Sound Waves Sound is a longitudinal wave It requires a medium to convey it e g a gas liquid or solid In a gas the amplitude of the sound wave is air pressure a series of slightly enhanced crest and reduced trough pressure or air density regions The speed that these pressure variations move the wave speed is the speed of sound A sound wave is longitudinal since for example the air molecules positions oscillate in the direction that the wave travels they oscillate from condensed regions crest to underdense regions trough Table 14 1 lists the sound speeds for various gases liquids and solids The sound speed in solids liquids gases Given some physical properties of the medium it is possible to calculate the sound speed For ideal gases low density gases for which the gas atoms or molecules do not interact discussed in Chap 17 the speed of sound is kb T v gas m m mass of a gas atom or molecule kg T temperature of the gas Kelvin K For temperature we must use the absolute scale of Kelvin T K T C 273 15 Chap 16 kb Boltzmann s constant 1 38x10 23 J K Think of kb as a conversion factor between temperature and energy adiabatic index of a gas a unitless constant which depends on the gas usually between 1 3 1 7 It is 1 4 for air Chap 18 Notice that the speed of sound increases with temperature It is also possible to calculate the speed of sound in liquids and solids We will not consider those expressions Just be aware of the trends e g vair 343 m s vwater 1482 m s vsteel 5960 m s Example Problem The wavelength of a sound wave in air is 2 74 m at 20 C What is the wavelength of this sound wave in fresh water at 20 C Hint the frequency is the same Solution Given air 2 74 m fair fwater v v vT f f f air f water v air v water v water water air air water v air 1482 m s water 2 74 m 11 8 m 343 m s How about for the sound wave in steel water 5960 m s 2 74 m 47 6 m 343 m s As a sound wave passes from one medium to another its speed and wavelength changes but not its frequency Example Problem A jet is flying horizontally as shown in the drawing When the jet is directly overhead at B a person on the ground hears the sound coming from A The air temperature is 20 C If the speed of the jet is 164 m s at A what is its speed at B assuming it has a constant acceleration A B Solution Given vA jet 164 m s ajet constant vair 343 m s Table 14 1 Find vB jet x R 36 0 P Let x be the distance between A and B and R the distance between A and the person P x R sin The time for the jet to travel from A to B is the same as the time for the sound wave to travel from A to P t jet t sound t R v sound t sound t sound From 1D kinematics R v sound x t v sound sin x x v A v B t jet v A v B v sound sin x v sound sin 12 v A v B x 1 2 1 2 2 v sound sin v A v B v B 2 v sound sin v A v B 2 343 m s sin 36 0 164 m s v B 239 m s Skip Section 14 5 The Doppler Effect of a Sound Wave When a car passes you at rest holding its horn the horn sound appears to have a higher pitch larger f as the car approaches and a lower pitch smaller f as the car recedes this is the Doppler Effect named for an Austrian physicist The effect occurs because the number of sound wave condensations crest changes from when the car is approaching to when the car is receding and is different if the car and you are both at rest The frequency of the car horn we call the source frequency fs Also called the rest frequency since it is the sound frequency you would hear if the car and you observer each had zero velocity f s v sound v sound T When both the source and observer are at rest a condensation wave crest passes the observer every T with the distance between each crest equal to the wavelength The frequency heard by the observer fo fs Now consider two different cases 1 the source moving with velocity vs and the observer at rest and 2 the source at rest and the observer moving with velocity vo Moving Source The car is moving toward you with vs It emits a wave A time T later it emits another wave but the car has traveled a distance d vsT The wavelength between each wave is reduced Therefore the frequency heard by the observer must increase d The reduced wavelength is d v sT The frequency heard by the observer is v sound v Let vsound v fo v sT v v v fo f s 1 v vs v vs v v s fs fs f s 1 For source moving towards observer fo fs f o f s 1 v s v For source moving away from observer wavelength increases v sT d Following the same procedures gives 1 f o f s For source moving away 1 vs v fo fs Observer hears lower pitch Moving Observer If the observer moves toward the source which is at rest with speed vo the emitted wavelength remains constant But the observer can run into more cycles wave crests than if she remained at rest The number of additional cycles encountered is vot Or the additional number of cycles second which is a frequency is v f o Therefore the frequency heard by the observer is vo vo fo fs f fs f s 1 fs vo f o f s 1 since v f s v Therefore if the observer is moving towards the source the frequency heard by the observer is increased fo fs Now for the observer moving away from the source she will encounter vot fewer wave crests than if she remained stationary The observed frequency will be vo vo fo fs f fs f s 1 fs vo f o f s 1 In this case f0 fs v To summarize 1 Moving source moving together 1 f o f s 1 vs v 2 Moving observer vo f o f s 1 v moving apart moving together moving apart Note that equations look similar but mechanisms for frequency shifts f fo fs are different Finally both observer and source can be moving 1 vo v f o f s 1 vs v Example Problem Suppose you are stopped for a traffic light and an ambulance …
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