Chapte 15 RAVELING WAVES 15.1 Simple Wave Motion Wave in which the ditubance i pependicula to the diection of popagation ae called the tanvee wave. Wave in which the ditubance i paallel to the diection of popagation ae called the longitudinal wave. 1
Wave pule t =, y = f(x) t >, y = f(x-vt) Fo wave pule on a ope, the peed of wave v = F / µ, whee F i the tenion, µ i the linea ma denity (ma pe unit length). Example 1. When will the pule each the left pole if the ma of the ope i.5 kg and the ma of the hanging weight i 1 kg?
he pule tavel with the peed v = F / µ. he tenion i F = mg = 1 9.81 N = 98.1 N. he linea denity of ma i µ = m ope / L ope =.5/ 5 kg / m =. kg / m hen the peed of pule i v = F / µ = 98.1/. m / = 7.4 m / and the popagation time i t = m / 7.4 m / =.9. 3
Fo ound wave in liquid and gae, v = B / ρ whee B i the bulk modulu and ρ i the denity. In ideal gae the bulk modulu i popotional to the peue and (Chapte 17) v = γ R / M whee i the abolute tempeatue in K, M i the mola ma (ma of 1 mol γ of the ga), R = 8.314 J/(mol K) i the univeal ga contant and = 1.4 fo diatomic gae ( γ = 1.67 fo monatomic gae). = Example. Calculate peed of ound in ai at o C. Fo ai, the mola ma M = 9x1-3 kg/mol, = (73 + ) K = 93 K, and 3 ( ) v = γ R / M = 1.4 8.314 93/ 9 1 m / = 343 m / (at o C, v = 331 m/). 4
he Wave Equation y = v y t x Any function y = y ( x vt ) y ( α ) atifie thi equation. Indeed, Similaly, = = v y dy α dy t d α t d α ( ) ( ), = v = v = v y dy d dy α d y t t d α d α d α t d α = = y dy α dy x d α x d α Combining, we get the wave equation ( ) ( ), = = = y dy d dy α d y x x d α d α d α x d α y = v y t x 5
15. Peiodic Wave Hamonic (inuoidal) wave he minimum ditance afte which the wave epeat i called the wavelength λ Duing the peiod = 1/f the wave move a ditance of one wavelength: λ v = = f λ = ωλ / π he hamonic (inuoidal) taveling wave ha the hape whee ( ) = ( π x ω ) ( ω ) ( ) y x t A t A kx t A k x vt, in in in ( ) k = π / λ = ω / v = π f / v i called the wave numbe. λ y he velocity of a point in the wave i v x = t = ω A co ( kx ω t ) y and the acceleation of thi point i a = = ω A in ( kx ω t ) x t 6
Example 3. he wave function fo a hamonic wave on a ting i Find paamete of the wave. 1 1 (, ) = (.1 ) in ( 4.1 ) ( 9. ) y x t m m x t he amplitude A =.1m. he wave numbe k = 4.1 m -1 and the wavelength λ = π / k = 1.53 m. ω = = π / ω =.74. he fequency 9. -1 and hen the peed of the wave λ ω 9. v = k = 4.1 m / =.4 m / 7
Enegy anfe via Wave he ate of enegy tanfe i powe, P = u F v t he vecto component ae u $ x y $, $, F = F i + F j v = v j t t P = F v in y t = F θ v t At mall angle, in θ tan θ = In the end, the powe u P = F v = F tan θ v = F y x y y t t x t ( ω ) ω ( ω ) = F ka co kx t A co kx t he tenion can be expeed via the wave peed v uing v = F / µ. hen and the aveage powe i co ( ) P = µ v ω A kx ω t P = µ v ω A av 1 8
he enegy tavel along the ting at the wave peed v he aveage enegy E t flowing pat a point in time x E = P t = µ v ω A t ( ) 1 av av x v t = hi enegy i ditibuted ove the length o the aveage enegy in the length i ( ) µω E = A x av 1 Example 4. A wave of wavelength 4 cm and amplitude.8 cm move along a 1 m egment of a 75 m ting with ma 3 g and tenion 1 N. What ae the paamete of the wave including the aveage total enegy? he peed of the wave i given by tenion and ma denity: ( ) v = F / µ = 1/.3/ 75 m / = 54.77 m / he angula fequency i ω = π v / λ = 6.8 54.77 /.4 ad / = 818.9 ad / hen the aveage total enegy i.3 ( E ) = 1 1 µω A x = 75 ( 818.9 ) (.8 ) 1 J =.86 J av 9
Hamonic Sound Wave he diplacement of molecule fom thei equilibium poition, ( ) ( ) x, t = in kx ω t, i out of phae by 9 o degee with the peue and denity wave, ( ) ( ) p x, t = p in kx ω t π /, whee p i the change in peue fom the equilibium. he amplitude p = ρω v he enegy and the enegy denity of ound wave ae E = ρω V u = ρω, av ( ) 1 1 av 1
15.3 Wave in hee Dimenion he motion of wavefont can be epeented by the ay pependicula to the wavefont. he aveage powe in the wave pe unit aea that i pependicula to the diection of popagation, i called the intenity, I P av / A W / m = If a point ouce emit unifomly in all diection, then the wavefont ae pheical, A = 4 π, and the intenity at a ditance fom the ouce i I = P / 4 π av 11
hee i a link between intenity and enegy denity: In the hell, whee we ued E = u V = u Av t av av av P = E / t = u Av, av av av I = P / A = u v = p / ρ v av av av 1, u = ρω = p ρω v 1, / 1
Example 5. A peake diaphagm 5 cm in diamete i vibating at 1 khz with an amplitude.15 mm. Find the paamete of the ound wave immediately in font of the peake and intenity at a ditance 5 m. he peue amplitude i ( )( )( )( ) p = ρω v = 1.9 kg / m π 1 34 m / 1.5 1 m = 413.1 N / m he intenity at the diaphagm i ρ 3 4 1 5 ( ) ( ) 1 1 I = p / v = 413.1 / 1.9 34 W / m = 194.8 W / m he powe of the peake i the intenity multiplied by the aea of the diaphagm, ( ) P = IA = 194.8 π.5 / 4 W =.38 W hen the intenity at a ditance 5 m, auming the unifom adiation in the fowad hemiphee, i ( ) ( ) I 5 m = P / π =.38 / 6.8 5 W / m =.43 mw / m 13
he intenity level of ound i meaued in decibel defined a β = 1log I I whee the efeence level I i uually taken a thehold of heaing, I = 1 W / m 1 Example 6. Find the intenity level of ound at a ditance 5 m fom the peake fom Example 5. he intenity I W m 3 =. hen the intenity level 5 m.43 1 / β ( 9 ) ( 3 1 ) I = 1 log = 1 log.43 1 /1 db I 1 log.43 1 db = 9 + 1.38 db = 93.8 db 14
15.4 Wave Encounteing Baie When a wave encounte an obtacle, pat of it i eflected and pat i tanmitted though: Hee the eflected pule i inveted; the tanmitted pule i not (the econd ting i heavie) Hee both the eflected and tanmitted pule ae not inveted (the econd ting i lighte) 15
Example 7. he incident wave with amplitude A encounte the junction of two wie with the ame tenion F. he wave peed in the fit wie i thee time that in the econd, and the amplitude of the eflected wave i a quate of the amplitude of the incident wave. What i the amplitude of the tanmitted wave? By enegy conevation, the incident powe i equal to the um of powe in eflected and tanmitted wave, 3 P P P P µ v ω A 1 i = + t, av =, 1 1 1 µ 1 v 1 ω A i = µ 1 v 1 ω A + µ v ω A t Since the wave peed v = F / µ, and the tenion i the ame in both wie, 1 1 1 1 v A = v A + v A = + A A F F F A i t v i v v t v v v We know that A = A i / 4, v 1 = 3 v. heefoe, 1 1 ( ) i i t i A A / 4 A 1 A 5 v = v + v /3 = v 16 + 3 A t A t = 16 A i 1 1 1 1 16
Diffaction Bending of a wavefont behind an obtacle i called diffaction. Almot all of the diffaction occu within eveal wavelength fom the edge of the obtacle Flat wavefont paing though a mall apetue bend, pead out, and become pheical. he lage i the wavelength in compaion to the apetue, the lage ae the diffaction effect. 17
he appoximation that wave popagate in taight line without diffaction i called the ay appoximation (zeo wavelength appoximation). Fo ound wave, λ 3 cm 3 m and fo viible light λ 7 7 4 1 7 1 m 18
15. 5 he Dopple Effect Change in fequency of the eceived ignal in compaion to the fequency of the emitted ignal, which eult fom a elative motion of the ouce and eceive, i called the Dopple effect. If the ouce and eceive move cloe togethe, the eceived fequency i geate than the ouce fequency. If the ouce and eceive move away fom each othe, the eceived fequency i lowe than the ouce fequency. 19
Example 1. Souce move to the ight with u ; eceive i tationay; peed of the wave i v; the time between emiion of two conecutive cet i. Between two conecutive event, the fit cet move v while the ouce move u. hu, the ditance between two cet (the wavelength) i hen the fequency in the eceive λ = ( v ± u ) ( ) = v ± u / f f = v / = f λ ± v v u (+ the eceive behind the ouce, - in font) Example. Souce doe not move; eceive move with u ; peed of the wave i v; the time between aival of two conecutive cet i. Between detecting two conecutive cet, the eceive move by u while the fit cet move by v. he ditance between two cet i the wavelength λ = v ± u (- eceive move away fom the ouce, + it move towad the ouce). hu the fequency i f 1 / ( ) / ( ) = = v ± u λ = f v ± u / v
Example 3. If both move, v u ± ± f = f v u he pope ign ae choen by emembeing that the fequency goe up when the ouce move towad the eceive and the eceive towad the ouce. If both u, u v ± f v f u Example 4. he fequency of ca hon i 5 Hz. Find the fequency and the wavelength of the eceived ignal if the ca move towad eceive with u = 9 km/h. he ca velocity i 9 km/h =9 (1 m)/(36 ) = 5 m/ he wavelength in font of the ca ( ) ( ) λ = v u / f = 34 5 / 5 m =.63 m and the eceived fequency i λ f = v / = f = 5 Hz = 54 Hz v v u 34 34 5 1
Example 5. he ame if the honking ca i tationay while the eceive move towad thi ca with u = 6 km/h 6 km/h = 6(1)/36 m/ =16.7 m/ he eceived fequency i ( ) ( ) f = f v ± u / v = 5 34 + 16.7 / 34 Hz = 55 Hz Example 6. he ca peed away fom the police ca. he ada unit emit the electomagnetic wave with fequency f. Find the peed of the ca u if the fequency of eflected wave, eceived by the ada unit, i f f. he ada wave tike the peeding ca at a fequency f = f v ± u v = f c u c = f ( ) / ( ) / ( 1 u ) c and ae eflected fom (emitted by) the ca with the ame fequency. he eceived ignal by the police i then ( ) ( ) ( ) ( ) f ' = f = f = f 1 f = 1 f 1 1 f v c 1 u u u u v ± u c + u 1 + u / c c c c c heefoe, f = f ' u f = c f and u = c f f
Dopple Shift and Relativity In non-elativitic ytem, the Dopple effect i detemined by the peed of both the ouce and the eceive elative to the medium which affected the peed of popagation of wave. Fo light and othe electomagnetic wave the peed of popagation c i a univeal contant which doe not depend on the peed of ouce o eceive (no wind ). Second, the time between the event uch a emiion of wave cet,, depend on the efeence fame and i diffeent in the efeence fame of ouce and eceive. In the end, the Dopple hift fo electomagnetic wave depend only on the elative velocity of the ouce and eceive u: f f c ± u = c m u whee ign i uch o that the fequency goe up when the ouce and eceive ae appoaching each othe. 3
Chapte 15 eview Wave on a ting: v = F / µ Sound wave in liquid and gae: v = B / ρ Sound wave in dilute ideal gae: v = γ R / M Electomagnetic wave: the peed i a univeal contant, c = 3x1 8 m/ Wave equation: = v y y t x Paamete of hamonic wave ( ) ( ) y x, t = A in kx ±ω t : k = π / λ = ω / v, ω = π f = π / = kv, v = f λ = ω / k = πωλ = λ / he enegy in hamonic wave i popotional to the quae of the amplitude P = 1/ µ v ω A Powe fo wave on the ting: ( ) av Enegy denity fo ound wave: ( ) av 1/ ρω, ρω u = p = v 4
I = P / A Intenity fo ound wave: av Sound intenity level: β ( ) I = u av v = 1log I / I, I = 1 W / m 1 Dopple effect v ± u f = v ± u f Moving ouce: λ = ( v ± u ) / f f = v ± u / λ = f v ± u / v Moving eceive: ( ) ( ) Small peed of ouce o eceive: f / f ± u / v 5