is equal to 1/J
and the quality of an oscillator Q measures the sharpness of damping near resonance:
Quality Q = ωo τ =
frequency at resonance
ωo
=
2 ∆ω full width at half −maximum power
Numerical Example of Harmonic Oscillator
Given oscillator data:
Then:
mass
m = 1 gram
spring constant k = 10 N/m
relaxation time J = 0.5 second
N
10
k
m = 102 radian = 16 cycles
ωo =
=
−3
m
second
second
10 kg
2
1
radian
ω = ω −   = 104 −1 = 102
 2τ
second
2
o
free oscillatio n frequency
Quality factor
Q = ωo τ = 102
1
radian
0 .5 second = 50
second
m
= 2 τ = 2 (0 .5 sec ) = 1 sec
β
c
1 gram
m
gram
the value of the damping constant is: c = =
=2
second
τ 0.5 second
The time for the amplitude to
damp to e&1 of its initial value is:
Given a driving force:
=2
F = 0.1 N sin ( 90 t ) thus
Fo = 0.1 N ;
Fo 0.1 N
N
N and the driving frequency is T = 90 rad/sec
=
= 0.1
= 100
gram
kg
m 1 gram
n = & 5.4E
The response of the oscillator is:
Amplitude A =
100 N / kg
(
104 − 902
)
2
 90 
+ 
 0 .5 
2
=
100
= 5 cm
1909
tan ϕ =
90 / 0 .5
= −0 .095
902 − 104
100 N / kg
Fo / m
=
= 50 cm
ω o / τ 100 rad / sec
0.5 second
01
. N
F
the amplitude at low frequency is A(ω = 0) = o =
= 1 cm
k 10 N / m
the amplitude response at resonance is A(ω = ω o ) =
1
2
The power absorption at resonance: Pres = 10 − 3 kg (100 N / kg)2 0.5 sec = 2.5 Watt
The full width of the resonance curve between half-power points is:
2 )T = To/Q = 100 rad/sec / 50 = 2 rad/sec.
Exercise 1
Show x=Asin(Tt+N) satisfies the equation of motion of a driven harmonic oscillator.
Equation of motion:
m x&& + c x& + k x = Fo sin ω t
&&
x+
c 1
≡
m τ
c
k
F sin ω t
x& + x = o
m
m
m
k
≡ ωo2
m
Substituting the first and second differentiation of the proposed solution x=Asin(Tt+N):
(ω
2
o
)
− ω 2 A sin (ω t + ϕ ) +
F
ω
A cos (ω t + ϕ ) = o sin (ω t )
τ
m
using: sin(Tt+N) = sinTt cosN + cosTt sinN
cos(Tt+N) = cosTt cosN - sinTt sinN
(
(
)
)
Fo
ω
ω
 2

 2

2
2
ϕ
ω
ϕ
ω
ω
ω
ϕ
ω
−
ω
cos
ϕ
−
sin
A
sin
t
+
−
sin
+
cos
A
cos
t
=
sin ω t
o
o




τ
τ
m
This equation can only be satisfied for all time t if the coefficients of
sinTt and cosTt are both zero, which establishes A and the phase angle N:
A=
Fo / m
(
)
ωo2 − ω 2 cosϕ −
tan ϕ =
ω
sin ϕ
τ
=
Fo / m
(ω
2
o
−ω
)
2 2
ω
+  
τ
2
ω /τ
sin ϕ
=− 2
cosϕ
ωo − ω 2
where we have used sin2N + cos2N / 1 in order to obtain:
sin ϕ =
−ω /τ
(ω
2
o
−ω
)
2 2
ω
+  
τ
2
cosϕ =
ωo2 − ω 2
(ω
2
o
−ω
)
2 2
ω
+  
τ
2
Exercise 2
Derive the power absorption P(T) for a driven harmonic oscillator.
For the motion of a forced or driven oscillator we have:
force F = Fo sin Tt ; position x = A sin ( Tt + N ) ; velocity v = TA cos ( Tt + N )
Note the phase angle N of the position x relative to the driving force F and
an additional 90E for the velocity v which is the derivative of the position.
The time averaged power is the time average of the force times the velocity
Power = + Fexternal @ velocity ,
+, / time average
Power = + ( Fo sinTt ) @ { TA cos( Tt + N ) } ,
cos( Tt + N ) = cosTt cosN - sinTt sinN
Power = Fo T A + sinTt cosTt cosN & sinTt sinTt sinN ,
The time average of sinTt cosTt is zero, that is + sinTt cosTt , = 0
and the time average of sin2Tt is one half, that is + sin2Tt , = ½
Power = Fo T A ( 0 & ½ sinN )
Fo / m
From exercise 1: A =
(ω
Power =
1
F
Fo ω o
2
m
2
o
2
ω
+  
τ
2
(ω
2
o
−ω
)
2 2
and2
ω
+ 
τ
ω/ τ
(ω
2
o
1  Fo 
m  τ
2 m 
)
2 2
−ω
ω
+
 
τ 
2
(ω / τ)
2
2
Power =
)
−ω2
−ω /τ
sin ϕ =
(ω
2
o
)
2 2
−ω
ω
+
 
τ 
Note the frequency dependence factor
containing T, damping J=c/m, spring
To2=
k/m
2
1
2
m (ω A )
= 2
τ
factor =
(ω / τ )2
(ω
2
o
−ω
)
2 2
ω
+  
τ
2
Exercise 3
Consider a series electromagnetic circuit consisting of: an inductor L,
a resistor R, a capacitor C, and an applied power supply õ = õosin Tt.
Show that such a circuit behaves identically as a harmonic oscillator.
Applying Kirchhoff’s Voltage Law to the series circuit loop, we obtain:
L
∆i
1
+ Ri + q = Eo sin (ω t )
C
∆t
where q is the time-varying quantity of electric charge stored in the capacitor and i is
the time varying intensity of current as the charge oscillates through the series circuit.
Substituting for i / )q/)t :
1
q = Eo sin (ω t )
C
m x&& + c x& + k x = Fo sin (ω t )
L q&& + R q& +
The form of the electric circuit equation is identical to that of the mechanical,
laminar-damped, driven forced oscillator. Mathematically, the two equations
describe the same behavior. It remains to identify the quantities that correspond
to each other and identify their physical significance and the role each plays.
Inductance
L
Resistance
R
elastance
1/C
õlectro Motive Force Eo
}|
}|
}|
}|
m mass
c damping factor
k spring’s elastic constant
Fo maximum strength of driving force
Thus:
Inductance plays the role of inertia, as expected since it takes time to establish a
current in an inductor when a voltage is applied to it.
Resistance plays the role of damping, as expected because it absorbs energy similar
to a dashpot or a shock absorber in a mechanical system.
The inverse of capacitance plays the role of elasticity, as expected because a capacitor
stores electric charge and energy similar to the elastic energy storage in a spring.
The natural frequency is therefore ωo 2 = elasticity = 1 / C = 1
inertia
L
LC
which is precisely the resonance frequency of the so-called electrical “tuning” circuit.
ωL − 1 / ωC
The phase angle between the applied voltage and the current is tan ϕ =
R
At resonance the current and voltage are in phase N=0, the circuit acts as a pure
resistance, and for a given impressed õmf the current in the circuit is a maximum.
The resonance full width at half-maximum power points is ( 2)T ) = 1/J = R/L .
The Quality factor of the electromagnetic oscillator is Q = To J = To L/R etc.
C. Deurzen 2005
Exercise 4
Consider the damping factor for a flat plate moving normal to its plane with
speed v through a low-pressure, rarefied gas. Assume the mean free path of the
gas molecules is large compared to the dimensions of the plate so that the
deceleration of the plate may be thought of in terms of individual collisions
between the plate and the molecules rather than hydrodynamic flow.
The drag force on the plate will be proportional to the rate at which molecules
strike the plate multiplied by the average momentum transfer per molecule.
The rate at which molecules strike the plate is proportional to the relative velocity.
Consider the relative velocity between the incoming molecules and the plate.
Suppose the molecules move in only one direction with speed vm or with rms speed.
On one side of the plate the relative velocity is vm + v; on the other side it is vm - v.
The average momentum transfer is itself proportional to the relative velocity.
The net drag force on the plate opposing the motion will thus be proportional to:
Fdamping = − F forward − Fbackward
∝−
(v m − v )2 − (v m + v )2
Case: v << molecular velocity
Fdamping % & 4 vm v % &v
The damping is proportional to the first power of the velocity.
For viscous hydrodynamic flow this is referred to as laminar flow.
Case: v >> molecular velocity
In this case there is no forward force as the molecules can’t catch up
and the molecular velocity is negligibly small and ignored.
Fdamping % & v2
The damping is proportional to the second power of the velocity.
For viscous hydrodynamic flow this is referred to as turbulent flow.
C. Deurzen 2005