Xem mẫu
- Oscillations
Pham Tan Thi, Ph.D.
Department of Biomedical Engineering
Faculty of Applied Science
Ho Chi Minh University of Technology
- What is an Oscillation?
• Any motion that repeats itself
• Described with reference to an
equilibrium position
where the net force is zero, and a
restoring force
which acts to return object to equilibrium
• Characterize by:
- Periodic (T) or frequency (f) or angular frequency (ω)
- Amplitude (A)
- 14.2 Model for periodic motion. When It’s simplest
the body is displaced from its equilibrium rium position,
position at x = 0, the spring exerts a x-component o
restoring force back toward the equilib-
Simple Harmonic Oscillation/Motion rium position. change in the
exerts on the
(a) ax = Fx>m.
Total force exerts to object given by Hooke’s law x . 0: glider displaced Fx , 0, so ax , 0: Figure 14.2
Whenever the
F = kx to the right from the
equilibrium position.
stretched spring
pulls glider toward tends to restor
equilibrium position.
restoring forc
Newton’s second law of motion: y y ax
ing to return th
d2 x Fx x Fx
n Let’s analyz
right to x = A
F =m 2 x
mg
x
(Fig. 14.2a). Th
dt When the body
then (b)
its motion it ov
rium position t
2 x 5 0: The relaxed spring exerts no force on the tion are to the r
d x k glider, so the glider has zero acceleration.
stop. We will s
2
= x= !2 x y y
The body then
dt m n the starting poi
O If there is no fr
General solution for motion x
mg
x
this motion re
toward the equ
x = Acos(!t + ) Phase of the motion In different
(c)
librium in diff
2 k x , 0: glider displaced Fx . 0, so ax . 0: force that tend
where ! = Angular frequency to the left from the compressed spring
m equilibrium position. pushes glider toward
equilibrium position.
Amplitude,
y ax y Here are some
Velocity v = A!sin(!t + ) The amplit
x Fx n Fx displacement f
x x positive. If the
2
Acceleration a= A! cos(!t + ) mg
motion is 2A. T
complete roun
through O to
- Period and Frequency
Period (T): is the time to complete one full cycle, or one oscillation [s]
Frequency (f): is the number of cycles per second [s-1]
The relation between Frequency and Period:
14.2 Simple Harmonic Motion 445
1 1
s A and gives an 14.13 How f = v and
x-velocity or T =
of SHM.
x
x
Tone cycle
x-acceleration a vary during Simple Harmonic
f Motion
x
x 5 2A x 5 0 x 5A
(14.18)
2A 2A/2 0 A/2 A
v0 x . We’ll sketch ax ! 2amax
14); then divide x
vx 5 0
). The right side ax
x
lt is vx
ax 5 0
x
vx 5 2vmax
(14.19) ax
vx x
ax 5 amax
x
a nonzero initial vx 5 0
ment. That’s rea- ax
x
vx
e velocity v0 x , it
ax 5 0
vx 5 vmax x
ax
vx x
ax 5 2amax
x
vx 5 0
I: Describing Motion
- Angular Frequency
The oscillation frequency is measured in cycles per second, Hertz.
We may also define an angular frequency ω, in radians per second, to
describe the oscillation
2⇡
!(in rad/s) = = 2⇡f
T
The position of an object oscillating with simple harmonic motion can then
be written as
x(t) = Acos!t
then, the maximum speed of this object is
2⇡A
vmax = = 2⇡f A = !A
T
- Displacement as function of time in SHM
2 k
x = Acos(!t + ) ! =
m
Changing m, A or k changes the graph of x versus t:
change f
- Mechanical Energy in Simple Harmonic Motion
Potential Energy, U
Consider the oscillation of a spring as a SHM, the potential energy
of the spring is given by
1 2
U = kx where k = m! 2
2
1 2 2
U = m! x
2
The potential energy in terms of time, t, is given by
1
U = m! 2 x2 where x = Asin(!t + )
2
1
U = m! 2 A2 sin2 (!t + )
2
- Mechanical Energy in Simple Harmonic Motion
Kinetic Energy, K
The kinetic energy of an object in SHM is given by
1 p
K = mv 2 where v2 = ! A2 x2
2
1
K = m! 2 (A2 x2 )
2
The kinetic energy in terms of time, t, is given by
1
K = m! 2 (A2 x2 ) where v = A!cos(!t + )
2
1 2 2 2
K = m! A cos (!t + )
2
- Mechanical Energy in Simple Harmonic Motion
Total Energy, E
The total energy of a body in SHM is the sum of its kinetic energy, K
and its potential energy, U
E =K +U
From the principle of conservation of energy, this total energy is
always constant in a closed system
E = K + U = constant
The equation of total energy in SHM is given by
1 1
E = m! 2 (A2 x ) + m! 2 x2
2
2 2
1 2 2 1 2
E = m! A OR E = kA
2 2
- vmax = A = vA (14.23)
Am
Mechanical
This agrees with Eq.Energy in Simple
(14.15): vx oscillates betweenHarmonic
-vA and +vA. Motion
Interpreting E,
Conservation of K, and U in SHM
Energy
Figure 14.14 shows the energy quantities E, K, and U at x = 0, x = ! A>2, and
1 1 1
x = ! A. Figure 14.15
1 2 is a graphical 2 display of Eq. 2 (14.21); energy 2 (kinetic,
E = mv + kx = kA = mv max
potential, and total) is plotted vertically and the coordinate x is plotted horizontally.
2 2 2 2
(a) The potential energy U and total mechanical (b) The same graph as in (a), showing 14.15 Kin
energy E for a body in SHM as a function of kinetic energy K as well energy U, and
displacement x At x 5 6A the energy is all potential; the kinetic as functions o
energy is zero. value of x the
The total mechanical energy E is constant. U equals the
Energy At x 5 0 the energy is all kinetic; show that the
E
the potential energy is zero. half potential
1
U5 2 kx2
K Energy
E5K1U
U
U K
x
2A O x A
x
2A O A
At these points the energy is half
kinetic and half potential.
- Thepoint massforce
restoring moveis in an arc.by
provided The restoring
Fu gravity;
= - mgsin
force isT proportional
the utension merely acts tonot make to uthebut to
(14.30)
sin mass
point u, so the
movemotion
in anisarc. simple
not The harmonic.
restoring forceHowever, if the angle
is proportional not tou isu 454
small,
but to sin u14 Periodic Motion
CHAPTER
The u,restoring
sinis very forceequal
so thenearly
motion is
is provided
u in by
nottosimple gravity;
radians (Fig.
harmonic. the tensionFor
14.22).
However, merely
ifT the angleacts
example, tosmall,
when
u is make = the
sin
u14.21 0.1 rad of a simple
point mass
(about
is very move
6°),
nearly sinin
equalu to =
anu0.0998,
arc.
in radians
The
Thea restoring Simple
difference force
(Fig. 14.22).of onlyFor
Pendulum
is proportional
0.2%. With
example, thisutoapproximation,
not
when butrad
=u 0.1 to
pendulum.
uThe dynamics The path of
line but the a
(Fig. 14.21b).
sin Eq.
(about
u, so(14.30)
the motion
6°), sin u is0.0998,
becomes
= not simple
a harmonic.
difference of However,
only 0.2%. if With
the angle
this is small,(a)sin
uapproximation, A realupendulum
The tangential component of the net force: the motion is
isEq.
very nearly
(14.30) equal to u in radians (Fig. 14.22). Forx example, when u = 0.1 rad
becomes to x or (becaus
In Fig. 14.2
(about 6°), sin u = F ✓ = a Fdifference
0.0998, mgsin✓
u = - mgu of =only - xmg0.2%. With or this approximation, radial compon
Fu = - mgu = - mg L or
Eq. (14.30) becomes force:
The restoring force is provided by gravity; L
mg the tension is
merely acts to make =mg- x(bob)
Fu mass x move in an arc. (14.31)
Fu =the point
- mgu
Fu = -= - mg x L or (14.31) The restoring
o be L L point mass m
and
e
The restoring
if the force is then
pendulum’s proportional
oscillation is
mgsmall to the(θcoordinate
< 10°) for small displace- sin u, so the m
Thements,
restoring
and force is then
the force Fuis=k - = to
proportional
constant xthe coordinate for small displace- (14.31) is very nearly
ble. mgLmg>L. From Eq. (14.10) the angular fre- (about 6°), sin
ments, andvtheof force
F constant is k== smallFrom
mg>L. Eq. (14.10) the angular fre-
.
deled quency ✓ = pendulum
a simple mg✓ with xamplitude is Eq. (14.30) be
ed
The
quencyrestoring
v of a force
simpleispendulum
then proportional
with smallto the coordinate
Lamplitude is for small displace-
mass.
ss. ments, and the force constant is k = mg>L. From Eq. (14.10) the angular fre-
(b) An idealized simple pendulum
The restoring force k is mg>L
then proportionalg to the coordinate
(simple pendulum,
quency v of a simple = k pendulum
v displacement, =mg>Lwith small
=the amplitude
g force (simple is
pendulum, (14.32)
ed for small A m B = mand A L constant
small is k
amplitude)= mg/L String is
s. v = = small amplitude) (14.32) assumed to be
The restoring
Am B m AL massless and
ments, and th
k mg>L g (simple pendulum, u unstretchable.
The corresponding
v = =frequency =and period small relationships are (14.32) T Bob is modeled quency v of a
The corresponding A mfrequency
B mand period A L relationships are amplitude) as a point mass.
L
v
mg cos u 1
v frequency g
cos u The corresponding
ƒ =v =
1 g and period relationships
(simple pendulum,are
small amplitude) (14.33) x m
ƒ = =
2p 2p A L (simple pendulum, small amplitude) (14.33) The correspon
2p 2p A L mg sin u
os u v 1 g The restoring force on the
u mg cos u v
ƒ = = (simple pendulum, small amplitude) (14.33)
bob is proportional to sin u, ƒ =
2
2p 12p A L L
2p not to u. However, for small
T 2p
= 1= = 2pL sin u ^ u, so the motion is
(simple pendulum, small amplitude) u,approximat (14.34)
T = = =ƒ 2p
v A g (simple pendulum, small amplitude) (14.34) ely simple harmonic.mg 2p
v ƒ Ag T =
v
=
ments 2p 1 L
nts T = Note =that =these
2p expressions
(simple
do pendulum,
not involvesmall amplitude)
the mass of particle.
the particle.
14.22This
(14.34) For smallisangular displacements Note that t
u on a Notevthat these
ƒ expressions
Ag do not involve the mass of the This is force
u, the restoring Fu = - mgsin u on a
because the re
- Fig. 14.2b). When the bob of an oscillating simple pendulum passes through its equilib-
rium position, is its acceleration zero? ❙
14.6 Dynamics
The of The
Physical Pendulum Physical Pendulum
A physical pendulum is any real pendulum that uses an extended body, as con- 14.23 Dynamics of a physical pendulum.
The weight
trasted (mg)
to the idealized causes
model a restoring
of the simple pendulum with torque
all the mass concen- The body is free to rotate
trated at a single point. For small oscillations, analyzing the motion of a real, Pivot around the z-axis.
⌧ =
physical pendulum
z (mg)(dsin✓)
is almost as easy as for a simple pendulum. Figure 14.23 Irregularly
shaped z
shows a body of irregular shape pivoted so that it can turn without friction about body O The gravitational force
an axis through point O. In the equilibrium position the center of gravity is acts on the body at
directly below thewhen
pivot; in the displacement
position shown in the figure, the body is displaced its center of
clockwise the u gravity (cg).
from equilibrium by an angle u, which we use as a coordinate for the system. The
is counterclockwise
distance from O to the center of and viceis versa
gravity d, the moment of inertia of the body
d
d sin u
ifabout
the the axis of rotation through O is I, and the total mass is m. When the body
pendulum’s oscillations is small (θ < 10°)
is displaced as shown, the weight mg causes a restoring torque
cg
mg sin u
⌧z = (mgd)✓
tz = -1mg21d sin u2 (14.36)
456 sign
The negative CHAPTER
shows14 thatPeriodic Motion
the restoring torque is clockwise when the displace-
The
ment equation of motion
is counterclockwise, is
and vice versa. ⌃⌧ = I↵
z z mg cos u
456
When the body is released, it
CHAPTER 14 Periodic Motion oscillates about its equilibrium position. The
Comparing this with Eq. (14.4), Thewe see that the role of 1k>m2 for the spring-mass
restoring
motion is not simple harmonic because the torque 2 system
tz is is
proportional to sin u
played here by the quantitythe
torque on body
d ✓
rather than to u itself. However, if u is small, we can approximate sin u by u in is proportional to mg
1mgd>I2. Thus the angular frequency is
(mgd)✓ = I↵ = I
radians, just as we did in analyzing z the simple Comparing
pendulum.
2 thisthe
Then with Eq. (14.4),
motion wenotsee
is sin u, to u.that the for
However, role ofu,1k>m2
small , u,for the spring-ma
sin u ,
dt
approximately simple harmonic. With this approximation,system is played
mgd so the motion is approximately simple harmonic.
v = here by the(physical 1mgd>I2. Thus
quantity pendulum, smallthe angular frequency
amplitude) is
(14.38)
A I
2 tz = -1mgd2u
The equation of motion is gtz = Iaz , so
d ✓ mgd The frequency
= ƒ
mgd
is (physical
1>2p times pendulum,
this, and small
the period T is amplitude)
2
= ✓ v
A I
(14.38
dt I d 2u I
-1mgd2u = Iaz = I
T = 2p
dt 2 frequency
The ƒ isA1>2p (physical
times this, andpendulum,
the periodsmall
T is amplitude) (14.39)
General solution mgd
d 2u mgd
= - u (14.37)
dt 2 I Equation
T = (14.39)
2p isI the basis of a common
(physical method
pendulum, for experimentally
small amplitude) determin-(14.39
✓ = ✓o cos(!t + ) ing the moment of A inertia
mgd of a body with a complicated shape. First locate the cen-
ter of gravity of the body by balancing. Then suspend the body so that it is free to
Equation
oscillate (14.39)
about is theand
an axis, basis of a common
measure method
the period for experimentally
T of small-amplitude determi
oscillations.
ing the moment
Finally, use Eq.of(14.39)
inertiatoofcalculate
a body with a complicated
the moment shape.
of inertia First
I of the locate
body aboutthethis
ce
- Damped Oscillation
Consider an additional force on a body due
to friction
dx
Fx = bvx where vx =
dt
The net force is then
⌃Fx = kx bvx
Newton’s second law for the system is
dx d2 x
kx bvx = max OR kx b =m 2
dt dt
General solution r s
p ✓ ◆2
k2 b2
x = Ae t
cos(! 0 t + ) !0 =
m2 4m 2
= !2 2 =! 1
!
r
The period k b
!= ; =
m 2m
2⇡ 2⇡
T = 0 =p
! !2 2
- time for an oscillator with little damping
[see Eq. (14.42)] and with phase angle
f = 0. The curves are for two values of
Damped Oscillation the damping constant b.
b! 0.1!km (weak damping force) You can
t 0 x b! 0.4!km (stronger damping force)
x = Ae cos(! t + ) A Ae2(b/2m)t
and sec
whethe
Decay smaller than ω
=0 tedious
The
O t ways. F
If b is large enough to be T0 2T0 3T0 4T0 5T0 because
k b2 p Eq. (14
=0 b = 2 km With stronger damping (larger b):
quickly
m 2
4m (Critical damping) 2A • The amplitude (shown by the dashed Seco
p curves) decreases more rapidly. v = 2
If b > 2 km (Overdamping) • The period T increases
(T0 ! period with zero damping).
p
If b < 2 km (Underdamping)
Overdamped
When E
Critically damped tem no
when it
If b
is no os
ical dam
where C
- Mechanical Waves
Pham Tan Thi, Ph.D.
Department of Biomedical Engineering
Faculty of Applied Sciences
Ho Chi Minh University of Technology
- What is Wave?
Wave motion, or abbreviated wave, is propagation of oscillation. Wave is capable of
accompanying energy and transferring in different media. Wave can be changed its
direction (refraction, reflection, scattering, diffraction,…) or its energy (by absorption,
emission,…) or even its characteristic (e.g., different frequency in non-linear medium)
- Mechanical Waves
All mechanical waves require:
(1) sources of disturbance,
(2) a medium that can be disturbed,
(3) physical medium through which elements of a medium can be
influenced from one another.
- Wave Properties
Wavelength
2π
k
Frequency (Hz) 1
f =
T
Angular
ω = 2πf = 2π
Frequency
T
ω
Velocity v= = λf
k
- The Function of a Wave
✓ ◆
2⇡ 2⇡
y(x, t) = Acos x t
T
= Acos (kx !t)
- The Equation of a Wave
The function of a wave:
y = Acos(!t kx)
Derivatives with respect to the time:
dy d2 y
= A!sin(!t kx) 2
= A! 2 cos(!t kx)
dt dt
Derivatives with respect to the position:
dy d2 y
= Aksin(!t kx) = Ak 2 cos(!t kx)
dx dx 2
where: d2 y
2⇡ 2⇡ ! dt2 !2
k= = = d2 y
= 2 = v2
vT v k
dx2
then:
d2 y 1 d2 y
=0 Wave Equation
dx2 2
v dt 2
nguon tai.lieu . vn
