Transcript Document

Chapter 13 periodic motion
Collapse of the Tacoma Narrows
suspension bridge in America in 1940
(p 415)
oscillation
Damped
oscillation
SHM
dynamics
Dynamic
equation
Energy
kinematics
Kinematics
equation
Superposition
of shm
Circle of
reference
Forced
oscillation
Key terms:
periodic motion / oscillation
restoring force
amplitude
cycle
period
frequency
angular frequency
simple harmonic motion
harmonic oscillator
circle of reference
phasor
phase angle
simple pendulum
physical pendulum
Damping
Damped oscillation
Critical damping
overdamping
underdamping
driving force
forced oscillation
natural angular frequency
resonance
chaotic motion
chaos
§1 Dynamic equation
1)
dynamic equation
Ideal model:
A) spring mass system



 F  ma  kx
2

d x
m 2  kx
dt
k

m
d2x k
 x0
2
m
dt
x  A cos( t   )
d2x
2


x0
2
dt
T
2

B) The Simple Pendulum
Ft  mat
d 2s
 mg sin   m 2
dt
s  L
d 2
 g sin   L 2
dt
Small angle approximation sin
d 2 g
  0
2
L
dt
g
 
l
2
   0 cos(t   )
C) physical pendulum
  I
d 2
 mgd sin   I 2
dt
d 2
mgd


2
dt
I
mgd
2
 
I
O
 d
mg
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
d2x
2


x0
2
dt
Solution:
x  A cos( t   )
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.
Solution:
GmM '
Fg  
r2
Fg  
GmM
r
3
R
GmM
d 2r

rm 2
3
R
dt
r  R,
4 3
r3
M 
r   3 M
4
R
R 3 3
3
'
M
2
d 2 r GM
 3 r 0
2
dt
R
Fg
r
O
M
R
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
M  (k
4
2
)
T
m
2
Example:the system is as follow,prove the block
will oscillate in SHM
Solution:
mg  T1  ma
(1)
(T1  T2 ) R  I
( 2)
T2  kx
( 3)
a  R
( 4)
2
We have
d y
ky

0
2
I
dt
m  R2
o
y
Alternative solution
1 2 1 2 1 2
mgy  ky  I  mv
2
2
2
(1)
v  R
(2)
Take a derivative of y with respect to x
§ 2 kinematic equation
d2x
2


x0
2
dt
2.1 Equation
Solution:
x  A cos( t   )
v
dx
  A sin( t   )
dt
d2x
a  2   A2 cos( t   )
dt
x  A cos( t   )
v   A sin( t   )
a   A 2 cos(t   )   2 x
2.2) the basic quantity——amplitude、period,phase
A) Basic quantity:
1) Amplitude (A): the maximum magnitude of displacement
from equilibrium.
2) Angular frequency():
2

T
Spring-mass:
Simple pendulum:


k
m
g
l
Caution:

is not angular frequency rather than velocity .it
depends on the system
3) Phase angle (  = t+ ): the status of the object.
0
   2  1  0
   2  1  0
Ahead in phase
Lag in phase
   2   1  ( 2 k  1 )
   2   1  2 k
k  0 ,1 ,2 
k  0 ,1 ,2 
In phase
Out of phase
B) The formula to solve: A, , 
1) 
is predetermined by the system.
d2x
2


x0
2
dt
x  A cos( t   )
2) A and  are determined by initial condition:
if t=0, x=x0, v=v0 ,
then, x0  A cos  , v0   A sin 
v0 2
A x ( )

2
0
Caution:

v0
tg  
x0
Is fixed by initial condition
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
Solution:
k
 
m
x  A cos(t  0 )
v02
A x  2

100

 5,
4
v0
 tg  
1
x0
 x  1.4 cos(5t 
2
0
and v0  0

4
)
 

4
 1.4m
Circle of reference method
v   A sin( t   )
x  A cos( t   )
a  A2 cos( t   )
Q
Compare SHM with UCM
A
x
A
O
SHM
UCM
Amplitude
Radius
Displacement
P A

x
Projection

Angular Frequency
Angular Velocity

Phase
Angle between OQ and
axis-x
In first quadrant
x(+), v(-), a(-)

Example:Find the initial phase of the two oscillation
x(m)
x(cm)
0.8
6
3
o
1
o
t(s)
1
t(s)
2
1
6
1
3
o
4
/3
x
0  
o
3
x

2
0  

3
SHM: x-t graph,find 0 , a , b , and the angular
frequency
Solution:
x (m)
From circle of reference
2
2

 0   ,
4

3
a 
b 
2
2
0
a
1
t (s)
-2
t t a
for   t   0
    t
  a   0
 


t
t
b
  (  )
2
4
1
2
3

4

2 2
t t b
t 0
§ 3 Energy in SHM
Kinetic energy:
1 2 1
1 2 2
2 2
2
KE  mv  m A sin (t   )  kA sin (t   )
2
2
2
Potential energy:
1 2 1 2
PE  kx  kA cos 2 (t   )
2
2
Total energy of the system:


1 2
1 2
2
2
KE  PE  kA sin (t   )  cos (t   )  kA
2
2
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
1
1 2
2
Mv 1  kA1 ,
2
2
Solution:
Mv 1  ( M  m )v2 ,
1
1 2
2
( M  m )v 2  kA2 ,
2
2
k

mM
0  
x  A2 cos(t  0 )
A2  A1
M
M m

2
k
M
O
X
v1
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
§ 4. Superposition of SHM
4.1 mathematics method
x1  A1 cos(t  1 )
x2  A2 cos(t  2 )
x  x1  x2  A cos(t   )
A1 sin 1  A2 sin  2
  arctg
A1 cos 1  A2 cos  2
A  A1  A2  2 A1 A2 cos( 2  1 )
2
2
  2  1  2k
  2  1  ( 2k  1)
A  A1  A2  2 A1 A2  A1  A2  Amax
2
2
A  A1  A2  2 A1 A2  A1  A2  Amin
2
2
B) circle of reference
x1 =A1cos( t+1 )
x2 =A2cos( t+2 )
ω
M
M2
x= x1+x2= Acos( t+ )
A
A2
A  A  A2  2 A1 A2 cos( 2  1 )
2
1
2
2
o
A1 sin 1  A2 sin  2
tg  
A1 cos 1  A2 cos  2

A1
A2
(2-1)
M1
1
x2
x1
x
x
Example:x1=3cos(2t+)cm, x2=3cos(2t+/2)cm,
find the superposition displacement of x1 and x2.
Solution:
Draw a circle of reference,
x  x1  x2  A cos( t   )
3
 3 2 cos(2t 
)cm
4

A

A1
3

A2
x
O
3
§ 5 Damped Oscillations
5.1 Phenomena
 F  ma  kx  bv
5.2 equation
d 2x
dx
m 2  kx  b
dt
dt
If damping force is relative small
x ( t )  Ae
k
b2
 

m 4m 2
 bt
2m
k
b2

m 4m 2
cos( t   )
0
0
underdamping
critical damping
0
overdamping ( x  c1e  a1t  c2e  a2 t )
x ( t )  Ae
1.0
 bt
2m
cos( t   )
0.8
0.6
overdamping
0.4
No oscillation
0.2
  0
0.0
-0.2
Critical damping
-0.4
underdamping
-0.6
Amplitude decrease
-0.8
0.0
0.5
1.0
1.5
2.0
:§6 Forced Oscillations
drive an oscillator with a sinusoidally varying force:
d 2x
dx
m 2  kx  b  F0 sin t
dt
dt
The steady-state solution is
x(t )  A cos(t   )
F0
A

2

m

2 2
0
 b 


 m 
2
where 0=(k/m)½ is the natural frequency of the system.
The amplitude has a large increase near 0 - resonance
1. Projectile motion with air resistance
Projectile motion with air resistance
(case study:p147)
2. Tracing problem
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
(X,Y)
(X0,h)
v
(x,y)
h
u
x
x( n  1)  x( n)  v x ( n)  t
X (n  1)  X (n)  V  t
y( n  1)  y( n)  v y ( n)  t
Y  Y0
dx
Xx

udt
(Y  y )2  ( X  x )2
dx
Xx
vx 
u
dt
(Y  y )2  ( X  x )2
dy
Yy

udt
(Y  y )2  ( X  x )2
dy
Yy
vy 
u
dt
(Y  y )2  ( X  x )2
(X,Y)
y
v
(x,y)
u
O
h
x
y(0)=0, x(0)=0
Y=h,X(0)=0
3.
Planets trajectory
Example:the orbits of satellites in the gravitational field
v
v
Solution:



r  mv  L
mrv  L
v  L
1
GmMe
2
mv 
E
2
r
v  v 
2
2
v r2
2
r
mr
1
L
GmMe
2
mvr 

E
2
2
r
2mr
dr
dr
v dr
vr 


dt
d
r d
me
ms
vr
We get:
d 
Ldr
r 2 Emr2  2Gme m 2r  L2
R
r
1  e cos
reference :《大学物理》吴锡珑 p 149
3.
Planets trajectory
U  GMm / r

F  U  GMm r̂ / r 2
Solution1: Newton’s laws
F ( r )  ma r  m( r  r 2 )
F (  )  ma   m( r  2 r )  0
 e  1 ellipse 
p


2
r
e

1
parabola
,
p

h
/ GM


1  e cos  

e

1
hyperbola


Solution2: conservation of mechanical energy
1
mv 2  U ( r )  E
2
r
p
2E
h
1 1
(
m ( GM )2
p  h 2 / GM
m( r  2r )  0
 E  0 ellipse 


E

0
parabola


) cos   E  0 hyperbola 
2