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