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9/Maio/2018 – Aula 17

17 Movimentos oscilatórios, amortecidos e forçados 17.1 Movimento na vertical 17.2 Pêndulo simples 17.3 Pêndulo físico 17.4 Oscilações amortecidas 17.5 Oscilações forçadas

16 Movimento periódico 16.1 Movimento harmónico simples (MHS) 16.2 Conservação da energia no MHS

1

It’s simplest to define our coordinate system so that the origin O is at the equilib-rium position, where the spring is neither stretched nor compressed. Then x is thex-component of the displacement of the body from equilibrium and is also thechange in the length of the spring. The x-component of the force that the springexerts on the body is and the x-component of acceleration is given by

Figure 14.2 shows the body for three different displacements of the spring.Whenever the body is displaced from its equilibrium position, the spring forcetends to restore it to the equilibrium position. We call a force with this character arestoring force. Oscillation can occur only when there is a restoring force tend-ing to return the system to equilibrium.

Let’s analyze how oscillation occurs in this system. If we displace the body to theright to and then let go, the net force and the acceleration are to the left (Fig. 14.2a). The speed increases as the body approaches the equilibrium position O.When the body is at O, the net force acting on it is zero (Fig. 14.2b), but because ofits motion it overshoots the equilibrium position. On the other side of the equilib-rium position the body is still moving to the left, but the net force and the accelera-tion are to the right (Fig. 14.2c); hence the speed decreases until the body comes to astop. We will show later that with an ideal spring, the stopping point is at The body then accelerates to the right, overshoots equilibrium again, and stops atthe starting point ready to repeat the whole process. The body is oscillating!If there is no friction or other force to remove mechanical energy from the system,this motion repeats forever; the restoring force perpetually draws the body backtoward the equilibrium position, only to have the body overshoot time after time.

In different situations the force may depend on the displacement x from equi-librium in different ways. But oscillation always occurs if the force is a restoringforce that tends to return the system to equilibrium.

Amplitude, Period, Frequency, and Angular FrequencyHere are some terms that we’ll use in discussing periodic motions of all kinds:

The amplitude of the motion, denoted by A, is the maximum magnitude ofdisplacement from equilibrium—that is, the maximum value of It is alwayspositive. If the spring in Fig. 14.2 is an ideal one, the total overall range of themotion is 2A. The SI unit of A is the meter. A complete vibration, or cycle, is onecomplete round trip—say, from A to and back to A, or from O to A, backthrough O to and back to O. Note that motion from one side to the other (say, to A) is a half-cycle, not a whole cycle.

The period, T, is the time for one cycle. It is always positive. The SI unit is thesecond, but it is sometimes expressed as “seconds per cycle.”

The frequency, is the number of cycles in a unit of time. It is always posi-tive. The SI unit of frequency is the hertz:

This unit is named in honor of the German physicist Heinrich Hertz(1857–1894), a pioneer in investigating electromagnetic waves.

The angular frequency, is times the frequency:

We’ll learn shortly why is a useful quantity. It represents the rate of change ofan angular quantity (not necessarily related to a rotational motion) that is alwaysmeasured in radians, so its units are Since is in we may regardthe number as having units

From the definitions of period T and frequency we see that each is the recip-rocal of the other:

(14.1)f = 1T T = 1

ƒ (relationships between frequency and period)

ƒrad>cycle.2p

cycle>s,ƒrad>s.

v

v = 2pƒ

2pv,

1 hertz = 1 Hz = 1 cycle>s = 1 s-1

ƒ,

-A-A,

-A

ƒx ƒ .

x = A,

x = -A.

x = A

ax = Fx>m.axFx,

438 CHAPTER 14 Periodic Motion

x , 0: glider displacedto the left from theequilibrium position.

Fx . 0, so ax . 0:compressed springpushes glider towardequilibrium position.

Fx

ax

Fx

x 5 0: The relaxed spring exerts no force on theglider, so the glider has zero acceleration.

(b)

O x

y

xn

mg

y

(a)

xx

y

xn

mg

y

x . 0: glider displacedto the right from theequilibrium position.

Fx , 0, so ax , 0:stretched springpulls glider towardequilibrium position.

Fx

ax

Fx

(c)

xx

y

x

n

mg

y

14.2 Model for periodic motion. Whenthe body is displaced from its equilibriumposition at the spring exerts arestoring force back toward the equilib-rium position.

x = 0,

Application Wing FrequenciesThe ruby-throated hummingbird (Archilochuscolubris) normally flaps its wings at about 50 Hz, producing the characteristic sound thatgives hummingbirds their name. Insects canflap their wings at even faster rates, from 330 Hz for a house fly and 600 Hz for a mos-quito to an amazing 1040 Hz for the tiny bitingmidge.

Um objeto que esteja ligado a uma mola, por exemplo, e que seja desviado da sua posição de equilíbrio, tende a voltar a essa posição: a mola exerce uma força de restituição, o que causa um movimento periódico (oscilação).

2

16. Movimento periódico

Massa-mola

simulação

Aula anterior

Quando a força de restituição é diretamente proporcional ao afastamento da posição de equilíbrio, como no caso de molas ideais, tem-se um movimento harmónico simples (MHS).

3

!F = −k

!x

16.1 Movimento harmónico simples

!Fm=!a = d

2 !x

dt2d2!x

dt2= −km!x

14.2 Simple Harmonic Motion 439

Also, from the definition of

(14.2)v = 2pƒ = 2pT (angular frequency)

v,

Example 14.1 Period, frequency, and angular frequency

An ultrasonic transducer used for medical diagnosis oscillates atHow long does each oscillation take,

and what is the angular frequency?

SOLUTION

IDENTIFY and SET UP: The target variables are the period T andthe angular frequency . We can find these using the given fre-quency ƒ in Eqs. (14.1) and (14.2).

v

6.7 MHz = 6.7 * 106 Hz.EXECUTE: From Eqs. (14.1) and (14.2),

EVALUATE: This is a very rapid vibration, with large and andsmall T. A slow vibration has small and and large T.vƒ

vƒ

= 4.2 * 107 rad>s= 12p rad>cycle216.7 * 106 cycle>s2v = 2pf = 2p16.7 * 106 Hz2T = 1ƒ

= 1

6.7 * 106 Hz= 1.5 * 10-7 s = 0.15 ms

Test Your Understanding of Section 14.1 A body like that shown inFig. 14.2 oscillates back and forth. For each of the following values of the body’sx-velocity and x-acceleration state whether its displacement x is positive,negative, or zero. (a) and (b) and (c) and (d) and (e) and (f) and ❙ax = 0.vx 7 0ax 6 0;vx = 0ax 6 0;vx 6 0

ax 7 0;vx 6 0ax 6 0;vx 7 0ax 7 0;vx 7 0ax,vx

14.2 Simple Harmonic MotionThe simplest kind of oscillation occurs when the restoring force is directlyproportional to the displacement from equilibrium x. This happens if the springin Figs. 14.1 and 14.2 is an ideal one that obeys Hooke’s law. The constant ofproportionality between and x is the force constant k. (You may want toreview Hooke’s law and the definition of the force constant in Section 6.3.) Oneither side of the equilibrium position, and x always have opposite signs. InSection 6.3 we represented the force acting on a stretched ideal spring as

The x-component of force the spring exerts on the body is the negativeof this, so the x-component of force on the body is

(14.3)

This equation gives the correct magnitude and sign of the force, whether x is pos-itive, negative, or zero (Fig. 14.3). The force constant k is always positive and hasunits of (a useful alternative set of units is We are assuming thatthere is no friction, so Eq. (14.3) gives the net force on the body.

When the restoring force is directly proportional to the displacement fromequilibrium, as given by Eq. (14.3), the oscillation is called simple harmonicmotion, abbreviated SHM. The acceleration of a body inSHM is given by

(14.4)

The minus sign means the acceleration and displacement always have oppositesigns. This acceleration is not constant, so don’t even think of using the constant-acceleration equations from Chapter 2. We’ll see shortly how to solve this equa-tion to find the displacement x as a function of time. A body that undergoessimple harmonic motion is called a harmonic oscillator.

ax = d2x

dt 2 = - km

x (simple harmonic motion)

ax = d2x>dt 2 = Fx>mkg>s2).N>m

Fx = -kx (restoring force exerted by an ideal spring)

Fx

Fx = kx.

Fx

Fx

Fx

The restoring force exerted by an idealizedspring is directly proportional to thedisplacement (Hooke’s law, Fx 5 2kx):the graph of Fx versus x is a straight line.

O

Displacement x

Restoring force Fx

x , 0Fx . 0

x . 0Fx , 0

14.3 An idealized spring exerts arestoring force that obeys Hooke’s law,

Oscillation with such a restoringforce is called simple harmonic motion.Fx = -kx.

Why is simple harmonic motion important? Keep in mind that not all periodicmotions are simple harmonic; in periodic motion in general, the restoring forcedepends on displacement in a more complicated way than in Eq. (14.3). But inmany systems the restoring force is approximately proportional to displacementif the displacement is sufficiently small (Fig. 14.4). That is, if the amplitude issmall enough, the oscillations of such systems are approximately simple har-monic and therefore approximately described by Eq. (14.4). Thus we can useSHM as an approximate model for many different periodic motions, such as thevibration of the quartz crystal in a watch, the motion of a tuning fork, the electriccurrent in an alternating-current circuit, and the oscillations of atoms in mole-cules and solids.

Circular Motion and the Equations of SHMTo explore the properties of simple harmonic motion, we must express the dis-placement x of the oscillating body as a function of time, The second deriv-ative of this function, must be equal to times the function itself,as required by Eq. (14.4). As we mentioned, the formulas for constant accelera-tion from Section 2.4 are no help because the acceleration changes constantly asthe displacement x changes. Instead, we’ll find by noticing a striking similar-ity between SHM and another form of motion that we’ve already studied.

Figure 14.5a shows a top view of a horizontal disk of radius A with a ballattached to its rim at point Q. The disk rotates with constant angular speed (measured in so the ball moves in uniform circular motion. A horizontallight beam shines on the rotating disk and casts a shadow of the ball on a screen.The shadow at point P oscillates back and forth as the ball moves in a circle. Wethen arrange a body attached to an ideal spring, like the combination shown inFigs. 14.1 and 14.2, so that the body oscillates parallel to the shadow. We willprove that the motion of the body and the motion of the ball’s shadow areidentical if the amplitude of the body’s oscillation is equal to the disk radius A,and if the angular frequency of the oscillating body is equal to the angularspeed of the rotating disk. That is, simple harmonic motion is the projection ofuniform circular motion onto a diameter.

We can verify this remarkable statement by finding the acceleration of theshadow at P and comparing it to the acceleration of a body undergoing SHM,given by Eq. (14.4). The circle in which the ball moves so that its projectionmatches the motion of the oscillating body is called the reference circle; wewill call the point Q the reference point. We take the reference circle to lie in the

v2pƒ

rad>s),v

x1t21-k>m2d2x>dt 2,

x1t2.


... but Fx 5 2kx can be agood approximation to the forceif the displacement x is sufficiently small.

Ideal case: The restoring force obeys Hooke’slaw (Fx 5 2kx), so the graph of Fx versus x is astraight line.

Typical real case: Therestoring force deviatesfrom Hooke’s law ...

O Displacement x

Restoring force Fx

14.4 In most real oscillations Hooke’slaw applies provided the body doesn’tmove too far from equilibrium. In such acase small-amplitude oscillations areapproximately simple harmonic.

u

Shadow of ballon screen

Ball’s shadow

Ball on rotatingturntable

While the ball Qon the turntablemoves in uniformcircular motion,its shadow P movesback and forth onthe screen in simpleharmonic motion.

Illuminatedvertical screen

Illumination

Table

Light beam

A

A

2A O P

Q

Ball moves in uniformcircular motion.

Shadow movesback and forth onx-axis in SHM.

(a) Apparatus for creating the reference circle (b) An abstract representation of the motion in (a)

OP

A

y

x

Q

x !A cos uv

14.5 (a) Relating uniform circular motion and simple harmonic motion. (b) The ball’s shadow moves exactly like a body oscillatingon an ideal spring.

Aula anterior

4

x = Acos ω t +δ( )d2x

dt2= −kmx

Massa-mola

simulação

δ é a fase do movimento e

14.2 Simple Harmonic Motion 443

Displacement, Velocity, and Acceleration in SHMWe still need to find the displacement x as a function of time for a harmonic oscil-lator. Equation (14.4) for a body in simple harmonic motion along the x-axisis identical to Eq. (14.8) for the x-coordinate of the reference point in uniformcircular motion with constant angular speed Hence Eq. (14.5),

describes the x-coordinate for both of these situations. If at the phasor OQ makes an angle (the Greek letter phi) with the positive x-axis,then at any later time t this angle is We substitute this into Eq. (14.5)to obtain

(14.13)

where Figure 14.9 shows a graph of Eq. (14.13) for the particularcase The displacement x is a periodic function of time, as expected forSHM. We could also have written Eq. (14.13) in terms of a sine function ratherthan a cosine by using the identity In simple harmonicmotion the position is a periodic, sinusoidal function of time. There are manyother periodic functions, but none so simple as a sine or cosine function.

The value of the cosine function is always between and 1, so in Eq. (14.13),x is always between and A. This confirms that A is the amplitude of the motion.

The period T is the time for one complete cycle of oscillation, as Fig. 14.9shows. The cosine function repeats itself whenever the quantity in parentheses inEq. (14.13) increases by radians. Thus, if we start at time the time T tocomplete one cycle is given by

which is just Eq. (14.12). Changing either m or k changes the period of oscilla-tion, as shown in Figs. 14.10a and 14.10b. The period does not depend on theamplitude A (Fig. 14.10c).

vT = A km

T = 2p or T = 2pAmk

t = 0,2p

-A-1

cos a = sin1a + p>22.f = 0.v = 2k>m .

x = Acos1vt + f2 (displacement in SHM)

u = vt + f.f

t = 0x = A cos u,v = 2k>m .

EXECUTE: (a) When the force the spring exerts onthe spring balance is From Eq. (14.3),

(b) From Eq. (14.10), with ,

T = 1ƒ

= 13.2 cycle>s = 0.31 s

ƒ = v

2p=

20 rad>s2p rad>cycle

= 3.2 cycle>s = 3.2 Hz

v = A km

= B200 kg>s2

0.50 kg= 20 rad>sm = 0.50 kg

k = -Fx

x= - -6.0 N

0.030 m= 200 N>m = 200 kg>s2

Fx = -6.0 N.x = 0.030 m, EVALUATE: The amplitude of the oscillation is 0.020 m, the dis-

tance that we pulled the glider before releasing it. In SHM theangular frequency, frequency, and period are all independent of theamplitude. Note that a period is usually stated in “seconds” ratherthan “seconds per cycle.”

12

x

2TtO

xmax 5 A

T

T 12 T

2xmax 5 2A

14.9 Graph of x versus t [see Eq. (14.13)]for simple harmonic motion. The caseshown has f = 0.

(a) Increasing m; same A and k

O t

x

1 2 3

Mass m increases from curve1 to 2 to 3. Increasing m aloneincreases the period.

14.10 Variations of simple harmonic motion. All cases shown have [see Eq. (14.13)].f = 0

(b) Increasing k; same A and m

tO

x

3 2 1

Force constant k increases fromcurve 1 to 2 to 3. Increasing k alonedecreases the period.

(c) Increasing A; same k and m

O t

x

32

1

Amplitude A increases from curve1 to 2 to 3. Changing A alone hasno effect on the period.

PhET: Motion in 2DActivPhysics 9.1: Position Graphs andEquationsActivPhysics 9.2: Describing VibrationalMotionActivPhysics 9.5: Age Drops Tarzan

ω= k m

T = 1f=2πω

16.1 Movimento harmónico simples Aula anterior

Solução da equação diferencial de 2ª ordem

Velocidade e aceleração no movimento harmónico simples:

5

16.1 Movimento harmónico simples

x t( ) = Acosωtvx t( ) = −Aω senωtax t( ) = −Aω2 cosωt

ax t( ) = −ω2 x

Aula anterior

A energia mecânica total conserva-se no movimento harmónico simples:

6

16.2 Energia do movimento harmónico simples

Wnc = ΔU +ΔEcin = ΔE = 014.3 Energy in Simple Harmonic Motion 447

(Recall that Hence our expressions for displacement andvelocity in SHM are consistent with energy conservation, as they must be.

We can use Eq. (14.21) to solve for the velocity of the body at a given dis-placement x:

(14.22)

The sign means that at a given value of x the body can be moving in eitherdirection. For example, when

Equation (14.22) also shows that the maximum speed occurs at Using Eq. (14.10), we find that

(14.23)

This agrees with Eq. (14.15): oscillates between and

Interpreting E, K, and U in SHMFigure 14.14 shows the energy quantities E, K, and U at and

Figure 14.15 is a graphical display of Eq. (14.21); energy (kinetic,potential, and total) is plotted vertically and the coordinate x is plotted horizontally.x = ! A.

x = ! A>2,x = 0,

+vA.-vAvx

vmax = A km

A = vA

v = 2k>m ,x = 0.vmax

vx = !A km BA2 - a!

A2b2

= !A34 A k

mA

x = ! A>2,!

vx = !A km2A2 - x2

vx

sin2a + cos2a = 1.)

E is all potentialenergy.

2A 2 A

E is all potentialenergy.

E is partly potential,partly kinetic

energy.

E is partly potential,partly kinetic

energy.

E is all kineticenergy.

ax 5 amax ax 5 2amax

vx 5 6vmax

ax 5 amax ax 5 0

vmaxvx 5 0 vx 5 0vx 5 6

12

Åx

34 vmaxvx 5 6Å 3

4

12 ax 5 2 amax

12

O AA12

zero

zero

zero

E 5 K 1 U E 5 K 1 U E 5 K 1 U E 5 K 1 U E 5 K 1 U

14.14 Graphs of E, K, and U versus displacement in SHM. The velocity of the body is not constant, so these images of the body atequally spaced positions are not equally spaced in time.

The total mechanical energy E is constant.

(a) The potential energy U and total mechanicalenergy E for a body in SHM as a function ofdisplacement x

Energy

x

U 5 kx212

E

K

U

O A2A x

At x 5 6A the energy is all potential; the kineticenergy is zero.

At x 5 0 the energy is all kinetic;the potential energy is zero.

At these points the energy is halfkinetic and half potential.

(b) The same graph as in (a), showingkinetic energy K as well

Energy

x

E 5 K 1 U

O A2A

UK

14.15 Kinetic energy K, potentialenergy U, and total mechanical energy Eas functions of position for SHM. At eachvalue of x the sum of the values of K andU equals the constant value of E. Can youshow that the energy is half kinetic and half potential at x = !21

2 A?

Energia MHS

simulação Ecin +U = constante

Aula anterior

A energia mecânica total conserva-se no movimento harmónico simples:

7

16.2 Energia do movimento harmónico simples

12mvx

2 +12k x2 = constante

x t( ) = Acos ωt +δ( )vx t( ) = −Aω sen ωt +δ( )

12mvx

2 =12mA2ω22sen2 ωt +δ( )

12k x2 = 1

2k A2 cos2 ωt +δ( )

ω=km

E = Ecin +U =12k A2

Aula anterior

E = Ecin +U = constante

Exemplo O bloco da figura é afastado 0,2 m da posição de equilíbrio e largado, no instante t = 0. Efetua 15 oscilações completas em 10 s. Determine: a)  o período T ; b) a velocidade máxima; c) a posição e a velocidade em t = 0,8s.

a) T = 1f=2πω

f = 15oscil10s

=1,5 Hz

b)

x = Acos ω t +δ( )

δ = 0( )

T =1/ f = 0,667 s

vx t( ) = −Aω senωt vx,max ⇒ senωt = −1 ⇒ vx,max = Aω

vx,max = A 2π f( ) = (0,2 m) 2π( ) 1,5s−1( ) =1,88 m/s

c) x t=0,8s( ) = (0,2 m)cos 2π (0,8 s)(0,667 s)

= 0,062 m

vx t( ) = −Aω senωt v t=0,8s( ) = −(1,88 m/s)sen 2π (0,8 s)(0,667 s)

= −1,79 m/s8

Quando um objeto se encontra pendurado na extremidade de uma mola vertical, fica sujeito ao seu peso, para além da força de restituição da mola.

9

17.1 Movimento oscilatório na vertical

y ' = y − y0

Fy = −k y +mg∑ = 0

Se o sistema estiver em equilíbrio:

Se se deslocar o sistema ligeiramente para fora do equilíbrio:

d2y '

dt2= −kmy '

y ' = Acos ωt +δ( ) , ω=km

U =12k y '( )2 +U0

The path of the point mass (sometimes called a pendulum bob) is not a straightline but the arc of a circle with radius L equal to the length of the string (Fig. 14.21b). We use as our coordinate the distance x measured along the arc. Ifthe motion is simple harmonic, the restoring force must be directly proportionalto x or (because to Is it?

In Fig. 14.21b we represent the forces on the mass in terms of tangential andradial components. The restoring force is the tangential component of the netforce:

(14.30)

The restoring force is provided by gravity; the tension T merely acts to make thepoint mass move in an arc. The restoring force is proportional not to but to

so the motion is not simple harmonic. However, if the angle is small,is very nearly equal to in radians (Fig. 14.22). For example, when rad(about a difference of only 0.2%. With this approximation,Eq. (14.30) becomes

or

(14.31)

The restoring force is then proportional to the coordinate for small displace-ments, and the force constant is From Eq. (14.10) the angular fre-quency of a simple pendulum with small amplitude is

(14.32)

The corresponding frequency and period relationships are

(14.33)

(14.34)

Note that these expressions do not involve the mass of the particle. This isbecause the restoring force, a component of the particle’s weight, is proportionalto m. Thus the mass appears on both sides of and cancels out. (This is thesame physics that explains why bodies of different masses fall with the sameacceleration in a vacuum.) For small oscillations, the period of a pendulum for agiven value of g is determined entirely by its length.

The dependence on L and g in Eqs. (14.32) through (14.34) is just what weshould expect. A long pendulum has a longer period than a shorter one. Increas-ing g increases the restoring force, causing the frequency to increase and theperiod to decrease.

We emphasize again that the motion of a pendulum is only approximately sim-ple harmonic. When the amplitude is not small, the departures from simple har-monic motion can be substantial. But how small is “small”? The period can beexpressed by an infinite series; when the maximum angular displacement is the period T is given by

(14.35)

We can compute the period to any desired degree of precision by taking enoughterms in the series. We invite you to check that when (on either side of™ = 15°

T = 2pALga1 + 12

22 sin2 ™2

+ 12 # 32

22 # 42 sin4 ™2

+ Áb™,

©FS

! maS

T = 2pv

= 1ƒ

= 2pALg (simple pendulum, small amplitude)

ƒ = v

2p= 1

2p A gL (simple pendulum, small amplitude)

(simple pendulum,small amplitude)v = A k

m= Bmg>L

m= A g

L

vk = mg>L.

Fu = -mgL

x

Fu = -mgu = -mgxL

sinu = 0.0998,6°),u = 0.1u

sinuusinu,u

Fu = -mg sinu

Fu

u.x = Lu)


The restoring force on thebob is proportional to sin u,not to u. However, for smallu, sin u ^ u, so the motion isapproximately simple harmonic.

Bob is modeledas a point mass.

(a) A real pendulum

(b) An idealized simple pendulum

L

T

x

mg sin u

mg

mg cos u

m

u

u

String isassumed to bemassless andunstretchable.

14.21 The dynamics of a simple pendulum.

2p/2 2p/4 p/4 p/2 u (rad)

FuFu 5 2mg sin u(actual)Fu 5 2mgu(approximate)

22mg

2mg

mg

2mg

O

14.22 For small angular displacementsthe restoring force on a

simple pendulum is approximately equalto that is, it is approximately pro-portional to the displacement Hence forsmall angles the oscillations are simpleharmonic.

u.-mgu;

Fu = -mg sinuu,

O chamado pêndulo simples (ou matemático) consiste numa massa m , de dimensões desprezáveis, suspensa na extremidade de uma corda ou de uma barra, de comprimento L e massa desprezável.

10

17.2 Pêndulo simples

O ângulo θ que o pêndulo faz com a vertical varia, no tempo, como um seno ou um cosseno.

Pêndulo simples

simulação



(14.30)



or

(14.31)


(14.32)


(14.33)

(14.34)




(14.35)


T = 2pALga1 + 12

22 sin2 ™2

+ 12 # 32

22 # 42 sin4 ™2

+ Áb™,

©FS

! maS

T = 2pv

= 1ƒ


ƒ = v

2p= 1



m= Bmg>L

m= A g

L

vk = mg>L.

Fu = -mgL

x

Fu = -mgu = -mgxL

sinu = 0.0998,6°),u = 0.1u

sinuusinu,u

Fu = -mg sinu

Fu

u.x = Lu)




(a) A real pendulum


L

T

x

mg sin u

mg

mg cos u

m

u

u



2p/2 2p/4 p/4 p/2 u (rad)


22mg

2mg

mg

2mg

O



u.-mgu;

Fu = -mg sinuu,

No pêndulo simples, a força de restituição é proporcional a sen θ, enquanto no sistema massa-mola essa força é proporcional ao deslocamento (θ, neste caso):

11


Para ângulos θ suficientemente pequenos, o seno é aproximadamente igual ao próprio ângulo:

F = −mg senθ ≠ −mgθ

F =ma∑ ⇒ −mg senθ =md2x

dt2, x = Lθ

senθ ≈θ

senθ =θ −θ3

3!+θ5

5!−!≈θ

se θ <<1



(14.30)



or

(14.31)


(14.32)


(14.33)

(14.34)




(14.35)


T = 2pALga1 + 12

22 sin2 ™2

+ 12 # 32

22 # 42 sin4 ™2

+ Áb™,

©FS

! maS

T = 2pv

= 1ƒ


ƒ = v

2p= 1



m= Bmg>L

m= A g

L

vk = mg>L.

Fu = -mgL

x

Fu = -mgu = -mgxL

sinu = 0.0998,6°),u = 0.1u

sinuusinu,u

Fu = -mg sinu

Fu

u.x = Lu)




(a) A real pendulum


L

T

x

mg sin u

mg

mg cos u

m

u

u



2p/2 2p/4 p/4 p/2 u (rad)


22mg

2mg

mg

2mg

O



u.-mgu;

Fu = -mg sinuu,

Na aproximação dos ângulos pequenos, , tem-se

12


Nestas condições, a força de restituição é proporcional ao deslocamento, pelo que

F = −mg senθ ≈ −mgθ = −mg xL

senθ ≈θ

ω=km=mg Lm

=gL

T = 2πω

= 2π mk= 2π L

g

Note-se que o período T não depende da massa m do pêndulo.

Pêndulo vs M.Mola

simulação



(14.30)



or

(14.31)


(14.32)


(14.33)

(14.34)




(14.35)


T = 2pALga1 + 12

22 sin2 ™2

+ 12 # 32

22 # 42 sin4 ™2

+ Áb™,

©FS

! maS

T = 2pv

= 1ƒ


ƒ = v

2p= 1



m= Bmg>L

m= A g

L

vk = mg>L.

Fu = -mgL

x

Fu = -mgu = -mgxL

sinu = 0.0998,6°),u = 0.1u

sinuusinu,u

Fu = -mg sinu

Fu

u.x = Lu)




(a) A real pendulum


L

T

x

mg sin u

mg

mg cos u

m

u

u



2p/2 2p/4 p/4 p/2 u (rad)


22mg

2mg

mg

2mg

O



u.-mgu;

Fu = -mg sinuu,

Para ângulos de oscilação grandes, o período já depende do ângulo inicial:

13


T =T0 1+1

22sen2

θ02+1

2234⎛

⎝⎜⎞

⎠⎟2sen4

θ02+!

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

θ0,

T0 = 2πLg

L senθ

θL

14

17.3 Pêndulo físico

O pêndulo físico consiste num objeto que pode rodar em torno de um eixo que não passa pelo centro de massa.

No exemplo da figura, o momento da força gravítica, em relação ao eixo de rotação, é

τ = −LsenθMg ≈ −LθMg(na aproximação dos ângulos pequenos)

O período de oscilação depende do momento de inércia I e da distância L entre o eixo de rotação e o centro de massa:

T = 2π IMg L

15

17.4 Oscilações amortecidas

Na maior parte dos sistemas físicos reais, existem forças não-conservativas que tendem a dissipar a energia. No caso dos movimentos oscilatórios, a presença dessas forças é visível na diminuição da amplitude das oscilações. Por exemplo, no caso de atrito viscoso, essas forças (ditas de arrastamento), são proporcionais à velocidade:

Então, a força total sobre o objeto será a soma da força de restituição com a força de arrastamento:

!D = −b !v

!F = −k

!x −b

!v = −k

!x −b d

!xdt

d2!x

dt2= −km!x − bmd!xdt

M.Mola amortecida

simulação

16


ω ' =ω0 1−b

2mω0

⎛

⎝⎜⎜

⎞

⎠⎟⎟

km!x + bmd!xdt

+d2!x

dt2= 0 x = A0 exp −

b2mt

⎛

⎝⎜

⎞

⎠⎟ cos ω 't +δ( )

ω0 =km

A resolução desta equação diferencial conduz a uma solução do tipo

com

The angular frequency of oscillation is given by

(14.43)

You can verify that Eq. (14.42) is a solution of Eq. (14.41) by calculating the firstand second derivatives of x, substituting them into Eq. (14.41), and checkingwhether the left and right sides are equal. This is a straightforward but slightlytedious procedure.

The motion described by Eq. (14.42) differs from the undamped case in twoways. First, the amplitude is not constant but decreases with timebecause of the decreasing exponential factor Figure 14.26 is a graph ofEq. (14.42) for the case it shows that the larger the value of b, the morequickly the amplitude decreases.

Second, the angular frequency given by Eq. (14.43), is no longer equal tobut is somewhat smaller. It becomes zero when b becomes so large that

(14.44)

When Eq. (14.44) is satisfied, the condition is called critical damping. The sys-tem no longer oscillates but returns to its equilibrium position without oscillationwhen it is displaced and released.

If b is greater than the condition is called overdamping. Again thereis no oscillation, but the system returns to equilibrium more slowly than with crit-ical damping. For the overdamped case the solutions of Eq. (14.41) have the form

where and are constants that depend on the initial conditions and and are constants determined by m, k, and b.

When b is less than the critical value, as in Eq. (14.42), the condition is calledunderdamping. The system oscillates with steadily decreasing amplitude.

In a vibrating tuning fork or guitar string, it is usually desirable to have as littledamping as possible. By contrast, damping plays a beneficial role in the oscillationsof an automobile’s suspension system. The shock absorbers provide a velocity-dependent damping force so that when the car goes over a bump, it doesn’t continuebouncing forever (Fig. 14.27). For optimal passenger comfort, the system should becritically damped or slightly underdamped. Too much damping would be counter-productive; if the suspension is overdamped and the car hits a second bump justafter the first one, the springs in the suspension will still be compressed somewhatfrom the first bump and will not be able to fully absorb the impact.

Energy in Damped OscillationsIn damped oscillations the damping force is nonconservative; the mechanicalenergy of the system is not constant but decreases continuously, approachingzero after a long time. To derive an expression for the rate of change of energy,we first write an expression for the total mechanical energy E at any instant:

To find the rate of change of this quantity, we take its time derivative:

But and so

dEdt

= vx1max + kx2dx>dt = vx,dvx>dt = ax

dEdt

= mvxdvx

dt+ kx

dxdt

E = 12 mvx

2 + 12 kx2

a2a1C2C1

x = C1e-a1t + C2e-a2t

21km ,

km

- b2

4m2 = 0 or b = 21km

v = 2k>m v¿,

f = 0;e-1b>2m2t.Ae-1b>2m2t

v¿ = B km

- b2

4m2 (oscillator with little damping)

v¿


O

2A

A

T0 2T0 3T0 4T0 5T0

Ae2(b/2m)t

xb ! 0.1!km (weak damping force)b ! 0.4!km (stronger damping force)

With stronger damping (larger b):• The amplitude (shown by the dashed curves) decreases more rapidly.• The period T increases (T0 ! period with zero damping).

t

14.26 Graph of displacement versustime for an oscillator with little damping[see Eq. (14.42)] and with phase angle

The curves are for two values ofthe damping constant b.f = 0.

Piston

Viscousfluid

Lower cylinderattached toaxle; moves upand down.

Pushed up

Pushed down

Upper cylinderattached to car’sframe; moves little.

14.27 An automobile shock absorber.The viscous fluid causes a damping forcethat depends on the relative velocity of thetwo ends of the unit.

17


x = A0 exp −b2mt

⎛

⎝⎜

⎞

⎠⎟ cos ω 't +δ( )

A amplitude das oscilações diminui exponencialmente com o tempo:

M.Mola amortecida

simulação

E = E0 exp −bmt

⎛

⎝⎜

⎞

⎠⎟

e a energia mecânica (total) também:

The angular frequency of oscillation is given by

(14.43)

You can verify that Eq. (14.42) is a solution of Eq. (14.41) by calculating the firstand second derivatives of x, substituting them into Eq. (14.41), and checkingwhether the left and right sides are equal. This is a straightforward but slightlytedious procedure.

The motion described by Eq. (14.42) differs from the undamped case in twoways. First, the amplitude is not constant but decreases with timebecause of the decreasing exponential factor Figure 14.26 is a graph ofEq. (14.42) for the case it shows that the larger the value of b, the morequickly the amplitude decreases.

Second, the angular frequency given by Eq. (14.43), is no longer equal tobut is somewhat smaller. It becomes zero when b becomes so large that

(14.44)

When Eq. (14.44) is satisfied, the condition is called critical damping. The sys-tem no longer oscillates but returns to its equilibrium position without oscillationwhen it is displaced and released.

If b is greater than the condition is called overdamping. Again thereis no oscillation, but the system returns to equilibrium more slowly than with crit-ical damping. For the overdamped case the solutions of Eq. (14.41) have the form

where and are constants that depend on the initial conditions and and are constants determined by m, k, and b.

When b is less than the critical value, as in Eq. (14.42), the condition is calledunderdamping. The system oscillates with steadily decreasing amplitude.

In a vibrating tuning fork or guitar string, it is usually desirable to have as littledamping as possible. By contrast, damping plays a beneficial role in the oscillationsof an automobile’s suspension system. The shock absorbers provide a velocity-dependent damping force so that when the car goes over a bump, it doesn’t continuebouncing forever (Fig. 14.27). For optimal passenger comfort, the system should becritically damped or slightly underdamped. Too much damping would be counter-productive; if the suspension is overdamped and the car hits a second bump justafter the first one, the springs in the suspension will still be compressed somewhatfrom the first bump and will not be able to fully absorb the impact.

Energy in Damped OscillationsIn damped oscillations the damping force is nonconservative; the mechanicalenergy of the system is not constant but decreases continuously, approachingzero after a long time. To derive an expression for the rate of change of energy,we first write an expression for the total mechanical energy E at any instant:

To find the rate of change of this quantity, we take its time derivative:

But and so

dEdt

= vx1max + kx2dx>dt = vx,dvx>dt = ax

dEdt

= mvxdvx

dt+ kx

dxdt

E = 12 mvx

2 + 12 kx2

a2a1C2C1

x = C1e-a1t + C2e-a2t

21km ,

km

- b2

4m2 = 0 or b = 21km

v = 2k>m v¿,

f = 0;e-1b>2m2t.Ae-1b>2m2t

v¿ = B km

- b2

4m2 (oscillator with little damping)

v¿


O

2A

A

T0 2T0 3T0 4T0 5T0

Ae2(b/2m)t

xb ! 0.1!km (weak damping force)b ! 0.4!km (stronger damping force)

With stronger damping (larger b):• The amplitude (shown by the dashed curves) decreases more rapidly.• The period T increases (T0 ! period with zero damping).

t

14.26 Graph of displacement versustime for an oscillator with little damping[see Eq. (14.42)] and with phase angle

The curves are for two values ofthe damping constant b.f = 0.

Piston

Viscousfluid

Lower cylinderattached toaxle; moves upand down.

Pushed up

Pushed down

Upper cylinderattached to car’sframe; moves little.

14.27 An automobile shock absorber.The viscous fluid causes a damping forcethat depends on the relative velocity of thetwo ends of the unit.

18


x = A0 exp −b2mt

⎛

⎝⎜

⎞

⎠⎟ cos ω 't +δ( )


bc = 2mω0 = 2 km

Valor crítico da constante de amortecimento:

19



20

17.5 Oscilações forçadas

Para se manter um sistema amortecido em oscilação, é necessário forçar o movimento, imprimindo uma força exterior, no sentido do movimento. Essa força deve oscilar com uma frequência próxima da frequência própria do sistema (ω0):

ω0 =km

!Fext =

!F0 cosωt

21


Para se manter um sistema amortecido em oscilação, é necessário forçar o movimento, imprimindo uma força exterior, no sentido do movimento. Essa força deve oscilar com uma frequência próxima da frequência própria do sistema (ω0):

ω0 =km

ω0 =gL

Então, a força total sobre o objeto será a soma da força de restituição com a força de arrastamento, com a força que obriga a que o movimento se mantenha:

!F = −k

!x −b d

!xdt

+ F0 cosωt

!Fext =

!F0 cosωt

maximum value That would involve more differential equations than we’reready for, but here is the result:

(14.46)

When the first term under the radical is zero, so A has a maximumnear The height of the curve at this point is proportional to the less damping, the higher the peak. At the low-frequency extreme, when

we get This corresponds to a constant force and aconstant displacement from equilibrium, as we might expect.

Resonance and Its ConsequencesThe fact that there is an amplitude peak at driving frequencies close to the naturalfrequency of the system is called resonance. Physics is full of examples of reso-nance; building up the oscillations of a child on a swing by pushing with a fre-quency equal to the swing’s natural frequency is one. A vibrating rattle in a carthat occurs only at a certain engine speed or wheel-rotation speed is an all-too-familiar example. Inexpensive loudspeakers often have an annoying boom orbuzz when a musical note happens to coincide with the resonant frequency of thespeaker cone or the speaker housing. In Chapter 16 we will study other examplesof resonance that involve sound. Resonance also occurs in electric circuits, as wewill see in Chapter 31; a tuned circuit in a radio or television receiver respondsstrongly to waves having frequencies near its resonant frequency, and this fact isused to select a particular station and reject the others.

Resonance in mechanical systems can be destructive. A company of soldiersonce destroyed a bridge by marching across it in step; the frequency of their stepswas close to a natural vibration frequency of the bridge, and the resulting oscilla-tion had large enough amplitude to tear the bridge apart. Ever since, marchingsoldiers have been ordered to break step before crossing a bridge. Some yearsago, vibrations of the engines of a particular airplane had just the right frequencyto resonate with the natural frequencies of its wings. Large oscillations built up,and occasionally the wings fell off.

A = Fmax>k FmaxA = Fmax>k.vd = 0,

1>b;vd = 2k>m .k - mvd

2 = 0,

A =Fmax21k - mvd

222 + b2vd2 (amplitude of a driven oscillator)

Fmax.


Each curve shows the amplitude A for an oscillator subjected to a driving forceat various angular frequencies vd. Successive curves from blue to gold representsuccessively greater damping.

A lightly damped oscillator exhibits a sharpresonance peak when vd is close to v (thenatural angular frequency of an undampedoscillator).

Driving frequency vd equals natural angular frequency v of an undamped oscillator.

Stronger damping reduces and broadens thepeak and shifts it to lower frequencies.

If b $ !2km, the peak disappears completely.Fmax/k

2Fmax/k

3Fmax/k

4Fmax/k

5Fmax/k

0.5 1.0 1.5 2.0

A

b 5 0.2!km

0vd /v

b 5 0.4!km

b 5 0.7!km

b 5 1.0!km

b 5 2.0!km

14.28 Graph of the amplitude A offorced oscillation as a function of theangular frequency of the driving force.The horizontal axis shows the ratio of to the angular frequency of an undamped oscillator. Each curve hasa different value of the damping constant b.

v = 2k>m vd

vd

Application Canine ResonanceUnlike humans, dogs have no sweat glandsand so must pant in order to cool down. Thefrequency at which a dog pants is very close tothe resonant frequency of its respiratory sys-tem. This causes the maximum amount of airto move in and out of the dog and so mini-mizes the effort that the dog must exert tocool itself.

Test Your Understanding of Section 14.8 When driven at a frequencynear its natural frequency, an oscillator with very little damping has a much greaterresponse than the same oscillator with more damping. When driven at a frequencythat is much higher or lower than the natural frequency, which oscillator will have thegreater response: (i) the one with very little damping or (ii) the one with more damping? ❙

22

k!x +b d

!xdt

+md2 !x

dt2= F0 cosωt

A=F0

m2 ω02 −ω2( )

2+b2ω2

Equação diferencial do movimento:


Amplitude das oscilações:

23

A=F0

m2 ω02 −ω2( )

2+b2ω2


Na ressonância, a frequência de oscilação é igual (ou muito próxima) da frequência própria do sistema: ω = ω0

Tacoma Bridge

filme

In November, 1940, the newly completed Tacoma Narrows Bridge, opened barely four months before, swayed and collapsed in a 42 mile-per-hour wind.

Download - 17 Movimentos oscilatórios, amortecidos e … – Aula 16 9/Maio/2018 – Aula 17 17 Movimentos oscilatórios, amortecidos e forçados 17.1 Movimento na vertical 17.2 Pêndulo simples

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