Chapter 19 Sound Waves• Molecules obey SHM

s(x,t)=smcos(kx-t)

• leads to variations in density and pressure in the air

• pressure is large where density is large but displacement s(x,t) is small

• hence pressure(or density) wave• is 900 out of phase with displacement

• p=p0 + pm cos(kx-t-/2) = p0 + pm sin(kx-t) (p is change in pressure relative to equilibrium)

= 0 + m sin(kx-t) “rho”

m = pm (0/B)

• B is bulk modulus

Speed of Sound• B=-p/(V/V)

• Water: B~2.2x109N/m2 Air: B~ 105 N/m2

• gases are more compressible (B is smaller)

• speed of sound in a fluid depends on the density and the elasticity-> i.e. the medium

• v = (B/0)1/2 wave speed in a fluid

• vair ~ 343 m/s at room temperature

• vwater ~ 1482 m/s ( B is larger!)

• vsteel ~ 5941 m/s

Intensity Level and Loudness• I=Pav/area

• area of sphere = 4r2 => I 1/r2

• sensation of loudness is not directly proportional to intensity

• varies as log(I) eg. log(100)=2log(10)

• define sound level SL=10 log(I/I0) in decibels (dB)

• I0 = 10-12 W/m2 is the hearing threshold

• hence for I=I0 , SL= 10 log(1) = 0

• pain threshold is I= 1 W/m2 or SL=10log(1012)=120 dB

Problem • When a pin of mass 0.1 g is dropped from a height of 1 m, 0.05% of

its energy is converted into a sound pulse with a duration of 0.1 s.

• (a) Estimate the range at which the dropped pin can be heard if the minimum audible intensity is 10 -11 W/m2 .

• (b) Your result in (a) is much too large in practice because of background noise. If you assume that the intensity must be at least 10-8 W/m2 for the sound to be heard, estimate the range at which the dropped pin can be heard. (In both parts, assume that the intensity is P/4r2 .)

• (a) Sound energy is 5 x 10-4 (mgh) = 5 x 10-4 (1m)(10-4 kg)(9.8 m/s2 ) = 4.9 x 10-7 J

• Pav = E/t = 4.9 x10-6 W = 4r2 x10-11 W => r ~ 200 m.

•

• (b) r ~ 200/ (1000)1/2 = 6.24 m.

Problem • The equation I = Pav / 4r2 is predicated on the assumption that the transmitting

medium does not absorb any energy.

• It is known that absorption of sound by dry air results in a decrease of intensity of approximately 8 dB/km.

• The intensity of sound at a distance of 120 m from a jet engine is 130 dB.

• Find the intensity at 2.4 km from the jet engine (a) assuming no absorption of sound by air, and (b) assuming a diminution of 8 dB/km.(Assume that the sound radiates uniformly in all directions.)

• (a) SL = 10 log(I1/I0)-10 log (I2/I0)= 10 log(I1/I2) = 10log(r22/r1

2)=20log(r2/r1)

• 20 log (20) = 26; SL at 2.4 km = (130 - 26) dB = 104 dB

• (b) Subtract 2.28 x 8 dB from result of (a)

• SL= (104 - 18.2) dB = 85.8 dB

ripple

Interference of Sound Waves

Interference andPhase Difference

• Consider shifting one of the waves by a distance relative to the other

• they add up constructively • corresponds to a phase shift by = 2• constructive interference for = m2 • or path difference L= m • a shift by = corresponds to a shift of one of

the waves by /2• destructive interference for = (2m+1) • or path difference L= (2m+1) /2

Musical Sounds

• oscillating strings• membranes• air columns• standing wave patterns correspond to

resonances• large amplitude oscillations push surrounding

air and generate sound waves at the same frequency

Pressure waves in open pipe

Pressure waves in pipeclosed at one end