Oscillations - University of Saskatchewanphysics.usask.ca/~hirose/ep225/PPT2 Oscillations.pdf ·...
Transcript of Oscillations - University of Saskatchewanphysics.usask.ca/~hirose/ep225/PPT2 Oscillations.pdf ·...
OscillationsMechanical• Mass-spring system• 2nd order differential eq.• Energy tossing between mass (kinetic energy) and
spring (potential energy)• Effect of friction, critical damping (shock absorber)• Simple pendulum, physical pendulum (sweet points)
• Tacoma Narrow Bridge: Example of Torsional Oscillation http://www.youtube.com/watch?v=3mclp9QmCGs
Mass-Spring System
( ) ( )
( )
2
2
2
2
by the spring: (N): spring force constant in N/m
Equation of motion for the mas
Restoring force
mass accelerations:
0
Try = co
=
F kxk
dv d dx d xM M M kxdt dt dt dt
d x t k x tdt M
x
F
t A
= −
= = = −
+ =
×
22
2
2 2
s .
cos sin , cos cos
Subsitution to the diff. eq. yields
cos cos 0
(rad/sec) oscillation angular frequency
t
d dA t A t A t A tdt dt
kMA t kA tM
kM
ω
ω ω ω ω ω ω
ω ω ω ω
ω
= − = −
− + = → =
=
Example: A spring 2 m longhanging from the ceiling elongates by 30 cm when a mass of 1.5 kg is attached to the end. If the mass is pulled down another5 cm and released, what is the oscillation frequen
2
cy?
The spring constant is from1.5 9.8 N 49 N/m
0.3 m49 N/m= 5.7 rad/s, 0.91 Hz1.5 kg
9.8 m/sAlternativley, 5.7 rad/s0.3 m
Example: Discuss energy exchange in the abov
F MgF k l kl l
k fM
Mg gM l l
ω
ω
×= ∆ → = = = =
∆ ∆
= = =
= = = =∆ ∆
( ) ( )
( )
2 2 2 20 0
22 2 2 2 2
0 0
2 2 20
e Example.Initial potential energy is1 1 149 0.05 0.061 J. At time t, cos2 2 2Kinetic energy of the mass is
1 1 1sin sin2 2 2
1 1The total enengy: cos sin2 2
k x kx t
dxM M x t kx tdt
kx t t k
ω
ω ω ω
ω ω
= × × =
= =
+ = 20 =const.x
Energy Tossing
( )
( )
0
22 2 2 2 2 2
0 0
2 2 2 20 0
Potential energy (red) + kinetic energy (green) = constant (black)If cos ,
1 1 1 1cos sin2 2 2 2
1 1cos sin2 2
x t x t
dxkx M kx t M x tdt
kx t t kx
ω
ω ω ω
ω ω
=
+ = +
= + =
0
0.1
0.2
0.3
0.4
0.5
0.5 1 1.5 2 2.5 3x
2 2
2 2
2
2
2 2
2 2
2
2
Energy conservation oscillation equation 1 1 const.2 2
1 1 0 02 2
,
Then 0 0
Effect of Friction
+ + 0 damping (dissipa
Mv kx
d dv dxMv kx Mv kxdt dt dt
dx dv d xvdt dt dt
d x d x kM kx xdt dt M
d x dxM f kxdt dt
→
+ =
+ = → + =
= =
+ = → + =
= →
2
2
tion of energy)
Compare with LCR circuit1+ + 0 d q R dq q
dt L dt LC=
2 4 6 8 10 12 14 16 18 20
-1.0
-0.5
0.0
0.5
1.0
t
x(t)
1 2 3 4 5 6 7 8 9 10
-1.0
-0.5
0.0
0.5
1.0
t
x(t)
0 1 2 3 4 5 6 7 8 9 100.0
0.5
1.0
t
x(t)( )
2 20 0
20
0
0
0
0
2 20
Damped oscillation
'' ' 0
" 2 ' 0
, 2
From top( ) 0.1( ) =0.5( )
Solution
cos sint
f kx x xM M
x x xf kM M
abc
x t x e t tγ
γ ω ω
γ ω
γ ωγ ωγ ω
γω ωω
ω ω γ
−
+ + =
+ + =
= =
=
=
= +
= −
Two springs
( )
1 1 2 22
1 2 1 12
21
1 1 122
21 2
1 1 121 2 1 2
Common force
=
1
1 1 1, effeff
F k x k xdM x x k xdtd kM x k xdt k
d k kM x x k xdt k k k k k
= − = −
+ −
+ = −
= − = − = ++
( )1 1 2 2 1 2 1
1 2
In this case, the force is additive,
eff
F k x k x k k xk k k= − − = − +
= +
Pendulum
2 2
2 2
Pendulm of length and mass Torque about the pivot sin (small )
0 0
Oscilaltion frequency independent of mass
Example: If 1 m,
9.8 rad/s =3.13 rad/s
0
L MMgL MgL
d d gI MgLdt dt L
gL
L
gL
f
θ θ θ
θ θθ θ
ω
ω
− −
+ = → + =
=
=
= =
=
.498 Hz
The oscillation frequency can be used for measuring .g
Nonlinear Oscillation
2
2
3 2
22
2
The pendulum equation
sin 0
is . The sine function can be expanded for small as1 1sin 16 6
1Then 1 0 and the oscillation frequency is modified a6
nonlinear
d gdt L
d gdt L
θ θ
θ
θ θ θ θ θ
θ θ θ
+ =
− + ⋅ ⋅ ⋅ = −
+ − =
( )
( )
20
0
20 0
s
1 / 6
where is the amplitude which reduces the oscillation frequency.
More advanced calsulation yields 1 / 8
gL
gL
ω θ θ
θ
ω θ θ
= −
= −
http://physics.usask.ca/~hirose/ep225/animation/pendulum/anim-pendulum.htm
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
-0.10-0.050.000.050.10
t
theta
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
-1.4-1.2-1.0-0.8-0.6-0.4-0.20.00.20.40.60.81.01.21.4
x
y
Numerical solutions small and large amplitude
Oscillation involving moment of inertia
2 2 2 2
2 2
A cylinder attached to a spring rolls on floor. Energy conservation is1 1 1 1const. where (moment of inertia of cylinder)2 2 2 2
3 1and . Then const.4 2
Oscillation frequency is =
Mv I kx I Ma
v a Mv kx
ω
ω
ω
+ + = =
= + =
23
kM
Stick pivoted at its end
2
2
2
2
2
A stick of length is pivoted at its end. The moment of inertia1about the end is . The restoring torque for small is3
1 1sin2 2
12
1 02
3 =2 2
If 1 m, =
L
I ML
LMg LMg
dI LMgdtdI LMgdt
LMg gI L
L
θ
θ θ
θ θ
θ θ
ω
ω
=
− −
= −
+ =
=
=
3.83 rad/sec, 0.61 Hzf =
Mg
-MgθL
θ
Circular loop
2 2 2 2
The moment of inertial of a circular loop about a point on itself is
+ =2Then the oascilaltion frequency
= 2
If fact, an incomplete loop also has the same oscillation frequency. (Ho
CMI I Ma Ma Ma Ma
ga
ω
= + =
mework)
pivot
Mg
θ
θ
a
a
CM
( )
( )
2/ 2
0
22 / 2
0
2 2/ 2
0
22
Half CircleThe moment of inertia about the pivot
2 2sin2
8 sin2
4 1 cos 4 12
2CM is at 1 from the pivot. Then
1 2 /
4 12
MI a ada
Ma d
Ma Mad
a
agMa
π
π
π
θ θπ
θ θπ
πθ θπ π
ππ
ωπ
π
=
=
= − = −
− −
=−
∫
∫
∫
2ga
=
θ/2θ/2
θ/2sin2a
a
θ
θ0
a(cosθ−cosθ0)
Sloshing Oscillation of Water in Pan, Lake, Bay (Seiche)
22
2
The center of mass follows the trajectory
6
where is the depth of water and is the channel length. The oscillation frequency is
= 12
Hy xaH a
H ga
ω
=
Sloshing Oscillation (cont.)
( )Let the water displacement at the edge be . The CM of the rectangle BDEC is at
10, ( )2
The CM of the triangle ABC is at2 1,
6 3 3Then the CM of the entire water is at
1 1 ,6 6
h t
x y H h
ax y h H h H h
a ax ah h yaH H
= = −
= = + − = −
= × = ( )2
2
22
2
2
1 1 12 3 2 6
Eliminating , we find 6
12
A ball rolling along a parabolic curve oscillates at
2
a H hH h ah H haH HHh y xa
Hga
y Ax
Ag
ω
ω
= − + − = +
=
=
=
= y = Ax2
Sweet Point
2
Sweet point of baseball bat, tennis racket, etc. canbe determined as follows. The grip point can be regardedas a pivot for physical pendulum. Its oscillation frequency is
=
Simple pendulum of
CM
MgdI Md
ω+
2
. Force actin
length : '=
g at the gri
. From '
p (pivot) becomes minimum if the ball is hit at thesweet point. (H
,
This determines the position of s
omework
weet poi
)
nt
CM CM
gLL
I Md IL dMd Md
ω ω ω=
+= = +
Electromagnetic Oscillations
2
2
2
2
Charged capacitor is suddenly connected to aninductor. Voltage balance is
10 But 0
The charge oscillates at1
If a resistance is inlcuded, 10
q di dq d qL i qC dt dt dt LC
LC
q di d q R dqL Ri qC dt dt L dt LC
ω
+ = = → + =
=
+ + = → + +
( ) -10
0
Weakly damped solution is
cos , (s )2
t Rq t q e tL
γ ω γ−
=
= =
( )
( )
22
2
2220
0
2 22 2 2 2 20
0
1 1The capacitor stores energy of 2 2
1The inductor stores energy of 2
1 1For cos , cos2 2
1 1 1 1sin sin2 2 2 2Total energy is constant as in spring-ma
qCVC
Li
qqq t q t tC C
dq t qLi L L q t tdt C
ω ω
ω ω ω
=
= =
= = =
ss mechanical oscillation.
The capacitor and inductor exchange energy periodically.
0
0.5
1
1.5
2
y
5e-05 0.0001 0.00015 0.0002 0.00025 0.0003t
33 -1
6
Example: 5 F capacitor charged to 1 kV is suddenlyconnected to a 2 H inductor. Assuming a 5 m circuit resistance, describe how the initial energy decays.
5 10 1.25 10 s is smaller t2 2 2 10RL
µµ
−
−
Ω
×= = ×
× ×
( ) ( )
( ) ( )
5 -1
3 1250 50
20
2500 2 5
han
1 =3.16 10 s . Weakly damped.
cos 10 cos 3.16 10
1Intial energy 2.5 J. Then2
2.5 cos 3.16 10
t t
t
LCV t V e t e t
CV
E t e t
γ
ω
ω− −
−
= ×
= = ×
=
= ×
Forced Oscillation
( )
0
02
2
0
If an circuit is driven by an oscillator with voltage cos , the amplitude of the current is
which exhibits a resonance1
1when .
Example: 1 mF, 1 mH, 0.1 1 V
1
10
LCRV t
Vi
L RC
LCC L R
V
i
ω
ωω
ω
ω
= − +
=
= = = Ω=
=2
33
1 0.0110
ωω
−−
− +
0
2
4
6
8
10
y
1000 2000 3000 4000 5000x