# 17 Movimentos oscilatórios, amortecidos e – Aula 16 9/Maio/2018 – Aula 17 17 Movimentos oscilatórios,…

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<p>7/Maio/2018 Aula 16 </p>
<p>9/Maio/2018 Aula 17 </p>
<p>17 Movimentos oscilatrios, amortecidos e forados 17.1 Movimento na vertical 17.2 Pndulo simples 17.3 Pndulo fsico 17.4 Oscilaes amortecidas 17.5 Oscilaes foradas </p>
<p>16 Movimento peridico 16.1 Movimento harmnico simples (MHS) 16.2 Conservao da energia no MHS </p>
<p>1 </p>
<p>Its 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</p>
<p>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.</p>
<p>Lets 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.</p>
<p>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.</p>
<p>Amplitude, Period, Frequency, and Angular FrequencyHere are some terms that well use in discussing periodic motions of all kinds:</p>
<p>The amplitude of the motion, denoted by A, is the maximum magnitude ofdisplacement from equilibriumthat 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 tripsay, 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.</p>
<p>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.</p>
<p>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:</p>
<p>This unit is named in honor of the German physicist Heinrich Hertz(18571894), a pioneer in investigating electromagnetic waves.</p>
<p>The angular frequency, is times the frequency:</p>
<p>Well 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 </p>
<p>From the definitions of period T and frequency we see that each is the recip-rocal of the other:</p>
<p>(14.1)f = 1T T = 1</p>
<p> (relationships between frequency and period)</p>
<p>rad>cycle.2p cycle>s,rad>s.</p>
<p>v</p>
<p>v = 2p2pv,</p>
<p>1 hertz = 1 Hz = 1 cycle>s = 1 s-1,</p>
<p>-A-A,</p>
<p>-A</p>
<p>x .</p>
<p>x = A,</p>
<p>x = -A.</p>
<p>x = A</p>
<p>ax = Fx>m. axFx,</p>
<p>438 CHAPTER 14 Periodic Motion</p>
<p>x , 0: glider displacedto the left from theequilibrium position.</p>
<p>Fx . 0, so ax . 0:compressed springpushes glider towardequilibrium position.</p>
<p>Fx</p>
<p>ax</p>
<p>Fx</p>
<p>x 5 0: The relaxed spring exerts no force on theglider, so the glider has zero acceleration.</p>
<p>(b)</p>
<p>O x</p>
<p>y</p>
<p>xn</p>
<p>mg</p>
<p>y</p>
<p>(a)</p>
<p>xx</p>
<p>y</p>
<p>xn</p>
<p>mg</p>
<p>y</p>
<p>x . 0: glider displacedto the right from theequilibrium position.</p>
<p>Fx , 0, so ax , 0:stretched springpulls glider towardequilibrium position.</p>
<p>Fx</p>
<p>ax</p>
<p>Fx</p>
<p>(c)</p>
<p>xx</p>
<p>y</p>
<p>x</p>
<p>n</p>
<p>mg</p>
<p>y</p>
<p>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.</p>
<p>x = 0,</p>
<p>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.</p>
<p>Um objeto que esteja ligado a uma mola, por exemplo, e que seja desviado da sua posio de equilbrio, tende a voltar a essa posio: a mola exerce uma fora de restituio, o que causa um movimento peridico (oscilao). </p>
<p>2 </p>
<p>16. Movimento peridico </p>
<p>Massa-mola </p>
<p>simulao </p>
<p>Aula anterior </p>
<p>Quando a fora de restituio diretamente proporcional ao afastamento da posio de equilbrio, como no caso de molas ideais, tem-se um movimento harmnico simples (MHS). </p>
<p>3 </p>
<p>!F = k</p>
<p>!x</p>
<p>16.1 Movimento harmnico simples </p>
<p>!Fm=!a = d</p>
<p>2 !x</p>
<p>dt2d2!x</p>
<p>dt2= km!x</p>
<p>14.2 Simple Harmonic Motion 439</p>
<p>Also, from the definition of </p>
<p>(14.2)v = 2p = 2pT (angular frequency)</p>
<p>v,</p>
<p>Example 14.1 Period, frequency, and angular frequency</p>
<p>An ultrasonic transducer used for medical diagnosis oscillates atHow long does each oscillation take,</p>
<p>and what is the angular frequency?</p>
<p>SOLUTION</p>
<p>IDENTIFY and SET UP: The target variables are the period T andthe angular frequency . We can find these using the given fre-quency in Eqs. (14.1) and (14.2).</p>
<p>v</p>
<p>6.7 MHz = 6.7 * 106 Hz.EXECUTE: From Eqs. (14.1) and (14.2),</p>
<p>EVALUATE: This is a very rapid vibration, with large and andsmall T. A slow vibration has small and and large T.v</p>
<p>v</p>
<p>= 4.2 * 107 rad>s= 12p rad>cycle216.7 * 106 cycle>s2v = 2pf = 2p16.7 * 106 Hz2T =</p>
<p>1</p>
<p>= 16.7 * 106 Hz</p>
<p>= 1.5 * 10-7 s = 0.15 ms</p>
<p>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 bodysx-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</p>
<p>ax 7 0;vx 6 0ax 6 0;vx 7 0ax 7 0;vx 7 0ax,vx</p>
<p>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 Hookes law. The constant ofproportionality between and x is the force constant k. (You may want toreview Hookes 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</p>
<p>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</p>
<p>(14.3)</p>
<p>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.</p>
<p>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</p>
<p>(14.4)</p>
<p>The minus sign means the acceleration and displacement always have oppositesigns. This acceleration is not constant, so dont even think of using the constant-acceleration equations from Chapter 2. Well 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.</p>
<p>ax =d2x</p>
<p>dt 2= - k</p>
<p>mx (simple harmonic motion)</p>
<p>ax = d2x>dt 2 = Fx>mkg>s2).N>m</p>
<p>Fx = -kx (restoring force exerted by an ideal spring)Fx</p>
<p>Fx = kx.</p>
<p>Fx</p>
<p>Fx</p>
<p>Fx</p>
<p>The restoring force exerted by an idealizedspring is directly proportional to thedisplacement (Hookes law, Fx 5 2kx):the graph of Fx versus x is a straight line.</p>
<p>O</p>
<p>Displacement x</p>
<p>Restoring force Fx</p>
<p>x , 0Fx . 0</p>
<p>x . 0Fx , 0</p>
<p>14.3 An idealized spring exerts arestoring force that obeys Hookes law,</p>
<p>Oscillation with such a restoringforce is called simple harmonic motion.Fx = -kx.</p>
<p>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.</p>
<p>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, well find by noticing a striking similar-ity between SHM and another form of motion that weve already studied.</p>
<p>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 balls shadow areidentical if the amplitude of the bodys 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.</p>
<p>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</p>
<p>v2p</p>
<p>rad>s), vx1t21-k>m2d2x>dt 2, x1t2.</p>
<p>440 CHAPTER 14 Periodic Motion</p>
<p>... but Fx 5 2kx can be agood approximation to the forceif the displacement x is sufficiently small.</p>
<p>Ideal case: The restoring force obeys Hookeslaw (Fx 5 2kx), so the graph of Fx versus x is astraight line.</p>
<p>Typical real case: Therestoring force deviatesfrom Hookes law ...</p>
<p>O Displacement x</p>
<p>Restoring force Fx</p>
<p>14.4 In most real oscillations Hookeslaw applies provided the body doesntmove too far from equilibrium. In such acase small-amplitude oscillations areapproximately simple harmonic.</p>
<p>u</p>
<p>Shadow of ballon screen</p>
<p>Balls shadow</p>
<p>Ball on rotatingturntable</p>
<p>While the ball Qon the turntablemoves in uniformcircular motion,its shadow P movesback and forth onthe screen in simpleharmonic motion.</p>
<p>Illuminatedvertical screen</p>
<p>Illumination</p>
<p>Table</p>
<p>Light beam</p>
<p>A</p>
<p>A</p>
<p>2A O P</p>
<p>Q</p>
<p>Ball moves in uniformcircular motion.</p>
<p>Shadow movesback and forth onx-axis in SHM.</p>
<p>(a) Apparatus for creating the reference circle (b) An abstract representation of the motion in (a)</p>
<p>OP</p>
<p>A</p>
<p>y</p>
<p>x</p>
<p>Q</p>
<p>x !A cos uv</p>
<p>14.5 (a) Relating uniform circular motion and simple harmonic motion. (b) The balls shadow moves exactly like a body oscillatingon an ideal spring.</p>
<p>Aula anterior </p>
<p>4 </p>
<p>x = Acos t +( )d2x</p>
<p>dt2= kmx</p>
<p>Massa-mola </p>
<p>simulao </p>
<p> a fase do movimento e </p>
<p>14.2 Simple Harmonic Motion 443</p>
<p>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),</p>
<p>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</p>
<p>(14.13)</p>
<p>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.</p>
<p>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.</p>
<p>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</p>
<p>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).</p>
<p>vT = A km T = 2p or T = 2pAmkt = 0,2p</p>
<p>-A-1</p>
<p>cos a = sin1a + p>22.f = 0.v = 2k>m .x = Acos1vt + f2 (displacement in SHM)</p>
<p>u = vt + f.f</p>
<p>t = 0...</p>