Oscilla'ons The knowledge structure o  Harmonic mo'on when the restoring force is linear F
( x ) = −kx
o  Damped harmonic mo'on when F
( x ) = −kx − cx!
Three damping condi'ons: over-­‐, cri'cal-­‐ and under-­‐damping. Energy dissipa'on dE
= −cx! 2
dt
The quality factor Q =
2π ⋅ energy in the oscillator ω d
=
energy loss in one period
2γ
o  Phase Space: in mechanics ( x,
p ) or ( x, x! )
o  Forced harmonic mo'on and resonance •  Idealized case F ( x ) = −kx + F0 cos (ω t )
•  More realis'cally F ( x ) = −kx − cx! + F0 eiωt
Harmonic mo'on when F(x) = −kx
Two idealized cases. x : displacement from equilibrium. F ( x ) = −kx Hook’s law. x = A cos (ω 0 t + φ0 ), ω 0 ≡
φ0
k
m
angular frequency 2π 1
=
ω 0 f0 frequency v = x! = −ω 0 Asin (ω 0 t + φ0 )
period T0
T0
t
−kx = m!!
x → m!!
x + kx = 0
x
x
A
y
t
ω 0t t = 0
=
φ0
x
x = A cos (ω 0 t + φ0 )
θ
displacement from equilibrium. l
θ!
θ!!
v = s! = , a = v! =
l
l
F ( x ) = −mgsin θ = ma → −gθ = a, when θ is small
s=
l
g
θ!! + θ = 0
l
θ = A cos (ω 0 t + φ0 ), ω 0 ≡
g
l
Both s and θ here are general coordinates. Harmonic mo'on when F(x) = −kx
Energy considera'ons. 2
1 2 1
mv = m (−ω 0 Asin (ω 0 t + φ0 ))
2
2
x
x
2
1 2 1
V
=
−W
=
−
F
dx
=
−
−kx
dx
=
kx
=
k
A
cos
ω
t
+
φ
( 0 0 ))
(
∫0
∫0 ( ) 2
x
2
1
1
E = T + V = mω 02 A 2 sin 2 (ω 0 t + φ0 ) + kA 2 cos2 (ω 0 t + φ0 )
2
2
1
1 2
= kA 2 = mvmax
2
2
T=
1 2 m !2
mv = 2 θ
2
2l
V = mg (l − l cosθ )
T=
l
E = T +V =
m !2
m !2
θ
+
mgl
1−
cos
θ
=
θ = mgl (1− cosθ max )
(
)
2
2 max
2l
2l
Ask the class to prove the above. Damped Harmonic mo'on when F(x) = −kx−cv
!
F ( x ) = −kx − cx,
− cx! is the retarding force, like that comes from fric'on or air resistance m!!
x + kx + cx! = 0 → !!
x + 2γ x! + ω 02 x = 0, with γ ≡
let D ≡
c
2m
d
, a differential operator
dt
!!
x + 2γ x! + ω 02 x = 0 → ( D + γ − q ) ( D + γ + q ) x = 0, with q ≡ γ 2 − ω 02
since the order of ( D + γ − q ) and ( D + γ + q ) can be changed, the only
possibility for the equation to hold is that
dx
− γ −q t
→ ( D + γ − q ) x = 0, or
= − (γ − q ) x → x = A1e ( ) , and
dt
dx
− γ +q t
→ ( D + γ + q ) x = 0, or
= − (γ + q ) x → x = A2 e ( )
dt
x = A1e
q ≡ γ 2 − ω 02
q real > 0
q real = 0
q imaginary
−(γ −q)t
+ A2 e
−(γ +q)t
can have three cases over-­‐damping cri'cal-­‐damping under-­‐damping Over-­‐ and cri'cal damping cases
over-­‐damping q real > 0, γ > q
→ (γ − q ) and (γ + q ) are real and positive
x = A1e
−(γ −q)t
+ A2 e
−(γ +q)t
exponential decay with two constants
cri'cal-­‐damping q = 0 → (D + γ )(D + γ ) x = 0
u ≡ (D + γ ) x → (D + γ ) u = 0
( D + γ ) u = 0 → u = Ae−γ t
u ≡ ( D + γ ) x → Ae−γ t = ( D + γ ) x → A = eγ t ( D + γ ) x = D ( xeγ t )
→ xeγ t = At + B
x = Ate−γ t + Be−γ t
In the case when cri'cal-­‐damping condi'on is met, the displacement (x) returns to equilibrium with the shortest 'me, faster than over-­‐damping. Under-­‐damping
under-­‐damping q imaginary, i.e., γ 2 − ω 02 < 0
q ≡ γ 2 − ω 02 = ω 02 − γ 2 −1 = ω 02 − γ 2 i ≡ ω d i, ω d ≡ ω 02 − γ 2
x = A1e
−(γ −q)t
+ A2 e
−(γ +q)t
→ x = A1e
−(γ −ω d i)t
+ A2 e
−(γ +ω d i)t
x = e−γ t ( A1eiωd t + A2 e−iωd t )
decay oscilla'ng But x as a coordinate of a par'cle with mass m has to be real, i.e., x*=x
x * = e−γ t ( A1*e−iωd t + A2*eiωd t ) = x = e−γ t ( A1eiωd t + A2 e−iωd t )
→ A1* = A2 and A2* = A1. assign C=A1* = A2 then C* = A1 = A2*
A
→ x = e−γ t (C *eiωd t + Ce−iωd t ), define C ≡ e−iφ0
2
x = e−γ t A cos (ω d t + φ0 )
decay oscilla'ng w/ a period Td =
2π
2π
=
ωd
ω 02 − γ 2
Energy considera'ons and the Q factor
E=
1 2 1 2
mx! + kx
2
2
dE
= mx!!!
x + kxx! = ( m!!
x + kx ) x! =
dt
energy dissipa'ng rate 0, as m!!
x + kx = 0, if harmonic, not dampted
! if harmonic dampted
−cx! 2 , as m!!
x + kx = −cx,
The quality factor is defined as Q =
2π ⋅ energy in the oscillator
energy loss in one period
Now let’s get the expression for Q when the oscilla'on is weakly damped. Why not also for other damping situa'on? Energy considera'ons and the Q factor
When weakly damped x = e−γ t A cos (ω d t + φ0 )
x! = −Ae−γ t [γ cosθ + ω d sin θ ], θ ≡ ω d t + φ0
E! = −cx! 2 = −cA 2 e−2γ t (γ 2 cos2 θ + ω d2 sin 2 θ + 2γω d cosθ sin θ )
Now find the energy loss in one period: Td
ΔE =
∫ E! dt,
use Td =
0
1
=
ωd
2π
ωd
2π
∫ E! dθ
2π
0
2
cA −2γ t
=−
e
ωd
2π
2
2π
∫ (γ
0
2
cos θ + ω d sin θ + 2γω d cosθ sin θ ) dθ
2
2
2
cA 2 −2γ t
cA 2 −2γ t 2
2
2
=−
e π (γ + ω d ) = −
e πω 0 , ← ω d2 ≡ ω 02 − γ 2
ωd
ωd
π
c
, ←γ ≡
ωd
2m
1
2π
1
2π
= − mA 2ω 02 e−2γ t
2γ , ← τ ≡ , and Td =
2
ωd
2γ
ωd
= −cA 2ω 02 e−2γ t
t
− T
1
= − mA 2ω 02 e τ d
2
τ
∫ cos θ dθ = ∫ sin
0
0
2π
∫ cosθ sinθ dθ = 0
0
2
θ dθ = π
Energy considera'ons and the Q factor
t
− T
1
ΔE = mA 2ω 02 e τ d
2
τ
t
−
1
1
k
2 2 τ
E = mA ω 0 e , harmonic case E = kA 2 , ω 02 ≡
2
2
m
1
2π E 2πτ 2π ⋅ 2γ ω d
Q≡
=
=
=
2π
ΔE
Td
2γ
ωd
Phase space
In mechanics, this is the collec'on of ( x,
p ) or ( x,
x! ) of the system. When F is a constant, we have dv F
= = a = constant
dt m
x! ≡ v = v0 + at, v0 ≡ v t=0
!!
x≡
1
x = x0 + v0 t + at 2
2
2a ( x − x0 ) = v 2 − v02
Plot ( x,
x ! ) with a = 1,
ß eliminate t from the above two equa'ons v
x0 = v0 = 0
One can see that with t → ∞
both x and v → ∞
This phase space is not contained. x
Phase space
Harmonic mo'on, no damping: x = A cos (ω 0 t + φ0 ), x! = −ω 0 Asin (ω 0 t + φ0 )
x2
x! 2
Eliminate t, one has + 2 2 = 1 , an ellipse. 2
A ω0 A
−γ t
−γ t
Under-­‐damping: x = e A cos (ω d t + φ 0 ), x! = −e A "#γ cos (ω d t + φ 0 ) + ω d sin (ω d t + φ 0 )$%
It will take some efforts to eliminate t. Let’s see the phase space plots first. π
Plot with A
=
1,
ω
0 = , φ 0 = 0.1
for the simple harmonic mo'on and 10
π
γ = 0.05,
ω d = A = 1,
, φ0 = 0.1 for the under damping case. 10
Phase space
harmonic no damping Harmonic under damping x
x
t
v
t
v
x
x
Phase space
Cri'cal damping: Over damping: x = ( At + B ) e−γ t , x! = −γ x + Ae−γ t
x = A1e
−(γ −q)t
+ A2 e
−(γ +q)t
, x! = −γ x + qe−γ t ( A1e qt − A2 e−qt )
To compare damping speed, let’s assume the same γ and x0 = 1, x! 0 = 0
B = 1, A = 0.5, x! = −γ x + 0.5e−γ t Decay faster than For under damping: A = 1.14, A = −0.14, q > 0,
1
2
For cri'cal damping: − γ −q t
x! = −γ x + qe−γ t (1.14e qt + 0.14e−qt ) ≅ −γ x + qe ( )
Plot with x 0 = 1,
x! 0 =
0,
and
γ =
0.5,
ω
0 =
0.314
for both cases. Phase space
Cri'cal damping Over damping x
x
t
v
t
v
x
x
Forced oscilla'on with F(x,t) = −kx−cv+F0cosωt
The equa'on of mo'on is F ( x, t ) = −kx − cx! + F0 cos (ω t ) = m!!
x or m!!
x + cx! + kx = F0 cos (ω t )
Let’s start without damping m!!
x + kx = F0 cos (ω t )
Assume a solu'on x = A
cos
( ω
t − φ
) , here A is the amplitude, φ is the phase difference between ω
0 and ω
. With this solu'on we have −mω
or 2
A cos (ω t − φ ) + kA cos (ω t − φ ) = F0 cos (ω t )
(k − mω ) A cos (ωt − φ ) = F cos (ωt )
2
0
For the equa'on to hold for any values of t: cos (ω t − φ ) = ±cos (ω t ),
or φ = 0, or π
φ = 0, A =
F0
F0 m
=
, A > 0 (amplitude), ω < ω 0
2
2
2
k − mω
ω0 − ω
φ = π, A =
F0
F0 m
=
, ω > ω0
mω 2 − k ω 2 − ω 02
ω = ω 0 → A = ∞ resonance A
φ
Plot A (ω )
with F0
= 1, m = 1
π
ω 0 = 1.5
ω0
0
ω
Forced oscilla'on with F(x,t) = −kx−cv+F0cosωt
Employing Euler’s iden'ty, e iθ =
cos
θ
+
isin
θ
, we express the driving force as F
0 e i ω t . Now with damping the equa'on of mo'on becomes m!!
x + cx! + kx = F0 eiωt
When the system reaches steady state, Assume a solu'on x =
Ae
, then x! = iω Ae
i(ω t−φ )
i(ω t−φ )
and !!
x = −ω 2 Ae (
Subs'tute back to the equa'on of mo'on, we have: −mω 2 Ae (
i ω t−φ )
+ icω Ae (
i ω t−φ )
+ kAe (
i ω t−φ )
= F0 eiωt ← cancel eiωt and move e−iφ
−mω 2 A + icω A + kA = F0 eiφ = F0 ( cos φ + isin φ )
A ( k − mω 2 ) = F0 cos φ
ß equate the real and imagery parts of the equa'on cω A = F0 sin φ
tan φ =
A=
cω
2γω
k
c
2
=
,
ω
≡
,
γ
≡
0
k − mω 2 ω 02 − ω 2
m
2m
F0
2
(k − mω 2 ) + c2ω 2
=
F0 m
2
(ω02 − ω 2 ) + 4γ 2ω 2
i ω t−φ )
Forced oscilla'on with F(x,t) = −kx−cv+F0cosωt
To find the resonant resonance from F0
A=
2
=
(k − mω 2 ) + c2ω 2
F0 m
2
(ω02 − ω 2 ) + 4γ 2ω 2
3
−
dA
F0 " 2
2 2
2 2% 2 "
We set =−
ω 0 − ω ) + 4γ ω ' #2 (ω 02 − ω 2 ) 2ω + 8γ 2ω %& = 0
(
$
&
dω
2m #
ω 02 − ω 2 + 2γ 2 = 0, or the resonant angular frequecy
ω r2 ≡ ω 02 − 2γ 2 = ω d2 − γ 2 ← ω d2 ≡ ω 02 − γ 2
F0 m
F0
At the resonance Amax =
=
2
2
2
2
2 2
2
2
2 2
2
γ
m
ω
−
γ
0
ω
−
ω
+
2
γ
+
4
γ
ω
−
2
γ
( 0 0
)
( 0
)
Amax ≅
F0
, when weak damping ω 0 >> γ
2γ mω 0
Forced oscilla'on with F(x,t) = −kx−cv+F0cosωt
A
φ
Plot A (ω ) and φ
π
with F0
= 1, m = 1
ω 0 = 1.5, γ = 0.02
ω
ω0
Check with Amax ≅
=
2γ
F0
2γ mω 0
1
= 16.7 (unit) 2 × 0.02 ×1×1.5
The quality or the sharpness of the peak is ωd ω0
≅
2γ 2γ
when weak damping ω 0 >> γ
Q=
ω0
Forced oscilla'on with F(x,t) = −kx−cv+F0cosωt
ωd ω0
≅ , when weak damping ω 0 >> γ
2γ 2γ
2
2
2
2
When damping is weak, near the resonance ω ≅ ω r , and ω r = ω 0 − 2γ ≅ ω 0 , so ω ≅ ω 0
Understand the Q factor: Q =
From A =
F0 m
=
2
(ω02 − ω 2 ) + 4γ 2ω 2
F0 m
2
(ω 0 + ω ) (ω 0 − ω )
2
+ 4γ 2ω 2
F0 m
≅
2
(2ω 0 ) (ω 0 − ω )
2mω 0γ
φ
+ 4γ 2ω 02
2
+γ 2
F0γ
=
A
2
(ω 0 − ω )
Amaxγ
=
(ω 0 − ω )
A2 =
2
+γ 2
, as Amax ≅
1 2
Amax , when ω = ω 0 ± γ
2
Half energy points as E ∝ A 2
2γ
More oYen Δf = 1
fr
Q
Larger Q, smaller frequency spread. ω0
ω
F0
2γ mω 0
Example
assume the energy loss during one period is ΔE
. We have Q ≡
The ra'o of two successive maxima in displacement is i−1
Amax
E − ΔE
ΔE
2π
4πγ
=
=
1−
=
1−
=
1−
i
Amax
E
E
Q
ωd
a constant 2π E ω d
=
ΔE
2γ