1、Chapter 19 Sound waves19-1 Properties of Sound waves19-2 Traveling sound waves19-4 Power and intensity of sound waves 19-3*The speed of sound19-5 Interference of sound waves19-6*Standing longitudinal waves19-7*Vibrating system and sources of sound19-8 Beats19-9 The Doppler effectWhen we discuss soun
2、d waves,we normally mean longitudinal wave in the frequency range 20 Hz to 20,000 Hz,the normal range of human hearing.For simplification,we will consider the sound wave in 1D case.19-1 Properites of sound wavesFig19-2In Fig19-2,as the piston(活塞)moves back and forth,it alternately compresses and exp
3、ands(使稀薄)the air next to it.This disturbance travels down the tube as a sound wave.xxxxvmu0P0mSmmpx),(),(txtx0+=),(),(txPPtxP0+=It is convenient to use density and pressure to describe the properties of fluids.Under certain conditions1)Let us assume that the pistol is driven so that the density and
4、pressure of air in the tube will vary as a sine function.(19-1)(19-2)2)Whats the relationship between and?From the definitions of bulk modulus(体模量/膨胀系数)(Eq(15-5)and density ,when m is fixed,we have)sin(tkxm=)sin(tkxPPm=)(vvPB=vm=BPBPvvvvvmvvm02)(=19-2 Traveling sound wavesP3)How to find the displace
5、ment of an element of gas inside the tube?BPmm0=(19-3)x=0 xs(x,t)s(x+,t)xxxxx xxAm=0The undisturbed density of isxA is the corss-sectional area.)()(tx,sxxtx,xsxx-xxx+=),(),(1xtxstxxsx+=.1),1(100+=xs/ifxs/-xs/xAmxs00=/(19-6)or s1xx+=(19-8)kBPksmmm=0)cos(),(tkxstxsm=)(sin)(mtkxutsx,tux=BPvumm=Combine
6、Eqs.(19-1)and(19-6),we have:)sin(),(tkxtxxs0m0=(19-9)v=/k is the velocity of sound wave.is the velocity of oscillation of an element in fluids.)(tx,x(19-4)0Bv=As in the case of the transverse mechanical wave,the speed of a sound wave depends on the ratio ofan elastic property of the medium and an in
7、ertial property.For a 3D fluid,Note:1)B is the bulk modulus,is the mass density.2)Use Newtons law for a system of particles.()xcmxextMaF,=19-3*The speed of soundCsmvair20,/343=019-4 Power and intensity of sound waves As the wave travels,each fluid element exerts a force on the fluid element ahead of
8、 it.If the pressure increase in the fluid element is ,P)sin(tkxPAPAFmx=The power delivered by the element is:)(sin2tkxuPAFuPmmxx=vPAPmav2)(2=(19-18)Average over any number of full cycles.)(sin)(mtkxux,tux=Intensity I:(19-19)The response of the ear to sound of increasing intensity is approximately lo
9、garithmic.One can define a logarithmic scale of intensity called the“sound level SL”(19-20)Where is a reference intensity,which is chosen to be (a typical value for the threshold of human hearing(听觉阈).0log10IISL=0I212/10mwvPAPImav2)(2=The unit of the sound level is“decibels”(dB).A sound of intensity
10、 (听觉阈)has a sound level of 0 dB.The sound at the upper range of human hearing,called the threshold of pain(痛觉阈)has an intensity of and a SL of 120 dB.0I2/1mw声源声源声强声强W/m2声强级声强级dB响度响度引起痛觉的声音引起痛觉的声音1120钻岩机或铆钉机钻岩机或铆钉机10-2100震耳震耳交通繁忙的街道交通繁忙的街道10-570响响通常的谈话通常的谈话10-660正常正常耳语耳语10-1020轻轻树叶的沙沙声树叶的沙沙声10-1110极轻
11、极轻引起听觉的最弱声音引起听觉的最弱声音10-120几种声音近似的声强、声强级和响度几种声音近似的声强、声强级和响度Sample problem 19-2Spherical sound waves are emitted uniformly in all directions from a point source,the radiated power P being 25 w.What are the intensity and the sound level of the sound wave at a distance r=2.5m from the source?Solution:22
12、2/32.0)5.2(4254mwmwrPI=dBmwmwIISL115/10/32.0log10log1021220=Fig19-6 shows two loudspeakersdriven from a common source.At point P the pressure variation due only to speaker is and that due to alone is .Thetotal pressure disturbance at point P is .sourceP1r1s2s2r2P1P21PPP+=1s2sFig 19-619-5 Interferenc
13、e of sound wavesThe type of interference that occurs at point P depends on the phase difference between the waves.L=2,2|2121Lrrkkrkr=(19-22)sin(1tkrPThe phase difference:21rrL=When (m=0,1,2,.)(19-23),The intensity reaches a maximum value,forming constructive interference.When destructive interferenc
14、e occurs.The intensity has a minimum value.mL=)21(+=mL)sin()2/cos(2),(),(),(21=+=tkxytxytxytxymWe assume a train of sine waves travels down a tube(Fig19-7).1)If the end is open,the wave at the end will behave as a pressure node(波节);2)If the end is closed,a pressure antinode(波腹)will form at the end.(
15、a)(b)(c)(d)close endopen endFig 19-719-6*Standing longitudinal wavesnLn2=,n=1,2,3nLn4=,n=1,3,5.See动画库波动与光学夹2-16纵驻波a).For open end,the longitudinal pressure wave is reflected with a phase change of ,because the pressure at the open end must at the value,same as the environments.In this case,it likes
16、the string fixed at both ends.b).For the closed end,the pressure can vary freely.c).The superposition of the original and reflectedwaves gives a pattern of standing waves.d).Resonance can happen,when the driving frequencymatches one of the natural frequency of the system,which are determined by the length of the tube(L).1800PNotes:We have already studied the propagation of the sound wave,and now to understand the nature of the sound we must study the vibration system that produces it.We can clas
