Sound waves. Content. Chapter 21. objectives. objectives. When we use Sound Waves. What are sound waves? How they work.

Transcription

1 Chapter 21. Sound wae Content 21.1 Propagation o ound wae 21.2 Source o ound 21.3 Intenity o ound 21.4 Beat 21.5 Doppler eect 1 2 objectie a) explain the propagation o ound wae in air in term o preure ariation and diplacement b) interpret the equation or diplacement, y = y o in ( t kx), and preure, p = p o in ( t kx + /2) c) ue the tanding wae equation to determine the poition o node and antinode o a tanding wae along a tretched tring d) ue the ormula = (T/ ) 1/2 to determine the requencie o the ound produced by dierent mode o ibration o the tanding wae along a tretched tring What are ound wae? A mechanical wae that ibrate a medium (like air or water) with dierent requencie. Thee requencie are then picked up by our ear. They are created through a ariety o interaction, but all are mechanical (Phyical). How they work 3 5 objectie decribe, with appropriate diagram, the dierent mode o ibration o tanding wae in air column, and calculate the requencie o ound produced, including the determination o end correction deine and calculate the intenity leel o ound ue the principle o uperpoition to explain the ormation o beat ue the ormula or beat requency, = 1 2 decribe the Doppler eect or ound, and ue the deried ormulae (or ource and/or oberer moing along the ame line) When we ue Sound Wae Muic tie into Sound wae and requencie. Each note ha a dierent requency. We talk through ound wae, and apply meaning to certain ound. Dolphin and bat ue ound wae (onar ). Dolphin ue it to communicate, like a language, and bat ue them to ly due to poor eye ight. How they work 4 6 Sound wae trael in a longitudinal way (ertical ahion), a hown by the tuning ork in the picture. The ound ibrate the medium between the whateer i traight in ront o it. 7 A ound wae i meaured in hertz (Hz) => ibration/econd Thee are High and Low requency wae, they how the dierence between the two. 8

2 How they work The period, T between the wae categorize their requencie, a low or high. 1/T The higher requency ha a maller amount o time between wae, while the lower requency ha a longer amount o time. Frequency 9 Frequencie Interal Frequency Ratio Example Octae 2:1 512 Hz and 256 Hz Third 5:4 32 Hz and 256 Hz Fourth 4:3 342 Hz and 256 Hz Fith 3:2 384 Hz and 256 Hz Thi chart explain ound wae pertain to muic make muic. For intance, raiing a note an octae would require multiplying the bae note by 2 (take a low c, with requency o 261.5, to raie it an octae: ha requency 523.) Sound 1 Note C C# D D# E F F# G G# A A# B C C# D Octae Hz Thi table how the alue in hertz o certain note (rounding applie). 11 A longitudinal traeling wae Produced by ibration in a medium The diturbance i the local change in preure generated by the ibrating object It trael becaue o the molecular interaction. The region o increaed preure (compared to the normal preure) i called condenation The region o lower preure i called rareaction. 12 Sound The maximum increae in preure (DP m ) i the amplitude o the preure wae. (meaurable) requency: 2Hz to 2kHz. Preure wae below 2 Hz are called inraonic wae Preure wae oer 2kHz are called ultraonic wae Propagation o ound wae 1. The propagation o ound wae occur due to the ocillation o indiidual particle with the medium producing traeling wae o preure luctuation 2. The general orm o particle ocillation y(x, t) = y o co(kx - t) or y = y o in ( t kx) where y o i the magnitude o the particle diplacement Propagation o ound wae 3. The general equation or the preure luctuation: P(x, t) = P o in(kx - t) or P = P o in ( t kx + /2) 21.2 Source o Sound Muical intrument produce ound in ariou way ibrating tring, ibrating membrane, ibrating metal or wood hape, ibrating air column. The ibration may be tarted by plucking, triking, bowing, or blowing. The ibration are tranmitted to the air and then to our ear

3 21.2 Source o Sound: Vibrating String The tring on a guitar can be eectiely hortened by ingering, raiing the undamental pitch. The pitch o a tring o a gien length can alo be altered by uing a tring o dierent denity Source o Sound: Vibrating String A piano ue both method to coer it more than eenoctae range: the lower tring (at bottom) are both much longer and much thicker than the higher one Source o Sound: Vibrating Air Column Source o Sound: Vibrating String and Air Column Wind intrument create ound through tanding wae in a column o air Source o Sound: Vibrating String and Air Column A tube open at both end (mot wind intrument) ha preure node, and thereore diplacement antinode, at the end Source o Sound: Vibrating Membrane A piece o elatic membrane can ibrate in the mode a hown in the igure below: Vibrating Membrane A tube cloed at one end (ome organ pipe) ha a diplacement node (and preure antinode) at the cloed end Intenity o ound Intenity o ound Wae tranport energy without tranporting ma. The amount o energy tranported per econd i the power (P) o the wae (in W) Intenity i a meaure o power tranmitted by a wae per unit area: I= 23 Power P = = Area A medium wae 2 2 m 24

4 21.3 Intenity o ound The energy tranmiion (power) i determined by the ource. The power i ditributed (pread) in all direction. Far away rom the ource, the power i pread oer a greater area. For a point ource, intenity decreae inerely with the quare o the ditance rom the ource: 21.4 Beat P I(r) = = A P 4 r 1. When 2 ound wae the reultant wae pattern exhibit both contructie and detructie intererence. 2. When the amplitude o the 2 wae are imilar but the requencie are lightly dierent then: a. The requency o the reultant wae i roughly the aerage requency o the 2 wae 2 25 Loudne & Decibel 1. The human doe not perceie ound intenity linearly but rather logarithmically Perceied Loudne, I perceied log (I actual ) 2. The aerage minimum perceiable ound intenity: I o -12 W/m 2 3. The decibel cale wa been deeloped to ear perception (intenity leel, ): = (1 db). log(i/i ) = (1 db). log(i + 12) 21.4 Beat 2. When the amplitude o the 2 wae are imilar but the requencie are lightly dierent then: a. The combined eect o intererence produce periodic rie and drop in loudne called beat b. The requency o the beat ( beat ) i equal to the dierence between the 2 ound requencie: beat = Muician oten tune their muical intrument by litening to beat requency Beat The uperpoition o 2 ound wae: wae1 =159.2 Hz wae1 =148. Hz The reulting beat requency: beat = wae1 - wae2 = Hz Hz = 21.2 Hz 21.4 Beat When two ound wae o dierent but nearly equal requency ( 1 and 2 ) uperimpoe, we an intenity ariation at the dierence requency The intenity ariation i called beat The beat requency i equal to the dierence requency beat Ued to tune muical intrument to ame pitch Beat Superimpoe ocillation o equal amplitude, but dierent requencie Ain(2 t) Ain(2 t) 1 2 ( ) ( ) 2 2 ( ) ( ) Ain(2 t) co(2 t) Aco(2 t) in(2 t) Modulation o amplitude requency o pule i 1-2 Not examinable Ocillation at the aerage requency 21.4 Beat intererence in time Conider two ound ource producing audible inuoidal wae at lightly dierent requencie 1 and 2. What will a peron hear? How can a piano tuner ue beat in tuning a piano? I the two wae at irt are in phae they will interere contructiely and a large amplitude reultant wae occur which will gie a loud ound. A time pae, the two wae become progreiely out o phae until they interere detructiely and it will be ery quite. The wae then gradually become in phae again and the pattern repeat itel. The reultant waeorm how rapid luctuation but with an enelope that ariou lowly. CP CP

5 21.4 Beat intererence in time The requency o the rapid luctuation i the aerage requencie = The requency o the lowly arying enelope = Beat intererence in time Since the enelope ha two extreme alue in a cycle, we hear a loud ound twice in one cycle ince the ear i enitie to the quare o the wae amplitude. The beat requency i beat CP =1 = 12 beat = 1 Hz 2 = 11 Hz rapid = 15 Hz T rapid = 9.5 m beat = 1 Hz T beat =.1 (loud pulation eery.1 ) time time 1 = 1 Hz 2 = 12 Hz rapid = 11 Hz T rapid = 9.1 m beat = 2 Hz T beat =.5 (loud pulation eery.5 ) CP 535 CP Doppler eect time 1 = 1 Hz 2 = 14 Hz rapid = 12 Hz T rapid = 9.8 m beat = 4 Hz T beat =.25 (loud pulation eery.25 ) =1 = 14 beat One might wonder why the iren on a moing ambulance eem to produce ound with a higher pitch when it pae an oberer and decreae when it recede the oberer. I thi imply becaue o the relatie ditance between the oberer and the ambulance (ound)? Or i it becaue o the loudne o the ound produced by the iren? CP Doppler eect Chritian Johann Doppler ( ) Studied motion related requency change (1842) o o 21.5 Doppler eect Doppler eect i the change in requency o a wae (or other periodic eent) or an oberer moing relatie to it ource. o o Source () Oberer (o) Source () Oberer (o) 39 4

6 21.5 Doppler eect o wae and the oberer are approaching each other, the ound heard by the oberer become higher in pitch, wherea i the ource and oberer are moing apart the pitch become lower. For the ound wae to propagate it require a medium uch a air, where it ere a a rame o reerence with repect to which motion o ource and oberer are meaured Doppler eect Application: police microwae peed unit peed o a tenni ball peed o blood lowing through an artery heart beat o a deeloping etou burglar alarm onar hip & ubmarine to detect ubmerged object detecting ditance planet obering the motion o ocillating tar Doppler eect Conider ource o ound at requency, moing peed, oberer at ret ( o = ) Speed o ound What i requency o heard by oberer? 21.5 Doppler eect On right - ource approaching ource catching up on wae waelength reduced requency increaed On let - ource receding ource moing away rom wae waelength increaed requency reduced SITUATION 1 Stationary Source and Oberer (NO DOPPLER EFFECT) A tationary ound ource S emit a pherical waeront o relatie to the medium air. In time t, the waeront moe a ditance t toward the oberer, O 1 & O 2. The number o waelength detected by the oberer inront and behind the ource are the ame and equal to t 21.5 Doppler eect What i both o the oberer in igure 1 are moing, i there any change in the requency and waelength o the ource? SITUATION 1 Stationary Source and Oberer (NO DOPPLER EFFECT) Thu, the requency heard by both tationary oberer i gien by, t / t - requency o ound ource - peed o ound wae t - time - waelength SITUATION 2 Stationary Source; Moing Oberer Oberer 1 moe a ditance o t toward the ource at peed o We had known earlier that waeront alo moe at peed toward O 1 in time t at ditance t. The ditance traeled by the waeront with repect to O 1 become t + O t. The number o waelength intercepted by O 1 at thi ditance i (t + t

7 SITUATION 2 Stationary Source; Moing Oberer Thi how that there i an increae in the requency heard by O 1 a it goe nearer to the ound ource a gien by, ' Since ( t t t) / = /, then ' SITUATION 2 Stationary Source; Moing Oberer ' (2) (3) In thee ituation only the requency heard by the oberer change due to there motion relatie to the ource. Howeer the waelength o ound remain contant. (5) 49 SITUATION 2 Stationary Source; Moing Oberer I oberer 2 moe away rom the ound ource, the ditance traeled by the waeront with repect to O 2 in time t, i t o t. Conequently, there would be a decreae in the requency heard by O2 a gien by, ' (4) SITUATION 3 Moing Source; Stationary Oberer A the ource moe a ditance S T (T=1/ period o wae) toward O 1 there i a decreae in the waelength o ound by a quantity o T. The hortened T SITUATION 3 Moing Source; Stationary Oberer The requency o ound wae heard by O 1 increae a gien by, ' ' ' T / (6) SITUATION 3 Moing Source; Stationary Oberer With repect to oberer 2, the waelength o ound increae, where T. The requency o ound wae heard by O 2 decreae a gien by, ' SITUATION 3 Moing Source; Stationary Oberer SITUATION 4 Moing Source and Oberer ' ' Combining Equation (6) and (7), we hae ' (6) (7) (8) 55 From the equation (5) and (8), we can now derie the equation o general Doppler Eect by replacing in equation (5) with o equation (8). Thi reult to, ' (Moing ource and oberer) (9) 56

8 SITUATION 4 Moing Source and Oberer ' (9) SITUATION 4 Moing Source and Oberer ' ' The ± ign correpond to the direction o the ource or oberer when they are moing relatie to the other. Thee would determine whether there i an increae or decreae on the requency heard by the oberer during the motion. Approaching oberer, receding ource I o >, obered requency increae I o <, obered requency decreae Receding oberer, receding ource Decreae in obered requency SITUATION 4 Moing Source and Oberer ' Approaching oberer, approaching ource Obered requency increae ' Receding oberer, approaching ource I o >, obered requency decreae I o <, obered requency increae Problem A train ha a whitle, which emit a 4 Hz ound. You are tationary and you hear the whitle, but the pitch i 44 Hz. How at i train moing toward or away rom you? 58 6 Solution: Summary: Sound wae The pitch i higher, o the train i moing toward you. It peed relatie to you i ound rom = /(- ). We hae (- ) = / = (4/)(33 m/)/(44/) = 3m/. Thereore = 33m/ 3m/ = 3m/ Propagation Source o ound Intenity leel Beat Doppler eect y = y o in ( t kx) p = p o in ( t kx + /2) Open end tube: L n = n /2, n = n/2l One Cloed End tube: L n = n /4, n = n/4l I = P/4 r 2 = (1 db). log(i/i ) beat = 1-2 [( )/( )] 61 62