Waves in cavities such as vehicle compartments, rooms or ducts

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1 7.1 Wavs in cavitis such as vhicl compartmnts, rooms or ducts Sound propagation from sourcs into th fr fild is rlativly simpl. At a rciving position in a distanc to th sourc th sound will arriv dlayd by th tims ndd for th sound to travl from sourc to rcivr and it will b attnuatd dpnding on th gomtrical distribution of th sound powr from th sourc into th spac. In th pictur on th right th situation is diffrnt. Th sourc might b tyr/road intraction. Th rcivr might b th drivr of th car. Sound, which coms into th car compartmnt ithr by vibration of th whl suspnsion or through xcitation of th car body by a prssur fild, will b modifid by th acoustic proprtis of th compartmnt. In a simplifid viw, ths proprtis ar dtrmind by th rsonancs of th compartmnt cavity and th damping du to absorption matrial insid th compartmnt. To study ths ffcts mor in dtail, howvr w hav to start with a mor simpl gomtry. Th most simpl cavity is tub at low frquncis, i.. th wavlngth is big in comparison to th diamtr of th tub. In this cas w only hav plan wavs insid th tub, w can considr th tub as a on-dimnsional wavguid. Th wav fild in th tub is dtrmind by th boundary conditions on both nds of th tub that ar charactrisd by thir impdanc and in this way by thir rflction cofficints. Imagin a sourc snd out a harmonic signal. This sourc is situatd at th lft nd of th tub x=). It is crating an initial wav front of th form p x,t ) = p! jkx jt. This wav front will b rflctd at th right nd of th tub x=l). It can b considrd as a wav, which startd at th sam tim as th initial wav front but twic th lngth of th tub away in x dirction and which travls in ngativ x dirction a imag sourc). Th amplitud of th rflctd wav is dtrmind by th rflction cofficint r L and can b writtn as p r1 x,t ) jt ) = p! jk 2L!x r L This wav travls back to th lft nd of th tub and is rflctd thr again. Th rflctd wav is p r 2 x,t ) jt ) = p! jk 2L +x r L

2 7.2 W can now continu this procdur and will gt all rflctd wavs. Th wav fild insid th tub is th sum of all rflctd wavs plus th initial wav: p x,t ) = p! jkx 1 + r L! jk 2L + r L ) 2! jk 4L + r L ) 3 % )! jk 6L... * # &! jk 2L!x r L ) 1 + rl! jk 2L + r L ) 2! jk 4L + r L ) 3 % +! jk 6L..., # &- On possibility to writ this quation is to sum up all trms in th brackts and rplac thm by th unknown amplituds A + and A!. Th xprssion obtaind in this way is calld th wav approach and is givn by p x,t ) = p A +! jkx! jk 2L!x + A! )% # & It xprsss th wav fild in th tub by two wavs propagating in th positiv and in th ngativ dirctions. Th unknown amplituds A + and A! of both wavs hav to b dtrmind by th boundary conditions. This dscription is widly usd whn discussing wav filds in coupld systms. Th quation consisting of th contribution of many rflctions can also b intrprtd in a diffrnt way. Th xprssion in th brackts has th form 1 +! +! 2 +! = 1 for! < 1 1! Assuming that th magnituds of th rflction cofficints on both sids ar smallr than 1 w can writ! jkx! jk 2L!x + r L ) p x,t ) = p! jk 2L 1! r L If th assumption! < 1is not fulfilld th sris dos not convrg. Or to say it diffrntly: If w xcit a systm without damping which rflcts th wavs at th boundaris compltly th wav fild will nvr bcom stabl, but incras continuously in amplitud undr crtain conditions. From th solution w can s that thr ar crtain frquncis whr th amplituds of th wavs bcom infinity if thr is no damping includd. Ths frquncis ar dtrmind by 1! r L! jk 2L = or [ ) )] = 1 r L cos k2l + j sin k2l Now w can assum crtain boundaris to find a solution to th quation. Whn assuming both nds of th duct as rigid w will find as rflction

3 7.3 cofficints = 1 and r L = 1. Th quation which has to b fulfilld is ) ) j sin k2l = and cos k2l = 1. Th solutions ar k n =!n L with n=,1,2,3,....this can b writtn a bit diffrntly in th way 2! =!n n L or! = 2L n n 11) Valus dtrmind by Eq. 1) ar oftn calld ignvalus for th wav propagation in a tub with rigid trmination on both sids. W gt vry high amplituds in th tub for frquncis whr th wavlngth quals twic th lngth, th lngth, or intgr fractions of th lngth. Th frquncis for ths wavlngths ar f n = c! n = c n 2L Ths frquncis ar oftn calld ignfrquncis or rsonanc frquncis. To ach of ths rsonanc frquncis blongs a crtain vibration pattrn in th tub. Th vibration pattrns can also b calculatd by th numrator of th solution! jkx! jk 2L!x + r L ) which givs p n x ) = cos!n % # & 1,5 Th vibration pattrn is shown in th Figur on px)/p 1 max th right for th cas that n qual 4, i.. that,5 two wav lngths fits to th lngth of th tub -,5 or 4 tims a half wavlngth). Th figurs - 1-1,5,2,4,6,8 1 show prssur and vlocity for th tim t= x/l 1,5 insid th tub compard with th ffctiv ux)/u 1 max valus of prssur and vlocity.,5 Each of ths vibration pattrns at th -,5-1 rsonanc frquncis is calld mods. Th -1,5,2,4,6,8 1 fantastic proprty of mods is, that w can x/l 1,5 built up any sound fild insid th tub by a p 2 1 ffx) / p2 max,5 summation ovr such mods. Th ida is vry similar to th ida that you can build up any -,5-1 tim signal by a summation of harmonic tons. -1,5 % p x,t ) = # p n x ) j!t = A n cos n,2,4,6,8 1 1,5 x/l # * j!t u 2 / 1 ffx) u2 max,5 & ) n = n = -,5-1 -1,5,2,4,6,8 1 x/l

4 7.4 Exrcis: driv th rsonanc frquncy for a tub whr on nd is closd and on nd is opn assum that an opn nd corrspond to prssur rlas condition). Intrprtation of th rsonanc phnomna Thr is a diffrnt introduction of th rsonanc phnomna, which is much mor appaling than th mathmatical drivation shown abov. It is basd on th concpt of th closd wav train. This concpt says that w will obtain rsonanc whn th rflctd wav fits in phas to th signal at th xcitation position. In th xampl of th tub, th rflctd wav has a phas shift, which is dtrmind by th dlay btwn starting at th xcitation sid and rturning to this point. Th dlay is twic th lngth of th tub dividd by th spd of sound!t = 2L 2L. This dlay corrsponds to th phas shift! = #t = c c. That th signal arrivs just in phas with th sourc signal, th phas has to b 2!, 4!, 6!,... This can only b fulfilld for crtain frquncis n2! = n #t = n 2L c which givs! n = c n2 2L This is idntical with th prvious rsults for rigid-rigid boundary conditions. Othr frquncis will not lad to a closd wav train. As a consqunc th rflctd wavs and th nwly xcitd wavs cannot add up to an incrasing amplitud. In th cas w includ damping th amplitud will b finit fortunatly). Howvr, structurs or cavitis can show clar rsonant bhaviour that also includs that th vibration or prssur amplituds bcom high vn for small xcitation strngth. Extnsion to thr-dimnsional cavitis Th modl of th tub allows only for on-dimnsional wav propagation. Similar drivation can b carrid out for th thr dimnsional room. W will find cass, whr multipl of half a wavlngth will fit to on or svral dirctions. Th condition can b formulatd as k nx = n!, k my = m! L y and k lz = l! L z.

5 7.5 From th wav propagation in th fr spac w know that th wav numbrs in th diffrnt dirctions ar not indpndnt but can b sn as vctors. Th magnitud of th vctor is th wav numbr in air. This mans that k 2 = k 2 2 nx + k my + k 2 lz has to b valid and th rsonanc frquncis ar combinations of th diffrnt cass whn th multipl of half a wavlngth fit into th diffrnt dirctions f n = ck 2! = c 2! k nx + k my + k lz Aftr substituting th wav numbrs on obtains an quation for th rsonanc frquncis which looks quit similar to th on-dimnsional cas Stting l= and m=, both quations will b idntical. f n = c 2! n! % # & 2 + m! % # L y & 2 + l! % # L z & 2 For th thr-dimnsional cas th concpt mods is applid in analogy to th on-dimnsional cas of th tub. Th shap functions i.. th function dscribing th vibration) at th vibration pattrn ar dtrmind by th boundary conditions in th diffrnt dirctions % p x,t) = # p n x) j!t = A n,m,l cos n % x* cos m % y* cos l ### z* j!t & ) & ) & L z ) n= n= m= l= L y Impdanc of th sound fild insid a cavity A main qustion of intrst is th impdanc of sound filds insid cavitis. Th impdanc dtrmins th transport of nrgy and th fficincy of sourcs. Placing a loudspakr insid a tub, it is not slfvidnt what th radiatd sound powr will b. It will dpnd on th impdanc of th fild in which th loudspakr is placd. For th on-dimnsional cas can driv th impdanc of th sound fild insid a tub. As so oftn Nwtons scond law is applid ) ) p x,t u x x,t! = #. This givs in this cas x t u x x,! ) = p j!# jk ) jkx r L jk 2Lx ) jk 2L 1 r L Th impdanc as a function of x and angular frquncy is Z x,! ) = c # jkx + r L # jk 2L#x ) = c # jkx + r L # jk 2L + jkx # jkx # jk 2L#x # r L ) # jkx # r L # jk 2L + jkx

6 7.6 By only considring rsonanc frquncis on can s that th function bcoms 1 for k = n! L and th othr xponntial function can b summarisd to ) = j#c cos Z x,! % n * & ) % sin n * & ) Th impdanc bcoms purly imaginary which mans that thr is no nrgy transport in th room. Howvr this will chang as soon as on introducs damping into th systm. In this cas th impdanc will b complx. Th simplst way to introduc damping would b by a complx ignfrquncy. On would obtain u x x,! ) = kp! + j! which givs Z x,! ) # ) = j#! k & cos n% ) + * & sin n% ) + * jkx jk 2Lx r L ) jk 2L 1 r L +! k & cos n% ) + * & sin n% ) + * Whil th first part is idntical to th cas without damping, th scond part is now ral and rprsnts th nrgy transport into th room. Whn placing th sourc into th middl of th room, th impdanc bcoms zro sinc th cosin function bcoms zro. Placing th sourc on th nds of th tub th impdanc bcoms infinity sinc th sin function bcoms zro. Th loudspakr can b considrd as a volum sourc, i.. th loudspakr mmbran vibrats with a crtain vlocity only dtrmind by th lctrical signal fd to th loudspakr. Zro impdanc mans that th prssur in front of th loudspakr will b zro as wll sinc th vlocity tim th impdanc will giv th prssur. With zro prssur no transport of nrgy into th room can occur. Infinit impdanc mans for a fixd vlocity an infinit prssur only in thory). Th product of vlocity and prssur will lad to a substantial transport of powr into th room

7 7.7 Numbr of ignfrquncis in a givn cavity Although th modal concpt is a quit powrful on, it is limitd by th incrasing numbr of mods on has to tak into account with incrasing frquncy. Additionally, a high numbr of mods pr frquncy band mans that th sound fild has rathr statistic thn dtrministic charactr. For th on-dimnsional cas th numbr of rsonanc frquncis N f blow a crtain frquncy f can asily b calculatd by using f n = c = c n. Th rsonanc frquncis ar situatd on lins as! n 2L shown in th figur. N f can b obtaind by dividing th frquncy f by th first intrval f 1 = c which lads 2L f 1 f 2 f 3 f 4 f 5 f 6 f to N f = f c 2L In th two-dimnsional cas i.. in a vry flat room) th rsonanc frquncis ar situatd on a plan surfac. Th frquncy f is rprsntd by a circl with th radius f. Th ara insid this circl, dividd by th rctangular ara A givn by th first rsonanc frquncis approximats th numbr of ignfrquncis. N f =!f 2 A whr A is A = c 2 c 2L y = c2 4 L y Th numbr of frquncis N f can b approximatly calculatd by N f =!f 2 c S, 2 whr S is th surfac L y. f,7 f, 5 f, 3 f, 1 f f 1, f 3, f 5, f 7, In thr dimnsions th procdur is similar and on obtains bsid that now, instad of a surfac, th volum will dtrmin how many rsonanc frquncis on can count up to a crtain frquncy. N f = 4 3!f 3 c 3 V

8 7.8 Bsid th numbr of rsonanc frquncis, th dnsity of th rsonancs is of importanc as mntiond abov. Th dnsity can b stimatd by th drivativ of th numbr of rsonanc frquncis with rspct to th frquncy on dimnsion!n f!f = 2L c two dimnsions!n f!f = 2f c S 2 thr dimnsions!n f!f = 4f 2 V + f c 3 2c S + L 2 8c Th last quation is somwhat modifid to corrct for rrors occurring whn only a fw rsonanc frquncis ar blow f. Exampl: A room with a volum of 72 m3. How many rsonanc frquncis ar locatd blow 1 khz?

9 7.9 Problms to sction A tub is opn in both nds and th diamtr of it is small compard to th wavlngth, s figur. l D a) Driv an xprssion for th input impdanc of th pip. b) What ar th first thr rsonanc frquncis for such a pip, having a lngth of 1 m? c) How high in frquncy, qualitativly, is th xprssion drivd in a) valid to us for calculations of th ignfrquncy? 7.2 A pan-pip panflöjt in Swdish) consists of a sris of tubs, which ar closd in on nd. Calculat th lngth of such tubs in ordr to covr on octav according to th tabl. Ton c d f g a b c f [Hz] Us c = 346 m/s corrsponds to 25 C). 7.3 How many ignmods ar thr always in a room? 7.4 a) Calculat th rsonanc frquncy for th 1-- mod in a room of dimnsions, L y, L z ) = 1 x 5 x 5 m 3. b) Draw th shap of th sound prssur distribution for th 1-- mod in th room whn ωt =, π/2 and π, by th us of th following xprssion: n! % m! % p n,m,l x, y,z,t ) = p! n,m,l cos x # # L cos y x & # L y & cos l! % ) z # L % jt n, m,l z & Part3 Part1! # c) Explain what th thr markd factors in th xprssion mans. 7.5 A corridor has th dimnsion W x H = 2 x 2 x 3 m 3. Mark th ignfrquncis on a frquncy scal from - 1 Hz. 7.6 Calculat th numbr of ignmods with ignfrquncis 2 Hz and 1 Hz rspctivly in a room of dimnsions 4 x 2.8 x 2.4 m 3. Part2

10 Th numbr of mods in a frquncy band insid a room has a grat influnc on th sound quality of th room and th choic of masurmnt mthod. Th dimnsions of th room ar 4.5 x 4. x 2.6 m 3. Calculat th numbr of ignmods in th third octav bands in th frquncy rang 1-1 Hz. 7.8 A prson living in an old hous has his bdroom right abov th furnac room Th nois from th furnac is vry annoying and its fundamntal frquncy 3 Hz) corrsponds to th lowst ignmod of th room. This mod dominats th sound prssur distribution in th room. Th prson has his pillow at th position x 1, y 1, z 1 ) =.3,.3,.3) m. Whn th authoritis masur th nois lvl thy do it at a position approximatly in th cntr of th room according to th rgulations at th position x 2, y 2, z 2 ) = 2.5, 1.5, 1.5) m. Th A- wightd sound prssur lvl was in this position masurd to 35 db, which didn t lad to any rmarks. Calculat th actual lvl at th pillow. 7.9 An lvator is drivn by an hydraulic piston pump. Earlir xprinc shows that thr is a larg risk for complains on th nois lvl in th lvator, if th fundamntal frquncy of th pump, du to pulss in th hydraulic systm, xcit som ignfrquncy insid th lvator. Th pulss can b rducd to accptabl lvls if a ractiv silncr is placd in th hydraulic systm nar th pump. Howvr, to dsign th silncr, thr is a nd to know th unwantd frquncis. Th rotational spd of th pump is assumd to vary btwn 13 to 153 rpm. Th numbr of pistons in th pump is 7, which lads to 7 similar pulss pr rotation. Th lvator compartmnt can b sn as a rctangular room with dimnsions W x H = 2. x 1.1 x 2.25 m 3. Th spd of sound in air is 343 m/s. a) Calculat th frquncy rang in which th fundamntal frquncy of th pump may vary. b) Is thr a risk that th fundamntal frquncy of th pump will xcit som of th lvators ignfrquncis? How many? What ar th ignfrquncis and corrsponding mod numbr? 7.1 Emilia livs in a singl room compartmnt whr th rooms dimnsions ar 5 x 4.75 x 2.5 m 3 according to th figur. Th rfrigrator is old and noisy and Emilia has placd hr bd in th opposit cornr, as far away from th rfrigrator as possibl. Unfortunatly, Emilia dosn t know about wav thory and room acoustics. Th nois has a vry strong 5 Hz componnt from th lctric powr supply), which almost xactly coinsid with on of th

11 7.11 rooms rsonanc frquncis, so th nois lvl is high in th room. 2,5 m 5 m 4,75 m a) Which mod, xprssd as n, m, l) has an ignfrquncy clos to 5 Hz? Th spd of sound is 34 m/s b) Assum that th sound fild is dominatd by th mod from a). What diffrnc in lvl can b found comparing th slping positions Th had in th cornr and Th ft in th cornr? Assum for simplicity that both positions ar xactly at th wall, 5 cm abov th floor. Emilia is 16 cm long.

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