Oscillations - Pham Tan Thi

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1. Oscillations Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Science Ho Chi Minh University of Technology
2. 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)
3. 438 CHAPTER 14 Periodic Motion 14.2 Model for periodic motion. When It’s simplest to define our coordinate system so that the origin O is at the equilib- the body is displaced from its equilibrium rium position, where the spring is neither stretched nor compressed. Then x is the position at x = 0, the spring exerts a x-component of the displacement of the body from equilibrium and is also the restoring force back toward the equilib- rium position. change in the length of the spring. The x-component of the force that the spring Simple Harmonic Oscillation/Motion exerts on the body is Fx, and the x-component of acceleration ax is given by (a) a F m. Total force exerts to object given by Hooke’s law x = x x . 0: glider displaced Fx , 0, so ax , 0: Figure 14.2 shows the body for three different displacements of the spring. to the right from the stretched spring Whenever> the body is displaced from its equilibrium position, the spring force F = kx equilibrium position. pulls glider toward tends to restore it to the equilibrium position. We call a force with this character a equilibrium position. restoring force. Oscillation can occur only when there is a restoring force tend- a Newton’s second law of motion: y y x ing to return the system to equilibrium. 2 n Let’s analyze how oscillation occurs in this system. If we displace the body to the F x d x x Fx F = m x x right to x = A and then let go, the net force and the acceleration are to the left 2 mg (Fig. 14.2a). The speed increases as the body approaches the equilibrium position O. dt When the body is at O, the net force acting on it is zero (Fig. 14.2b), but because of its motion it overshoots the equilibrium position. On the other side of the equilib- then (b) rium position the body is still moving to the left, but the net force and the accelera- 2 x 5 0: The relaxed spring exerts no force on the tion are to the right (Fig. 14.2c); hence the speed decreases until the body comes to a glider, so the glider has zero acceleration. d x k 2 stop. We will show later that with an ideal spring, the stopping point is at x =-A. y y 2 = x = ! x The body then accelerates to the right, overshoots equilibrium again, and stops at dt m n the starting point x = A, ready to repeat the whole process. The body is oscillating! O General solution for motion x x If there is no friction or other force to remove mechanical energy from the system, mg this motion repeats forever; the restoring force perpetually draws the body back toward the equilibrium position, only to have the body overshoot time after time. x = Acos(!t + ) Phase of the motion In different situations the force may depend on the displacement x from equi- (c) librium in different ways. But oscillation always occurs if the force is a restoring 2 k x , 0: glider displaced Fx . 0, so ax . 0: force that tends to return the system to equilibrium. where ! = Angular frequency to the left from the compressed spring m equilibrium position. pushes glider toward equilibrium position. Amplitude, Period, Frequency, and Angular Frequency y ax y Here are some terms that we’ll use in discussing periodic motions of all kinds: Velocity v = A!sin(!t + ) The amplitude of the motion, denoted by A, is the maximum magnitude of x F n x Fx displacement from equilibrium—that is, the maximum value of ƒxƒ. It is always 2 x x positive. If the spring in Fig. 14.2 is an ideal one, the total overall range of the Acceleration a = A! cos(!t + ) mg motion is 2A. The SI unit of A is the meter. A complete vibration, or cycle, is one complete round trip—say, from A to -A and back to A, or from O to A, back through O to -A, and back to O. Note that motion from one side to the other (say, -A 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 the second, but it is sometimes expressed as “seconds per cycle.” The frequency, ƒ, is the number of cycles in a unit of time. It is always posi- Application Wing Frequencies The ruby-throated hummingbird (Archilochus tive. The SI unit of frequency is the hertz: colubris) normally flaps its wings at about -1 50 Hz, producing the characteristic sound that 1 hertz = 1 Hz = 1 cycle s = 1 s gives hummingbirds their name. Insects can flap their wings at even faster rates, from This unit is named in honor of the German> physicist Heinrich Hertz 330 Hz for a house fly and 600 Hz for a mos- (1857–1894), a pioneer in investigating electromagnetic waves. quito to an amazing 1040 Hz for the tiny biting midge. The angular frequency, v, is 2p times the frequency: v = 2pƒ We’ll learn shortly why v is a useful quantity. It represents the rate of change of an angular quantity (not necessarily related to a rotational motion) that is always measured in radians, so its units are rad s. Since ƒ is in cycle s, we may regard the number 2p as having units rad cycle. From the definitions of period T and frequency> ƒ we see that> each is the recip- rocal of the other: > 1 1 f = T = (relationships between frequency and period) (14.1) T ƒ
4. 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 1Motion 445 1 To find f, we divide Eq. (14.17) by Eq. (14.14). This eliminates A and gives an 14.13 Howfx-velocity= vx and or T = equation that we can solve for f: x-acceleration ax vary during one cycle Simple Harmonicf Motion of SHM. T x v0 x -vAsinf = =-vtanf x 0 Acosf v0 x x 52Axx 5 0 5A f = arctan - (phase angle in SHM) (14.18) vx 0 a b 2A 2A/2 0 A/2 A It is also easy to find the amplitude A if we are given x 0 and v0 x. We’ll sketch a ϭ 2a x max x the derivation, and you can fill in the details. Square Eq. (14.14); then divide vx 5 0 Eq. (14.17) by v, square it, and add to the square of Eq. (14.14). The right side ax 2 2 2 2 x will be A sin f + cos f , which is equal to A . The final result is vx a 5 0 x x 1 2 2 vx 52vmax 2 v0 x (14.19) ax A = x 0 + 2 (amplitude in SHM) x v vx a 5 a B x max x Note that when the body has both an initial displacement x 0 and a nonzero initial vx 5 0 a velocity v0 x, the amplitude A is not equal to the initial displacement. That’s rea- x x vx sonable; if you start the body at a positive x 0 but give it a positive velocity v0 x, it ax 5 0 will go farther than x 0 before it turns and comes back. x vx 5 vmax a x x vx a 52a x max x vx 5 0 Problem-Solving Strategy 14.1 Simple Harmonic Motion I: Describing Motion IDENTIFY the relevant concepts: An oscillating system undergoes EXECUTE the solution as follows: simple harmonic motion (SHM) only if the restoring force is 1. Use the equations given in Sections 14.1Copyright and 14.2 © to 2008 solve Pearson for Education, Inc., publishing as Pearson Addison-Wesley. directly proportional to the displacement. the target variables. 2. To find the values of x, v , and a at particular times, use Eqs. SET UP the problem using the following steps: x x (14.13), (14.15), and (14.16), respectively. If the initial position 1. Identify the known and unknown quantities, and determine x and initial velocity v are both given, determine f and A which are the target variables. 0 0x from Eqs. (14.18) and (14.19). If the body has an initial posi- 2. Distinguish between two kinds of quantities. Properties of the tive displacement x but zero initial velocity v 0 , then system include the mass m, the force constant k, and quantities 0 0x = the amplitude is A x and the phase angle is f 0. If it has derived from m and k, such as the period T, frequency ƒ, and = 0 = an initial positive velocity v but no initial1 displacement2 angular frequency v. These are independent of properties of the 0x x 0 , the amplitude is A v v and the phase angle is motion, which describe how the system behaves when it is set 0 = = 0x f =-p 2. Express all phase angles in radians. into motion in a particular way; they include the amplitude A, 1 2 > maximum velocity vmax, and phase angle f , and values of x, vx, EVALUATE your> answer: Make sure that your results are consistent. and ax at particular times. For example, suppose you used x 0 and v0x to find general expres- 3. If necessary, define an x-axis as in Fig. 14.13, with the equilib- sions for x and vx at time t. If you substitute t = 0 into these expres- rium position at x = 0. sions, you should get back the given values of x 0 and v0x. Example 14.3 Describing SHM We give the glider of Example 14.2 an initial displacement x 0 = SOLUTION +0.015 m and an initial velocity v0x ϭ +0.40 m s. (a) Find the IDENTIFY and SET UP: As in Example 14.2, the oscillations are period, amplitude, and phase angle of the resulting motion. (b) Write SHM. We use equations from this section and the given values k ϭ equations for the displacement, velocity, and acceleration> as func- 200 N m, m ϭ 0.50 kg, x 0, and v0x to calculate the target variables A tions of time. and f and to obtain expressions for x, vx, and ax . > Continued
5. 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 v = =2⇡fA = !A max T
6. Displacement as function of time in SHM k x = Acos(!t + ) !2 = m Changing m, A or k changes the graph of x versus t: change f
7. 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 U = kx2 where k = m!2 2 1 U = m!2x2 2 The potential energy in terms of time, t, is given by 1 U = m!2x2 where x = Asin(!t + ) 2 1 U = m!2A2sin2(!t + ) 2
8. Mechanical Energy in Simple Harmonic Motion Kinetic Energy, K The kinetic energy of an object in SHM is given by 1 K = mv2 where v2 = ! A2 x2 2 1 p 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 K = m!2A2cos2(!t + ) 2
9. 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 x2)+ m!2x2 2 2 1 2 2 1 2 E = m! A OR E = kA 2 2
10. 14.3 Energy in Simple Harmonic Motion 447 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 at equally spaced positions are not equally spaced in time. 1 1 ax 5 amax ax 5 2 amax ax 5 0 ax 5 2 2 amax ax 52amax 6 3 v 56v 6 3 vx 5 0 vx 5 4 vmax x max vx 5 4 vmax vx 5 0 Å Å x 2 1 1 A 2 2 A OA2 A 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 E is all potential E is partly potential, E is all kinetic E is partly potential, E is all potential energy. partly kinetic energy. partly kinetic energy. energy. energy. 2 2 (Recall that sin a + cos a = 1.) Hence our expressions for displacement and velocity in SHM are consistent with energy conservation, as they must be. We can use Eq. (14.21) to solve for the velocity vx of the body at a given dis- placement x: k 2 2 vx = Ϯ A - x (14.22) m 2 TheϮ sign means that at a given valueA of x the body can be moving in either direction. For example, when x = Ϯ A 2, 2 k 2 > A 3 k vx = Ϯ A - Ϯ = Ϯ A m 2 4 m A B a b A A Equation (14.22) also shows that the maximum speed vmax occurs at x = 0. Using Eq. (14.10), v = k m, we find that 2 > k vmax = A = vA (14.23) m A MechanicalThis agrees with Eq. Energy(14.15): vx oscillatesin Simple between Harmonic -vA and +vA. Motion ConservationInterpreting Eof, KEnergy, and U in SHM Figure 14.141 shows the energy1 quantities E, 1K, and U at x = 10, x = Ϯ A 2, and x =EϮ A=. Figuremv 14.152 + is a graphicalkx2 = display kA of Eq.2 (14.21);= mv energy2 (kinetic, potential, and2 total) is plotted vertically2 and the2 coordinate x is plotted2 horizonmax> tally. (a) The potential energy U and total mechanical (b) The same graph as in (a), showing 14.15 Kinetic energy K, potential energy E for a body in SHM as a function of kinetic energy K as well energy U, and total mechanical energy E displacement x At x 56A the energy is all potential; the kinetic as functions of position for SHM. At each energy is zero. value of x the sum of the values of K and The total mechanical energy E is constant. U equals the constant value of E. Can you Energy E At x 5 0 the energy is all kinetic; show that the energy is half kinetic and the potential energy is zero. half potential at x Ϯ 1 A? U 5 1 kx2 = 2 2 Energy K E 5 K 1 U 2 U U K x 2AxOA x 2A OA At these points the energy is half kinetic and half potential.
11. 454 CHAPTER 14 Periodic Motion 454 CHAPTER 14 Periodic Motion 14.21 The dynamics of a simple The path of the point mass (sometimes called a pendulum bob) is not a straight 454pendulum.CHAPTER 14 Periodic Motion line but the arc of a circle with radius L equal to the length of the string The dynamics of a simple The path of the point mass (sometimes called a pendulum bob) is not a straight 14.21 (Fig. 14.21b). We use as our coordinate the distance x measured along the arc. If pendulum.(a) A real pendulum line but the arc of a circle with radius L equal to the length of the string Thethe motionpath of theis simple point mass harmonic, (sometimes the restoring called a pendulumforce must bob) be directlyis not a straightproportional 14.21(a) A realThe pendulum dynamics of a simple (Fig. 14.21b). We use as our coordinate the distance x measured along the arc. If pendulum. linetheto motion butx or the (because is arcsimple of x aharmonic, = circleLu) to with uthe. Is radiusrestoring it? L equalforce must to the be lengthdirectly of proportional the string In Fig. 14.21b we represent the forces on the mass in terms of tangential and (a) A real pendulum (Fig.to x or14.21b). (because We x use= Lasu) our to ucoordinate. Is it? the distance x measured along the arc. If the radialInmotion Fig. components. 14.21bis simple we harmonic, represent The restoring the forcesrestoring force on F forceuthe is massthe must tangential in be terms directly ofcomponent tangential proportional of and the net force: toradialx or components.(because x = TheLu) restoring to u. Is it? force Fu is the tangential component of the net In Fig. 14.21b we represent the forces on the mass in terms of tangential and force: Fu =-mgsinu (14.30) radial components. The restoring force F is the tangential component of the net F mgu sinu (14.30) force:The restoring force is providedu by=- gravity; the tension T merely acts to make the Thepoint restoring mass force move is inprovided an arc. by The gravity; restoring the tensionforce is T proportionalmerely acts to not maketo uthe but to Fu =-mgsinu (14.30) pointsin massu, so themove motion in an is arc. not Thesimple restoring harmonic. force However, is proportional if the angle not to u isu 454 small, but toCHAPTERsinu 14 Periodic Motion Thesinisu restoring, veryso the nearly motion force equal isis providednot tosimple u in byradians harmonic. gravity; (Fig. the However, 14.22). tension For ifT themerely example, angle acts u iswhen to small, make u =sin the0.1u rad The Simple Pendulum 14.21 The dynamics of a simple The path of the point mass (sometimes called a pendulum bob) is not a straight pointis very(about mass nearly 6°), move equalsin inu to=an u 0.0998,arc. in radians The a restoringdifference (Fig. 14.22). force of only Foris proportional example,0.2%. With when notthis uto approximation,= u0.1pendulum. but radto line but the arc of a circle with radius L equal to the length of the string sinu, so the motion is not simple harmonic. However, if the angle u is small,(a) sinA realu pendulum (Fig. 14.21b). We use as our coordinate the distance x measured along the arc. If (aboutEq.The (14.30)6°), tangentialsin ubecomes= 0.0998, component a difference of of the only net 0.2%. force: With this approximation, the motion is simple harmonic, the restoring force must be directly proportional isEq. very (14.30) nearly becomes equal to u in radians (Fig. 14.22). For example, when u = 0.1 rad to x or (because x = Lu) to u. Is it? x In Fig. 14.21b we represent the forces on the mass in terms of tangential and (about 6°), sinu = F0.0998,✓ = a Fdifferencemgu =-sinmg✓ ofu =-onlymg 0.2%. Withor this approximation, (b) An idealized simple pendulum x L radial components. The restoring force Fu is the tangential component of the net Eq. (14.30) becomes Fu=-mgu =-mg or force: (b) An idealized simple pendulum The restoring force is provided by gravity;mgL the tension is Fu =-mgsinu (14.30) Fu =-mg x x (14.31) String is merely acts to makeF themg pointu massmg (bob)L ormove in an arc. The restoring force is provided by gravity; the tension T merely acts to make the (b) An idealized simple pendulum u =- Fu =-=- x (14.31) Stringassumed is to be L L point mass move in an arc. The restoring force is proportional not to u but to sinu, so the motion is not simple harmonic. However, if the angle u is small, sinu assumedmassless to be and Theif the restoring pendulum’s force is thenoscillation proportional ismg small to the (θ coordinate coordinate is for small displace- x masslessBobas is a modeledpoint and mass. L F mgu mg or (b) An idealized simple pendulum u =- =- u unstretchable.as a point mass. ments, and the force constant is k = mg>L. From Eq. (14.10) the angular fre- L L The restoring forcek is mgthenL proportionalg (simple to the coordinatependulum, mg L T quency v of a simple pendulum with small amplitude is Fu =- x (14.31) Bob is modeled for smallv displacement,= k =mg L and =theg force(simple constantsmall pendulum, amplitude) is k = mg/L (14.32) String is L m m > L assumed to be as a point mass. v = = = (14.32) The restoring force is then proportional to the coordinate for small displace- m m > L small amplitude) massless and L x m kA mgB>L g A (simple pendulum, u unstretchable. ments, and the force constant is k = mg L. From Eq. (14.10) the angular fre- x m A A T Bob is modeled quency v of a simple pendulum with small amplitude is The correspondingv = = frequencyB =and period relationships are (14.32) > mg sin u The correspondingm frequencym and periodL relationshipssmall amplitude) are as a point mass. mg sin u > L k mg L g (simple pendulum, v = = = (14.32) The restoring force onx the m mg cos u Av 1B g A m m L small amplitude) The restoring force on the u The corresponding frequencyg and period relationships are > bob is proportional to sin u, u mg cos u ƒ =v =1 (simple pendulum, small amplitude) (14.33) x m A B A bobnot is to proportional u. However, tomg forsinsin smallu,u ƒ = 2=p 2p L (simple pendulum, small amplitude) (14.33) The corresponding frequency and period relationships are not to u. However, for small 2p 2p L mg sin u Theu ,restoring sin u ^ uforce, so theon themotion is The restoring force on the mg cos u v 1 gA u mg cos u v 1 g u, sin u ^ u, so the motion is u A bob is proportional to sin u, ƒ = = (simple pendulum, small amplitude) (14.33) boba ppis proportionalroximately simple to sin harmonic.u, ƒ = = (simple pendulum, small amplitude) (14.33) 2p 2p L approximately simple harmonic. mg 2p2p 12p L L not to u. However, for small not to u. However, for small mg 2p 1 L u, sin u ^ u, so the motion is A u u u T = = = 2p (simple pendulum, small amplitude) approxima(14.34)tely simple harmonic. , sin ^ , so the motion is T = v= =ƒ 2pA g (simple pendulum, small amplitude) (14.34) mg 2p 1 L approximately simple harmonic. v ƒ g T = = = 2p (simple pendulum, small amplitude) (14.34) mg 2p 1 LA v ƒ g 14.22 For small angular displacements A A 14.22 For small angular displacements T = Note =that =these2p expressions (simple do pendulum,not involve small the massamplitude)of the particle.(14.34)14.22 ThisFor small is angular displacements Note that these expressions do not involve the mass of the particle. This is Notev that theseƒ expressionsg do not involve the mass of the particle. Thisu, the restoring is force Fu =-mgsinu on a u,uthe, the restoring restoring force force F Fu =-mgmgsinsinu onu ona a because the restoring force, a component of the particle’s weight, is proportional u =- because the restoring force, a component of the particle’s weight, is proportionalsimple pendulum is approximately equal S S ma and cancels out. (This is the ؍ simple pendulum is approximately equal because the restoring force, a component of the particle’sS weight, is proportionalto mgu; that is, it is approximately pro- to m. Thus the mass appears on both sides of ©F simple pendulum is approximately equal A S S - S maof and the cancels particle. out. Thisportional (This is to is the the displacement u. Hence for same physics that explains why bodies of different masses fall with the same؍14.22to mgForu; thatsmall is, angular it is approximately displacements pro- Noteto m. thatThus these the mass expressions appears ondo both not sidesinvolve of ©theF mass ma and cancels out. (Thissmall is anglesthe the oscillations are simple ؍ to mg- u; that is, it is approximately pro- to m. Thus the mass appears on both sides of ©F u, the- restoring force Fu =-mgsinu on a acceleration in a vacuum.) For small oscillations, the period of a pendulum for a portionalportional to tothe the displacement displacement u. Henceu. Hence for for becausesamesame physics the physics restoring that that explains force, explains a why component why bodies bodies of the different of particle’s different masses weight, masses fall is with fall proportional with theharmonic. same the same given value of g is determined entirely by its length. simple pendulum is approximately equal S small angles the oscillations are simple S F The dependence on L and g in Eqs. (14.32) through (14.34) is just what we mthea andperiodthe cancelsperiod of a out.pendulumof a (This pendulum is for the a for ua ؍ small angles the oscillations are simple toaccelerationmacceleration. Thus the in mass ain vacuum.) aappears vacuum.) onFor both Forsmall sidessmall oscillations, of oscillations, F to - mgu; that is, it is approximately pro- © 2 mg Fu 52mg sin u should expect. A long pendulum has a longer period than a shorter one. Increas- harmonic.harmonic. (actual) portional to the displacement u. Hence for samegivengiven physicsvalue value of that gofis g explainsdeterminedis determined why entirely bodies entirely by of its by different length. its length. masses fall with the same ing g increases the restoring force, causing the frequency to increase and the mg F 52mgu u period to decrease. small angles the oscillationsF Fu are simple accelerationTheThe dependence dependence in a vacuum.) on Lonand ForL and gsmalling Eqs.in oscillations, Eqs. (14.32) (14.32) through the throughperiod (14.34) of (14.34) a pendulumis just is whatjust for whatwe a we (approximate) u We emphasize again that the motion of a pendulum is only approximately sim- harmonic. F F52u 52mg mgsin sinu u u (rad) 2 mg2 mg u givenshouldshould value expect. expect. of g Ais long Adetermined long pendulum pendulum entirely has hasa bylonger aits longer length. period period than thana shorter a shorter one. one.Increas-2p/ 2 Increas-2p/4 O p/4 p/2 ple harmonic. When the amplitude is not small, the departures from simple har- (actual)(actual) 2mg ing g increases the restoring force, causing the frequency to increase and the monic motion can be substantial. But how small is “small”? The period can be Fu F 52mgu ingTheg increasesdependence the on restoring L and g force,in Eqs. causing (14.32) the through frequency (14.34) to increaseis just what and we the mgmg FFu5252u mgmg sinu u 22 mg expressed by an infinite series; when the maximum angular displacement is ™, 2 mg u (approximate) shouldperiodperiod expect.to decrease.to decrease. A long pendulum has a longer period than a shorter one. Increas- the period T is given by (actual)(approximate) 2 2 2 u u (rad) ing Weg increasesWe emphasize emphasize the again restoring again that that the force, motionthe motion causing of a ofpendulum thea pendulum frequency is only is to onlyapproximatel increase approximatel andy sim- they sim- L 1 2 ™ 1 # 3 4 ™ F 52mgu (rad) Á (14.35) 2p2p2 22p2p4 4mgO O p 4pu4 p 2p 2 T = 2p 1 + 2 sin + 2 2 sin + / / / / / (approximate)/ / / periodpleple harmonic. toharmonic. decrease. When When the theamplitude amplitude is not is small,not small, the departuresthe departures from from simple simple har- har- g 2 2 2 # 4 2 2mg2mg We can compute the periodA a to any desired degree of precision byb taking enough u (rad) monicWemonic emphasizemotion motion can again can be substantial.be that substantial. the motion But Buthowof a how pendulumsmall small is “small”? isis only “small”? approximatel The periodThe period cany sim- becan be 2p/2 2p/4 O p/4 p/2 terms in the series. We invite you to check that when ™=15° (on either side of 222mg2 mg pleexpressed expressedharmonic. by byanWhen infinitean infinitethe amplitudeseries; series; when iswhen notthe small,maximumthe maximum the departures angular angular displacement from displacement simple is har- ™ ,is ™, 2mg monicthethe period motionperiod T is T cangivenis givenbe bysubstantial. by But how small is “small”? The period can be 2 mg expressed by an infinite series; when2 the maximum2 2 angular displacement is ™, 2 L L 1 12 ™ 1 # 312 32 ™ the period T is given by 2 2 ™ # 4 4 ™Á (14.35) T =T 2=p 2p 1 + 1 +2 sin sin + 2+ 2 sin sin+ + Á (14.35) g g 2 2 2 2 4 2 2 2 2 2 2 2 # 22 # 4 2 L 1 2 ™ 1 # 3 4 ™ WeWe can can compute computeT = the2p theperiodA periodAa 1to+ aany to anydesiredsin desired +degree degree ofsin precision of precision+ Ábyb taking byb taking(14.35) enough enough g 22 2 22 # 42 2 termsterms in thein theseries. series. We Weinvite invite you youto check to check that thatwhen when ™= ™=15° (on15° either (on either side ofside of We can compute the periodA a to any desired degree of precision byb taking enough terms in the series. We invite you to check that when ™=15° (on either side of
12. 14.6 The Physical Pendulum 455 the central position), the true period is longer than that given by the approximate Eq. (14.34) by less than 0.5%. The usefulness of the pendulum as a timekeeper depends on the period being very nearly independent of amplitude, provided that the amplitude is small. Thus, as a pendulum clock runs down and the amplitude of the swings decreases a little, the clock still keeps very nearly correct time. Example 14.8 A simple pendulum Find the period and frequency of a simple pendulum 1.000 m long EXECUTE: From Eqs. (14.34) and (14.1), 2 at a location where g = 9.800 m s . L 1.000 m T = 2p = 2p = 2.007 s > g 9.800 m s2 SOLUTION 1 A 1 A ƒ = = = 0.4983 Hz> IDENTIFY and SET UP: This is a simple pendulum, so we can use T 2.007 s the ideas of this section. We use Eq. (14.34) to determine the pen- dulum’s period T from its length, and Eq. (14.1) to find the fre- EVALUATE: The period is almost exactly 2 s. When the metric sys- quency ƒ from T. tem was established, the second was defined as half the period of a 1-m simple pendulum. This was a poor standard, however, because the value of g varies from place to place. We discussed more mod- ern time standards in Section 1.3. Test Your Understanding of Section 14.5 When a body oscillating on a horizontal spring passes through its equilibrium position, its acceleration is zero (see Fig. 14.2b). When the bob of an oscillating simple pendulum passes through its equilib- rium position, is its acceleration zero? ❙ 14.6 TheDynamics Physical Pendulum of The Physical Pendulum A physical pendulum is any real pendulum that uses an extended body, as con- 14.23 Dynamics of a physical pendulum. Thetrasted weight to the idealized (mg model) causes of the simple a restoringpendulum with torqueall 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 is almost as easy as for a simple pendulum. Figure 14.23 Irregularly ⌧z = (mg)(dsin✓) 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 the pivot; in the position shown in the figure, the body is displaced its center of clockwise when the displacement gravity (cg). from equilibrium by an angle u, which we use as a coordinate for the system. The u is counterclockwise and vice versa d distance from O to the center of gravity is d, the moment of inertia of the body d sin u about the axis of rotation through O is I, and the total mass is m. When the body cg ifis thedisplaced pendulum’s as shown, the weight oscillations mg causes a restoring is small torque (θ 2 radians, just(mgd as we ) did✓ in= analyzingI↵z the= simpleI Comparing pendulum. Then this thewith motion Eq. (14.4), is sin uwe, not see to u .that However, the forrole small of u , sink um, u for, the spring-mass 2 mgd so the motion1 is app> roxima2 tely simple harmonic. approximatel y simple harmonic. With this approximation,dsystemt is playedv = here by the (physical quantity pendulum,mgd I . Thussmall the amplitude) angular frequency(14.38) is I 1 > 2 2 tz =-mgd u d ✓ mgd Amgd 1 > 2 The equation of motion is gtz Iaz, so The frequencyv = ƒ is 1 2p times(physical this, and pendulum, the period small T is amplitude) (14.38) == 1 ✓2 I 2 2 dt I d u > - mgd u = Iaz = I A I dt 2 T = 2p p (physical pendulum, small amplitude) (14.39) The frequency ƒ is 1mgd2 times this, and the period T is General solution 1 d2u mgd =- u A(14.37)> 2 I I dt EquationT = (14.39)2p is the basis(physical of a common pendulum, method small for experimentally amplitude) determin-(14.39) ✓ = ✓ocos(!t + ) ing the moment of inertiamgd of a body with a complicated shape. First locate the cen- ter of gravity of Athe body by balancing. Then suspend the body so that it is free to oscillateEquation about (14.39) an axis, is the and basis measure of a common the period method T of small-amplitude for experimentally oscillations. determin- ingFinally, the moment use Eq. of (14.39) inertia to of calculate a body withthe moment a complicated of inertia shape. I of the First body locate about the this cen- teraxis of gravityfrom T , ofthe the body’s body mass by balancing. m, and the Then distance suspend d from the the body axis so to that the itcenter is free of to oscillategravity about(see Exercise an axis, 14.53). and measure Biomechanics the period researchers T of small-amplitude use this method oscillations.to find the Finally,moments use of Eq. inertia (14.39) of an to animal’s calculate limbs. the moment This information of inertia is I importantof the body for about analyz- this axising from how anT, animalthe body’s walks, mass as we’ll m, and see thein the distance second doffrom the two the following axis to the examples. center of gravity (see Exercise 14.53). Biomechanics researchers use this method to find the moments of inertia of an animal’s limbs. This information is important for analyz- Example 14.9 Physical pendulum ingversus how ansimple animal pendulum walks, as we’ll see in the second of the two following examples. If the body in Fig. 14.23 is a uniform rod with length L, pivoted at EVALUATE: If the rod is a meter stick L = 1.00 m and 2 one end, what is the period of its motion as a pendulum? g = 9.80 m s , then Example 14.9 Physical pendulum versus simple pendulum 1 2 > 2 1.00 m SOLUTION T = 2p = 1.64 s If the body in Fig. 14.23 is a uniform rod with length L, pivoted at EVALUATE: If the rod is3 9.80 a meter m s2 stick L = 1.00 m and IDENTIFY and SET UP: Our target variable is the oscillation period T 2 1 2 one end, what is the period of its motion as a pendulum? g = 9.80 m s , then B 2 of a rod that acts as a physical pendulum. We find the rod’s moment The period is smaller by a 1factor of> 2 3 = 0.8161 than that 2of a of inertia in Table 9.2, and then determine T using Eq. (14.39). simple pendulum> of the same2 length1.00 (see m2 Example 14.8). The rod’s SOLUTION T 2p 1 21.64 s moment of inertia =around one end, I =23 ML= , is one-third that of EXECUTE: The moment of inertia of a uniform rod about an axis 3 9.801 m s2 IDENTIFY and SET UP: Our1 target2 variable is the oscillation period T the simple pendulum, and the rod’s cg is half as far from the pivot through one end is I = 3 ML . The distance from the pivot to B ? Theas that period of the is simplesmaller pendulum. by a factor You of can show2 that,0.816 taken than together that of a of a therod rod’s that actscenter as of a gravityphysical is pendulum.d L 2. Then We findfrom the Eq. rod’s (14.39), moment 1 > 2 3 = = in Eq. (14.39), these two differences account for the factor 2 by of inertia in Table 9.2, and then determine T using Eq. (14.39). simple pendulum of the same length (see2 Example 14.8). The3 rod’s 1 2 which the periods differ. 1 2 I > 3 ML 2L moment of inertia around one end, I = ML , is one-third2 that of EXECUTE: The momentT = 2p of inertia= of2p a uniform rod= 2aboutp an axis 3 1mgd2 MgL 2 3g the simple pendulum, and the rod’s cg is half as far from the pivot through one end is I = ML . The distance from the pivot to A3 B A ? as that of the simple pendulum. You can show that, taken together the rod’s center of gravity is d = L 2. Then >from Eq. (14.39), 2 in Eq. (14.39), these two differences account for the factor 3 by 1 2 which the periods differ. I > 3 ML 2L 2 T = 2p = 2p = 2p Example 14.10 mgdTyrannosaurusMgL 2rex and the3g physical pendulum A B A All walking animals, including humans,> have a natural walking 14.24 The walking speed of Tyrannosaurus rex can be estimated pace—a number of steps per minute that is more comfortable than from leg length L and stride length S. a faster or slower pace. Suppose that this pace corresponds to the Exampleoscillation 14.10 of the Tyrannosaurusleg as a physical pendulum. rex and (a) Howthe doesphysical this pace pendulum depend on the length L of the leg from hip to foot? Treat the leg as a All walkinguniform rod animals, pivoted including at the hip humans,joint. (b) Fossil have aevidence natural shows walking that 14.24 The walking speed of Tyrannosaurus rex can be estimated pace—aT. rex, numbera two-legged of steps dinosaur per minute that livedthat isabout more 65 comfortable million years than ago, from leg length L and stride length S. a fasterhad or a legslower length pace. L =Suppose3.1 m andthat athis stride pace length corresponds S = 4.0 to m the (the oscillationdistance of fromthe leg one as footprint a physical to pendulum. the next print (a) Howof the does same this foot; pace see dependFig. on 14.24). the length Estimate L of the the walking leg from speed hip toof foot?T. rex. Treat the leg as a uniform rod pivoted at the hip joint. (b) Fossil evidence shows that T. rex, a two-legged dinosaur that lived about 65 million years ago, Leg SOLUTION Stride length S had a leg length L = 3.1 m and a stride length S = 4.0 m (the length distanceIDENTIFY from andone SETfootprint UP: Our to targetthe next variables print of are the (a) same the relationship foot; see L Fig. between14.24). Estimate walking thepace walking and leg speed length of L T.and rex. (b) the walking speed of T. rex. We treat the leg as a physical pendulum, with a period of Leg SOLUTION Stride length S length IDENTIFY and SET UP: Our target variables are (a) the relationship L between walking pace and leg length L and (b) the walking speed of T. rex. We treat the leg as a physical pendulum, with a period of
13. Damped Oscillation Consider an additional force on a body due to friction dx Fx = bvx where v = x dt The net force is then ⌃F = kx bv x x Newton’s second law for the system is dx d2x kx bvx = max OR kx b = m dt dt2 General solution k2 b2 2 t ! = = !2 2 = ! 1 x = Ae cos(!0t + ) 0 2 2 rm 4m s ! p ✓ ◆ k b The period ! = ; = m 2m 2⇡ 2⇡ r T = = !0 !2 2 p
14. 458 CHAPTER 14 Periodic Motion 14.26 Graph of displacement versus The angular frequency of oscillation v¿ is given by time for an oscillator with little damping [see Eq. (14.42)] and with phase angle k b2 f 0. The curves are for two values of (14.43) = v¿= - 2 (oscillator with little damping) Damped Oscillationthe damping constant b. m 4m B bϭ 0.1Ίkm (weak damping force) You can verify that Eq. (14.42) is a solution of Eq. (14.41) by calculating the first bϭ 0.4Ίkm (stronger damping force) t x and second derivatives of x, substituting them into Eq. (14.41), and checking A 2(b/2m)t x = Ae cos(!0t + ) Ae whether the left and right sides are equal. This is a straightforward but slightly =0 Decay smaller than ω tedious procedure. The motion described by Eq. (14.42) differs from the undamped case in two - b 2m t O t ways. First, the amplitude Ae is not constant but decreases with time b 2m t If b is large enough to be T0 2T0 3T0 4T0 5T0 because of the decreasing exponential1 > 2 factor e- . Figure 14.26 is a graph of k b2 Eq. (14.42) for the case f = 0; it shows that the1 > larger2 the value of b, the more ⇒ p quickly the amplitude decreases. 2 =0 b =2 km With stronger damping (larger b): m 4m (Critical damping) 2A • The amplitude (shown by the dashed Second, the angular frequency v¿, given by Eq. (14.43), is no longer equal to curves) decreases more rapidly. v = k m but is somewhat smaller. It becomes zero when bbecomes so large that • The period T increases If b>2pkm (Overdamping) 2 2 (T0 ϭ period with zero damping). > k b - = 0 or b = 2 km (14.44) If p (Underdamping) m 4m 2 b > dE = vx max + kx Pushed down dt 1 2
15. Mechanical Waves Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Sciences Ho Chi Minh University of Technology
16. 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)
17. 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.
18. Wave Properties Wavelength 2π k 1 Frequency (Hz) f = T Angular 2π ω = 2πf = Frequency T ω Velocity v = = λf k
19. The Function of a Wave 2⇡ 2⇡ y(x, t)=Acos x t T ✓ ◆ = Acos (kx !t)
20. The Equation of a Wave The function of a wave: y = Acos(!t kx) Derivatives with respect to the time: dy d2y = A!sin(!t kx) = A!2cos(!t kx) dt dt2 Derivatives with respect to the position: dy d2y = Aksin(!t kx) = Ak2cos(!t kx) dx dx2 where: 2 d y 2 2⇡ 2⇡ ! 2 ! k = = = dt = = v2 vT v d2y k2 dx2 then: d2y 1 d2y =0 Wave Equation dx2 v2 dt2
21. Sound Wave Sound wave is a longitudinal Threshold of hearing at 1000 Hz: wave with a speed of 343 m/s I = Io I Threshold of pain I = 1 W/m2, L = 10log (dB) 10 I L = 120 dB ✓ o ◆ 12 2 Io = 10 W/m
22. Boundary Conditions • When a wave reflects from a fixed end, the pulse inverts as it reflects (Figure (a)). • When a wave reflects from a free end, the pulse reflects without inverting (Figure (b)).
23. Wave Interference and Superposition • Interference is the result of overlapping waves. • Principle of superposition: when two or more waves overlap, the total displacement is the sum of the displacements of the individuals.
24. Doppler Effect The change in frequency heard by an observer whenever there is a relative motion between a source of sound waves and the observer is called Doppler effect Let’s call the observed frequency is f’, and the source frequency is f (1) When the observer and the source are moving toward each other, the observed frequency is greater than the source frequency: f’ > f (2) When the observer and the source are moving way from each other, the observed frequency is less than the source frequency: f’ < f v vo v: speed of sound in the medium f 0 = f ± vs: speed of source v v vo: speed of observer ✓ ± s ◆