Transcript Chapter 2

Review: The rules to distinguish simple harmonic motion
1) The restoring force of the oscillator is:
f  kx
2) The dynamical equation of the oscillator is:
d 2x
2


x0
2
dt
k
 
m
2
3) The kinetic equation of the oscillator is:
x  A cos t   
k

m
Amplitude A, Period T/ Frequency f, phase constant φ
2
m
T
 2

k
1
1
f  
T 2
k
m
The period and the frequency depend only on the mass of the
particle and the force constant of the system.
m
Determine the frequency of the following
simple harmonic vibration:
1
1
f  
T 2
k1
k
m
Fnet    k1  k2  x  kx
1
1
f  
T 2
k
1

m 2
k1  k2
m
k  k1  k2
k2
The forces on the two springs are:
f1  k1 x1
f 2  k2 x2
k1
k2
m
The forces on the vibrator is:
f  kx
The displacement of the vibrator is:
f
f1 f 2
We have:
 
k k1 k 2
x  x1  x2
The forces on the spring 1, spring 2
and the vibrator are equal.
k1
k2
m
k1k2
k
k1  k2
The frequency of the simple harmonic vibrator is:
1
f 
2
m
1

k 2
m  k1  k2 
k1k2
The characters of simple harmonic motion
1) Simple harmonic vibration is periodic motion
2) The state of the oscillator is determined by the parameters
of amplitude A, angular frequency ωand phase angle φ.
3) ω is determined by the natural quantities of the system.
A andφare determined by the system and the initial condition
of the system.
Example 1-1-1
A particle moves along x axis in simple harmonic motion. Its
amplitude A=0.12m, period T=2s. When t=0, its displacement
is x(0)=0.06m, moving to the positive direction of the
equilibrium position. Find:
1) The kinetic equation of the simple harmonic motion.
2) t=T/4, the position, velocity and acceleration of the particle.
3) The time to arrive the equilibrium position for the
first time.
solution 1) the kinetic equation of the particle is:
x  A cos t   
T  2s
A  0.12m
2

  rad / s
T
x (0)  0.06m  A cos   0.12 cos 
cos   1/ 2
1
 
3
The velocity of the vibrator:
dx
v
  A sin t   
dt
v (0)   A sin   0
1
 
3
sin   0
1
 
3
Thus, the kinetic equation of the particle is:


x (t )  0.12 cos   t  
3

2)t=T/4, the position, velocity and acceleration of the particle
the kinetic equation of the particle is: x (t )  0.12 cos   t   
3

dx


the velocity of the particle is: v 
 0.12 sin   t  
dt
3

dv


2
 0.12 cos   t  
the acceleration of the particle is: a 
dt
3

Take t=T/4=0.5 s into the above equations, we get:
x  0.104m
v  0.189m / s
a  1.03m / s
2
3) When the vibrator arrives the equilibrium position, the
displacement is 0. We get:
0  0.12 cos(t 
t 

3

3
)
 ( 2k  1)

2
For the first time, t is minimum.
t  5 6  0.83s
, k  1, 2, 
x (cm )
Example 1-1-2
The simple harmonic vibration
curve (cosine form) of a
vibrator is shown in the right
diagram. Try to get its
kinetic equation.
2
1
t (s )
0
1
2
1s
solution the kinetic equation of the particle is:
x  A cos t   
A  0.02m
x (0)  A cos   0.02 cos   0.01
cos   0.5
2

3
x (cm )
2
1
??
t (s )
0
1
2
2

x (1)  A cos t     0.02 cos   
3

2

 2
3
4

3
1s

  0.02m

4
2
x  2 cos( t   )
3
3
The energy consideration simple harmonic motion
1 2 1
1 2 2
2 2
2
Ek  mv  mA  sin t     kA sin t   
2
2
2
1 2 1 2
E p  kx  kA cos2 t   
2
2
1 2
E  Ek  E p  kA
2
Ek
E
(  0)
E  Ek  E p
E
E
p
Ek
A
0
T
4
T
2
3T
4
T
o
Ep
A
t
1. The amplitude and frequency of Ek and Ep are identical,
but the phase angle is opposite.
2. The mechanical energy of the vibrator is proportional to
A2.
The average kinetic energy of simple harmonic vibration
1
Ek 
T

T
0
1
2
mv dt
2
1 T1
1 2 1
2 2
2
  mA  sin (t   )dt  kA  E
T 0 2
4
2
The average potential energy：
1
EP 
T

T
0
1 2
1
kx dt 
2
T
1
1
2
 kA  E
4
2

T
0
1
kA2 cos 2 ( t   )dt
2
§1-2 represent simple harmonic vibration
by rotating vector method
The kinetic equation of the simple harmonic vibrator
x  A cos t   
The simple harmonic vibration can be expressed by
Rotating vector.
X axis is vibration axis
M

O is the equilibrium position
The magnitude of vector A
is identical with the amplitude A.
The vector A is rotating vector.
o

t A


x
A cos(t   )
The angular speed ωis the angular
frequency of the vibration.
Φ: the initial phase / phase angle.
The displacement of the vibrator away from the
equilibrium position is the projection of vector A on x axis.
The three parameters of simple harmonic vibration are
expressed by rotating vector method
M

o


t A

A cos(t   )
x
the velocity of the vibrator:
v  R  A

the direction of the velocity is
negative when 0<ωt+φ<π
the acceleration of the vibrator:
an  R    A  
2
2
a x  an cos(t   )
  A cos(t   )
  2 A cos(t     )
2

v
 
a x an 
A
O

x
Example 1-1-3
A particle moves along x axis in simple harmonic motion. Its
amplitude A=0.12m, period T=2s. When t=0, its displacement
is x(0)=0.06m, moving to the positive direction of the
equilibrium position. Find: the kinetic equation of the simple
harmonic motion.
Solve it with rotating vector method 
2
x0  0.06m


T
v (0)  0
1
 
3


x (t )  0.12 cos   t  
3


O
A  0.12m 
x
Rotating vector method
M

The three parameters of simple
harmonic vibration are
expressed by rotating vector
method
o


t A

A cos(t   )
x

the velocity and the
acceleration of the vibrator in
rotating vector method:

v
 
a x an 
A
O

x
a>0
X>0
v>0
§1-3 damped oscillations
The simple harmonic vibration is the simplification model of
an ideal frictionless system.
In realistic systems, resistive forces, such as friction and
viscous force, are present and retard the motion of the system.
The mechanical energy of the system diminishes in time.
The restoring force of simple harmonic vibration is:
F  kx
The restoring force of damped oscillation is:
F  kx  Fresis t ive
If the resistive force is proportional to the velocity and acts in
the direction opposite the velocity:
Fresis t ive  bv
2
dx
d x
F  kx  bv  kx  b  m 2
dt
dt
d 2x
dx
m 2  b  kx  0
dt
dt
If the resistive force is small, b  4mk
Solve the differential equation

x(t )   Ae


b
t
2m

 cos t   

The oscillatory character of
the motion preserved.
  2bm t 
x(t )   Ae
 cos t   


The amplitude of the vibration decreases with time.
The vibration ultimately ceases.
Underdamped oscillator
x
Ae

b
t
2m
t
The angular frequency of the underdamped oscillation:
 b 
   

 2m 
2
2
0
k
 
m
2
0
If b=0, there is no resistive force and the system is simple
harmonic vibration.
If b=2mω0, ω=0. The system does not oscillate.
Critically damped
x
The vibrator returns to equilibrium
in an exponential manner.
t
If b>2mω0, the medium is highly viscous. x
The system does not oscillate but simply
returns to its equilibrium position.
overdamped
As the damping increases, the time it takes the particle to
approach the equilibrium increases.
When a resistive force is present, the mechanical energy
of the oscillator decreases, and falls to zero eventually.
t
§1-4 forced oscillations
The mechanical energy of a damped oscillator decreases
in time as a result of resistive force.
The diminished mechanical energy is possible to be
compensated by applying an external force that does positive
work on the system.
If the external force varies periodically F (t )  F sin t
0
Note:
The angular frequency of the external force / driven force is
different from the natural frequency ω0 of the oscillator
The net force on the oscillator:
Fnet  Fdriven  kx  bv
Connecting with Newton’s law:
2
d x
dx
m 2  F0 sin t  b  kx
dt
dt
Solve the differential equation:
The amplitude A:
x(t )  A cos t   
F0 / m
A

2

2 2
0

 b 


m


2
ω0 natural frequency / Resonance frequency
When   0 , the amplitude A arrives its maximum
Resonance
Resonance: the oscillating system exhibits its maximum
response to a periodic driving force when the frequency of
the driving force matches the natural frequency of the
oscillator.
The kinetic equation of the forced oscillation / steady forced
vibration is:
x(t )  A cos t   
It is a harmonic vibration.
§1-5 superposition of harmonic vibrations
superposition
Superposition of motions
Superposition of vibrations
We will focus on Case : superposition of harmonic
vibrations with identical frequency and common vibration
axis
1) Superposition of two harmonic vibrations with identical
frequency and common vibration axis
x1  A1 cos t  1 
The net vibration:
x2  A2 cos t  2 
x  x1  x2
A: rotating vector method
A
The amplitude A of net vibration:
A
A  A  2 A1 A2 cos2  1 
2
1
2
2

A2

))φ
2)ψ
)φ
A1 y
1
The initial phase angle of the net vibration: O
 A1 sin 1  A2 sin 2 
  tg 

 A1 cos 1  A2 cos 2 
1
x
x
The angular frequency of net vibration is identical with that
of the two components.

Thus, the kinetic equation of the
net vibration is:

A2
x  A cos t   
A
A  A  2 A1 A2 cos2  1 
2
1
2
2
 A1 sin 1  A2 sin 2 
  tg 

A
cos


A
cos

1
2
2
 1
1
)ψ
O
A

A1
x
Resume
simple harmonic motion
1) The restoring force
f  kx
2
2) The dynamical equation
d x
2
 x  0
2
dt
3) The kinetic equation
x  A cos t   
k

m
x  A cos t   
dx
v
  A sin t   
dt
dv
2
a
  A cos(t   )
dt
xmax  A
vmax  A
amax  A
2
x  A cos t   
Simple harmonic motion:
Damped oscillation, the resistive force:

x(t )   Ae

 b 
   

2
m


2
0
2
b
 t
2m
Fresis t ive  bv

 cos t   

k
 
m
2
0
Underdamped oscillator b is very small
x
Ae
Critically damped b=2mω0, ω=0
x

b
t
2m
t
overdamped b>2mω0
t
x
t
Steady forced oscillations
The external periodic force
F (t )  F0 sin t
The net force on the oscillator:
Fnet  Fdriven  kx  bv
x(t )  A cos t   
F0 / m
A

2

2 2
0

When   0
 b  the amplitude A arrives its maximum


m


Resonance
2