Physics 2 - Lecture 1: Oscillations - Pham Tan Thi

Nội dung text: Physics 2 - Lecture 1: Oscillations - Pham Tan Thi

1. 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
2. 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
3. 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
4. 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
5. 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
6. 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
7. Mechanical Waves Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Sciences Ho Chi Minh University of Technology
8. 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.
9. The Function of a Wave 2⇡ 2⇡ y(x, t)=Acos x t T ✓ ◆ = Acos (kx !t)
10. 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
11. 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.