11.4 SoundProducing sound wavesSpeed of soundEnergy and IntensitySpherical and Plane waves.Interference of sound wavesProducing sound waves•Produced by compression and rarefaction of media (air) •Sound waves are longitudinal resulting in displacement in the direction of propagation.• The displacements result in oscillations in density and pressure.DensityPressureDisplacementFrequencies of sound waveAudible SoundFrequency (Hz)10 20,000ultra- sonicinfra-sonicWavelength (m) in air30 0.015Speed of soundSpeed of sound in a fluidBv =ρPBV/V∆=−∆Bulk modulusmVρ=DensitySimilarity to speed of a transverse wave on a stringelastic _propertyvintertial_property=Bv =ρDensity is higher in water than in air. Why is the speed of sound higher in water than in air?Speed of sound in airPvγ=ργ is a constant that depends on the nature of the gas γ =7/5 for air.P - Pressureρ -DensitySince P is proportional to the absolute temperature T by the ideal gas law. PV=nRTTv 331273=(m/s)v is dependent on T2Find the speed of sound in air at 20oC.Tv331273=273 20v 331 343m/s273+==For calculations use v=340 m/sEnergy and Intensity of sound wavespowerenergyPtime=Intensity power PIarea A==area A(units W/m2)Sound intensity levelThe decibel is a measure of the sound intensity leveloII10log⎛⎞β=⎜⎟⎝⎠Io= 10-12 W/m2the threshold of hearingdecibels (dB)note - decibel is a logarithmic unit. It covers a wide rangeof intensities. The ear is capable ofdistinguishing a widerange of sound intensities.What is the intensityof sound at a rock concert? (W/m2)oII10log 120⎛⎞β= =⎜⎟⎝⎠0II120log 1210⎛⎞==⎜⎟⎝⎠0II1210=120I = 10 I12 1210 10 1−=⋅=W/m2Spherical and plane wavesAs sound spreads out uniformly from a point sourceThe intensity decreases as 1/r22A4r=πarea of sphere2PI4r=πSuppose you are standing near a loudspeaker that can is blasting away with 100 W of audio power. How far away from the speaker should you stand if you want to hear a sound level of 120 dB. ( assume that the sound is emitted uniformly in all directions.)2PPIA4r==πPr4I=π2100W2.8m4(1W/m)==π3Interference of sound wavesConstructive InterferenceDestructive InterferenceTwo sound waves superimposed Noise canceling headphonesNoiseAnti-noiseWave 1Wave 2Interference due to path differencepath difference =δ =r2–r1In phase at x=0Condition for constructive interferencemδλ=Condition for destructive interference1()2mδλ=+m = 0 + 1, + 2,….r1r2Superposition of wavesat A shows interferencedue to path differencesAwhere m is any integerWave 1Wave 2Interference of sound wavesPhase shift due to path differencesxWhen r2–r1=nλConstructive InterferenceWhen r2–r1= (m+½) λDestructive InterferenceDetermining the wavelength of a sound wave – determine the speed of soundFind wavelength of the sound from interference.Example 14.6 Path difference for two sources.At position P the listener hears the first minimum in sound intensity. Find the frequency of the oscillation.r2-r1=0.13 mAt position P the path difference is equal to λ/2. (firstminimum) destructive interference.21r r 0.13m2λ=−=2(0.13) 0.26mλ==3v 340m/sf1.31x10Hz0.26m==
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