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Transcript
```Chapter 9 Oscillation
9-0 Basic Requirements of the Chapter Teaching
9-1 Simple Harmonic Motion, Amplitude, Period and
Frequency, Phase
9-2 Rotating Vector
9-3 Single Pendulum and the Physical Pendulum
9-4 Energy in Simple Harmonic Motion
9-5 Superposition of Simple Harmonic Motions
*9-6 Damped Oscillation , Forced Oscillation and
Resonance
9-7 Electromagnetic Oscillation
Brief Summary of this Chapter
Basic Requirements of the Chapter Teaching
1、To master physical significance of physical quantities
that describe simple harmonic motion and the relations
between them.
2 、 To master the representation of rotating vector for
simple harmonic motion, and be able to analysis and resolve
problems of simple harmonic motion.
3、To master the basic characteristics of simple harmonic
motion, be able to establish the differential equation of
simple harmonic motion, to obtain the motion function of
the simple harmonic motion determined by the initial
condition, and to understand its physical significance.
Basic Requirements of the Chapter Teaching
4、to understand the energy converting process of simple
harmonic motion, and be able to calculate the mechanical
energy of a simple harmonic motion.
5、To understand the regulation of superposition of simple
harmonic motions with same frequency in the same direction,
to grasp the characteristics of beat and superposition of two
simple harmonic motions with the same frequency in
perpendicular direction.
6、To grasp the physical principle of electromagnetic
oscillation.
§ 9-1 Simple Harmonic Motion, Amplitude,
Period and Frequency, Phase
Oscillation is a periodical motion.
If a periodical motion is only the back-and-forth
motion of the position, which is called Mechanical
Oscillation.
Examples for Oscillation：
1. The back-and forth motion of objects
(ex. The spring oscillation and single pendulum)
2. The cycle of current and voltage
3. Atoms vibration within a molecule.
§ 9-1 Simple Harmonic Motion, Amplitude,
Period and Frequency, Phase
Reason of Mechanical Oscillation
•Restoring force
•Inertia
The Characteristics of Simple Harmonic Motion
1. The relation between displacement and time：
When an object in the simple harmonic motion, its
displacement is a cosine function of time.
x
x  A cos( t   )
t
The Characteristics of Simple Harmonic Motion
2. Kinematics descriptions
A object goes into a back and forth motion on
both side of the equilibrium position under
elastic force that is directly proportional to the
displacement x to the equilibrium position.

F
o
m
x
x
The Characteristics of Simple Harmonic Motion
3. Dynamics descriptions
 F   kx  ma
k
a
x
m
The acceleration of the spring oscillation is
proportional to the magnitude of the displacement
and has opposite direction to the displacement.

F
o
m
x
x
Simple Harmonic Motion
Every complicated
oscillation is the
superposition of several
simple harmonic
motions.
Simple Harmonic
Motions
Superposition
Decomposition
Complicated
Oscillation
The Regulation of Simple Harmonic Motion
(Take the spring oscillator as an example)
The oscillation system including the spring and the
object is called the spring oscillation.
F
o
x
x


According to Hooke’s law： F   k x
（where k is spring constant)
(1) The elastic force that the object subjected is directly proportional to
the displacement x to the equilibrium position.
(2) The direction of the elastic force is opposite to the direction of the
displacement, and orients to the equilibrium position.
The Regulation of Simple Harmonic Motion
According to Newton’s second law：
2
d x
F  m 2  k x
dt
We obtain:
2
d x
k


x
2
m
dt
k
Assume  
m
2
d x
2
  x
2
dt
The Regulation of Simple Harmonic Motion
Equation of simple harmonic motion and its solution
d 2x
k
 x
2
dt
m
Assume：
2
k

m
2
d x
k

x0
2
dt
m
2
d x
2
 x  0
2
dt
The differential equation of the simple harmonic motion
Its solution：
where

x  A cos  t   
and A are the integral constants
Equation of simple harmonic motion
The Regulation of Simple Harmonic Motion
Results：
（1）The vibration of the spring oscillator is a
simple harmonic motion.
k
（2）Angular frequency：  
m
Period： T 
2

 2
m
k
（3）The angular frequency and period of the spring oscillator
are only depend on the physical properties of the oscillating
system.
These kinds of period and frequency that only depended on
the properties of the oscillating system are called natural period
and natural frequency.
The Regulation of Simple Harmonic Motion
Velocity：
dx

v
  A sin(  t   )  v m cos( t    )
dt
2
Acceleration：
dv
a
  2 A cos( t   )  am cos( t     )
dt
ω2 A
A
x, v, a
a
ωA
O
x
O
v
t
A
T
Physical Quantities for Simple Harmonic Motion Description
x  A cos  t   
A ：Amplitude，（the maximum displacement，x =±A ）
The absolute value A of the maximum position of the
object to its equilibrium position is the amplitude.
x
A
o
A
xt
T
T
2
t
Physical Quantities for Simple Harmonic Motion Description
2
Angular Frequency  ：   2  
T
Frequency ：the number of complete cycles or vibrations
per unit of time.
Period T：the time that the object takes to finish one
complete cycle of motion.
A cos  t     A cos t  T    
cos  t     cos  t     T 
2
 T  2 , T 

1 
 

 2 
T 2
Physical Quantities for Simple Harmonic Motion Description
Phase ： （ t +  )
Initial Phase ： 
(t = 0)
Phase is a physical quantity, which decides the motion state
of an object in a simple harmonic motion.
x
A
o
A
xt
T
T
2
t
Physical Quantities for Simple Harmonic Motion Description
Velocity Expression
dx

v
  A sin(  t   )  v m cos( t    )
dt
2
vm   A
Velocity is leading to displacement of /2。
Acceleration Expression
dv
2
a
  A cos( t   )  am cos( t     )
dt
am   A
2
Acceleration is out of phase with displacement.
Methods of Characteristics for Simple Harmonic Motion
i、Mathematic Equation： x  A cos(t   )
ii、Diagram of Rotating Vector：
1. To determine the phase and
initial phase of a simple harmonic
motion by rotating vector.
2. To discuss the phase difference
of two simple harmonic motions
by rotating vector.
iii、Diagram x-t (or v-t, a-t)
x  2 cos(
4π 2π
t  ) cm
3
3
x/cm
0
1
2
1
t/s
The solution of constant A and 
x  A cos( t   )
v   A  sin(  t   )
Initial condition:
A  x02 
v

2
0
 v0
tan  
x0
2
t  0 x  x0 v  v0
For given oscillating system, the
period
(or
frequency)
is
determined by the oscillating
system itself, while the amplitude
and the initial phase are
determined by the initial condition.
Example
When t  0, x  0, v 0  0, Try to resolve

π
v
0  A cos    
2
 v 0   A  sin   0
 sin   0
π
Therefore  
2
π
x  A cos( t  )
2

x
o
x
A
o
A
xt
T
T
2
t
§9-2 Rotating Vector
The projection component
of the vector A on Ox axis is ：
x  A cos( t   )
Conclusion：
The motion of the projecting
point P of the vector A can
represent the simple harmonic
motion along x axis.
Rotating Vector
• Circular frequency：the angular velocity  of rotating vector
• Amplitude：the absolute value of rotating vector A
• Phase： (  t+  )
y
• Initial Phase： (t = 0)

A
P

 t  o
• After one period, the
phase
varies
2π,
everything will go on
repeatedly.
M
Period：
T 
2

x
Rotating Vector

vm
y
t  
O

an
π
t   
 2
 v
a
A
x  A cos(t   )
v m  A
 v   A sin( t   )
x
an  A
2
a   A 2 cos(t   )
Rotating Vector Diagram
Rotating vector diagram and graph of x versus t
for a simple harmonic motion.
The graph of x versus t for simple harmonic motion
(1) The graph
represents the
regulation of the simple
harmonic motion. The
direction of velocity is
shown in the graph.
(2) The graph reflects
period T, amplitude A,
initial phase and phase.
The graph of x versus t for simple harmonic motion
(3) The graph shows the relation between displacement and time:
x  A cos  t   
x
t
(4) The graph shows the relation between phase difference and
time:
   t
x
The graph of x versus t for simple harmonic motion
x
(4) The graph indicates the direction of the force that the
object subjected and the direction of the acceleration.
F = - kx，The direction of the force is opposite to the direction of the
displacement, and the acceleration has same direction to the force.
(5) The graph indicates the position that object has maximum
and minimum kinetic energy and potential energy.
When the displacement is zero, kinetic energy is the maximum,
while potential energy is zero.
When the displacement reaches maximum, potential energy is the
maximum, while kinetic energy is zero.
Methods of Resolution
According to initial condition, try to
resolve Amplitude and Initial phase
（1）Assume time t = 0, displacement x = x0,
and velocity v = v0
From Eq.
x  A cos ( t   ) , we have
xo  A cos 
From Eq. v   A sin ( t   ) , we have
v o   A sin 
Methods of Resolution
xo  A cos 

vo

 A sin 
v
2
2
2
2
x 
 A (sin   cos  )  A

2
o
2
o
2
A
x0
2
 v0 


 
vo
tg   
 xo
2
Methods of Resolution
Analysis：
xo  A cos 
vo   A sin 
x 0  0,
v0  0
x 0  0,
v0  0
x 0  0,
v0  0
x 0  0,
v0  0
 In the fourth quadrant
 In the first quadrant
 In the third quadrant
 In the second quadrant
Or analysis using rotating vector diagram:
After confirming x0, we draw
the rotating vector diagram,
and then confirm the direction
of velocity, v0  0, or v0  0
at last we could obtain

x
Methods of Resolution
（2）Assume time t = 0，xo = kA, where k is a constant，
given the direction of velocity, we obtain:
x  A cos t   
k  cos 
kA  A cos 
 has two values that can be determined by the
direction of velocity from rotating vector diagram.
If given initial condition, we can obtain the integral
constants A and  by equation (displacement or velocity).
（3）If given xt, vt at time t, we can obtain the integral
constants A and  by equation (displacement , velocity or
acceleration), especially using the following expressions:
v max   A
a max   A
2
Examples
An object is in simple harmonic motion along x axis，its
amplitude is 12 cm，the period is 2 s, at the initial moment
(t = 0) the object is at the position of 6 cm and moves
toward the positive direction of Ox axis. Try to resolve
(1) the equation of oscillation；
(2) At t = 0.5 s，where is the object located、and what is
the velocity and acceleration；
(3) Assume the object was located at x = - 6 cm and moves
toward the negative direction of x axis, what is the
minimum time when the object moves from the position to
equilibrium position.
Examples
Solution：(1) First we need to obtain the equation of
oscillation. Assume
x  A cos ( t   )
According to this sample：A =0.12 m , T = 2 s, we have
2

  s 1
T
Substitute t = 0 ， x0 = 0.06 m into equation of simple
harmonic motion with v0 > 0, we have
0.06 =0.12 cos 
Examples
0.06 =0.12 cos 
1

 cos     
2
3
y

3
v 0   A sin   0
 sin   0


 
3

x
3

Therefore, the equation of oscillation is
x  0.12 cos( t 


3
)
Examples
(2) At t = 0.5 s，we have the object located、the
velocity and acceleration:
x t  0 .5

 0.12 cos( t  ) t  0.5  0.104 m
3
dx

v t  0 .5 
 0.12 sin(  t  ) t  0.5  0.189 m  s 1
dt t 0.5
3
dv

2
2
a t  0 .5 
 0.12 cos( t  ) t  0.5  1.023 m  s
dt t 0.5
3
Examples
(3) Assume at t1， x = - 0.06 m
Substitute them into the motion equation：
 0 .06  0 .12 cos ( t1   3 )
 2
4
 t1  
or
3 3
3
 2
 t1  
 t1  1 s
3
3
 3
11
 t2  
 t2  s
3
2
6
11
5
 t  t 2  t1   1  s
6
6
2 3
4 3
x
Examples
Method 2：to resolve the problem using rotating vector
3 2 5
 


2
3
6
   t

5 5
t 



6 6
2 3
3 / 2
x
Examples
Sample: Two objects are in simple harmonic motion with same
frequency and amplitude along the same direction. At the initial
moment the object is at the position of x1= A/2 and moves
toward the negative direction, the other object is at the position
of x 2= -A/2 and moves toward the positive direction. Try to
resolve the phase difference between the two objects.
x1  A cos ( t   1 )
A 2  A cos ( t   1 )   t   1    3
Solution：
v1   A sin ( t  1 )  0
sin ( t   1 )  0
 t  1   3
Examples
-A
-A/2
o
A/2
A
 A 2  A cos( t   2 )
  t   2   2 3
v 2   A sin ( t   2 )  0
 sin  t     0
 t   2   2 3

2
   ( t   1 )  ( t   2 )   (  )  
3
3
Examples
Method 2：to resolve the problem using rotating vector
  
x
Examples
Sample: An object is in simple harmonic motion along x
axis，at the initial moment (t = 0) the object is at the
position of 0.04 m along the positive direction of Ox axis.
Try to resolve
(1) the equation of oscillation；
(2) The velocity where the object was located at x = A/2 at
the first time when the object moves from the initial
position to the position.
Examples
Solution: (1) According to this sample,
x 0  0.04m
, v0  0
,   6 .0 s
Substitute them into the following expression, we have
Amplitude： A 
x0 2 
v0 2
0
2
 x0  0 . 04 m
 v0
  0
Initial phase   arctg
 x0
then：
x  0.04 cos (6.0t ) m
Examples
(2) First we need to obtain phase
x  A cos( t )

x
 t  arccos
A
A2
1 
5
 t  arccos
 arccos  (or
)
A
2 3
3
A

x  A x   , t 
2
3
Therefore,

v   A sin  t  0.04  6.0(sin )
3
1
 0.208 m  s
Examples
Method 2：to resolve the problem using rotating vector
Solution:
According to this sample

   t 
3
A
x  A x 
2
v   A sin  t  0.04  6.0(sin
 0.208 m  s 1

3
)
Examples
Sample：An object is in simple harmonic motion as
shown in the following figure, what the period is
x（m）
（A）2.62 s
4
2
（B）2.40 s
t（s）
1
2  4 cos 
  
（D）0.382 s

3
0  4 cos      
5

6
（C）0.42 s

3
2 12
T


5


2
§9-3 Single Pendulum and the Physical Pendulum
1. Single pendulum
O

Dynamics：
d s
 mg sin   m 2
dt
2

l
T
s  l
d 
 mg sin   ml
dt 2
2
  5

mg
 sin   
Single Pendulum
g
d 2
 
2
dt
l
g
d 2
 0
2
dt
l
Conclusion：single pendulum is a simple harmonic motion.
   0 cos  t   

g
l
T  2
l
g
Application:
We can measure the period of single pendulum
to obtain the gravity acceleration at this place.
Physical pendulum
 

M l F
M   mgl sin 
d 2
 J  J
dt 2
d 2
 mgl  J
dt 2
ω
O
 l
*C

P
（C is the center of mass）
mgl
Assume  
J 2
d 
mgl

(


5
)


2
dt
J
2
Physical pendulum
d 2
2




2
dt
ω
O
   m cos( t   )

 l
mgl
J
T 
2π
*C
 2π

Applications
J
mgl

P
（C is the center of mass）
• To calculate gravity acceleration
• To calculate moment of inertia
§9-4 Energy in Simple Harmonic Motion
Take the spring oscillator as an example
v
o
x
x
1
1
2
Kinetic energy：E  mv  m  2 A 2 sin 2 ( t   )
k
2
2
Potential energy：E  1 kx 2  1 kA2 cos 2 ( t   )
p
2
m  k
2
2
Energy in Simple Harmonic Motion
x  A cos ( t   )
1
E k  m  2 A 2 sin 2  t   
2
1 2
E p  kA cos 2  t   
2
x
E
m 2  k
Ep
The total mechanical energy：
E  Ek  E p
t
O
Ek
O
t
1 2 1
1
2 The restoring force is
2 2
E  kA  m  A  mv m
a conservative forces
2
2
2
Summary
(1) Both the kinetic energy and potential energy of the
system vary in period with time, while the total energy
remains unchanged.
When the displacement is zero, kinetic energy is the maximum,
while potential energy is zero.
When the displacement reaches maximum, potential energy is the
maximum, while kinetic energy is zero.
(2) The frequency of kinetic energy and potential energy
variation is two times of the frequency of vibration of the
spring oscillator.
(3) The total mechanical energy of a simple harmonic
oscillator is proportional to the square of the amplitude.
Graph of potential energy for a simple harmonic motion
Ep
1 2
Potential Energy E p  kx
2
Kinetic Energy
E
Ek
Ek  E  E p
Ep
A
Simple harmonic motion
is a motion with
invariable amplitude.
O
1
E  kA2
2
A
x
Example
When the displacement is 1/2A,what are the kinetic
energy and potential energy? At which position
kinetic energy is equal to potential energy?
1 2
Solution： E  E p  E k  kA
2
2
1 2 1  A
1
At x  A 2：E p  kx  k    E
2
2 2
4
3
Ek  E  E p  E
4
1
1
1 1
2
2
A  0.707 A
kx0   kA x0  
2
2
2 2
§9-5 Superposition of Simple Harmonic Motions
1. Superposition of two simple harmonic motions
with same frequency in the same direction
There are two simple harmonic motions in the same
direction, both of their angular frequency are ω, the
amplitudes and initial phases are A1, A2, and φ1, φ2,
respectively. Their motion equations are：

A2
x1  A1 cos( t   1 )
x 2  A2 cos( t   2 )
2
O
x2
1

A1
x1
The phase difference     2  1 = Constant
x
Superposition of two simple harmonic motions with
same frequency in the same direction
  
A  A1  A2
x  x1  x2
x  A cos( t   )

A2
2
x2


A



A1
1
x1
x
x
    ( 2   1 )
Superposition of two simple harmonic motions with
same frequency in the same direction
A  A  A  2 A1 A2 cos 
2
2
1
2
2
 A12  A22  2 A1 A2 cos 2  1 
A  A12  A22  2 A1 A2 cos 
y A1 sin 1  A2 sin  2
tan   
x A1 cos 1  A2 cos  2
Conclusion：
The combined oscillation is still a simple
harmonic motion, whose angular frequency is the
same as the frequency of the component oscillation,
and its amplitude of the combined oscillation is A.
Superposition of two simple harmonic motions with
same frequency in the same direction
A
A1  A2  2 A1 A2 cos( 2   1)
2
  tg
(1)
1
A1 sin  1 A2 sin  2
A1 cos  1 A2 cos  2
Assume : 2  1  2k
obtain :
(2)
2
A
k  0,1,2,
A  A  2 A1 A2  A1  A2
2
1
2
2
Assume : 2  1  (2k  1)
obtain :
A
k  0,1,2,
A12  A21  2 A1 A2  A1  A2
Superposition of two simple harmonic motions with
same frequency in the same direction
(1) If the phase difference is
x

oA
1
A2

 2  1  2 k
k  0,1, 2,  
x
o
T
t
A
x  ( A1  A2 ) cos(t   )
A  A1  A2
   2  1  2k π
Superposition of two simple harmonic motions with
same frequency in the same direction
(2) If the phase difference is  2  1  ( 2 k  1)
x
x
o
o
k  0,1,2,  
A1
2
T

t
A
A2
x  ( A2  A1 ) cos(t   )
A  A1  A2
   2  1  (2k  1)π
Superposition of two simple harmonic motions with
same frequency in the same direction
Summary
（1）If the phase difference is
 2   1  2 k π ( k  0 ， 1，) then
A  A1  A2
Enhance
（2）If the phase difference is
 2   1  ( 2 k  1) π ( k  0 ， 1，)
A  A1  A2
Weaken
（3）The phase difference can be arbitrary value
A1  A2  A  A1  A2
Example
There are two objects in simple harmonic motions in the same
direction as shown in the following figure: try to resolve:
1、the amplitude of the combined oscillation
2、the motion equation of superposition
x x1 (t )
A
1
2
Resolution  
T
A  A2  A1
A2
x 2 (t )
T

A2


A1 cos  1  0 1     1  

2
2
A



A2 cos  2  0  2     2 
A1
2
2

2

  
x  A2  A1 cos(
t )
2
T
2
t
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
Simple harmonic motion along x direction
x  A1 cos( t   1 )
y  A 2 cos( t   2 )
Simple harmonic
motion along y direction
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
x  A1 cos( t   1 )
y  A2  cos( t   2 )
y
x
 cos  t cos  1 sin  t sin  1
A1
y
 cos  t cos  2  sin  t sin  2
A2
2
x
2
x
y
2 xy
2
 2
cos( 2  1 )  sin ( 2  1 )
2
A1 A2 A1 A2
Conclusion：the superposition of two simple harmonic
motions in perpendicular directions with the same frequency is
ellipse motion, which is determined by phase difference.
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
Discussion：
（a）If 2  1  0 (or 2k )
2
2
x
y
2 xy
 2
0
2
A1 A2 A1 A2
y
2
 x
y 
    0
 A1 A2 
A2
A2
y  x Slop  0
A1
A1
Conclusion：the trajectory of the
combined oscillation is a straight line.
x
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
x 2 y 2 2 xy
2


cos(



)

sin
( 2  1 )
2
1
2
2
A1 A2 A1 A2

（b）If： 2  1 
2
2




or
2
k

1

2 
2
x
y
 2 1
2
A1 A2
Conclusion：the trajectory of the
combined oscillation is a perfect ellipse.
y
x
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
2
2
x
y
2 xy
2
 2
cos( 2  1 )  sin ( 2  1 )
2
A1 A2 A1 A2
（c）If  2   1  2 k  1
2
y
2
x
y
2 xy
 2
0
2
A1 A2 A1 A2
2
 x
y 
    0
 A1 A2 
x
A2
A2
y   x , Slop :   0
A1
A1
The trajectory of the combined oscillation is a straight line.
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
The
superposition of
two simple
harmonic
motions in
perpendicular
directions with
the same
frequency but
different phase
difference.
Superposition of two simple harmonic motions with
same frequency in perpendicular direction
Lissajou Figure
*Superposition of several simple harmonic motions
with same frequency in the same direction
x1  A1 cos t  1 
x2  A2 cos t   2 
x3  A3 cos t   3 
A3
A
A2
A1
A
A  A , tan  
2
y
2
1

x  A cos  t   
2
x
3
Ay
Ax
A x  A1 cos  1  A2 cos  2  A3 cos  3   
A y  A1 sin  1  A2 sin  2  A3 sin  3   
*Superposition of several simple harmonic motions
with same frequency in the same direction

x1  A0 cos  t
A
x 2  A0 cos( t    )
    
o A1 A2 A3 A A5 x
x3  A0 cos( t  2   )
A   Ai  NA0

i
x N  A0 cos[ t  ( N  1)   ]

4
(1)    2k π
Discussion ( k  0,  1,  2, )
(2) N    2k ' π
( k '  kN , k '   1,  2 , )

A4  
A3


A
2
 O



A6  A
x
1

A5
A0
Superposition of two simple harmonic motions with
different frequency in the same direction beat
The rotating angular
speed of


A1 with respect to A2 is
 2  1
1

A

A1

A2

A2
O
2

A
1
A1
x
2
For the superposition of two
simple harmonic motions in
the same direction when the frequencies of two component
oscillations are relatively large while difference of the two
frequencies is very small, the phenomenon of the resultant
amplitude ever and agah strengthening and weakening is called
the beat.
Superposition of two simple harmonic motions with
different frequency in the same direction beat
Superposition of two simple harmonic motions with
different frequency in the same direction beat
The period of the beat：
Beat frequency：
Assume :
2
T
 2  1
 2  1

  2  1
2
A1  A2  A
x1  A cos(1t   )
x 2  A cos( 2 t   )
x  x1  x 2  A cos(1t   )  A cos( 2 t   )
 2  1
 2  1
 2 A cos
t cos(
t  )
2
2
Superposition of two simple harmonic motions with
different frequency in the same direction beat
If
 2   1   2   1
 2  1
  2  1

t cos
t  
We obtain： x  2 A cos
2
 2

Superposition of two simple harmonic motions with
different frequency in the same direction beat
 2  1
  2  1

x  2 A cos
t cos
t  
2
 2

Assume :   0
 2  1
 2 1
x  2 A cos 2
t  cos 2
t
2
Amplitude：2 A cos 2
 2  1
2
2
t
The amplitude of the combined oscillation
various slowly and periodically with the time.
 2   1 is a harmonic oscillator.
cos 2
t
2
Superposition of two simple harmonic motions with
different frequency in the same direction beat
The second method：Superposition of rotating vector

( 2  1 )t  ( 2  1 )

A
2 2
 2t   2
1t  1
o
1   2  0
x2

A
 2  1
1 
A1
x1
x
  2 π ( 2   1 )t
x
Superposition of two simple harmonic motions with
different frequency in the same direction beat
Amplitude A  A1 2(1  cos  )
 2  1
 2 A1 cos(
Beat frequency
Angular frequency
x1  x2
cos t 
A
2
2
t)
   2  1
( 2   1 )t

A2
o
 
A1 1
x2
1t  2t
1   2

2


A
x
x
 2  1
x1
Superposition of two simple harmonic motions with
different frequency in the same direction beat
Applications：
• Acoustics
• Speed measurement;
• Radio-technology;
• Satellite tracking
§9-7 Electromagnetic Oscillation
The phenomenon that charges and current, electric field
energy and magnetic field energy vary with time
periodically is called Electromagnetic Oscillation.
L
L
+
C
E
ε
C
I0
Q0
Q0
L
S
LC Circuit
I0
L
C
B
+ Q0
C
A

B
C

E
Q0
L

B
C
D
The free electromagnetic oscillation without damping
The equation of the free electromagnetic oscillation without damping
The self induction
electromotive force is ：
dI q
L

dt C
q
VC 
C
I
L
dI q
Equation of Circuit：  L

dt C
2
d q
1
dq
 I


q
2
dt
dt
LC
C
The equation of the free electromagnetic oscillation without damping
Charge： q  q 0 cos  t   

1
LC

1

Frequency：  
Period：T  2 LC
2 2 LC
dq
  q 0 sin  t   
Current： I 
dt
I   I max sin  t    I max  q 0
q0
Voltage： V 
cos  t   
C
Conclusion：In the circuit of LC，the current, voltage, charges
are in simple harmonic motions.
The variation of charges and current with time for a
free electromagnetic oscillation without damping
q i
Q0 I
0
O
π
2
﹡
π
﹡
2π
(t   )
π
q  Q0 cos( t   ) i  I 0 cos(t    )
2
The energy of the free electromagnetic oscillation without damping
2
q
q
Electric field energy： W e 
 0 cos 2  t   
2C 2 C
2
2 2
L

q
1
2
0
Magnetic field energy：W m  LI 
sin 2  t   
2
2
q 02
2

sin  t   
2C
The total energy of LC oscillation circuit：
q 02
W  We  W m 
2C
In the process of free electromagnetic oscillating without
damping, the electric field energy and the magnetic field energy
converts into each other constantly, while at any moment the
sum of them remains unchanged.
The energy of the free electromagnetic oscillation without damping
The conditions of conservation of electric and
magnetic field energy in LC oscillation circuit:
1. the resistances in the circuit must be zero ；
2. there are not any electromotive forces existing in
the circuit；
3. the electromagnetic energy can not be radiated in
the form of electromagnetic wave.
Therefore, LC electromagnetic oscillation
circuit is an ideal model of oscillation circuit.
LC electromagnetic oscillation
compared with simple harmonic motion
Equation of LC
electromagnetic oscillation
d 2q
2



q
2
dt
q  Q0 cos( t   )
dq
i
  Q0 sin(  t   )
dt
Equation of simple
harmonic motion
d2x
2
  x
2
dt
x  A cos( t   )
dx
v
  A sin(  t   )
dt
Example
In a LC circuit, it is known L = 260 μH，C = 120 pF, at the
beginning the electric potential difference between the two
plates of the condenser is U0 = 1V and the current is i0 = 0.
Try to resolve
Solution（1）Oscillation frequency： 
（2）The maximum current
At t = 0

1
2 π LC
  9 .01  10 5 Hz
q0  Q0 cos   CU 0
i0  Q0 sin   0
C
I 0  Q0  CU 0 
U 0  0.679 mA
L
Example
（3）the relation that the electric field energy between the two
plates of the condenser varies with time:
1
2
2
10
2
E e  CU 0 cos  t  (0.60  10 J ) cos  t
2
（4）the relation that the magnetic field energy in the self
induction coil varies with time:
1 2 2
10
2
E m  LI 0 sin  t  (0.60  10 J ) sin  t
2
1
10
（5） E e  E m  0.60  10 J  E e 0  CU 02
2
Therefore at any time the sum of the electric field energy and
the magnetic field energy equals the electric energy at the
initial time.
Brief Summary of this Chapter
1、 Characteristics of simple harmonic motion
(1) Definition of simple harmonic motion
(2) Kinematics and dynamics descriptions
(3) The velocity and acceleration of the oscillator
(4) The period, frequency and angular frequency of the oscillator
(5) The relation between amplitude, initial phase and initial
displacement and initial velocity
(6) The characteristics of mechanical energy
Brief Summary of this Chapter
2、Rotating vector
(1) to determine the phase and initial phase for simple harmonic
motion using rotating vector.
(2) to discuss the phase difference of two simple harmonic
motions.
3、The superposition of simple harmonic motions
(1) to calculate the amplitude and phase difference of the
combined oscillation, the component oscillations have same
frequency and same direction.
(2) Superposition of two simple harmonic motions with same
frequency in perpendicular direction，the corresponding chart in
the special phase difference.
(3) The engender of beat and the calculation of beat frequency.
Brief Summary of this Chapter
4、 Electromagnetic Oscillation
(1) to understand the physical diagram of electric
field and magnetic field oscillation in LC oscillation
circuit.
(2) to understand the differential equation of LC
oscillation circuit and to master the formula
of oscillation frequency for LC oscillation circuit.
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