Chapter 17 Oscillations
description
Transcript of Chapter 17 Oscillations
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Chapter 17 Oscillations
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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.
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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.
(17-1)
Unit: period (s)
frequency(Hz, SI unit), 1 Hz = 1 cycle/s
Tf
1
4. The magnitude of the maximum displacement from equilibrium is called the amplitude of the motion.
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17-2/3 The simple harmonic oscillator and its motion
1. Simple harmonic motionAn oscillating system which can be described in terms of sine and cosine functionssine 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
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a body of mass m that slides on a frictionless horizontal surface. The body moves in x direction.
Fig 17-5
x
xm
mF
o
o
Relaxed state
origin is chosen at here
kxFx 2
2
dt
xdax
2
2
dt
xdmkx
02
2
xm
k
dt
xd (17-4)
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Eq(17-4) is called the “equation of motion of the simple harmonic oscillator”. It is the basis of many complex oscillator problems.
Rewrite Eq(17-4) as
(17-5)
We write a tentative solution to Eq(17-5) as
(17-6)
xm
k
dt
xd)(
2
2
3. Find the solution of Eq. (17-4)
)cos( txx m
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We differentiate Eq(17-6) twice with respect to the Time.
Putting this into Eq(17-5) we obtain
Therefore, if we choose the constant such that
(17-7)
Eq(17-6) is in fact a solution of the equation of motion of a simple harmonic oscillator.
)cos(22
2
txdt
xdm
)cos()cos(2 txm
ktx mm
m
k2
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a) :
If we increase the time by in Eq(17-6), then
Therefore is the period of the motion T.
2
2
k
mT
22
(17-8)(17-9)
)cos(])2
([cos txtxx mm
m
k
Tf
2
11
The quantity is called the angular frequency.f 2
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b) : is the maximum value of displacement. We call it
the amplitude of the motion.
c) and : The quantity is called phase of the motion.
is called “phase constant (常相位 )”.
and are determined by the initial position and
velocity of the particle. is determined by the system.
mxmx
t t
mx
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How to understand ?)cos( txx m
2
0
T
xtmx
o
tx 图
mx
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0
x
to
同相
How to compare the phases of two SHOs with same ?
)cos( 111 txx m
)cos( 222 txx m
)()( 12 tt
12
x
to
为其它超前
落后
t
x
o
π 反相
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(c) same: , different:
1 2 3 4 5 6t
-1
-0.5
0.5
1
xt
2 4 6 8t
-1
-0.5
0.5
1
xt a
1 2 3 4 5 6t
-1
-0.5
0.5
1
xt
Fig 17-6 shows several simple harmonic motions.
(a) (b)
(c) (a) same: , different:
mx
(b) same: , different:
mx
Fig 17-6
mx
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d). Displacement, velocity, and acceleration
Displacement
Velocity
Acceleration
When the displacement is a maximum in either direction, the speed is zero, because the velocity must now change its direction.
)cos( txx m
)sin( txdt
dxv mx
)cos(22
2
txdt
xda mx
)2
cos( txm
)cos(2 txm (17-11)
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tx 图
tv 图
ta 图
T
mx
mx
2mx
2mx
x
v
a
t
t
t
mx
mxo
o
o T
T
)cos( txx m
0取π2
T
)2
πcos( txm
)sin( txmv
)πcos(2 txm
)cos(2 txa m
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17-4 Energy in simple harmonic motion
1.The potential energy
(17-12)
2.The kinetic energy
1 2 3 4 5 6
0.2
0.4
0.6
0.8
1
)(cos2
1
2
1 222 tkxkxU m
U(t)
K(t)
T/2 T)(sin2
1
)(sin2
1
2
1
22
2222
tkx
txmmvK
m
m
)sin( txmv
0
(17-13)
Fig 17-8(a)
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• Fig17-8(a), both potential and kinetic energies oscillate with time t and vary between zero and maximum value of .
• Both U and K vary with twice the frequency of the displacement and velocity.
3. The total mechanical energy E is
2
2
1mkx
2
2
1mkxUKE (17-14)
mxmx
K(x)
U(x)
x
Fig 17-8 (b)
2
2
1)( kxxU
)()( xUExK
E
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At the maximum displacement , .
At the equilibrium position , .
Eq(17-14) can be written quite generally as
2
2
1mkxK
2
2
1mkxU 0K
0U
222
2
1
2
1
2
1mx kxkxmvUK
)( 222 xxm
kv mx
)( 22 xxm
kv mx (17-16)
(17-15)
then
or
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Sample problem 17-2
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?
xmo
Fig 17-5
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Solution:
(a)
(b)
(c) The maximum acceleration occurs just at the
instant of release, when the force is greatest
(d)
JmmNkxE m 49.1)116.0)(/221(2
1
2
1 22
smkg
J
m
E
m
Kv /11.1
43.2
)49.1(222 maxmax
2maxmax /6.10
43.2
)116.0)(/221(sm
kg
mmN
m
kx
m
Fa m
sradm
k/9536.0
![Page 20: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/20.jpg)
Since at t=0, then
So at t=0.215s
)cos()( txtx m
mxx m 116.0 0)536.9cos(116.0cos)( ttxtx m
msx 0535.0)215.0)(536.9cos(116.0
smtxv mx /981.0sin 222 /87.4)0535.0()/536.9( smmsradxax
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Sample problem 17-3
In Fig17-5, m=2.43kg, k=221N/m, when the block
m is pushed from equilibrium to x=0.0624m, and
its velocity , 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.
smvx /847.0
xm
Fig 17-5v
0
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Solution:
:
Setting this equal to , we have
To find the phase constant , we still need to use the information give for t=0:
???,,)()( φω,xφωtcosxtx mm
JJJ
mmNsmkg
kxmvE
302.1430.0872.0
)0624.0)(/221(2
1)/847.0)(43.2(
2
12
1
2
1
22
22
2
2
1mkx m
k
Exm 1085.0
2
mx
sradm
k/54.9
At t=0
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tx 图
T
xto
mxx m 0624.0cos)0(
5751.0)0(
cos mx
x 9.54
)9.54()(1 ωtcosxtx m
)9.54360()9.54()( -ωtcosx-ωtcosxtx mm2 So only will give the correct initial velocity.
smvx /847.0
9.54
★(1)
)()( φωtcosxtx m
![Page 24: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/24.jpg)
)sin()( m φωtωxdt
dxtvx Or
sm
sm
smxv mx
/847.0
/847.0
sin)/035.1(sin)0(
9.541 1.3052
forfor
radians33.51.3052
]33.5)/(54.9cos[)109.0()( radtsradmtx
is correct.
★(2)
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p
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
O’Fixed clamp
Fig 17-9
R
Q
m
m2
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For small twist, the restoring torque is
(17-17)
Here is constant ( the Greek letter Kappa ), and
is called torsional constant.
The equation of motion for such a system is
(17-18)
where is the rotational inertia of the disk about z
axis. Using Eq(17-17) we have
I
z
2
2
dt
dII zz
![Page 27: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/27.jpg)
or (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) or (17-21)
2
2
dt
dIk
I
k
dt
d
2
2
I
k2
)cos( tm
k
IT 2
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.
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The restoring force is:
(17-22)
If the is small,
2. The simple pendulum(单摆 )
Fig(17-10) shows a simple pendulum of length L and particle mass m.
sinmgF
L
xmgmgF
sin LT
m
mg
Fig(17-10)
(17-23)
x
xm
g
L
Lmg
mmT 2
/22 (17-24)
k
k
![Page 29: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/29.jpg)
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
C
d
Mgx
y
Fig(17-11)
In Fig. 17-11 a body of irregular shape is pivoted about a horizontal frictionless axis through P
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The restoring torque for an angular displacement is (17-26)For small angular displacement . (17-27) then (17-28)
(a)The rotational inertia can be found from Eq(17-28).
(17-29)
sinMgdz
kmgdz
Mgd
I
k
IT 22
I
2
2
4MgdT
I
sin
![Page 31: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/31.jpg)
Suppose the mass of the physical pendulum were concentrated at one point with distance L from the pivot, it will form a simple pendulum.
(b)
Mgd
I
g
LT 22
Md
IL (17-30)
Center of oscillation ( 振动中心 )”
The resulting simple pendulum would have same period as the original physical pendulum.
P
C
d
Fig(17-11)
O
L
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• 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.
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.
P
C
d
Fig(17-11)
O
LP
C
O
![Page 33: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/33.jpg)
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.
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Solution:
The rotational inertia about its Cm is
from Eq(17-21)
252
1060.812
mkgml
I rod
or 242 1038.6)( mkgIT
TI rod
rod
triamgletriamgle
2
1
)(triamgle
rod
triamgle
rod
I
I
T
T
![Page 35: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/35.jpg)
Sample problem 17-5
A uniform disk is pivoted at
its rim ( Fig17-12). Find its
period for small oscillations
and the length of the
equivalent simple pendulum.
Solution:
The rotational inertia about
pivot at the rim is
O
C
R
P
R2
3R
2
3
Fig 17-12
![Page 36: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/36.jpg)
From Eq(17-28) with d=R, then
The simple pendulum having the same period has
a length (Eq(17-30))
The center of oscillation of the disk pivoted at P is
therefore at o, a distance below the point of support.
222
2
3
2
1MRMRMRI
g
R
MgR
IT
2
322
RMR
IL
2
3
R2
3
You may check that the period of the pendulum pivoted at O is the same as that pivoted at P.
![Page 37: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/37.jpg)
Torsional Pendulum
(扭摆 )
Torsional pendulum clock
![Page 38: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/38.jpg)
17-6 Simple harmonic motion and uniform circular motion
Simple harmonic motion can be described as a
projection of uniform circular motion along a
diameter of the circle.
1.1. Fig17-14 shows a particle
P in uniform circular motion. 0
QP
'p t
v
r
Fig 17-14
x
y
![Page 39: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/39.jpg)
At a time t, the vector ( ) makes an angle with x axis, and the x component of is
(17-31)
This is of course identical to Eq(17-6) for the displacement of the simple harmonic oscillator. • The radius r corresponding to ;• Point , which is the projection of P on the x axis, executes simple harmonic motion along the x axis.
r
r
op t
)cos()( trtx
mx'p
![Page 40: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/40.jpg)
2.2. The magnitude of the tangent velocity of the point P is , the x component of is
(17-32)
3.3. The centripetal acceleration is , and its x component is (17-33)
Eqs(17-32) and (17-33) are identical with Eqs(17-11) for simple harmonic motion, again with replace by r.
v
)sin()( trtvx
)cos()( 2 trtax
vr
r2
mx
See动画库 \波动与光学夹 \1-07辅助圆
![Page 41: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/41.jpg)
4.4. In Fig17-14, the y projection OQ of at time t, is
(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.
)2
3cos( tr
r)sin()( trty
See动画库 \波动与光学夹 \1-10垂直振动的合成
![Page 42: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/42.jpg)
17-7 Damped (阻尼 ) harmonic motion
Up to this point we have assumed that no frictional
force act on the system.
For real oscillator, there may be friction, air resistance act on the system, the amplitude will decrease.
1.1. This loss in amplitude is called “damping” and
the motion is called “damped harmonic motion”.
xm v
0 f
![Page 43: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/43.jpg)
1 2 3 4 5 6
-1
-0.5
0.5
1
1 2 3 4 5 6
-1.5
-1
-0.5
0.5
1
1.5
x
x
t
t
(a)
(b)
t
e
Fig 17-16
Fig17-16 compare the motion of undamped and damped oscillators.
![Page 44: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/44.jpg)
(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
(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.
)cos()( / textx tm
![Page 45: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/45.jpg)
(b) When the damping force is not large, the mechanism energy is
(17-37)
Eq(17-37) shows that the mechanical energy of the oscillator decreases exponentially with time.
/22
2
1)( t
m ekxtE
The energy decreases twice as rapidly as the amplitude.
![Page 46: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/46.jpg)
2*. Mathematical analysis
Assume the damping force is , where b is a
positive constant called the “damping constant”.
With , Newton’s second law gives
or
(17-38)
xx mabvkx
02
2
kxdt
dxb
dt
xdm
xx bvkxF
xbv
![Page 47: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/47.jpg)
The solution is (17-39) where (17-40)
(a) If b is negligible, . It is ideal simple harmonic oscillation.
)'cos()( 2
textx m
bt
m
2' )2
(m
b
m
k
./2 bm
'
If , that is , damping slows downthe motion. This case is called underdamping (欠阻尼 )
Comparing Eqs(17-39) and (17-36) we have
kmb 2 '
![Page 48: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/48.jpg)
(b) When , , 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, .
kmb 2
0'
/1
ot
x 三种阻尼的比较
)(b
(a)
(c)
(c) When , the motion also decays exponentially to zero with no oscillation, called overdamping (过阻尼) .
kmb 2
![Page 49: Chapter 17 Oscillations](https://reader035.fdocuments.us/reader035/viewer/2022081419/56815001550346895dbdcbde/html5/thumbnails/49.jpg)
Forced oscillations:
Oscillations of a system carried out under the action of an external periodical force, such as
or a successive action of an external non-periodical force.
17-8 Forced oscillations and resonance
tFtF mx''sin)(
and , ,Which frequency will the forced oscillation system take?
''
Forced oscillation system takes the frequencyof the external force, namely .''
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Resonance: The amplitude of the forced oscillation can increase much as approaches .
'' This condition is known as “resonance” and is called “resonant angular frequency”.
''
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Am
plit
ud
e
Fig 17-19
Small damping
Large damping
5.0 25.1''
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.
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String theory: Theory of everything (万物之理 )
Theory of anything (任意之理 )
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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!