当前位置:高等教育资讯网  >  中国高校课件下载中心  >  大学文库  >  浏览文档

复旦大学:《大学物理》课程教学资源(PPT课件,英文)Chapter 19 Sound waves

资源类别:文库,文档格式:PPT,文档页数:38,文件大小:1.09MB,团购合买
19-1 Properties of Sound waves 19-2 Traveling sound waves 19-4 Power and intensity of sound waves 19-3* The speed of sound 19-5 Interference of sound waves 19-6* Standing longitudinal waves 19-7* Vibrating system and sources of sound 19-8 Beats 19-9 The Doppler effect
点击下载完整版文档(PPT)

Chapter 19 Sound waves 19-1 Properties of Sound waves 19-2 Traveling sound waves 19-3 The speed of sound 19-4 Power and intensity of sound waves 19-5 nterference of sound waves 19-6 Standing longitudinal waves 19-7 Vibrating system and sources of sound 19-8 Beats 19-9 The Doppler effect

Chapter 19 Sound waves 19-1 Properties of Sound waves 19-2 Traveling sound waves 19-4 Power and intensity of sound waves 19-3* The speed of sound 19-5 Interference of sound waves 19-6* Standing longitudinal waves 19-7* Vibrating system and sources of sound 19-8 Beats 19-9 The Doppler effect

19-1 Properites of sound waves When we discuss sound 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

When we discuss sound 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 waves

In Fig19-2, as the piston (活塞) moves back and forth, it alternately compresses and expands certain (使稀薄)) the air next to it conditions X This disturbance travels down the tube as a sound AP wave 0 X p(x,t)=po+△p(x,D) △x ipm (x,D)=P0+△P(x,D) It is convenient to use density and pressure to describe the V X properties of fluids Fig 19-2

Fig19-2 In Fig19-2, as the piston (活塞) moves back and forth, it alternately compresses and expands (使稀薄)) the air next to it. This disturbance travels down the tube as a sound wave. x x x x v m u P0  0 m S  m pm x (x,t) (x,t)  = 0 +  P(x,t) P P(x,t) = 0 +  It is convenient to use density and pressure to describe the properties of fluids. Under certain conditions

19-2 Traveling sound waves 1)Let us assume that the pistol is driven so that the density and pressure of air in the tube will vary as a sine function △p=△OnSn(kx-Ot) (19-1) △P=△PnSi(kx-ot) (192) 2)What's the relationship between Ap and AP From the definitions of bulk modulus(体模量/膨胀系数) (Eq(15-5)B=A and density p=m, when m is fixed,We have △ν △v △P △ν n=△phn B

1) Let us assume that the pistol is driven so that the density and pressure of air in the tube will vary as a sine function. (19-1) (19-2) 2) What’s 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( k x t)  =  m − P P sin( kx t)  =  m − ( ) v v P B   = − v m  = B ρ P B P ρ v v ρ v v v m v v m ρ 0 2 Δ Δ ) Δ ( Δ Δ = − Δ = − = − = = 19-2 Traveling sound waves  P

or△ △P (193) B 3How to find the displacement of an element of gas inside the tube? The undisturbed density of Oxis OX ox A is the corss- iS(x+Sxt) Aδx sectional area SIX 8x=x"x′=[x+x+(x+8,)-x+8x+(x: xx 6x[1+ S(+ar, t-s(x 6x[1+ Sm ASx 1+osar P(1-as/Ox), if as/ax <<1 △p=p-po=-pOs/Ox (196)

3) How to find the displacement of an element of gas inside the tube? B m Pm  0  =  (19-3) x=0 x s(x,t) s(x+ ,t x ) x' x x’ x’’ x A x m   0 = The undisturbed density of is x A is the corss￾sectional area. x ' = x ''-x' = [x + x + s(x + x ,t)]−[x + x + s(x ,t)] ] ( , ) ( , ) δ [1 x s x x t s x t x  + − = + (1 ), 1. ' 1 0 0       +   = = ρ - s/ x if s/ x s/ x ρ A x δm ρ  s x  = −0 = −0  /  (19-6) or ] s δ [1 x x   = +

Combine Egs(19-1)and(19-6), we have Os△p(x,t △Pnsm(k x-ot s(x, t=Sm cos(kx-@t) (198) k kB I >u,(x, t)=2=um sin (kx-at)(19-9) v△ B u(r, t)is the velocity of oscillation of an element in fluids v=@/k is the velocity of sound wave

(19-8) kB P k s m m m  =  =  0  s(x,t) s cos(k x t) = m − ( ) sin ( ) um k x ωt t s u x,t x = −   = B v P u m m  = Combine Eqs. (19-1) and (19-6), we have: sin( ) ( , ) k x t x t x s 0 m 0 −   = −   = −   (19-9) v= /k is the velocity of sound wave. μ (x,t) is the velocity of oscillation of an element in fluids. x 

19-3* The speed of sound as in the case of the transverse mechanical wave the speed of a sound wave depends on the ratio of an elastic property of the medium and an inertial property. For a 3D fluid BP (19-4 v=343m/S.20°C Note: 1)B is the bulk modulus, 2a_is the mass density 2)Use Newton's law for a system of particles (∑Fanx=Mm

(19-4)  0 B v = As in the case of the transverse mechanical wave, the speed of a sound wave depends on the ratio of an elastic property of the medium and an inertial property. For a 3D fluid, Note:1) B is the bulk modulus, is the mass density. 2) Use Newton’s law for a system of particles. ( ) Fext,x = Macm,x 19-3* The speed of sound vair m s C  = 343 / , 20 0

19-4 Power and intensity of sound waves u,(x, t)=um sin(kx-ot) as the wave travels each fluid element exerts a force on the fluid element ahead of it. If the pressure increase in the fluid element is AP F=A·AP=A·_Pnsn(kx-ot The power delivered by the element is P=u,F=A·△ Pu sin2(kx-ot) A(△Pn) (19-18) Average over any number of full cycles

19-4 Power and intensity of sound waves As the wave travels, each fluid element exerts a force on the fluid element ahead of it. If the pressure increase in the fluid element is , P F A P A P sin( k x t) x =  =  m − The power delivered by the element is: sin ( ) 2 P u F A P u k x t = x x =  m m − v A P P m a v 2 ( ) 2  = (19-18) Average over any number of full cycles. ( ) sin ( ) u x,t um k x ωt x = −

Pn,(△Pn) Intensity I: 1=av (19-19) The response of the ear to sound of increasing intensity is approximately logarithmic. One can define a logarithmic scale of intensity called the sound level sl SL=10 log (1920) Where I is a reference intensity which is chosen to be 10 -12w/m2(a typical value for the threshold of human hearing〔听觉阈)

Intensity I: (19-19) The response of the ear to sound of increasing intensity is approximately logarithmic. 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)). 0 10log I I SL = 0 I 12 2 10 w/ m − v P A P I av m 2 ( ) 2  = =

The unit of the sound level is decibels(dB) A sound of intensity/0〔听觉阈) has a sound level of0 dB The sound at the upper range of human hearing called the threshold of pain(痛觉阈) has an intensity of lw/m2 and a sL of 120 dB

The unit of the sound level is “decibels” (dB). A sound of intensity (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. 0 I 2 1w/ m

点击下载完整版文档(PPT)VIP每日下载上限内不扣除下载券和下载次数;
按次数下载不扣除下载券;
24小时内重复下载只扣除一次;
顺序:VIP每日次数-->可用次数-->下载券;
共38页,可试读13页,点击继续阅读 ↓↓
相关文档

关于我们|帮助中心|下载说明|相关软件|意见反馈|联系我们

Copyright © 2008-现在 cucdc.com 高等教育资讯网 版权所有