Solution Set #2

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Θεοδόσιος Αρβανίτης
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1 . For the followng two harmon waves: (a) Show on a phasor dagram: Soluton Set # phasor s the omplex vetor evaluated at t 0: f [t] os[ω 0 t] h f [t] 7os ω 0 t π f [t] exp[ 0] + 0 h f [t] 7exp π ³ h 7 os π h + sn π ³ h 7 os + π h π sn 7 ( ) As tme nreases, the two phasors rotate about the orgn at the rate ω 0. (b) Fnd the mathematal expresson for the superposton f [t]+f [t] n the form of a osne. Re {f [0] + f [0]} Im {f [0] + f [0]} s f [0] + f [0] ( ) Im Φ {f [0] + f [0]} tan {f [0] + f [0]} 7 tan Re {f [0] + f [0]} radans 0.7π radans 5.5 The snusodal expresson for ths osllaton s: f [t]+f [t] 8.5 os [ω 0 t 0.68]

2 . Two plane waves of the same frequeny that vbrate along the z dreton are: x f [x, t] A os π ν t + π ; A 0mm, X 0mm, ν 0Hz X y f [y, t] A os π ν t + π ; A 0mm, Y 0mm, ν 0Hz Y Evaluate the resultant waveform f [x, t]+f [y, t] at [x, y] [50mm, 0 mm] full redt f done ether way: (a) f [50 mm,t]+f [0 mm,t] 50 mm 0mm os π 0 Hz t + π 0 mm 5 0mm os π 0 Hz t +0mm 0π 0 mm os 0πt 0 mm os [π 0πt] 0π 0 mm os 0πt +0mm os [ 0πt] 0 mm os 0πt 0π +0mm os [0πt] ½ Re 0 mm exp [+0πt] 0 mm exp + +0mm os os 0πt 0π 0 mm π 0 mm π ¾ 0 Hz t 0 Hz t + π Beause the temporal frequenes are the same, we an evaluate the sum at t 0by usng phasors: f [50 mm, 0] + f [0 mm, 0] 0mm exp [+0π 0] 0 mm exp + 0π 0 0π 0mm 0 mm exp 0π Re {f [50 mm, 0] + f [0 mm, 0]} 0mm 0 mm os Im {f [50 mm, 0] + f [0 mm, 0]} 0 mm sn 0π 0π +0mm 0 mm.6 mm q f [50 mm, 0] + f [0 mm, 0] (+0 mm) +(.6 mm) 5.5 mm.6 Φ {f [50 mm, 0] + f [0 mm, 0]} tan radans 0.7 8π radans 0.8

3 (b) f [50 mm,t]+f [0 mm,t] 50 mm 0 mm 0mm os π 0 Hz t + π +0mm os π 0 Hz t + π 0 mm 0 mm 5 0mm os π 0 Hz t 0 mm os π 0 Hz t +π 0π 0 mm os +π 0πt 0 mm os π +π 0πt 0π 0 mm os 0πt +π 0 mm os 0πt π +π ½ 0π Re 0 mm exp + 0πt +π 0 mm exp + 0πt π +π ¾ Beause the temporal frequenes are the same, we an evaluate the sum at t 0by usng phasors: 0π Re {f [50 mm, 0] + f [0 mm, 0]} 0 mm os +π 0 mm os π +π 0π +π 0 mm os π +π 0 mm os.77 mm Re {f [50 mm, 0] + f [0 mm, 0]} 0 mm sn mm 0π +π 0 mm sn π +π q f [50 mm, 0] + f [0 mm, 0] (.77 mm) +( mm) 5.5 mm ½ ¾ Φ [f [50 mm, 0] + f [0 mm, 0]] tan.58 radans

4 . Consder the superposton of two snusodal travelng waves: f [z, t] A os [k z ω t], A 0 mm, ν 00 Hz, v 50 m s f [z, t] A os [k z ω t], A mm,ν 50 Hz, v 500 m s (a) Fnd an expresson for the resultng wave n terms of the average wave, the modulaton wave, plus any remanng ampltude. We have an expresson for the sum of two waves wth dfferent frequenes bu the same magntude: A + B A B os [A]+ os[b] os os so we have: A os [k z ω t]+a os [k z ω t] [A +(A A )] os [k z ω t]+a os [k z ω t] A (os [k z ω t]+os[k z ω t]) + (A A )os[k z ω t] k + k ω + ω k k ω ω A os z t os z t +(A A )os[k z ω t] whh s the sum of a travelng wave and the produt of the average and modulaton waves (b) Calulate the wavelengths of the average and modulaton waves. In ths ase: ν v v 50 m s ν 00 Hz 5 m ν v v 500 m s ν 50 Hz 0 m > and v < v k avg π k + k π avg Ã! avg + + 5m + 0 m 0 7 m avg k mod π k k π mod mod Ã! 5m 0 m 0 m mod n.b., mod > avg

5 () Fnd the velotes of the average and modulaton waves. ω avg ω + ω ω mod ω ω ν + ν 00 Hz + 50 Hz π π π 5 ν ν 00 Hz 50 Hz π π π 5 v avg ω avg avg ν avg m m 5 Hz k avg 7 7 s v avg v mod ω mod k mod mod ν mod 0m 5 Hz 500 m s v mod sothemodulatonwavetravelsntheoppostedreton (d) Does ths system exhbt normal or anomalous dsperson? > and v < v longer wave travels slower anomalous dsperson 5

6 . The phase veloty of waves n some medum s proportonal to ω +. Fnd an expresson for the modulaton veloty and determne whether the waves exhbt normal or anomalous dsperson. ω v φ k αω+,whereα s some onstant k ω αk ω + αω + ω (αk) ths s the dsperson relaton ω [k] Ã! dω v mod dk d h (αk) α k α ω + αω + dk α v mod αω + vavg anomalous dsperson 6

7 5. Plot and wrte the equaton of the superposton of the followng harmon waves: h π E sn 8 ωt 5π E os ωt h π E sn 6 ωt where the perod of eah s s. h π E sn 8 ωt h³ π ωt os 8 h³ π os 8 π ωt os 5π E os ωt h π E sn 6 ωt ωt π os 8π 8 ωt 5π h³ π ωt os 6 π os h π h ωt os ωt + π os ωt + π sne all three omponents have the same frequeny ω πν π se π radans se,thesumhasthesame frequeny. Evaluate the ampltude for t 0: E [t 0]+E [t 0]+E [t 0] exp + π + exp 5π h + exp + π Re {E [t 0]+E [t 0]+E [t 0]} os + π os + os 5π π h + os + π Im {E [t 0]+E [t 0]+E [t 0]} sn + π sn + sn π 5π h + sn + π 7

8 s E [t 0]+E [t 0]+E [t 0] os π + sn π " sn Φ {E [t 0]+E [t 0]+E [t 0]} tan π # os π π radans 0 E + E + E os h ωt π 0.65 os [ωt 0.] (a) the three omponent snusods; (b) the sum usng both the sum of the three funtons n (a) and the omputaton the two results are dental 8

9 6. A laser emts a monohromat beam of wavelength 0,whhsrefleted normally from a plane mrror that reedes from the lght soure at veloty v. (a) Determne the beat frequeny between the ndent and refleted lght. Frst try t wthout for v 0;therefleted lght dffers from the ndent lght only n the dreton (assumng no losses or phase hanges n the mrror): The sum reates standng waves: f [z, t] A 0 os [k z ω t] f [z, t] A 0 os [k z + ω t] f [z,t]+f [z,t] A 0 os [k z ω t]+a 0 os [k z + ω t] A 0 os [k z]os[ω t] Note that f the mrror moves wth veloty v, the frequeny and the wavelength of the refleted lght hange due to the Doppler effet. From Eq.(-) n Pedrott(s): In the lmt v<<, then Eq.(-5) apples: after before after before s v + v s v + v r v r ³ + v r r r ³ v ³ + v v +( ) v r r v v v ³ 0 0 v 0 so the wavelength of the refleted beam s longer f v reedes from the soure (v < 0, red shft) and shorter f t approahes the soure (v > 0, blue shft). In our ase, the refleted beam exhbts twe the Doppler shft beause t has to hange dreton, so the orrespondng wavevetors are: k π 0 k 0 π 0 0 unhanged where the dentty has been used: π v k 0 v k v k + v α X α n α + α α f α << α n0 The orrespondng osllaton frequenes are easy to fnd beause the veloty of lght s unhanged: 0 ν ν ν 0 0 ν 0 v 0

10 : ν 0 ν s unhanged ν v + v ν So f v s postve, the wavelength dereases and the frequeny nreases. Thesumofthetwowavess: f [z,t]+f [z,t] A 0 os [k z ω t]+a 0 os [kz 0 + ω 0 t] A 0 os [k avg z ω avg t] os [k mod z ω mod t] where: k avg k + k 0 k mod k k 0 ω avg ω + ω 0 ω mod ω ω 0 ν mod vν k + k + v k k + v k v ω + + v ³ ³ ω + v ω vω ω ω + vω v k + k k + k v f [z,t]+f [z,t] A 0 os [k z ω t]+a 0 os [kz 0 + ω 0 t] A 0 os k + k ³ v z ω + vω A 0 os k + k ³ v z ω + vω t so the beat frequeny n the equaton s: ν mod ω mod π vω ν π v t os k v t z vω h os v (k z + ω t) but we atually see the squared magntude, whh osllates twe as fast (Pedrott Eq.(5-)) ν mod ν v 0

11 (b) Determne the beat frequeny between the ndent and refleted lght f the lght s ndent on the plane mrror at angle θ. Here the refleted beam travels at angle θ, sothek vetors are dfferent before and after the refleton. Agan onsder the ase v 0: Thesumofthetwowavess: f [x, y, z, t]+f [x, y, z, t] os k k π π π k avg k + k0 k mod k k 0 π sn [θ] π 0 0 sn [θ] 0 π os [θ] sn[θ] π 0 +os[θ] π sn[θ] 0 os[θ] sn [θ] 0 os [θ] x + +os[θ] π z ωt os sn [θ] x + so the beat frequeny s zero (standng waves). If we add the movement, we get the same wavelength shfts, but now they depend on the angle: v z v os [θ] v x v sn [θ] ν mod ν v os [θ] os [θ] z

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