testAnnouncementPAL #9 SoundPAL #9 Sound (cont.)Intensity of SoundIntensity and DistanceInverse Square LawThe Decibel ScaleSound LevelsMusicSound Waves in a TubeHarmonicsHarmonics in Closed and Open TubesMusical InstrumentsBeat FrequencyBeatsBeats and TuningtestPhysics 202Professor Lee CarknerLecture 10AnnouncementTuesday help session moved to SC120 (physics discussion room)Practice problems on WebAssign and web pageTest Friday½ conceptual multiple choice½ problemsEquations given, but not labeled or explainedBring calculator and pencilPAL #9 SoundChanging medium to get max vv = (B/)½Want large B and small A low density fluid that is hard to compressChanging medium to get max  pmpm = vsm = (B/)½sm = B½½sm Large v and large Want to increase v by increasing B, not decreasing Want medium with large B and large Such a fluid would be hard to moveHeavy and hard to compressPAL #9 Sound (cont.)Interference from two loudspeakersTo get destructive interference you want the received waves to be out of phase by ½ wavelength Want the difference in path length to be ½ f = 1150 Hz, v = 343 m/s (for room temperature air)v = f,  = v/f = 343/1150 = 0.3 mWant L to be 0.15 mIf L1 is 4m, make L2 4.15 mConstructive interference occurs when L = 0, 1, 2 …L2 = 4 m (or 4.3 m or 3.7 m etc.)Intensity of SoundThe loudness of sound depends on its intensity, which is the power the wave delivers per unit area:I = P/AThe units of intensity are W/m2The intensity can be expressed as:I = ½v2sm2Compare to expression for power in a transverse waveDepends directly on  and v (medium properties)Depends on the square of the amplitude and the frequency (wave properties)Intensity and DistanceConsider a source that produces a sound of initial power PsAs you get further away from the source the intensity decreases because the area over which the power is distributed increasesThe total area over which the power is distributed depends on the distance from the source, rI = P/A = Ps/(4r2)Sounds get fainter as you get further away because the energy is spread out over a larger areaI falls off as 1/r2 (inverse square law)Inverse Square LawSourcer2rA1=4r2I1 = Ps/A1A2=4(2r)2 = 16r2 = 4A1I2 = Ps/A2 = ¼ I1The Decibel ScaleThe human ear is sensitive to sounds over a wide range of intensities (12 orders of magnitude)To conveniently handle such a large range a logarithmic scale is used known as the decibel scale = (10 dB) log (I/I0)I0 = 10-12 W/m2 (at the threshold of human hearing)log is base 10 log (not natural log, ln)There is an increase of 10 dB for every factor of 10 increase in intensitySound LevelsHearing Threshold0 dBWhisper10 dBTalking60 dBRock Concert110 dBPain120 dBMusicA musical instrument is a device for setting up standing waves of known frequencyA standing wave oscillates with large amplitude and so is loudWe shall consider an generalized instrument consisting of a pipe which may be open at one or both endsLike a pipe organ or a saxophoneThere will always be a node at the closed end and an anti-node at the open endCan have other nodes or antinodes in between, but this rule must be followedClosed end is like a tied end of string, open end is like a string end fixed to a freely moving ringSound Waves in a TubeHarmonicsPipe open at both endsFor resonance need a integer number of ½ wavelengths to fit in the pipeAntinode at both endsL = ½  n v = ff = nv/2Ln = 1,2,3,4 …Pipe open at one endFor resonance need an integer number of ¼ wavelengths to fit in the pipeNode at one end, antinode at otherL = ¼ n v = ff = nv/4Ln = 1,3,5,7 … (only have odd harmonics)Harmonics in Closed and Open TubesMusical InstrumentsWhen playing a musical instrument you change n, v or L to produce a sound at the desired frequencyMusical notes are related to a specific frequency For example: A = 440 Hz Music is the superposition of all of the notes being played at one timeSmaller instruments generally produce high frequency soundf is inversely proportional to LBeat FrequencyYou generally cannot tell the difference between 2 sounds of similar frequencyIf you listen to them simultaneously you hear variations in the sound at a frequency equal to the difference in frequency of the original two sounds called beatsfbeat = f1 –f2BeatsBeats and TuningThe beat phenomenon can be used to tune instrumentsCompare the instrument to a standard frequency and adjust so that the frequency of the beats decrease and then disappearOrchestras generally tune from “A” (440 Hz) acquired from the lead oboe or a tuning