Download Superposition of two simple harmonic motions with same frequency

Chapter 9 Oscillation
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
Chapter 9
Oscillation
§ 9-1 Simple Harmonic Motion, Amplitude,
Period and Frequency, Phase
Reason of Mechanical Oscillation
•Restoring force
•Inertia
Simple Harmonic Motion
1. The relationship 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
Simple Harmonic Motion
2. Dynamics 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
Simple Harmonic Motion
3. kinematics 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
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
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
Simple Harmonic Motion
Results：
（1）The vibration of the spring oscillator is
simple harmonic motion.
k
Angular
frequency：
（2）

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.
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 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.
Method of Characteristics of Simple Harmonic Motion
i、Mathematic Equation： x  A cos(t   )
ii、Diagram of Rotating Vector：
x/cm
iii、Diagram x-t (or v-t, a-t)
0
x  2 cos(
4π 2π
t  ) cm
3
3
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 and graph of x versus t
for simple harmonic motion.
Method 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 
Method 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
Method 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, draw the
rotating vector diagram, and
then confirm the direction of
velocity, v0  0, or v0  0
at last we could obtain

x
Example
An object is in simple harmonic motion along x axis，its
amplitude is 12cm，the period is 2s, 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 form the position to
equilibrium position.
Example
Solution：(1) First we need to obtain the equation of
oscillation. Assume
x  A cos ( t   )
Aaccording 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 
Example
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
)
Example
(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
Example
(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
Example
Method 2：to resolve using rotating vector
3 2 5
 


2
3
6
   t

5 5
t 



6 6
2 3
3 / 2
x
Example
Sample: Two objects are in simple harmonic motion with same
frequency and amplitude along same direction. At the initial
moment the object is at the position of x1= A/2 处 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 difference of phase 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
Example
-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
Example
Method 2：to resolve using rotating vector
  
x
Example
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 quality）
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

Application
*C
 2π
J
mgl

P
（C is the center of quality）
• 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
t
O
E
m 2  k
Ep
The total mechanical energy：
E  Ek  E p
Ek
O
1 2 1
1
2
2 2
E  kA  m  A  mv m
2
2
2
t
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 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
当 x  A 2 时： E p  kx  k    E
2
2 2
4
1
2
kx0
2
3
Ek  E  E p  E
4
1
1 1
2
A  0.707 A
  kA x0  
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 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)
Phase difference  2   1  2 k 
x

oA
1
A2

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）Phase difference  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
Conclusion
（1）Phase difference  2   1  2 k π ( k  0 ， 1，)
A  A1  A2
Enhance
)
（2）Phase difference  2   1  ( 2 k  1) π ( k  0 ， 1，
A  A1  A2
Weaken
（3）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 equations 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
Example

A
A
A2

sin  sin  6
 6
A
 20 
sin  
sin  sin  1
A2
6 10
6


2

A2

A1

    2   1    
2

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 the same 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 the same direction
Discussion：
（a）
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 the same direction
x 2 y 2 2 xy
2


cos(



)

sin
( 2  1 )
2
1
2
2
A1 A2 A1 A2

（b）当：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 the same direction
2
2
x
y
2 xy
2
 2
cos( 2  1 )  sin ( 2  1 )
2
A1 A2 A1 A2
（c） 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 the same 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 the same 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
2
The period of the beat：T 
 2  1
 2  1
  2  1
Beat frequency：  
2
Assume :
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
When  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
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
1. The free electromagnetic oscillation without damping
2. 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
2. 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
3. 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：
2
0
q
W  We  W m 
2
C
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.
3. 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 oscillation motion
Equation of LC
electromagnetic oscillation
Equation of simple
oscillation motion
d 2q
2



q
2
dt
d2x
2
  x
2
dt
q  Q0 cos( t   )
dq
i
  Q0 sin(  t   )
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
当t = 0 时

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 in 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.