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Chapter 19 Sound Waves

Chapter 19 Sound Waves. Molecules obey SHM s(x,t)=s m cos(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 90 0 out of phase with displacement

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Chapter 19 Sound Waves

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  1. 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

  2. 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)1/2wave speed in a fluid • vair ~ 343 m/s at room temperature • vwater ~ 1482 m/s ( B is larger!) • vsteel ~ 5941 m/s

  3. Intensity Level and Loudness • I=Pav/area • area of sphere = 4r2 => I  1/r2 • sensation of loudness is not directly proportional to intensity • varies as log(I) eg. log(100)=2log(10) • define sound levelSL=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

  4. 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/4r2 .) • (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 = 4r2 x10-11 W => r ~ 200 m. • (b) r ~ 200/ (1000)1/2 = 6.24 m.

  5. Problem • The equation I = Pav / 4r2 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/r12)=20log(r2/r1) • 20 log (20) = 26; SL at 2.4 km = (130 - 26) dB = 104 dB • (b) Subtract 2.28 x8 dB from result of (a) • SL= (104 - 18.2) dB = 85.8 dB

  6. Interference of Sound Waves ripple

  7. 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

  8. 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

  9. Pressure waves in open pipe Pressure waves in pipe closed at one end

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