the rules to distinguish simple harmonic motion
DESCRIPTION
Review:. The rules to distinguish simple harmonic motion. 1) The restoring force of the oscillator is:. 2) The dynamical equation of the oscillator is:. 3) The kinetic equation of the oscillator is:. Amplitude A , Period T / Frequency f , phase constant φ. - PowerPoint PPT PresentationTRANSCRIPT
The rules to distinguish simple harmonic motion
1) The restoring force of the oscillator is: f kx
22
20
d xx
dt
2) The dynamical equation of the oscillator is:
3) The kinetic equation of the oscillator is:
cosAx t k
m
Review:
2 k
m
22
mT
k
1 1
2
kf
T m
The period and the frequency depend only on the mass of theparticle and the force constant of the system.
Amplitude A, Period T/ Frequency f, phase constant φ
m
k1 k2
Determine the frequency of the following simple harmonic vibration:
1 1
2
kf
T m
1 2netF k k x kx 1 2k k k
1 21 1 1
2 2
k k kf
T m m
m
k1 k2
The forces on the two springs are:
1 1 1f k x2 2 2f k x
The forces on the vibrator is:
f kxThe displacement of the vibrator is:
1 2x x x We have: 1 2
1 2
f f f
k k k
The frequency of the simple harmonic vibrator is:
m
k1 k2
1 2
1 2
1 1
2 2
m k kmf
k k k
1 2
1 2
k kk
k k
The forces on the spring 1, spring 2 and the vibrator are equal.
The characters of simple harmonic motion
1) Simple harmonic vibration is periodic motion
2) The state of the oscillator is determined by the parametersof 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. Itsamplitude A=0.12m, period T=2s. When t=0, its displacementis x(0)=0.06m, moving to the positive direction of theequilibrium 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.
0.12A m
solution 1) the kinetic equation of the particle is:
cosAx t
2T s 2/rad s
T
(0) 0.06 cos 0.12cosx m A
cos 1/ 2 1
3
sindx
v A tdt
The velocity of the vibrator:
(0) sinv A 0 sin 0
1
3
1
3
Thus, the kinetic equation of the particle is:
( ) 0.12cos3
x t t
2)t=T/4, the position, velocity and acceleration of the particle
the kinetic equation of the particle is: ( ) 0.12cos3
x t t
the velocity of the particle is: 0.12 sin3
dxv t
dt
the acceleration of the particle is:20.12 cos
3
dva t
dt
Take t=T/4=0.5 s into the above equations, we get:
mx 104.0 smv /189.0 2/03.1 sma
3) When the vibrator arrives the equilibrium position, the displacement is 0. We get:
)3
cos(12.00 t
,2 ,1 ,2
)12(3
kkt
st 83.065
For the first time, t is minimum.
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.
0
12
1
2
)(st
)(cmx
s1
solution the kinetic equation of the particle is:
cosAx t 0.02A m
(0) cos 0.02cos 0.01x A
0
12
1
2
)(st
)(cmx
s1
cos 0.5 2
3
2(1) cos 0.02cos 0.02
3x A t m
22
3
4
3
)
3
2
3
4cos(2 tx
??
The energy consideration simple harmonic motion
2 2 2 2 2 21 1 1sin sin
2 2 2kE mv mA t kA t
2 2 21 1cos
2 2pE kx kA t
21
2k pE E E kA
pk EEE
t0 T2
T
4
T
4
3T
E
)0(
pEkE
A
kE pEE
A o
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.
EkA2
1
4
1 2
The average potential energy：
T
dttkAT 0
22 ) (cos2
11
The average kinetic energy of simple harmonic vibration
T
k dtmvT
E0
2
2
11
EkAdttmAT
T
2
1
4
1)(sin
2
11 2
0
222
T
P dtkxT
E0
2
2
11
§1-2 represent simple harmonic vibration by rotating vector method
The kinetic equation of the simple harmonic vibrator
cosAx t
The simple harmonic vibration can be expressed by Rotating vector.
)cos( tA
M
t
xo
A
X axis is vibration axis
O is the equilibrium position
The magnitude of vector Ais identical with the amplitude A.
The angular speed ωis the angular frequency of the vibration.
Φ: the initial phase / phase angle.
The displacement of the vibrator away from theequilibrium position is the projection of vector A on x axis.
The vector A is rotating vector.
The three parameters of simple harmonic vibration areexpressed by rotating vector method
)cos( tA
M
t
xo
A
)cos(2 tA
ARv
22 ARan
)cos(
)cos(2
tA
taa nx
v
na
A
x
O
xa
the velocity of the vibrator:
the direction of the velocity is negative when 0<ωt+φ<π
the acceleration of the vibrator:
Example 1-1-3A particle moves along x axis in simple harmonic motion. Itsamplitude A=0.12m, period T=2s. When t=0, its displacementis x(0)=0.06m, moving to the positive direction of theequilibrium position. Find: the kinetic equation of the simple harmonic motion.
mx 06.00
Solve it with rotating vector method2
T
0.12A m
(0) 0v 1
3
( ) 0.12cos3
x t t
Ox
The three parameters of simple harmonic vibration areexpressed by rotating vectormethod )cos( tA
M
t
xo
A
Rotating vector method
v
na
A
x
O
xa
the velocity and the acceleration of the vibrator in rotating vector method:
X>0
v>0
a>0
§1-3 damped oscillationsThe 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 kxThe restoring force of damped oscillation is:
tresis iveF kx F
If the resistive force is proportional to the velocity and acts in
the direction opposite the velocity: tresis iveF bv
2
2
dx d xF kx bv kx b m
dt dt
2
20
d x dxm b kx
dt dt
If the resistive force is small, 4b mk
Solve the differential equation
2( ) cosb
tmx Aet t
2( ) cosb
tmx Aet t
The oscillatory character of the motion preserved.
The amplitude of the vibration decreases with time.
t
x
2
bt
mAe
The vibration ultimately ceases.
Underdamped oscillator
220 2
b
m
20
k
m
The angular frequency of the underdamped oscillation:
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
t
x
The vibrator returns to equilibrium
in an exponential manner.
If b>2mω0, the medium is highly viscous. The system does not oscillate but simply returns to its equilibrium position.
t
x
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.
§1-4 forced oscillationsThe 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 periodically0( ) sinF t F t
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: net drivenF F kx bv
2
02sin
d x dxm F t b kx
dt dt
Connecting with Newton’s law:
Solve the differential equation: ( ) cosAx t t
0
222 2
0
/F mA
bm
The amplitude A:
ω0 natural frequency
When 0 , the amplitude A arrives its maximum
Resonance
/ Resonance frequency
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: ( ) cosAx t t
It is a harmonic vibration.
§1-5 superposition of harmonic vibrationssuperposition
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
1 1 1cosx A t 2 2 2cosx A t
A: rotating vector method
The net vibration: 1 2x x x
O x
1A��������������
)φ1
2A��������������
)φ2)
A��������������
122122
21 cos2 AAAAA
The amplitude A of net vibration:
The initial phase angle of the net vibration:)ψ
2211
22111
coscos
sinsin
AA
AAtg
y
x
The angular frequency of net vibration is identical with that of the two components.
A��������������
)ψO x
1A��������������
2A��������������
Thus, the kinetic equation of the net vibration is:
cosAx t
122122
21 cos2 AAAAA
2211
22111
coscos
sinsin
AA
AAtg
Resume
simple harmonic motion
1) The restoring force f kx
22
20
d xx
dt 2) The dynamical equation
3) The kinetic equation
cosAx t k
m
cosAx t
sinAdx
v tdt
maxx A
maxv A
2 cos( )dv
a tdt
A 2
maxa A
2( ) cosb
tmx Aet t
Simple harmonic motion: cosAx t
Damped oscillation, the resistive force: tresis iveF bv
220 2
b
m
20
k
m
t
x
2
bt
mAe
Underdamped oscillator b is very small
Critically damped b=2mω0, ω=0
t
xoverdamped b>2mω0
t
x
Steady forced oscillations
The external periodic force 0( ) sinF t F t
( ) cosAx t t
The net force on the oscillator: net drivenF F kx bv
0
222 2
0
/F mA
bm
When 0
the amplitude A arrives its maximum
Resonance