chapter 16oscillations 16.1 simple harmonic motion 16.1.1linear restoring force linear restoring...
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Chapter 16 OSCILLATIONS
16.1 Simple Harmonic Motion
16.1.1 Linear restoring force
Linear restoring force:
xf
The magnitude of force is proportional to displacement respect to equilibrium position with opposite direction of displacement.
Linear restoring torque may be defined in the same way.
16.1.2 Simple harmonic motion
Spring oscillation analysis:
2
2
dt
xdmkx
022
2
2
2
xdt
xdx
m
k
dt
xd
kxf
The equation can be solved as:
tCtCx sincos 21
We prefer using the form of cosine function
)cos( tAx
Angular frequency:
Amplitude: A
Phase constant:
m
k
The torsional oscillator
For small twists,
z
From rotational law:
Iz
2
2
dt
dI
)cos( tm
The solution is
A simple pendulum
sinmgF
For small angle swings
sin
We obtain
xL
mg
L
xmgmgF
The solution is
t
L
gAx cos
The physical pendulum
sinMgdz
For small angular displacement
2
2
dt
dIMgd
We get
tAcos
whereI
mgd
The definition of simple harmonic motion
When the resultant force or resultant torque exerted on a body is of linear restoring, the body undergoes a oscillation called simple harmonic motion.
When a oscillation is described by a cosine function, the oscillation is called simple harmonic motion.
To determine a SHM, three factors are needed, that is, the angular frequency, the amplitude, the phase constant.
)cos( tAx
16.1.3 Period, frequency, and angular frequency
Period is a time when one complete oscillation undergoing.
o
T2
Frequency is numbers of oscillation in unit time.
2
1
T
2
16.1.4 Amplitude, phase, and phase constant
Amplitude A is the maximum distance of an oscillator from its equilibrium position.
1)cos( t
AxA
Since
We get
Phase: t
Phase constant (or initial phase):
: angular frequency determined by the oscillation system;
Amplitude A and phase constant are determined from the initial conditions:
o
o
vv
xxt ,0
We get
sin
cos
Av
Ax
o
o
Therefore,
2
22
o
ov
xA
o
o
x
vtan
16.1.5 Oscillation diagram
16.2 Uniform Circular Motion and SHM
16.2.1 Rotational vector
16.2.2 Phase difference
Compare the phase difference of two oscillations with:
i) If 2 1 > 0, SHM-2 is in
before SHM-1;
ii) If 2 1 < 0, SHM-2 is in
after SHM-1;
iii) If 2 1 = 0, SHM-2 is in
synchronization with SHM-1 (or in synchronous phase);
iv) If 2 1 = , SHM-2 is in
anti-phase with SHM-1.
16.2.3 x, v, and a in SHM
)cos( tAx
)2
cos(
)sin(
tv
tAv
m
)cos(
)cos(2
ta
tAa
m
16.3 Oscillation Energy
The kinetic energy of a mass in simple harmonic motion
)(sin2
1 22 tkAEk
The potential energy associated with the force kx
)(cos2
1 22 tkAEp
The total energy
2
2
1kAE
The total energy is conserved for all SHMs.
By means of total energy
kEAx /2max
mEAv /2max
16.4 The Composition of SHM
16.4.1 Two SHMs with the same direction and frequency
)cos( 111 tAx o
)cos( 222 tAx o
)cos(
)cos()cos( 221121
tA
tAtAxxx
o
oo
)cos(2 122122
21 AAAAA
2211
2211
coscos
sinsintan
AA
AA
where
DISCUSSION:
(i) If two SHMs are synchronous,
(ii) If two SHMs are in antiphase,
(iii) In general case,
21 AAA
21 AAA
2121 AAAAA
16.4.2 Two SHMs with the same direction but different frequency
)cos( 11 tAx
)cos( 22 tAx
ttA
tAtAxxx
2cos
2cos2
)cos()cos(
2112
2121
m
2
12 : modulating frequency
a
2
12 : average frequency
If two component frequencies are very close and their difference is small. Therefore, the average frequency is much larger than modulating frequency. The phenomenon that the composite amplitude will change periodically is named a beat.
16.4.3 Two SHMs with the same frequency but perpendicular directions
)cos( 11 tAx o
)cos( 22 tAy o
)(sin)cos(2 122
1221
22
2
21
2
A
y
A
x
A
y
A
x
(i) If ,
(ii) If ,
(iii) If ,
012 xA
Ay
1
2
12 xA
Ay
1
2
212
122
2
21
2
A
y
A
x
16.4.4 Two SHMs with different directions and frequencies
Generally, the composite motion is rather complicated and sometimes is hard to get a periodical oscillation. However, if two frequencies have a ratio of two simple integers, we still get a periodical motion of which the trajectory is called a Lissajous figure.
16.5 Damped Harmonic Motion
A frictional force may be expressed as
vf
mavkxF
The Newton’s equation
02
2
kxdt
dx
dt
xdm
Three situations:
(i) For a small values of , satisfying mo 2
22
)'cos( teAx to
where oo 22'
(ii) If damping is large, mo 2
22
tt oo eCeCx )(2
)(1
2222
(iii) Critically damped motion:
tetCCx )( 21
16.6 Forced Harmonic Motion: Resonance
tFkxdt
dx
dt
xdm o cos
2
2
tfxm
k
dt
dx
mdt
xdo
cos2
2
)cos()'cos( tAtAex ot
16.6.1 Characterization of forced oscillation
22222 4)(
o
oo
fA
22
2tan
o
16.6.2 Resonance
222 2 oIf , the resonant amplitude
This is called a displacement resonance. 222
o
or
fA
Problems:
1. 16-26 (on page 367), 2. 16-28, 3. 16-36, 4. 16-46, 5. 16-58, 6. 16-60,
7. A cart consists of a body and four wheels on frictionless axles. The body has a mass m. The wheels are uniform disks of mass M and radius R. The cart rolls, without slipping, back and forth on a horizontal plane under the influence of a spring attached to one end of the cart. The spring constant is k. Taking into account the moment of inertia of the wheels, find a formula for the frequency of the back-and-forth motion of the cart.
8. Assume a particle joins two perpendicular SHMs of
Draw the trajectory of combined oscillation of the particle by using rotational vector method.
)4
cos(
cos2
ty
tx