复旦大学《大学物理》课件（英文）-第17章 Oscillations(1).pdf

Chapter 17 Oscillations17-1 Oscillating SystemsEach day we encounter many kinds of oscillatory motion,such as swinging pendulum of a clock,a person bouncing on a trampoline,a vibrating guitar string,and a mass on a spring.They have common properties:1.The particle oscillates back and forth about aequilibrium position.The time necessary for one complete cycle(a complete repetition of the motion)is called the period T.2.No matter what the direction of the displacement,the force always acts in a direction to restore the system to its equilibrium position.Such a force is called a“restoring force(恢复力)”.3.The number of cycles per unit time is called the“frequency”f.(17-1)Unit:period(s)frequency(Hz,SI unit),1 Hz=1 cycle/sTf1=4.The magnitude of the maximum displacement from equilibrium is called the amplitude of the motion.17-2/3 The simple harmonic oscillator and its motion1.Simple harmonic motionAn oscillating system which can be described in terms of sine and cosine functions is called a“simple harmonic oscillator”and its motion is called“simple harmonic motion”.2.Equation of motion of the simple harmonic oscillatorFig 17-5 shows a simple harmonic oscillator,consisting of a spring of force constant K acting ona body of mass m that slides on a frictionless horizontal surface.The body moves in x direction.Fig 17-5xxmmFooRelaxed stateorigin is chosen at herekxFx=22dtxdax=22dtxdmkx=022=+xmkdtxd(17-4)Eq(17-4)is called the“equation of motion of thesimple harmonic oscillator”.It is the basis of manycomplex oscillator problems.Rewrite Eq(17-4)as(17-5)We write a tentative solution to Eq(17-5)as(17-6)xmkdtxd)(22=3.Find the solution of Eq.(17-4)cos(+=txxmWe differentiate Eq(17-6)twice with respect to the Time.Putting this into Eq(17-5)we obtain Therefore,if we choose the constant such that(17-7)Eq(17-6)is in fact a solution of the equation of motion of a simple harmonic oscillator.)cos(222+=txdtxdm)cos()cos(2+=+txmktxmmmk=2a):If we increase the time by in Eq(17-6),then Therefore is the period of the motion T.22kmT22=(17-8)(17-9)cos()2(cos+=+=txtxxmmmkTf211=The quantity is called the angular frequency.f2=b):is the maximum value of displacement.We call itthe amplitude of the motion.c)and :The quantity is called phase of the motion.is called“phase constant（常相位)”.and are determined by the initial position and velocity of the particle.is determined by the system.mxmx+t+tmxHow to understand?)cos(+=txxm2=0=Txtmxotx图图mx0=xto同相同相How to compare the phases of two SHOs with same?)cos(111+=txxm)cos(222+=txxm)()(12+=tt12=xto为其它为其它超前超前落后落后txo=反相反相(c)same:,different:123456t-1-0.50.51x t2468t-1-0.50.51x ta123456t-1-0.50.51x tFig 17-6 shows several simple harmonic motions.(a)(b)(c)(a)same:,different:mx(b)same:,different:mxFig 17-6mxd).Displacement,velocity,and accelerationDisplacement VelocityAccelerationWhen the displacement is a maximum in either direction,the speed is zero,because the velocity must now change its direction.)cos(+=txxm)sin(+=txdtdxvmx)cos(222+=txdtxdamx)2cos(+=txm)cos(2+=txm(17-11)tx图图tv图图ta图图Tmxmx2mx2mxxvatttmxmxoooTT)cos(+=txxm0=取取2=T)2cos(+=txm)sin(+=txmv)cos(2+=txm)cos(2+=txam17-4 Energy in simple harmonic motion1.The potential energy(17-12)2.The kinetic energy1234560.20.40.60.81)(cos2121222+=tkxkxUmU(t)K(t)T/2T)(sin21)(sin2121222222+=+=tkxtxmmvKmm)sin(+=txmv0=(17-13)Fig 17-8(a)Fig17-8(a),both potential and kinetic energiesoscillate with time t and vary between zero and maximum value of .Both U and K vary with twice the frequency of the displacement and velocity.3.The total mechanical energy E is221mkx221mkxUKE=+=(17-14)mxmxK(x)U(x)xFig 17-8(b)221)(kxxU=)()(xUExK=EAt the maximum displacement ,.At the equilibrium position ,.Eq(17-14)can be written quite generally as 221mkxK=221mkxU=0=K0=U222212121mxkxkxmvUK=+=+)(222xxmkvmx=)(22xxmkvmx=(17-16)(17-15)thenorSample problem 17-2In Fig 17-5,m=2.43kg,k=221N/m,the block is stretched in the positive x direction a distance of 11.6 cm from equilibrium and released.Take time t=0 when the block is released,the horizontal surface is frictionless.(a)What is the total energy?(b)What is the maximum speed of the block?(c)What is the maximum acceleration?(d)What is the position,velocity,and acceleration at t=0.215s?xmoFig 17-5Solution:(a)(b)(c)The maximum acceleration occurs just at the instant of release,when the force is greatest(d)JmmNkxEm49.1)116.0)(/221(212122=smkgJmEmKv/11.143.2)49.1(222maxmax=2maxmax/6.1043.2)116.0)(/221(smkgmmNmkxmFam=sradmk/9536.0=Since at t=0,thenSo at t=0.215s)cos()(+=txtxmmxxm116.0=0=)536.9cos(116.0cos)(ttxtxm=msx0535.0)215.0)(536.9cos(116.0=smtxvmx/981.0sin=222/87.4)0535.0()/536.9(smmsradxax=Sample problem 17-3 In Fig17-5,m=2.43kg,k=221N/m,when the block m is pushed from equilibrium to x=0.0624m,and its velocity ,the external force is removed and the block begins to oscillate on the horizontal frictionless surface.Write an equation for x(t)during the oscillation.smvx/847.0=xmFig 17-5v0Solution:Setting this equal to ,we haveTo find the phase constant ,we stil