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Chapter 13 periodic motion. Collapse of the Tacoma Narrows suspension bridge in America in 1940 (p 415). oscillation. SHM. Damped oscillation. Forced oscillation. kinematics. dynamics. Kinematics equation. Dynamic equation. Circle of reference. Energy. Superposition of shm. - PowerPoint PPT Presentation
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Chapter 13 periodic motion
Collapse of the Tacoma Narrowssuspension bridge in America in 1940(p 415)
SHMSHM
dynamicsdynamics kinematicskinematics
Dynamic equation
Dynamic equation
Circle of reference
Circle of reference
Kinematicsequation
Kinematicsequation
oscillationoscillation
EnergyEnergy Superposition of shm
Superposition of shm
Damped oscillation
Damped oscillation
Forcedoscillation
Forcedoscillation
periodic motion / oscillationrestoring forceamplitude
cycleperiod
frequencyangular frequencysimple harmonic motionharmonic oscillator
circle of referencephasor
phase anglesimple pendulum
Key terms:
physical pendulumDampingDamped oscillationCritical dampingoverdampingunderdampingdriving forceforced oscillationnatural angular frequencyresonancechaotic motionchaos
Ideal model:
A) spring mass system
1) dynamic equation
xkdt
xdm
xkamF
2
2
02
2
xm
k
dt
xd
)tcos(Ax
022
2
xdt
xd
m
k
2
T
§1 Dynamic equation§1 Dynamic equation
Small angle approximation sin
02
2
L
g
dt
d
B) The Simple Pendulum
2
2
2
2
sin
sin
dt
dLg
Lsdt
sdmmg
maF tt
)cos(0 t
l
g2
mg
d
O
I
mgdI
mgd
dt
d
dt
dImgd
I
2
2
2
2
2
sin
C) physical pendulum
Example: Tyrannosaurus rex and physical pendulum
the walking speed of tyrannosaurus rex can be estimated from its leg length L and its stride length s
Conclusion:Equation of SHM
0xdt
xd 22
2
Solution: )tcos(Ax
Solution:
2
'
g r
GmMF ,Rr M
R
rr
3
4
R34
MM
3
33
3
'
RO
gF r
M
rR
GmMF
3g
2
2
3 dt
rdmr
R
GmM 0r
R
GM
dt
rd32
2
2
Example:A particle dropped down a hole that extends from one side of the earth, through its center, to the other side. Prove that the motion is SHM and find the period.
mTkM 22 )
4(
Example:An astronaut on a body mass measuring device (BMMD),designed for use on orbiting space vehicles,its purpose is to allow astronauts to measure their mass in the ‘weight-less’ condition in earth orbit.
The BMMD is a spring mounted chair,if M is mass of astronaut and m effective mass of the BMMD,which also oscillate, show that
02
2
2
RIm
kydt
ydWe have
Example:the system is as follow,prove the block
will oscillate in SHM
)4(
)3(
)2()(
)1(
2
21
1
Ra
kxT
IRTT
maTmg
Solution:
o
y
Alternative solution
222
2
1
2
1
2
1mvIkymgy
Rv
(1)
(2)
Take a derivative of y with respect to x
0xdt
xd 22
2
Solution: )tcos(Ax
)tsin(Adt
dxv
)tcos(Adt
xda 2
2
2
2.1 Equation
§ 2 kinematic equation§ 2 kinematic equation
)tcos(Ax )tsin(Av
xtAa 22 )cos(
1) Amplitude (A): the maximum magnitude of displacement from equilibrium.
2) Angular frequency(): T
2
Spring-mass:m
k
Simple pendulum:l
g
is not angular frequency rather than velocity .it depends on the system
Caution:
A) Basic quantity:
2.2) the basic quantity——amplitude、 period,phase
In phase
2,1,0k
k212
2,1,0k
)1k2(12
Out of phase0
012 012
Lag in phaseAhead in phase
3) Phase angle ( = t+ ): the status of the object.
)tcos(Ax
sinAv,cosAx,then 00
2020 )
v(xA
0
0
x
vtg
0xdt
xd 22
2
Caution:Caution:
B) The formula to solve: A, ,
1) is predetermined by the system.
2) A and are determined by initial condition:
if t=0, x=x0, v=v0 ,
Is fixed by initial condition
Solution:
m
k ,5
4
100 2
202
0
vxA
m4.1
10
0 x
vtg
00 vand
4
)4
5cos(4.1 tx
An object of mass 4kg is attached to a spring of k=100N.m-1. The object is given an initial velocity of v0=-5m.s-1 and an initial displacement of x0=1. Find the kinematics equation
)cos( 0 tAx
Circle of reference method
)tcos(Ax
)tsin(Av
)tcos(Aa 2
Compare SHM with UCM
x(+), v(-), a(-)In first quadrant
Angle between OQ and axis-x
Phase
Angular VelocityAngular Frequency
ProjectionDisplacement
x
Radius
UCM
Amplitude
SHM
A
Ox
AA
Q
P
20
30
o xo x /3
6
3
1
x(m)
o t(s)
0.8
1
x(cm)
o t(s)
63
Example:Find the initial phase of the two oscillation
1
2
3
4
1
2
a b
10
-2
x (m)
t (s)
2
2
,40
22 2
att
btt
2
3
2
ba
t
tfor
0
4
3
1
)4(20
tt
a 0t
SHM: x-t graph,find 0 , a , b , and the angular frequency
Solution:
From circle of reference
§ 3 Energy in SHM§ 3 Energy in SHM
)(sin2
1)(sin
2
1
2
1 222222 tkAtAmmvKE
)(cos2
1
2
1 222 tkAkxPE
2222
2
1)(cos)(sin
2
1kAttkAPEKE
Kinetic energy:
Potential energy:
Total energy of the system:
Mk
X
O
Solution:
1v
,kA2
1Mv
2
1 21
21 ,v)mM(Mv 21
,kA2
1v)mM(
2
1 22
22
mM
MAA 12
Mm
k
Example:Spring mass system.particle move from left to right, amplitude A1. when the block passes through its equilibrium position, a lump of putty dropped vertically on to the block and stick to it. Find the kinetic equation suppose t=0 when putty dropped on to the block
)cos( 02 tAx
20
Example:A wheel is free to rotate about its fixed axle,a spring is attached to one of its spokes a distance r from axle.assuming that the wheel is a hoop of mass m and radius R,spring constant k. a) obtain the angular frequency of small oscillations of this system b) find angular frequency and how about r=R and r=0
)cos( 111 tAx )cos( 222 tAx
)cos(21 tAxxx
)cos(2 1221
2
2
2
1 AAAAA2211
2211
coscossinsin
AAAA
arctg
k212 max2121
2
2
2
1 2 AAAAAAAA
)12(12 kmin2121
2
2
2
1 2 AAAAAAAA
4.1 mathematics method
§ 4. Superposition of SHM§ 4. Superposition of SHM
2211
2211
coscos
sinsintg
AA
AA
(2-1)M1
A1
1
M
A2
xo
xx1 x2
ω
A
A2
2
M2
)cos(2 12212
22
1 AAAAA
B) circle of reference
x= x1+x2= Acos( t+ )
x1 =A1cos( t+1 )
x2 =A2cos( t+2 )
Solution:
x3 3O1A
2AA
Draw a circle of reference,
)tcos(Axxx 21
cmt )4
32cos(23
Example:x1=3cos(2t+)cm, x2=3cos(2t+/2)cm, find the superposition displacement of x1 and x2.
dt
dxbkx
dt
xdm
bvkxmaF
2
2
5.1 Phenomena
)cos()( 2
tAetx mbt
5.2 equation
If damping force is relative small
2
2
4m
b
m
k
2
2
4m
b
m
k
)(0
0
0
22
11
tata ececxgoverdampin
dampingcritical
ngunderdampi
§ 5 Damped Oscillations§ 5 Damped Oscillations
0.0 0.5 1.0 1.5 2.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
underdamping
overdamping
No oscillation
Amplitude decrease
0
)cos()( 2
tAetx mbt
Critical damping
tFdt
dxbkx
dt
xdm sin02
2
The steady-state solution is
2
220
2
0
)cos()(
mb
mF
A
tAtx
where 0=(k/m)½ is the natural frequency of the system.
The amplitude has a large increase near 0 - resonance
:§6 Forced Oscillations:§6 Forced Oscillations
drive an oscillator with a sinusoidally varying force:
Projectile motion with air resistance
(case study:p147)
1. Projectile motion with air resistance
A plane moves in constant velocity due eastward,a missile trace it,suppose at anytime the missile direct to plane,speed is u,u>v,draw the path of missile
h
x
(x,y)
(X,Y)(X0,h) v
u
2. Tracing problem
vy
O
u
y(0)=0, x(0)=0
Y=h,X(0)=0
h
x
(x,y)
(X,Y)
22 )()( xXyY
yY
udt
dy
22 )()( xXyY
xX
udt
dx
tnvnyny
tnvnxnx
y
x
)()()1(
)()()1(
22 )()( xXyY
xXu
dt
dxvx
22 )()( xXyY
yYu
dt
dyv y
tVnXnX )()1(
0YY
Solution:
Example:the orbits of satellites in the gravitational field
d
dr
r
v
d
dr
dt
drv
Er
GmM
mr
Lmv
vvv
Er
GmMmv
mrLvLmrv
Lvmr
r
er
r
e
2
22
222
2
22
1
2
1
v
rvv
r
me
ms
3. Planets trajectory
We get:222 22 LrmGmEmrr
Ldrd
e
cos1 e
Rr
reference : 《大学物理》吴锡珑 p 149
2r/r̂GMmUF
Solution1: Newton’s laws
)rr(mma)r(F 2r
0)r2r(mma)(F
GM/hp,
hyperbola1e
parabola1e
ellipse1e
cose1
pr 2
3. Planets trajectory
r/GMmU
Solution2: conservation of mechanical energy
E)r(Umv2
1 2 0)r2r(m
GM/hp
hyperbola0E
parabola0E
ellipse0E
cos))GM(
h(
mE2
11
pr
2
2
2