09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
Linear Harmonic oscillator
Mechanical model: mass m on a spring characterized by a spring constant k
Elastic restoring force Fs = − kx is balanced according to Newton's second law
Fs = ma
acceleration
mx = − kx
mx + kx = 0
2
x + ω0 x = 0
where
ω0 =
k
m
free natural angular frequency
We will show that such system oscillates with amplitude A and angular frequency ω0 .
Basic facts about second order linear differential equations:
1. Solutions x = x (t; C1 , C2 ) will have two constants dependent on initial conditions
2. If x1 (t ) is a solution then Cx1 (t ) is also a solution.
3. If x1 (t ) and x2 (t ) are solutions then x1 (t ) + x2 (t ) as well as any linear combination
C1 x1 (t ) + C2 x2 (t ) is also a solution.
To solve the equation of motion
2
x + ω0 x = 0
we multiply both sides by 2x
= −2ω 0 2 xx
2 xx
which allows us to immediately carry out the first integration:
2
x 2 = −ω 0 x 2 + C
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
We determine the first integration constant C by observing that at a full "swing", the
oscillator position is equal to its amplitude, x=A, and its velocity is zero (turning point of
2
oscillation) x = 0 . Thus C = ω 0 A2 and the equation to be integrated further becomes:
2
x 2 = ω0 ( A2 − x 2 )
Separation of variables gives
∫
dx
A2 − x 2
= ω0 ∫ dt
and after integration:
x
sin −1   = ω0t + φ
 A
or in a more familiar form:
x = A sin(ω 0t + φ )
ω0 is the natural angular frequency.
ω0 = 2π f 0 , where f0 is the natural frequency.
Oscillation period T0 =
1
.
f0
Alternative solution method uses trial exponential function
x = exp( λ t ) , and our task is to determine the constant λ .
Substituting into the equation to be solved we obtain
λ 2 exp( λt ) + ω0 2 exp( λt ) = 0
from which it is immediately clear that
λ 2 = −ω0 2
and thus
λ = ±i ω0 where i = −1
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
Thus the general solution has to be of the form
x = A1 exp( +iω0t ) + A2 exp( −iω 0t )
(remember: we need two constants).
Recall Euler's formula
e ± iθ = cos θ ± i sin θ
Since A1 and A2 are complex quantities, they may be inconvenient to use, and it is more
informative to write the solution as:
x = A exp[i (ω0t + φ )]
= A cos(ω0t + φ ) + iA sin(ω0t + φ )
Energy considerations
Consider solution
x = A sin(ω 0t + φ )
The velocity v is equal to
x = Aω0 cos(ω0t + φ )
and thus the kinetic energy
1 2 1
2
mx = mA2ω 0 cos 2 (ω 0t + φ ) = K 0 cos 2 (ω0t + φ )
2
2
where the maximum kinetic energy is equal to
K=
K0 =
1
1
2
mA2ω 0 = kA2
2
2
The potential energy – work done by applied force displacing the system from 0 to x
x
U ( x ) = ∫ kx dx =
0
1 2
kx
2
Substituting x
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
1 2 2
kA sin (ω0t + φ ) = U 0 sin 2 (ω 0t + φ )
2
U ( x) =
where U0 is the maximum potential energy (for x=A)
U0 =
1 2
kA
2
The average values over one oscillation period are calculated using the definition
t
1 2
f =
f (t )dt
t2 − t1 ∫t1
Thus:
T
U =
T
∫ Udt ∫ U
0
T
=
0
sin 2 (ω0t + φ )
0
∫ dt
T
1
1
= U 0 = kA2
2
2
0
and
K =
1
1
K 0 = kA2 .
2
2
In conclusion, the average values of kinetic and potential energy per oscillation cycle are
equal to
U = K =
1
E
2
Damped harmonic oscillator
This time, we introduce the additional force, which will dissipate the energy.
Fd = −bv = −bx
The equation of motion gains one more term:
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
mx + bx + kx = 0
Denote:
γ =
b
k
2
, ω0 =
2m
m
2
x + 2γ x + ω 0 x = 0
Use exponential trial function
x = exp( λ t )
x = λ exp(λ t )
x = λ 2 exp( λt )
2
exp( λt )[λ 2 + 2γλ + ω 0 ] = 0
since
exp(λ t ) ≠ 0
λ 2 + 2γλ + ω0 2 = 0
λ1 = −γ + γ 2 + ω 02
λ2 = −γ − γ 2 + ω 02
x (t ) = A1 exp( λ1t ) + A2 exp(λ2t )
2
2
x (t ) = exp( −γ t )[ A1 exp( +t γ 2 − ω0 ) + A2 exp( −t γ 2 − ω0 )]
Depending on the sign of the expression under the root, there are three
possible cases:
Underdamped
γ 2 − ω0 2 < 0
λ1 and λ2 are imaginary: oscillating solutions.
Overdamped
γ 2 − ω0 2 > 0
λ1 and λ2 are both real
Critically damped
γ 2 = ω02
Underdamped oscillator
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09-723 Proximal Probe Techniques
ω1 = ω0 2 − γ 2 =
Harmonic Oscillator Summary
k
b2
−
m 4m 2
x (t ) = exp( −γ t )[ A1 exp(iω1t ) + A2 exp(iω1t )]
which recalling Euler's formula becomes
x (t ) = exp( −γ t ) [i ( A1 − A2 )sin(ω1t ) + ( A1 + A2 ) cos(ω1t )]
Substitute
i ( A1 − A2 ) = B and A1 + A2 = C
x (t ) = exp( −γ t ) [ B sin(ω1t ) + C cos(ω1t )]
Introduce
A = B 2 + C 2 and tan(φ ) = −
C
B
x (t ) = A exp( −γ t ) cos(ω1t + φ )
Damped oscillator moves at "frequency" smaller than undamped:
γ2
ω1 = ω0 − γ = ω0 1 − 2
ω0
For small damping γ expand in binomial series and retain only the first two terms
γ2
γ2
+
.....)
≈
(1
−
)
ω1 = ω 0 (1 −
ω
0
2
2
2ω0
2ω 0
and for small damping γ ω0 and ω1 ≈ ω0
2
2
Critically damped
λ1 = λ2 = −γ
Solution
x (t ) = ( A1 + A2 ) exp( −γ t ) = ( B1 ) exp( −γ t )
This is not a general solution (it contains just one constant).
We can show that if
x (t ) = exp( −γ t ) than x (t ) = t exp( −γ t ) is also a solution.
Substitute
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
x (t ) = exp( −γ t ) − γ t exp( −γ t ) = (1 − γ t ) exp( −γ t )
x (t ) = (1 − γ t )( −γ ) exp( −γ t ) − γ exp( −γ t ) = (γ 2t − 2γ ) exp( −γ t )
2
[γ 2t − 2γ + 2γ (1 − γ t ) + ωo t ]exp( −γ t ) = 0
2
[(ωo − γ 2 )t ]exp( −γ t ) = 0 Always satisfied.
Thus the general solution for a critically damped oscillator is:
x (t ) = ( B1 + B2t ) exp( −γ t )
Overdamped
γ 2 − ω0 2 < ω 2
x (t ) = exp( −γ t )[ A1 exp(ω 2t ) + A2 exp( −ω 2t )]
Energy considerations
Total energy:
E (t ) = E (0) + W f
work performed by friction
Frictional force:
f = −bx = −bv
t
Wf = ∫
0
t
t
t
dx
fdx = ∫ f
dt = ∫ fvdt = ∫ −bv 2dt
dt
0
0
0
The rate of energy loss:
dE dW f
=
= −bv 2
dt
dt
E (t ) = K (t ) + U (t ) =
1 2 1 2
mx + kx
2
2
recall that
x (t ) = A exp( −γ t ) cos(ω1t + φ )
γ
cos(ω1t + φ )]
ω1
γ
Assume that the system is lightly damped ( 1 ), so we can neglect the second term.
ω1
x (t ) = −ω1 A exp( −γ t )[ sin(ω1t + φ ) +
Then:
E (t ) =
1 2
A exp( −2γ t )[m ω1 sin 2 (ω1t + φ ) + k cos 2 (ω1t + φ )]
2
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
2
2
Since light damping was assumed, ω1 ≈ ω 0 =
E (t ) =
k
and E(t) becomes
m
1 2
kA exp( −2γ t )
2
The initial energy
1
E0 = kA2
2
and thus
E (t ) = E0 exp( −2γ t )
(notice that the energy decays twice as fast as amplitude!)
The characteristic decay time (E decreases to E/e)
E0
= E0 exp( −2γτ )
e
2γτ = 1
1 2m m
τ=
=
=
2γ 2b b
Quality factor
Q = 2π
energy stored in the oscillator
energy dissipated in one time period
Define P = power loss = rate of energy dissipation
2π
since one time period T1 =
ω1
the denominator can be written as
2π
PT1 = P
ω1
Thus
Q = 2π
E
2π
P
ω1
1
- time necessary to complete 1 radian of oscillation, we can redefine Q as
ω1
energy stored in the oscillator
Q=
average energy dissipated per radian
Since
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
For a lightly damped oscillator Q can be calculated as follows:
E (t ) = E0 exp( −2γ t )
dE
= −2γ E
dt
Thus the energy dissipated in time ∆t will be equal to
dE
∆t = 2γ E ∆t
dT
1
If we choose ∆t =
(time necessary to complete 1 radian)
ω1
∆E =
Q=
ω
E
E
=
= 1
∆E 2γ E / ω1 2γ
For light damping ω1 ≈ ω0 and thus
ω
Q= 0
2γ
Forced (driven) damped harmonic oscillator
Net force
Fnet = Fs + F f + Fd
restoring
elastic force
dissipative
friction force
driving
force
where: Fs = − kx ; F f = −bx
Fnet = mx
mx + bx + kx = Fd
assume harmonic driving force
Fd = F0 cos(ω t + θ 0 )
mx + bx + kx = F0 cos(ω t + θ 0 )
This is an inhomogeneous 2-nd order linear differential equation. Its solution is the sum
of two parts, according to the following theorem:
If Xi is a particular solution of an inhomogeneous differential equation, and
Xn is a solution of a complementary homogeneous equation, then
X(t)=Xi(t)+Xh(t)
is a general solution.
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
From our previous considerations, the solution of the complementary homogeneous
equation is given by one of the three equivalent forms:
xh (t ) = exp( −γ t ) [ A1 exp( +iω1t ) + A2 exp( −iω1t )]
xh (t ) = exp( −γ t ) [ B sin(ω1t ) + C cos(ω1t )]
xh (t ) = Ah exp( −γ t ) cos(ω1t + φh )
For an inhomogeneous equation, let's postulate the following particular solution
xi (t ) = A cos(ω t ± φ )
and focus on the – sign solution.
xi (t ) = A cos(ω t − φ )
xi (t ) = − Aω sin(ω t − φ )
xi (t ) = − Aω 2 cos(ω t − φ )
Upon substitution
− mAω 2 cos(ω t − φ ) − bAω sin(ω t − φ ) + kA cos(ω t − φ ) = F0 cos(ω t )
(for simplicity we assumed that θ 0 = 0 ).
Recall that:
cos(ω t − φ ) = cos ω t cos φ + sin ω t sin φ
sin(ω t − φ ) = sin ω t cos φ − cos ω t sin φ
thus
− mAω 2 [cos ω t cos φ + sin ω t sin φ ]
−bAω [sin ω t cos φ − cos ω t sin φ ]
+ kA [cos ω t cos φ + sin ω t sin φ ] =
= F0 cos(ω t )
This can be regrouped as
cos ω t[− mAω 2 cos φ + bAω sin φ + kA cos φ ]
− sin ω t[− mAω 2 sin φ − bAω cos φ + kA sin φ ]
= F0 cos(ω t ) + 0 sin(ω t )
Since the cos(ω t ) and sin(ω t ) coefficients on both sides of the equation have to be equal,
we obtain the system of two equations:
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
F0
A
2
( k − mω )sin φ − bω cos φ = 0
( k − mω 2 ) cos φ + bω sin φ =
From the above it follows that:
tan φ =
b
ω
m
sin φ
bω
2γω
=
=
= 2
2
2
k
cos φ k − mω
− ω 2 ω0 − ω
m
thus
sin φ =
cos φ =
2γω
2
(ω 0 − ω 2 ) 2 + 4γ 2ω 2
ω02 − ω 2
2
(ω 0 − ω 2 ) 2 + 4γ 2ω 2
Substistute these back to obtain
F0 / m
A=
2
ω 02 − ω 2 + 4γ 2ω 2
(
)
Thus a particular solution of the inhomogeneous equation is:
xi (t ) =
F0 / m
(ω
2
0
−ω
2
)
2
+ 4γ ω
2
2
cos(ω t − φ )
where
φ = tan −1
2γω
ω0 2 − ω 2
The general solution is:
x (t ) = xh (t ) + xi (t ) = Ah exp( −γ t ) cos(ω1t + φh ) +
F0 / m
(ω
2
0
−ω
2
)
2
+ 4γ ω
transient term
steady state term
11
2
2
cos(ω t − φ )
09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
Amplitude resonance
The amplitude of the particular solution reaches maximum when the driving force is
equal to
ω = ω r = ω 02 − 2γ 2
On resonance, the phase shift
π
φ=
2
Far below resonance
ω ωr , φ → 0
Far above resonance
ω ωr , φ → π
Energy resonance
x (t ) = A cos(ω t − φ )
v = x (t ) = −ω A sin(ω t − φ )
1
1
K (t ) = mv 2 = mω 2 A2 sin 2 (ω t − φ )
2
2
1
1
U (t ) = kx 2 = kA2 cos2 (ω t − φ )
2
2
1 2
A [mω 2 sin 2 (ω t − φ ) + k cos2 (ω t − φ )]
2
Recall that per period:
1
cos 2 (ω t − φ ) = sin 2 (ω t − φ ) =
2
Substitute A (ω ) to K(t)
E (t ) = K (t ) + U ( t ) =
2
K (t ) =
F0 / m 2
1
mω 2
sin 2 (ω t − φ )
2
2
2
2 2
2
ω0 − ω + 4γ ω
(
)
2
K (t ) =
1 F0
ω2
4 m ω 2 − ω 2 2 + 4γ 2ω 2
0
(
)
Kinetic energy resonance:
d K (t )
= 0 for ω = ω 0
dω
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09-723 Proximal Probe Techniques
Harmonic Oscillator Summary
Potential energy resonance
since U (t ) =
1 2
2
kx , occurs at the same frequency as amplitude resonance ω = ω0 − 2γ 2
2
Total energy resonance. Resonance peak width and Q
E (t ) =
1 2 2
1
A mω sin 2 (ω t − φ ) + A2k cos2 (ω t − φ )
2
2
and after substituting A
2
E =
2
1 F0
ω 2 + ω0
4 m ω 2 − ω 2 2 + 4γ 2ω 2
0
(
)
For very weak damping γ ω0
ω 2 + ω 02 ≈ 2ω0 2
ω 2 − ω 02 = (ω − ω 0 )(ω + ω 0 ) ≈ 2ω0 (ω − ω0 )
2
E (ω ) =
1 F0
1
8 m (ω 0 − ω )2 + γ 2
Lorentzian
Maximum at ω = ω 0
2
1 F0 1
8 m γ2
The energy is equal ½ of its value at
(ω − ω 0 )2 = γ 2
or ω − ω 0 = ±γ
Resonance peak width at half height
E (ω 0 ) =
∆ω = 2γ
Recall from previous considerations that
ω
Q= 0
2γ
This is the basis for determining Q from the energy resonance peak width:
Q=
ω0
.
∆ω
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