Transcript Ch17 Oscillations

Chapter 17 Oscillations
17-1 Oscillating Systems
Each day we encounter many kinds of oscillatory
motion, such as swinging pendulum of a clock, a
person bouncing on a trampoline, a vibrating guitar
string, and a mass on a spring.
They have common properties:
1. The particle oscillates back and forth about a
equilibrium position. The time necessary for one
complete cycle (a complete repetition of the motion) is
called the period T.
2. No matter what the direction of the displacement,
the force always acts in a direction to restore the
system to its equilibrium position. Such a force is
called a “restoring force(恢复力)”.
3. The number of cycles per unit time is called the
“frequency” f.
1
f 
(17-1)
T
Unit: period (s)
frequency(Hz, SI unit), 1 Hz = 1 cycle/s
4. The magnitude of the maximum displacement
from equilibrium is called the amplitude of the motion.
17-2/3 The simple harmonic
oscillator and its motion
1. Simple harmonic motion
An oscillating system which can be described in
terms of sine and cosine functions is called a
“simple harmonic oscillator” and its motion is called
“simple harmonic motion”.
2. Equation of motion of the simple harmonic oscillator
Fig 17-5 shows a simple harmonic oscillator,
consisting of a spring of force constant K acting on
a body of mass m that slides on a frictionless horizontal
surface. The body moves in x direction.
Fig 17-5
origin is chosen at here
m

o
m
F

o
F
 kx
d 2x
 kx  m 2
dt
d 2x k
 x0
2
m
dt
x
x
Relaxed state
2
d x
ax  2
dt
x
(17-4)
Eq(17-4) is called the “equation of motion of the
simple harmonic oscillator”. It is the basis of many
complex oscillator problems.
3. Find the solution of Eq. (17-4)
Rewrite Eq(17-4) as
d 2x
k
 ( ) x
2
m
dt
(17-5)
We write a tentative solution to Eq(17-5) as
(17-6)
x  x cos(t   )
m
We differentiate Eq(17-6) twice with respect to the
Time.
2
d x
2



xm cos(t   )
2
dt
Putting this into Eq(17-5) we obtain
k
2
  x m cos(t   )   x m cos(t   )
m
Therefore, if we choose the constant  such that
k
2 
(17-7)
m
Eq(17-6) is in fact a solution of the equation of
motion of a simple harmonic oscillator.
a)  :
If we increase the time by 2 in Eq(17-6), then
x  xm [cos  (t 
2


)   ]  xm cos(t   )
Therefore 2 is the period of the motion T.

2
m
T
 2
(17-8)

k
1
1 k
f  
(17-9)
T 2 m
The quantity  is called the angular frequency.
  2f
b) x m :
x m is the maximum value of displacement. We call it
the amplitude of the motion.
c) t   and  :
The quantity t   is called phase of the motion.
 is called “phase constant （常相位)”.
x m and  are determined by the initial position and
velocity of the particle.  is determined by the system.
How to understand

?
x  xm cos(t   )
xm
o
 xm
x
x t 图
T
t
 0


2
 
How to compare the phases of two SHOs with
same  ?
x1  xm1 cos(t  1 )
x  x
2
m2
cos(t  2 )
  (t   2 )  (t  1 )
   2  1
为其它
  0同相
   π 反相
x
x
x
o
o
o
t
t
超前
落后
t
Fig 17-6 shows several simple harmonic motions.
x t
Fig 17-6
a
(a)
1
x t
(b)
1
0.5
0.5
t
2
4
6
t
8
1
-0.5
-0.5
-1
-1
x t
(c)
1
0.5
t
1
-0.5
-1
2
3
4
5
6
2
3
(a) same: x m , 
different: 
(b) same:  , 
different: x m
(c) same: x m , 
different: 
4
5
6
d). Displacement, velocity, and acceleration
Displacement x  xm cos(t   )

dx
Velocity v x 
 x m sin( t   )  xm cos(t    )
2
dt
2
d x
Acceleration a x  2   2 x m cos(t   )
dt
  2 xm cos(t     )
(17-11)
When the displacement is a maximum in either
direction, the speed is zero, because the velocity
must now change its direction.
x  xm cos(t   )
T
2π
取

 0
v   xm sin( t   )
xm
o
 xm
xm
2
x

a   xm cos(t   ) m
2
 xm cos(t    π)
o
 xm 2
x t图
T
v
a
t
v t 图
T
o
π
 xm cos(t    )  x 
m
2
2
x
t
a  t图
T
t
17-4 Energy in simple harmonic
motion
1.The potential energy
U 
1 2 1
2
kx  kxm cos 2 (t   )
2
2
(17-12)
2.The kinetic energy
1
1
2
K  mv 2  m 2 xm sin 2 (t   )
2
2
1
2
(17-13)
 kxm sin 2 (t   )
2
 0
U(t)
1
0.8
0.6
0.4
0.2
K(t)
1
2
3
4
T/2
Fig 17-8(a)
v   xm sin( t   )
5
6
T
• Fig17-8(a), both potential and kinetic energies
oscillate with time t and vary between zero and
maximum value of 1 kxm .2
2
• Both U and K vary with twice the frequency of the
displacement and velocity.
3. The total mechanical energy E is
1
2
E  K  U  kxm
2
E
(17-14)
U ( x) 
U(x)
K ( x)  E  U ( x)
K(x)
 xm
xm
Fig 17-8 (b)
1 2
kx
2
x
At the maximum displacement K  0 , U  1 kxm. 2
2
At the equilibrium position U  0 , K  1 kxm. 2
2
Eq(17-14) can be written quite generally as
1
1 2 1
2
2
K  U  mv x  kx  kxm
2
2
2
k
2
2
then v x  ( x m  x )
m
k
2
or
vx  
( xm  x 2 )
m
(17-15)
2
(17-16)
Sample problem 17-2
Fig 17-5
m

o
In Fig 17-5, m=2.43kg, k=221N/m, the block is
stretched in the positive x direction a distance of
11.6 cm from equilibrium and released. Take time
t=0 when the block is released, the horizontal
surface is frictionless.
(a) What is the total energy?
(b) What is the maximum speed of the block?
(c) What is the maximum acceleration?
(d) What is the position, velocity, and acceleration
at t=0.215s?
x
Solution:
(a) E  1 kxm 2  1 (221N / m)(0.116m) 2  1.49 J
2
2
(b)
2 K max
2E
2(1.49 J )
vmax 
m

m

2.43kg
 1.11m / s
(c) The maximum acceleration occurs just at the
instant of release, when the force is greatest
a max
Fmax kxm (221N / m)(0.116m)



 10.6m / s 2
m
m
2.43kg
(d)  
k
 0.9536rad / s
m
x(t )  xm cos(t   )
Since x  xm  0.116m at t=0, then   0
x(t )  xm cos t  0.116 cos(9.536t )
So at t=0.215s

x  0.116 cos(9.536)(0.215s )  0.0535m
v x  xm sin t  0.981m / s
a x   2 x  (9.536rad / s ) 2 (0.0535m)  4.87m / s 2
Sample problem 17-3
Fig 17-5
m
0
v
x
In Fig17-5, m=2.43kg, k=221N/m, when the block
m is pushed from equilibrium to x=0.0624m, and
its velocity v x  0.847m / s , the external force is
removed and the block begins to oscillate on the
horizontal frictionless surface. Write an equation
for x (t) during the oscillation.
Solution: x(t)  xmcos(ωt  φ), xm , ω, φ???
k

 9.54rad / s
m
x m : At t=0
E
1
1
mv 2  kx 2
2
2
1
1
2
 ( 2.43kg)( 0.847 m / s )  (221N / m)( 0.0624m) 2
2
2
 0.872 J  0.430 J  1.302 J
2E
1
2
Setting this equal to kxm , we have xm  k  0.1085m
2
To find the phase constant
information give for t=0:
 , we still need to use the
x(t)  xm cos(ωt  φ)
v x  0.847m / s
x(0)  xm cos   0.0624m
★(1)
x ( 0)
cos  
 0.5751
xm
  54.9 
x1(t)  xm cos(ωt  54.9 )
x
o
x t图
T
t
x2 (t)  xm cos(ωt - 54.9 )  xm cos(ωt  360 - 54.9 )
So only   54.9 will give the correct initial velocity.
★(2)
dx
 ωxm sin( ωt  φ)
Or vx (t ) 
dt
v x (0)  x m sin   (1.035m / s ) sin 
 0.847m / s
 0.847m / s
for
for
1  54.9 
2  305.1
2  305.1  5.33radians
is correct.
x(t )  (0.109m) cos[9.54(rad / s)t  5.33rad ]
17-5 Applications of simple
harmonic motion
1. The torsional oscillator(扭转振子)
Fig 17-9 shows a torsional oscillator.
If the disk is rotated in the horizontal (xy)
plane, the reference line op will move
to the OQ, and the wire oo’ will be
twisted. The twisted wire will exert a
restoring torque on the disk, tending to
return the system to its equilibrium.
O’
Fixed
clamp
m
o
p
R
2 m
Fig 17-9
Q
For small twist, the restoring torque is
 z  
(17-17)
Here  is constant ( the Greek letter Kappa ), and
is called torsional constant.
The equation of motion for such a system is
d 2
 z  I z  I dt 2
(17-18)
where I is the rotational inertia of the disk about z
axis. Using Eq(17-17) we have
d 
 k  I 2
dt
2
or
d 2
k
 
2
I
dt
(17-19)
Eq(17-19) and (17-5) are mathematically identical.
The solution should be a simple harmonic oscillation
in the angle coordinate  ,
(17-20)
   m cos(t   )
k
 
I
2
or T  2
I
k
(17-21)
A torsional oscillator is also called torsional pendulum(扭
摆). The Cavendish balance, used to measure the
gravitational force constant G, is a torsional pendulum.
2. The simple pendulum(单摆)
Fig(17-10) shows a simple pendulum of length L
and particle mass m.
The restoring force is:
F  mg sin 
(17-22)
If the  is small, sin   
k x
F  mg  mg
T  2
m
k
L
 mx

T
L
m
(17-23)
m
L
 2
 2
mg / L
g

(17-24)
x
mg
Fig(17-10)
3. The physical pendulum(复摆)
Any rigid body mounted so that it can swing in
vertical plane about some axis passing through it
is called “physical pendulum”.
P

d
y
C

x
Fig(17-11)
Mg
In Fig. 17-11 a body of
irregular shape is pivoted
about a horizontal
frictionless axis through P
The restoring torque for an angular displacement  is
 z  Mgd sin 
(17-26)
For small angular displacement sin    .
(17-27)
 z  mgd  k
then
(17-28)
I
I
T  2
 2
k
Mgd
(a)The rotational inertia I can be found from Eq(17-28).
T 2 Mgd
I
4 2
(17-29)
(b) Center of oscillation (振动中心 )”
Suppose the mass of the physical pendulum were
concentrated at one point with distance L from the pivot,
it will form a simple pendulum.
L
I
T  2
 2
g
Mgd
P
I
d L
L
(17-30)

Md
C
The resulting simple pendulum
O
would have same period as
the original physical pendulum.
Fig(17-11)
The point O is called the “center of oscillation” of the
physical pendulum.
• If we pivot the original physical pendulum from point O, it
will have the same period as it does when pivoted from point P.
• The center of oscillation is often also
called the “center of percussion（撞
击中心）”.
If an impulsive force in the plane of
oscillation acts at the center of
oscillation, no effect of this force is
felt at the pivot point P.
P

Pd L
CC
OO
Fig(17-11)
Sample problem 17-4
A thin uniform rod of mass M=0.112kg and
length L=0.096m is suspended by a wire that
passes through its center and is perpendicular
to its length. The wire is twisted and the rod
set oscillating. The period is found to be 2.14s.
When a flat body in the shape of an equilateral
triangle is suspended similarly through its
center of mass, the period is found to be 5.83m.
Find the rotational inertia of the triangle about this
axis.
Solution:
The rotational inertia about its Cm is
ml 2
I rod 
 8.60  10 5 kg  m 2
12
from Eq(17-21)
1
Trod
I rod 2
(
)
Ttriamgle
I triamgle
Ttriamgle 2
or I triamgle  (
) I rod  6.38 10 4 kg  m 2
Trod
Sample problem 17-5
A uniform disk is pivoted at
P
its rim ( Fig17-12). Find its

period for small oscillations 3
R R
2 C
and the length of the

O
equivalent simple pendulum.
Solution:
Fig 17-12
The rotational inertia about
pivot at the rim is
3
R
2
1
3
2
2
I  MR  MR  MR 2
2
2
From Eq(17-28) with d=R, then
T  2
I
 2
MgR
3R
2 g
The simple pendulum having the same period has
a length (Eq(17-30))
I
3
L
 R
MR 2
The center of oscillation of the disk pivoted at P is
3
therefore at o, a distance R below the point of support.
2
You may check that the period of the pendulum
pivoted at O is the same as that pivoted at P.
Torsional Pendulum
(扭摆)
Torsional pendulum clock
17-6 Simple harmonic motion
and uniform circular motion
Simple harmonic motion can be described as a
projection of uniform circular motion along a
y 
diameter of the circle.
v
1. Fig17-14 shows a particle
P in uniform circular motion.
Q
0
P
r
t  
p'
Fig 17-14
x


At a time t, the vector r ( op ) makes an angle

t   with x axis, and the x component of r is
x(t )  r cos(t   )
(17-31)
This is of course identical to Eq(17-6) for the
displacement of the simple harmonic oscillator.
• The radius r corresponding tox m ;
'
p
• Point
, which is the projection of P on the x axis,
executes simple harmonic motion along the x axis.

2. The magnitude of the tangent velocity
v of the

point P is r , the x component of v is
v x (t )  r sin( t   )
(17-32)
3. The centripetal acceleration is  2 r, and its x
component is a x (t )   2 r cos(t   ) (17-33)
Eqs(17-32) and (17-33) are identical with Eqs(1711) for simple harmonic motion, again with x m
replace by r.
See动画库\波动与光学夹\1-07辅助圆

4. In Fig17-14，the y projection OQ of r at time t, is
y (t )  r sin( t   )
3
 r cos(t 
)
2
(17-34)
So the projection of uniform circular motion along the y
direction also gives simple harmonic motion.
On the contrary, the combination of two simple harmonic
motions at right angles, with identical amplitude and
frequencies, can form a uniform circular motion.
See动画库\波动与光学夹\1-10垂直振动的合成
17-7 Damped (阻尼) harmonic motion
m
v
x
f
Up to this point we have assumed that no frictional
force act on the system.
0
For real oscillator, there may be friction, air resistance
act on the system, the amplitude will decrease.
1. This loss in amplitude is called “damping” and
the motion is called “damped harmonic motion”.
Fig17-16 compare the motion of undamped and
damped oscillators.
x
1
0.5
1
2
3
4
5
6
3
4
5
6
t
(a)
t
(b)
-0.5
-1
x
1.5
e
1
t

0.5
1
2
-0.5
-1
-1.5
Fig 17-16
(a) When we add a small damping force, the amplitude
gradually decreases to zero but the frequency changes
by a negligible amount. In this case Eq(17-6) becomes
x(t )  x m e  t /  cos(t   )
(17-36)
where  is called the “damping time constant” or
the “mean lifetime” of the oscillator.

is the time necessary for the amplitude to drop to 1/e
of its initial value.
(b) When the damping force is not large, the
mechanism energy is
1
2  2t / 
E (t )  kxm e
2
(17-37)
Eq(17-37) shows that the mechanical energy of
the oscillator decreases exponentially with time.
The energy decreases twice as rapidly as the
amplitude.
2*. Mathematical analysis
Assume the damping force is  bv x , where b is a
positive constant called the “damping constant”.
With Fx  kx  bv x , Newton’s second law gives

or  kx  bv x  ma x
d 2x
dx
m 2  b  kx  0
dt
dt
(17-38)
The solution is
x(t )  xm e

bt
2m
cos( ' t   )
(17-39)
where
k
b 2
(17-40)
 
( )
m 2m
(a) If b is negligible,  '   . It is ideal simple harmonic
oscillation.
'
If b  2 km ,  '   that is , damping slows down
the motion. This case is called underdamping (欠阻
尼)
Comparing Eqs(17-39) and (17-36) we have
  2m / b.
(b) When b  2 km , ' 0 , the motion decays
exponentially to zero with no oscillation at all.
This condition is called “critical damping (临界阻尼)”.
The lifetime  has its smallest possible value, 1 /  .
(c) When b  2 km , the motion also decays
exponentially to zero with no oscillation, called
overdamping （过阻尼）.
x
三种阻尼的比较
(c)
o
(b)
(a)
t
17-8 Forced oscillations and
resonance
Forced oscillations:
Oscillations of a system carried out under the action of
an external periodical force, such as
Fx (t )  Fm sin  t
''
or a successive action of an external non-periodical force.
 ' ' and 
,
,
Which frequency will the forced oscillation system take?
Forced oscillation system takes the frequency
of the external force, namely  ' ' .
Resonance:
The amplitude of the forced
oscillation can increase much
''

as
approaches  .
This condition is known as
“resonance” and  '' is called
“resonant angular frequency”.
Amplitude
In the case with damping, the rate at which energy
is provided by the driving force exactly matches the
rate at which energy is dissipated by the damping
force.
Small
damping
Large
damping
0.5

1.5 2
 ''
Fig 17-19
String theory:
Theory of everything (万物之理)
Theory of anything (任意之理)
Everything or nothing?
Where string theory stands today
弦理的根本问题是
它很难或不可能得到实验证明
Peter Woit (Princeton)
弦论根本不是理论
Not even wrong!
简直比错还差！