Một số phương pháp giải bài tập trắc nghiệm cơ học 12: Phần 2

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Nối tiếp nội dung phần 1 tài liệu Phương pháp giải bài tập trắc nghiệm cơ học 12, phần 2 cung cấp cho người đọc phương pháp giải bài tập trắc nghiệm sóng cơ và sóng âm. Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Một số phương pháp giải bài tập trắc nghiệm cơ học 12: Phần 2

PtiLiOng ph^p giai bai tgp tr^c nghigm Co hqc 12 - Trgn Trgng Hang Cty TIMHH MTV DVVH Khang Vigt BAI TAP C d BAN dy 2. T r e n mat nu-dc dang c6 song, ta thay khoang each giiJa hai gdn song lien tiep la 10cm va chiec la nho tren mat nuTdc nho l e n lien tiep 3 Ian trong D a n g 1. C H U K I - T A N S O S 6 P 1 G thdi gian 5s. Toe do truyen song tren mat nu-dc la: - Chu k i T cua song la chu k i dao dong cua mot phan tuf cua m o i trUdng c6 \ . . \ 4cm/s B . V = 6cm/s C. v = 8cm/s D . v = 2cm/s song truyen qua. > Giai - Tan so cua song: f = •'•r•;^'•^•^r^''''r. • ' Khoang each each giffa hai gdn song lien tiep la bude song: X = 10cm La nho nho l e n lien tiep 3 Ian iJng vdi 2 chu k i dao dong, vay 2T = 5 Vi du: M o t ngiTdi quan sat tha'y chiec phao tren mat niidc bien nho l e n cao - > T = ^ = 2,5 (s) .... I,.'..I.•••• 10 Ian trong 36s. Chu k i cua song bien la: ' • ^ • i - A . T = 3,6s B . T = 4,0s C. T = 2,0s ' D . T = 1,5s Toe do truyen song: v = — = ^ = 4,0 (em/s) T 2,5 Giai !]:hpn dap an A . Nu^dc bien ngay tai phao dao dong khi c6 song truyen qua nen phao dao dong theo. Phao nho len cao 10 Ian tufc nu-dc bien tai diem do dao dong dat ciTc dai Dang 3. VIET P H U O N G TRIPOl s6riG T A I M O T D I E M 10 Ian. Thdi gian giiJa 10 Ian ciTc dai lien tiep la 9 chu k i (9T). V a y : 9T = 36 B i e t phu^dng trinh song tai O la: Uo = Aeoscot - Phu-dng trinh song tai M each O khoang O M = x do song truyen tiT O tdi la: => T = = 4 (s): do cung la chu k i cua song bien. Uf^^ =: Acoso) t - -f = A c o s 0)t - 2 7 1 . - £-i ^ Chon dap an IJ. ^ x) - Phu-dng trinh song tai N tru'de O vdi O N = x (sdng truyen tijr N tdi O) la: Dang 2. BU6c S6NG N_v_ O u ^ = A c o s cot + 2 ; : . - - Bu'dc song X la quang diTcJng ma song truyen di diTdc trong mot chu k i . X - Cong thufc: A. = v.T = y , vdi v la van toe truyen song Vi du 1. M o t sdi day O A rS't dai cang ngang. Cho dau O dao dong theo phU'dng - Hai phan ttf tren cting phUdng truyen song cac nhau m o t biTdc song X thi vuong goc vdi day. PhiTdng trinh dao dong tai O la: U Q = 2eos 57Ct + - (em). dong pha 4 - Hai dinh song lien tiep each nhau mot bi/dc song X. Biet toe do truyen song tren day la 2m/s i • • — • ' Phu-dng trinh song tai M each dau O khoang 50cm l a : Vi du 1, M o t day dan hoi cang ngang. Cho mot dau day dao dong theo phu'dng A. u ^ = 2 cos 57rt + - (cm) B . U w = - 2 c o s 5 j i t (cm) thang du-ng vdi chu k i 2s thi tren day c6 song truyen d i . Sau thdi gian 0,3s dao 6J dong truyen di du^dc 1,5m. Bu-dc song l a : 7t C. u ^ = 2 cos 57tt - - (cm) D. u^^ = 2 cos 57tt+ — (cm) ' ' A.X = 2,5m B.X=\ C. = 5,0m D.X = 4,0m 4 2 —- • • Giai Giai Toe do truyen song: v = — = - ^ = 5(m/s) Cd: U Q= 2COS 57tt + - M. At 0,3 4J Birdc song: X^v.T = 5.2 = 10 (m) 05^ Nen: UM = 2 cos 57t t - - + — = 2eos 57t t- + — Chon dap an B . 4 4 2 J = 2cos(57it - 7t) = - 2cos57tt (cm) 195 1Q4 •••••
Chon dap an B . Phu'dng irinh dao dpng truyen den M : V i du 2. Song truyen tren phUdng x'x tiY M N each nhau 40cm vdi hiidc son 0,3^ U M = 2cos4Tr t - • = 2cos47r t - = 2cos 47rt (em) SOTTt-- (em). Phu-dng 4J X = 0,8m. Bie't phUdng trinh song tai N la u ^ = 4eos 4 Chpn dap an D. trinh song tai M la: Yi du 4. M o t song cd hpe du'de truyen theo phu'dng Ox vdi toe dp 20em/s. Cho 371 rang khi truyen di bien dp song khong doi. Biet phu'dng song tai O la A. = 4 cos 50711 (cm) B. U M = 4 e o s 507tt - (em) 4 ) -t "o = 4 c o s (em) l i dp dao dpng tai M each O 40cm liic l i dp dao dpng tai v6 y C. U M = 4cos507tt (cm) D . U M = 4 e o s 507:t - (cm) O dat cifc dai la: A. UM = 4em B. UM = 0 :, • C. UM = -2em D. UM = 2cm Giai Giai N —1— • UQ = 4C0S -t 27tx V6 ^ / UN = 4 cos 507rt - - UM =4 COS 507rt - - + { 4 X x 0,4 U M =4eos- = 4cos — t - = 4cos (cm) t - - 6 0,2 (dau + V I song truyen den M trUdc) '•*f. 7C 2Tt.0,4 ( 371^ ^71 ^ n U M = 4 e o s 507ct - - + — = 4cos 507it + — (cm) • UQ: max cos - t = 1 ^ - t = k27t 4 0,8 I 4 J v6 y 6 Chpn dap an A . 7t V a y : U M = 4 c o s k 2 7 r - — = 4 cos = 2 (cm) - V i d u 3. Sdi day dan hoi Ox ra't dai Qang ggang. Cho dau O dao dong dieu hoa theo ph^dng thang duTng vdi bien dp 2cm, chu k i 0,5s. L a y t = 0 la liic dau 0 Chpn dap an D. dat l i dp cifc dai. Toe dp truyen song tren day la 8m/s. Phu'dng trinh song tai Dang 4. B I E T F H U O M G T R I N H SOING. M each O 50cm la: T I M B U 6 C SOIMG X V A V A N T O C V A . U M = 2cos4TCt ( c m ) B. U M = 2cos 47tt + — (cm) So sanh phu'dng trinh da cho vdi phu'dng trinh tdng quat: ) 2 27tX^ u = Acos cot - • bu'dc song X 7t X C. UM = 2 cos 47Ct-- (cm) D. U M =2COS 47tt- (em) 6 • T i m van toe v: T i m X:=^v - — Giai T • Phu'dng trinh dao dpng tai dau O: U Q = Aeos(cot + cp) hoac so sanh phu'dng trinh da cho vdi phu'dng trinh tong quat: 271 271 Vdi: + (0 = • = 471 (rad/s) u = Acos CO t - - => van toe v T 0,5 V + A 2 (cm) r TtX ^> d u 1. M o t song ngang truyen tren day eo phu'dng trinh u = 2cos lOOTTt-- + t = 0, u = A A = Acoscp coscp = 1 ->(p = 0 (cm) trong do x tinh bang cm, t bang giay. Bu'dc song la: V a y : u,, = 2eos(47it) (cm) A . A,= lOcm B.X = 5cm C.X = 20cm D. A, = 40cm 196 I 197
Phuang phdp gii'i b^i t$p tr^c nghigm Cp, hgc 12 - Tian !i:in(j Hifng Giai Dang 5 . TIM V A N T O C D A O D O N G C U A M O T D I E M 7TX TREN FHUONG TRUYEN S6NQ. PhLTcfng t r i n h d a c h o : u = 2cos lOOnt- To ^ , . r 27rx^ PhU'dng t r i n h t r u y e n s o n g : u = A c o s cot X So s a n h v d i p h i f d n g t r i n h s o n g t d n g q u a t : i i = A c o s cot - t r o n g d o x va ^ J V a n t d c c u a m o t d i e m c 6 t o a d p x l a d a o h a m c u a u do'i v d i t: X cijng ddn vj. '• i > » , du u = — = -coAsm cot-- dt X T a diTdc: — = — ^X = 20.Vi\h b h n g c m n e n A. = 2 0 ( c m ) 10 X V i d u . PhU'dng t r i n h s o n g t a i d i e m M v d i O M - x l a : ChondapanC. ' ' " v:: u = 6cos - t - 0 , 2 7 i x ( c m ) t r o n g d o t t i n h b h n g g i a y , x t i n h b h n g c m . V a n tdc Vi d u 2. PhiTdng t r i n h s o n g t a i m o t d i e m c 6 t o a d p x t r e n phU'dng t r u y e n soim 2 v a o i u c t diTdc c h o b d i : u = 5 c o s ( 2 0 t - 0 , 0 5 x ) ( c m ) t r o n g d o t t i n h b a n g giay, d a o d p n g t a i d i e m M c 6 x = 10cm l u c t = I s l a : X b a n g c m . V a n toe t r u y e n s o n g l a : , | I'feP: : A. 0 ,; ' B . - 9 , 4 2 m/s C. 9,42 m/s D . 6 m/s A . V = 4m/s ., B . V = 2m/s C. v = 5m/s D . v = 400m/s Giai V a n tdc dao d p n g l a dao h a m cua li dp u theo t: PhU'dng t r i n h s o n g da c h o : u = 5 c o s ( 2 0 t - 0 , 0 5 x ) u = — = -37isin -t-0,2KX Cdcit 1: S o s a n h v d i phU'dng t r i n h s o n g t d n g q u a t : dt 2 27rx V d i t = Is, X = 1 0 c m , ta c 6 : u = A c o s (Ot - • ta du'dc u' = -37:sin - . 1 - 0 , 2 7 1 . 1 0 = -37rsin - - 2 7 1 = -37rsin- = - 9 , 4 2 (cm/s) 2 2 (0 20 10 C h p n d a p a n li. + 0,05 = ~ - ^ l =^ = 40n ( c m ) Dang 6 . DO L E C H P H A G I O A H A I D I E M 0,05 TREIN P H U O N G T R U Y E N S61NG. V a n t d c t r u y e n s o n g : V = — - "^^"^^^"^^ = 4 0 0 (cm/s) = 4 (m/s) g PhU'dng t r i n h s o n g t a i 2 d i e m c 6 tpa d p X | v a X 2 c u n g l u c t l a : ( o ^2 Ui = Acos cot-271-!- ; U 2 = Acos c o t - 2 7 1 ^ X ^ 0,05 ^ Cdcli 2: V i e t l a i phU'dng t r i n h s o n g d a c h o : u = 5cos 2 0 t - XT - X 20 D p l e c h p h a giffa h a i d i e m : Acp = 2 T C . ^ ^ — — ; Acp = 2 7 t . — X X X So s a n h v d i phU'dng t r i n h s o n g t d n g q u d t : u = A c o s t - - Vf d u 1. S o n g t r u y e n t r e n d a y d a i v 6 h a n v d i b i f d c s o n g X. K h o a n g e a c h giiJa hai d i e m t r e n day dao d p n g c u n g pha each nhau Tad.«c:M5a^v =^ = 400 A . d = 2kA. B. d = ( 2 k + l ) ? . C. d = k . - D . d = k?^ 20 V 0,05 2 VI x CO d d n v i la c m n e n v c 6 d d n vj l a cm/s (vdi k = 1,2,3,...) e V a y : v = 4 0 0 (cm/s) = 4 (m/s) Giai ChpndapanA. H a i d i e m tren day each nhau k h o a n g d dao d p n g c u n g pha k h i :
Phuohg phap giai bai t^p trSc nghigm Co^bpc 12 - Trai HKng Cty TNHH MTV DVVH Khang'Vift Acp = 2n.— = 2kn d = kX (vd\k= 1,2,3,...) y,' du 4. Hai diem M , N each nhau 28cm tren day c6 song truyen qua luon luon X lech pha vdi nhau mot goe Acp =(2k + l ) ^ vdi k = 0; ±1; ±2; ... Chon dap an D. Vi du 2. Song truyen tren day dai v6 han c6 bu'dc song X. Mai diem tren day dac To'e do truyen song tren day la 4m/s va tan so cua song eo gia tri trong dong ngu'dc pha each nhau Ichoang ttr 22Hz de'n 26Hz. Tan so cua song la A. f = 2 5 H z B.f=20Hz C. f = 45Hz D. f = 1 5 H z A. d = k . - B. d = 2 U C. d = U D. d = k + - X 2 V 2y Giai ( V d i k = 0, 1,2, ...) ;,. ,.,:vl 27rd ,,7t , 4d Giai V 2 k + 1 , 2k + l 2 k + 1 , Hai diem tren day each nhau khoang d dao dong ngUdc pha khi: ' f = _ = v. = 4. = (*) f 1 ^ X 4d 4.0,28 0,28 A(p =271.- = ( 2 k + 1)71 d= k+i X vdi k = 0, 1,2, . . . 22
nap g\ai oai isin(ntj - 0 , 2 7 i x ) = - V6 A. chi phu thupc vao bien dp cua song. , •i V 3 , B. khong d o i . L i i c sau 0 C. chi phu thupc vao tan so cua sdng. D. phu thupc vao bien dp va tan so cua sdng. 203
Phi/Ong phap gi^i bai t j p trSc nghi§m Cc hpc 12 - T d n Trpng Hung j2.14. Phircfng trinh cua mot song ngang truyen tren sdi day dan hoi ra't dai la: 1 2 . 6 . Song truyen tur O tren true Ox, toe do truyen song la v, bu'dc song X. Ne\ bieu thiJe dao dong tai O la: Uo = Acoscot thl phu'dng trinh song truyen theo 507rt - — X (cm) trong do t tinh bang giay, x tinh bang em. Toe do I u = 4cos 10 true Ox la: X 2nx truyen song tren day la: ' ' - • A. = Acos cot - - B. u ^ = Acos cot - • A. V = 5m/s I T . B. v = 5cm/s C. v = lOm/s D. v = lOem/s j2 15. Phu'dng trinh dao dong tai mot diem O tren phifdng Ox la U(, = 2coslOOTtt 2 TUX X C. = Acos cot-- D. u^^ = A c o s CO ' (cm) trong do t tinh bang giay, sau do dao dong truyen den M tren Ox each O 50cm vdi to do 4m/s. Phu'cJng trinh dao dong cua M la: 12.7. Cong thiJc dung la: A. U M = 2cos(1007tt-7i)(cm) B. U M = 2eosl007it (em) A. ;^ = v.f B.v = X.T C. f = v . ^ V C. U M = 2 cos l O O n t - - (cm) D. khac vdi A, B, C. 2 12.8. Cho mot song cd la song ngang. Vao luc t song 12.16. Tren sdi day dan hoi ra't dai c6 song hinh sin truyen di. Phu'dng tnnh dao C O hinh dang nhu* hinh ve, trong do phan tiir A dao dong vcti van toe v theo hu'dng chi tren hinh ve. dong tai diem M tren day la: u ^ = 5cos 257:t-^ (cm) trong do t tinh bang 4 Song dang truyen ve: giay, sau do dao dong truyen den N each M 1 met vdi toe do 5m/s. Phu'dng A. phia ben phai ' \ B. phia ben trai trinh dao dong tai N la: C. hu'c^ng cua v D. ngu'dc hU'dng cua v A. U N = 5 e o s ( 2 5 T c t - 7 i ) (cm) B. U N =5COS 2 5 7 r t - ^ (cm) 12.9. Mot ngirdi ngoi d bien nhan thay c6 20 ngon song di qua trU'dc mat minh trong 76 giay. Chu ki cua song la: :;nn J M , ? , i?'Vv;r;i / i ; C. U N =5eos(257it) (cm) D. U N = -5 cos 25Kt-^ (cm). A. T = 4s B . T = 3,8s C.T=15s D. T = —s 4 38 76 12.17. Tren day dan hoi kha dai eo song truyen qua. Song truyen theo ehieu tif 12.10. Mot ngUdi ngoi d bien thay khoang each giiJa ba ngon song lien tiep la • M -> N nhu' hinh ve vdi toe do Im/s. Biet phiTdng trinh dao dong cua diem N 15m va ngu'di do de'm du'pc 10 ngon song di qua trUde mat mlnh trong th5i gian 27s. Toe do truyen song bien la: la: Ufq = 4eos 507it-- (cm), trong do t tinh bang giay. M each N 65cm, A. V = 2,5m/s B. V = 2,78m/s C. v = 0,4m/s D. v = 0,28m/s _M v_ N I phu'dng trinh dao dong cua M la: 12.11. Mot ngu'di go mot nhat bua tren du'dng sat va each day 1530m c6 mot i A. U M = 4cos(507tt - n) (em) ; B. U M = 4cos50;ct (em) ngufdi khac ap tai len du'dng sat thi nghe du'dc hai tie'ng biia each nhau 4,2s. Biet toe do truyen am trong khong khi la 340m/s va nho hdn toe do truyen am C. =4cos 5 0 m - ^ (cm) D. U M =4eos 507tt + - (em) 4 2y trong sat. Toe do truyen am trong sat la: 12.18. Day dan hoi Ox kha dai cSng thang nam ngang. Cho dau O dao dong dieu A. V = 3600m/s B. v = 2700m/s C. v = 8400m/s D. v = 51 OOm/s hoa theo phu'dng thang diirng vdi chu ki 0,2s va bien do 3em. Toe do truyen dao 12.12. PhUc^ng trinh song truyen theo true Ox cho bdi: u = 5eos(207it - 0,47tx) I dong tren day la 2m/s. Lay t = 0 la lue dau O dao dong qua vj tri can bhng theo (cm) trong do t tinh bhng giay. Chu kl cua song la: ehieu du'dng. Phu'dng trinh dao dong tai diem M tren day each O 1 m la: A. T = 0,2s B. T = 0,4s C. T = 0,ls D. T = 0,3s 71 A. U M =3 COS 107:t + - (em) B. U M = 3cos(107it) (em) 12.13. Phu'cfng trinh song tren true Ox la: u = 3cos 5 7 C t - - X (em) trong do 2 2 X tinh bang met. Bi/de song la: C. UM = 3cos(107tt + 7t) (cm) D. U M =3cos ' l O T T t - ^ ' (em). 2j A.X^lm B. >i = 4m C. X = 2em D. = 4em 205 204
Hhd'dng ph^p giit bit tjp trie nghigm Do hgc 12 - Tran Trgng Hung 12.19. PhiTcfng trinh dao dong cua mot diem M tren phiTdng Ox khi c6 song truyej, j2 26. PhU'dng tnnh song ngang truyen tren phiTdng Ox nhU' sau: qua cho bdi: U M = 2cos(47it - 0,057tx) (cm), trong do t tinh bang giay va x V\n\y ^2n ^ • - - u = Acos t + STIX , trong do x tinh bang met. Do thi cua li do u theo t tai bang cm. L i do dao dong cua diem M c6 toa do x = 2cm vao luc t = 0,5s la; •-s A . u = 2cm B. u = -2cm C. u = l c m D. u = - l c m = 10cm la du'dng sin ed dang. 12.20. D a u O cua day dan hoi raft dai dao dong theo phiTdng vuong goc vdi d a y '""f"/T\'" , , A , CO phu'dng trinh: U Q = A C O S — t . D i e m M tren day each dau O —bu'dc sono "• ' : T 6 2 / i \ "\ T ' T I \ \ j ^; _ _/4 3_ _T T \ A / 4 vao luc t = — C O li do 1,5cm. B i e n do dao dong la: -A 4 •• -A 2 4 V / 4 A B A . A = 2cm B. A = 2 V 2 c m C. A = % / 3 c m D . A = 2^3 cm u i ,• 12.21. D i e m O tren mat nu'dc dao dong vcti phu'dng trinh: Uo = Acos47it, trong do t A A r 3 T T tinh bang giay. To'c dp truyen song tren mat nu'dc la 2m/s. Khoang each giL7a 1 \ 4 I i \^ O \ / ^ 0 hai d i e m gan nhau nhat tren du'dng thang qua O each nhau T A : / T \ 4 -A . \ i Z . . . ? - A . . . A. d = l m B . d = 2m C. d = 0,5m D. d = l , 5 m -A 2 Viy 12.22. M o t day dan hoi kha dai cang ngang. D d u O cua day dao dong vdi tan so D 36Hz thi tren day c6 song ngang truyen di vdi van toe trong khoang tu" 4m/s 12.27. L i i c t = 0 dau O cua day dan hoi rat dai cang ngang dao dpng theo phufdng 5m/s. Quan sat tha'y hai d i e m tren day each nhau 40cm dao dong cung pha. t h i n g dilng v d i phufdng trinh: U Q =3eos 2 . t - ^ (cm), trong do t tinh bang Toe do truyen song la: 2 A . v = 4,8m/s B . V = 7,2m/s C. v = 4,1 m/s D . v = 5,6m/s giay. T r e n day ed song truyen di vdi biTde sdng~K= 4cm. Dang cua day vao 12.23. M o t day dan hoi ra't dai c6 song ngang truyen qua vdi van toe v = lOm/s, luc t = 1,5s la: t^n so song trong khoang tiJf 40Hz - > 60Hz. Hai d i e m tren day each nhau (cm) > * u (cm) 6,25cm se dao dong lech pha vdi nhau goc Acp = k + - . - v d i k = 1 , 2 , 3 , ... 3 3 2 4 tan so song la: X (cm) A.f=20Hz B. f=IOOHz C. f = 50Hz D. f = 3 2 H z 12.24. M o t song ngang tren day dan hoi rat dai c6 phU'dng trinh dao dpng tai d i e m ed tpa dp x vao luc t la: u = Acos(47it - a.x), trong do t tinh bang giay, ^ la hang so diTdng. Bie't toe dp truyen song tren day la 16m/s. Gia trj cua hang so a la: A . a = 0,25n(m) ^/:,:r.^i.::^:'ry B. a = 0,257t (1/m) X (cm) C. a = 0,57t (m) • V , D . a = 0,57t (1/m) v 12.25. M o t song ngang ed phiTdng trinh song la: u = 0,3cos(314t - 5x) (cm), trong do t tinh bhng giay. V a n toe dao dpng cifc dai eua mot phan tuT vat chat khi c6 12.28. PhiTdng trinh sdng tren phiTdng Ox la: u = 3cos(57it + 0,4x) (cm), trong do t song truyen qua la: tinh bang giay. T a i mot diem nhat djnh tren day vao liic nao do no cd l i dp A . v„,ax = 0,3 em/s B . v,„ax = 0,6 cm/s 2cm thi sau 1 s d i e m dd cd l i d6: C. v,„ax = 94,2 cm/s ' D . v„„x = 15,8 cm/s A. u = l c m B . u = 2cm C. u = - 2 e m D. u = - I c m 207
Cty TIMHH MTV DVVH Khang Vi^l 12.29. Hinh ve ben bieu dien hlnh dang scfi day .9. Dap an A. , ,^ r , - chieu truyen song dan hoi ngay thcJi diem t khi c6 song truyen Trong 76 giay trifc^c mat mlnh c6 20 ngon song di qua tiJc cho do c6 20 Ian qua theo chieu tij" trai qua phai trong do diem ni/dc len cao nhat. Thdi gian giffa 20 ian lien tiep la 19 chu ki. Vay: P d vj tri can bang va Q d vj tri cao nhat. 19.T = 76 = > T - — = 4 (s) Hu'dng chuyen dong cua P va Q ngay thcJi 19 diem t la: - • f.- 12.10. Dap an A. .H") 1 A. Pdi len, Q d i xuo'ng B. P di len, Q diJng yen • Khoang each giifa ba ngon song lien tiep la < h—^ ^ ^ ^ C. P di xuong, Q di xuong D. P di xuong, Q diJng yen 2:^ (Hinh ve) nen: X = — = l,5 (m) \ \ HI/6NG DAN GIAI 12.1. Dap an C. • 10 ngon st5ng lien tiep di qua trifdc mat muih ti'rc C\\Q do vat chat da thifc 27 C. Sai vl: song ngang la song trong do cac phan tuf cua moi tru'dng dao dony hien 9 dao d()ng lien tiep nen chu ki song: T = — 3 (s) theo phu'dng vuong goc vdi phu"clng truyen song. \5 12.2. Dap an D . Song doc truyen du'pc ca trong chat ran, long va khi. Toe do truyen song: ^ = — = = 2,5 (m/s) 12.3. Dap an A. Song ngang truyen du'dc trong cha't ran (va tren mat niTdc). 12.11. Dap an D . 12.4. Dap an D . Chii y rang phat bieu III dung la: Ngirdi nghe se nghe hai tieng bila: tieng do am truyen trong du'dng sa't (nghc Bu"dc song la khoang each giiJa hai diem GAN NHAU NHAT tren phu'dng trUde) va tieng do am truyen trong khong khi. truyen song ma dao dong tai hai diem do dong pha. Goi V va t la toe do va thdi gian truyen am trong du'dng sat, ta ed: 12.5. Dap an l i . Doi vdi mot moi tru'dng nhat djnh thi toe do truyen song khong ddi S = v . t - 3 4 0 ( t + 4,2)= 1530-^t = - 5 ^ - 4 , 2 - 0 , 3 ( s ) -^^-^ ^ x^ CO 271 2K 340 12.6. Dap an C. u^^Acosco A c o s CO t - - = A COS cot vdi — = — = • nen V v.t X Vav= ^ = 1 ^ . 5 1 0 0 (m/s) 2 Tlx t 0,3 . U M = Aeos cot-- 12.12. Dap an C. : 27rx 12.7. Dap an D. c6 ;i = v.T = - hay: T = - . Phu'dng trlnh song eo dang long quat: u = Acos cot-- f V 12.8. Dap an B. , - So vdi phu'dng trlnh da cho: u = 5cos(20jit - 0,47ix) Ta dirdc: co = 207i (rad/s) T =^ = - ^ = o,I(s) CO 20n '2.13. Dap an B. Vao luc t diem A di len theo hu'dng v nen cac diem cung pha C, E, ... cung i-'' ^ S So c phu'dng trlnh da cho vdi phu'dng trlnh tdng quat: u = Acos cot -271 — ?; len va liic t + — cac diem nay len cao nhat. NgUde lai liic t cac diem B, 1^' , i ''J'*,';/ , i : K 4 T f Ta ladirdc: — = — — -> A = 4 (m) ... dao dong ngu'dc pha vdi A nen di xuong va luc t + — chi'mg xuong thfip 4 ' T , '2.14. Dap an A. nhat. Vay luc t + — song la du'dng cong khong lien tuc nhu' tren hinh ve. Q^''^ 4 2n 2nx) So phu'dng trlnh da cho vdi phirdng trlnh tdng quat: u = A cos ket qua do ta thay song truyen ve ben trai. 208
Phuong ph^p giai bai tjp trie nghigm Co hpc 12 - TrSn Trpng Hung oiy ii\]HH M i v u v v H T ' T i a n g v i e i T a diTdc: — = 507i T = — = 0 , 0 4 (s) + CO = — = — = 107t (rad/s) T 50 T 0,2 1" • 1 + A = 3cm • / •i X, = 2 0 ( c m ) X 10 cm + t = 0 : Uo = 0 va U Q > 0 0 = Acoscp coscp = 0 7t Toe do t r u y e n song: v = — = = 5 0 0 (cm/s) = 5 (m/s) ^ (p = — U(, = - C L ) A s i n ( p > 0 [sm(p v = 5 0 0 (cm/s) = 5 (m/s) • V 500 = 3 c o s ( 1 0 7 t t - 5 , 5 7 i ) = 3 c o s 107rt + - (cm). 2 12.19. D a p a n B . , , , . 12.15. D a p a n C . U M = 2cos(47:t - 0,057rx) ( c m ) r ^ x^ 05 UM=2cosl007r t - - = 2cosl007t t - V 4 j T a i X = 2 0 c m va t = 0,5s t h i : U M = 2cos(47i.0,5 - 0,0571.20) = 2cos7: = - 2 ( c m ) . 12.20. D a p a n C . = 2 c o s ( 1 0 0 j t t - 1 2 , 5 7 1 ) = 2 cos 1 0 0 7 : 1 - - (cm) 27t 27tX X T 2j UM = ACOS — t vdi X = - , t = — thi U M = 1,5 ( c m ) n e n 6 4 12.16. D a p a n D . r27i T 271.X/6 Phi/dng t r i n h d a o d o n g t a i N : 1,5 = A cos = Acos- = A.—-> A = V3 (cm) \1 A X 6 2 X = 5 C O S 257t ' 0 UN = 5 C O S 2571 t — 4 t — |2.21.DapanC. '^'^''^v ' »,4; >n V . 5J (cm) T a c 6 : T = — = — = 0,5 ( s ) ; ? i = v . T = 2.0,5 = 1 ( m ) ^^«''',' = 5 COS 2 5 7 t t - 5 7 r - - = - 5 COS 2 5 7 : t - ^ 4J 4j (0 471 12.17. D a p a n B . H a i d i e m g a n n h a u nha't t r e n phiTdng t r u y e n s o n g va d a o d o n g ngu'dc p h a V i s o n g t r u y e n t d i M triTdc n e n d a o d o n g t a i M s d m h d n d a o d o n g t a i N tht'i e a c h n h a u - J = 0,5 ( m ) . * gian At = — . V a y : 12.22. D a p a n A . '' ' ' > ' V H a i d i e m dao d o n g c i j n g pha each nhau: 0,65 u^^ = 4 cos 5071 t + - = 4 cos 507t t - ^ 1 r V X,- 1 1 -1 o d.f 0,4.36 14,4 , , , d = k . ? i = k . - v d i k = 1, 2, 3, ... => v = -— = — = ——{mis) f k k k = 4cos(507it + 327t) = 4cos507tt ( c m ) 12.18. D a p a n A . - , ,u 4 < V = < 5
PhUdng phap giSi bai tgip t r i e Co !i i(, 12 - TrSn Trpng Hang n 2nx n 2nx^ 7t 2KX 2nd 2nd ^ ^. . 71 hue t = 1,5s: u = 3cos 3n = 3 cos n-- = 3cos A(p = -^;— = .1 vdi A(p = k + - lien 2 X V 2 4 2 X 2 X 2nd • n V 10 0 Tai X = 0 u = 3cos- = 0 vOil I .f = k + - k + - = 20 k + - (Hz) 2 2 4 2 8d 2 8.0,0625 2 Ma: 4 0 < f < 6 0 ( H z ) • T a i X = — = 1cm: u = 3cosO = 3cm I v ' ' / 4 , •' . 40 < 20 k + - < 60 1,5 < k < 2,5 - » k = 2 Dang cua day la do thi cua u theo x: do la du'dng sin lap lai sau X = 4cm 2, • Chu y rang liic t = 0 dau O bat dau dao dong nen de'n liic t = 1,5s song Vay: f = 20 2.1 = 50 (Hz). truyen di xa nhal la: x,„„ = v.t = —.t vdi T = ^ = ^ _ \ n^,,. , 2) M i t l :U 12.26. D a p an D . PhiTdng tnnh dao dong tai d i e m x = 10cm = 0,1m ^ - Bai 13. G I A O T H O A S O N G *' * ' -' '2n '2n n u = Acos t + 57r.0,l - Acos — 1 + - ' ^ * TOMTATLITHUYET T IT 2J 1. Hien tU'dnjj jjiao thoa ciia hai .son u = AcosTi = - A - K e t qua: tren mat nu'dc xuat hien cac gdn song on 4 dinh CO hinh la cac du'dng hypebol vdi tieu d i e m la «^ Vay do thi la hinh D . 12.27.Dapan A . ; . - Hien tu'dng hai song gap nhau tao nen cac gdn song Phu^dng tnnh dao dong cua song: on dinh goi la hien tu'dng giao thoa cua hai song. u = 3cos 2Kt vdi ?i = 4 cm Cac gdn song c6 hlnh hypebol goi la cac van giao thoa. 2i2 •I , 2 X ) I 213
Phuong phap giai bai tgp trSc nghigm Co hqc 12 - I r a n l:oi).j Hung Vjiy i N n n i v i i v u v v n r .diiy V151 2. Cue d a i va ci/c t i e u . j ,, yj' d u 1: T a i hai d i e m Si, S2 tren mat nifdc c6 hai nguon ket hdp gio'ng nhau. M - Goi phifdng trinh dao dpng cua hai nguon S|, S2 la: u^^ = u^^ = Acos(a)t) la mot d i e m trong viing giao thoa each Si, S2 Ian lu'dt d i , d2. BiTdc song tren mat nu'dc la X. Cho rang bien dp song khong ddi va bang bien thi piiiTdng tnnh dao dong tdng hdp tai d i e m M trong vung giao thoa la: dp dao dpng cua nguon. { = 2 A c o s 7 i - ^ 7 — - . cos C0t-7t p p lech pha Acp = (pi - (p2 cua hai dao dpng tai M do hai song truyen tiT Si, S2 i; ;,>•,,;• ^-i den la: .•• • 1-).•..:.j^:,;..J:; , - Dao dong tong hdp tai M la dao dgng dieu hoa cung phifdng, ciing chu kl d, + d . A. Acp =7t B. Acp = 2 n ^ ^ d2-d, X X vdi hai nguon va c6 bien dp: A^^ = 2 A cos 71 d2-d, -d C. A(p =7X D. A ^ = n ^ ^ ' - V i t r i cac cifc dai giao thoa ifng v d i : 2X .d2-d, Giai cos 71- = 1 => d j - d i =kA. (k = 0 ; ± I ; ± 2 ; . . . ) M X S, S, Gpi phu'dng trinh dao dpng cua hai nguon Si, S2 la: II d Quy tich cac d i e m nay la nhiJng di/dng hypebol c6 hai tieu d i e m la Si, S. "S| =^s, =Acoscot thi: 'd: - chung du'dc gpi la van giao thoa ciTc dai. > > siii 'ji.*' ' - PhiTdng trinh dao dpng tai M do song truyen tiT S| den: - V j t r i cac ci/c tieu giao thoa ilng v d i : , i d.^ , ,,l = A c o s cot-2K-!- COSJt k + - X (k = 0 ; ± l ; . . . ) X 2 - Phu'dng trinh dao dpng tai M do song truyen tif S2 den: Quy tich cac d i e m nay la nhifng du'dng hypebol c6 hai tieu d i e m la Si, S2 ^ C17 UjM = A c o s va gpi la van giao thoa circ tieu. X 3. D i e u k i e n giao t h o a . Song k e t hdp. — D p lech pha giffa hai dao dpng tai M : Acp =cp, -(P2 = De CO van giao thoa on dinh Iren mat nu'dc thi hai nguon song Si, S2 phai: X a. Dao dpng cung phiTdng, cung chu k i . Chpn dap an B . b. Co hieu so'pha khong ddi theo thdi gian Vi d u 2: Can rung trong thi nghiem giao thoa song tren mat nu'dc dao dpng vdi ' - Hai nguon nhu" tren gpi la hai nguon ket hdp. Hai song do hai nguon k e l chu k i T = 0,5s. M la mot d i e m tren mat n\idc each hai mui nhpn Si, S2 ; hdp phat ra la hai nguon ket hdp. Ian lu'cJt di = 20cm va da = 15cm. B i e t tdc dp truyen song tren mat niidc la ^- H i e n tu'dng giao thoa la hien tu'dng dac triTng cua s6ng. V = 0,8m/s. Dao dpng tai M do Si truyen idi. . BAITAPCdBAN A. Sdm pha hdn dao dpng tai M do S2 truyen t d i goc - Dang 1. D O L E C H P H A C U A H A I D A O D O N G T A I M O T D I E M B. Tre pha hdn dao dpng tai M do S2 truyen tdi goc T R O N G VUNG G I A O THOA D O HAI S6NG TRUYEN T6I - T i m phiTdng trinh dao dpng tai M do song truyen tu" nguon C. Sdm pha hdn dao dpng tai M do S2 truyen td'i goc ^ ^|«' S| den (UIM): ' - - M - T i m phUdng trinh dao dpng tai M do song truyen til" nguon d| 'd2 D . Tre pha hdn dao dpng tai M do S2 truyen tdi goc ^ ''"-'^
vdi X = v.T = 80.0,5 = 40 (cm) A. Acp = + - B. Acp = - — C . Acp = + — D . Acp — X 5 4 4 Giiii • ;•. Nen: Acp = (p,-(p-, = ZTT = —
. _ nung V i d u 2: T a i S|, S2 tren mat niidc c6 hai nguon ket hdp cCing bien do A. M y/jdy 4: T a i hai d i e m 81,82 tren mat nU'dc c6 hai nguon dao dong ket hdp ngU'dc mot d i e m trong viing giao thoa cua hai song. Hai dao dong tai M do hai song pha, C L i n g bien dp V3 cm. nU'dc song cua song truyen tren mat nU'dc la 0 , 3 n . truyen tdi lech pha nhau gdc A(p. Cho rhng bien do song khong d o i . B i e n do dao dpng tdng hdp tai M : Coi bien dp song truyen di khong doi. Bien dp dao dpng tdng hdp tai d i e m M trdn mat nU'dc each S|, S2 Ian lu'dt di = 50cm, d2 = 40cm la: A. A M = 2 A cos-• - B. A M = 2 A cosAcp A. AM = 1cm B. A M = 2cm C. A M = 3cm D. A M = 4cm Giai C . A M =2A|cos2Acp| Acp D. A M = A cos S I , S2 la hai nguon ket hdp ngUdc pha nen dp lech pha cua hai dao dpng tai diemM: " ' J.^•.••ji 1 . . v, Giai ^2 - d | ^ 0,4-0,5 7C , . 57t A p dung cong thu-c tinh bien do dao dong tdng hdp: " * Acp =cpi-cp2 = 2 7 t - ^ i - ± 7 t = 27t-^ ^±7i=^-hoac 2 2 2 X \ 0,3 3 3 A M = A , + A 2 +2AiA2Cos((p2 - ( p i ) [Cong thilc 5.1 - trang 2 3 - S G K 1 2 ) 7t A, = A 2 =A 2.^/3 COS — = 3 ( c m ) ;? , , (1,1 Vdi Acp 6 N e n : A ^ = A^ + A^ + 2A^ cos Acp = 2A^(1 + cos A(p) A M = 2 A cos MM. ••, ;;;('••" ' III t I t I l i )( ( (P2 -cp, = Acp 2.^/3 COS = 3 (cm) ' M a : I + c o s A ( p = 2 c o s 2 ^ nen: A ^ = 4 A ^ c o s 2 ^ . f ,1;. 1 , 2 2 Chpn dap an C. Acp I A M = 2 A cos . Chon dap an A. D a n g 3 . FHUONQ TRINH D A O DONG TONG HOP C U A V i d u 3 : T a i hai d i e m O i , O2 tren mat niTdc c6 hai nguon ke't hdp v d i phu'dng HAI SONG K E T HOP trinh: UQ^ =yl2cosSnt (cm); U Q , = \/2cos(8jit-7r) (cm) Ne'u phu'dng trinh dao dpng cua hai nguon ket hdp la: Ui = U2 = Acoscot thi Song tren mat nu'dc truyen di v d i van tdc 2m/s. Coi b i e n do song khong doi. phu'dng trinh dao dpng tdng hdp do hai song tao ra tai mot d i e m M : M la mot d i e m tren mat niTdc trong vung c6 giao thoa each 0|, O2 Ian lu'dt d.-d, UM =2Acosji-^ ^-cos cot -71 di = 36cm, d2 = 23,5cm. Bien do dao dong tdng hdp tai M la: A. AM=2%/2cm B. A M = 2cm C. A M = 3cm D. A M = V 3 c m Vf d u : T r e n mat nU'dc c6 hai nguon ket hdp gio'ng het nhau vdi phu'dng trinh dao Giai dpng: U | = U2 = 2cosl07it (cm). Tdc dp truyen song tren mat niTdc la 5m/s. Cho H a i phu'dng trinh dao dpng tai M do hai song truyen t d i : rhng bien dp song khong thay d d i . PhiTdng trinh dao dpng tdng hdp tai M tren " l M = V 2 cos COS 87Ct-7C-27t^ mat ni/dc each hai nguon khoang di = 40cm va d2 = 15cm la: ^2M A. U M = 2 c o s (cm) B. U M = 2 c o s (cm) d, - d , 2) : 4J D p lech pha ciia hai dao dpng tai M : A(p = cpj - ( p j = 2TI^^ ^' + n C. u^,=272 COS ' l O ; : t - ^ ' (cm) D. U M = 2 V 2 c o s ( l 0 7 i t - 0 , 5 5 7 c ) (cm) T.g.0.25,s) 2J Vdi , n e n : Acp = 2 7 t . ^ ^ ' ^ ~ ^ ^ + = 50 2 Giai - v.T = 2.0,25 - 0,5 (m) = 50 (cm) 271 B i e n dp dao dpng tdng hdp tai M : A ^ = 2A COS Acp = 2 ^ COS — = 2 (cm) T = — = 0,2s ; = v.T = 5.0,2 = 1 (m) . ... 4 1071 d|+d2 ChpndapanB. = 2Acos7r-^ '-.cos cot -71 218
"luy i i u u i i i j i i i v i i i u u m/'j—re iidii iiviiynung" 0,15-0,4 0,4 + 0,15^ = 2.2 cos Tt ——.cos lOTtt - 7 1 ^ — C. d 2 - d | = k ^ D. d j - d , = : ( 2 k + DA, = 4cos(-0,257r).cos( 1 OTII - 0,557t) = 2>/2 cos( 1 OTit - 0,55;:) (cm) (vdik = 0 ; ± l ; ...) Chpn dap an D . ' Giai Bien dp dao dpng to'ng hdp tai M : D a n g 4. VI TRI C ^ C D A I - C y C T I E U G I A O T H O A A(p Neu hai nguon ket hdp Si, S 2 dao dpng cung pha: u^^ = u^^ = Acos((i)t) thi: 2 M A M =2A COS- a. V i tri cac ciTc dai giao thoa: d i - d | = k.A, (k = 0; ± ! ; . . . ) = 1o — = k 7 i - > Aq) = 2kn 2 , • b. V i t n cac cifc tieu giao thoa: d 2 - d| = k + - ^ ( k = 0 ; ± l ; . . . ) 5* A M : max o eos- 2 I V d i A(p =2n^- ^ ' - 7 t X V i du 1: T r e n mat nU'dc c6 hai nguon ket hdp ciing pha ciing chu k i T = 0,4s, t o c f 1 ^ dp truyen song tren mat nu'dc la v = 0,8m/s. G p i M , N , P, Q la bon d i e m trcii Nen: 2n^- - 7 t = 2k7i: d , - d , = k + - .X . Chpn dap an B . A. ^ 2j mat nu'ctc nam trong vimg giao thoa c6 hieu khoang each tiT m o i d i e m den hai Vidu 3: Hai nguon ket hdp ngUdc pha: U i = Acoscot nguon (tiJc d : - d|) cho bdi bang sau: U2 = Acos(Q)t - n) "' ' " Diem M N P Q .:':} nl mA) m>n Bu'dc song tren mat nUdc la A. - • » . d2-d|{cm) 20 40 60 80 Vi tri cac ciTc tieu giao thoa difdc xac djnh bang bieu thifc: D i e m tai do c6 ciTc tieu giao thoa la: f 1 'SI V A. M :, . r,,-- :, ' B. N > 'r-' • C. P D. Q A. d . - d i = k?. B. d 2 - d | = k +- X '^'^ '••••••»•'•. - Giai X I = v.T = 0,8.0,4 = 0,32 (m) = 32 (cm) D. d . - d , = ( 2 k + \).X C. d2-d|=k.- Xet bang sau: i (k = 0 ; ± l ; . . . ) Diem M P Q Giai d2 - d | (cm) 20 40 60 80 A(p A M = 2 A cos 0,625 1,25 1,875 2,5 X Qua bang ta tha'y d i e m Q c6: AM:min o c o s ^ = 0 « ^ = - + k7i (k = 0; ± 1 ; . . . ) - > Acp = 7t + 2k7t d.-d, 2 2 2 - = 2 , 5 - > d 2 - d , =2,5:V = 2 Ma: A(p = 2n^^ f ^ ' + 7 1 , nen: 2 7 1 ^ - ^ ^ ^ +Jt = Tt-f 2k7i: dj - d , = k.A. X •""""""" ^.k Vay tai Q c6 ciTc tieu giao thoa. Chpn dap an D . * Chpn dap an A . V i du 2: T r e n mat niTotc c6 hai nguon ket hdp ngiTdc pha: U | = Acoswt VI du 4: T r e n mat nu^dc c6 hai nguon ket hdp: u, = Acos(407it) \ . _ ;< ^ ^2 = Acos((ot + 71) U2 = Acos(407it - J i ) Bu-dc song truyen tren mat nu-dc la X. Vj tri cac ciCc dai giao thoa du'dc xac djnh bang bieu thii'c: Toe dp truyen song tren mat ni/dc la 30cm/s. M la mot d i e m tren mat niTdc c6 ciTc tieu giao thoa khi hieu khoang each: k + - X A. d 2 - d | = 6 , 7 5 ( c m ) B. d 2 - d, = - 4 , 5 e m A. d 2 - d , =kA. B. d 2 - d i = 2 C. d 2 - d , = - 3 , 5 c m D . d . - d, = l,8em 221 220
Giai X Si S7 271 (3) va (4) cho: dj = k + - ma 0 — < k < I • Vj tri ciTc dai giao thoa: d2 - di = k+ - 2 2 2 X X M • Vj tri ciTc tieu giao thoa: d2 - di = :; Vdi ;t = v.T = 3 0 . 0 , l = 3 ( c m ) V D a n g 5 . T I M s6 V A N GIAO THOA Xet tru'dng hdp hai nguon ke't hdp 81,82 cung pha. Nen: < k < — =>-3,3
Cty TIMHH MTV DVVH Khang Viet S|S. 1 V _ 50 '"^ - < k < ^ - - vdi ^ = - = — = 5 ( c m ) j
PhtiOng phip g\i\ tap trSc nghigm Co hgc 12 - TrSn Trpng Hiing Cty TNHH MTV DVVH Khang Vi§t C A U HOI VA BAI TAP TRAC NGHIEM 13.7. Tren mat nu'dc c6 hai nguon ke't hpp dao dong ngu'dc pha: 13.1. Hai ke't hdp la hai nguon dao dong U| = Acos207it U 2 = Acos(207xt + 71) A. CLing bien do, cijng chu ki, cung phu"c(ng. To'c dp truyen song tren mat nu'dc la 0,5m/s. M mot diem tren mat nifdc each B. cijng phu'Ong, cCing tan so', hieu so' pha khong doi theo thdi gian. hai nguon Ian lu'dt d| = 15cm va d2 - 18cm. Dp lech pha ciia hai song tai M la: C. cung bien do, ciing tan so', hieu so pha khong doi theo thdi gian. A. Acp = ^ R a d B. Acp = y R a d C. Acp = y R a d D. Acp = y R a d D. Cling bien do, ciing phu'dng, hieu so' pha khong doi theo thdi gian. 13.2. Hai nguon dong bo la hai nguon ke't hdp • 13.8. Trong thi nghiem giao thoa tren mat nu'dc, bien dp song truyen di khong A. cijng bien do B. cung tan so' C. ciingpha D. ciing phu'cJng doi va bang icm. tai diem M tren mat nirdc hai song truyen tiT hai nguon den 13.3. Neu hai nguon ke't hdp ciing pha, thi nhiJng diem tai do dao dong c6 bien lech pha nhau y - r a d . Bien dp song tong hdp tai M la: " ' ' ' ' dp ciTc dai la nhffng diem ma hieu du'dng di cua hai song tif nguon truyen tdi thoa man: . .M • .^ • A. A M = 2cm B. A M = 0,5cm C. A M = l , 5 c i n D. A M = l c m . 13.9. Hai nguon ke't hdp tren mat nu'dc cd phiTdng trinh Ui = U 2 = 2cos(207rt)(cm). A. d , - d i = kX B. d , - d | = k- To'c dp truyen song la 0,4in/s. M la mot diem tren mat nifdc each hai nguon Ian lu'dt 15cm va 12 cm. Bien dp song tong hdp tai M la C.d2-d, - k + - D. d 2 - d , = 2 2 A. A M = V3 cm B. A M = 2>y2 cm C. A M = V2 cm D. A M = 3^3 cm (vdik = 0 ; ± l ; ...) 13.10. Tren mat nu'dc cd hai nguon ke't hdp ngu'dc pha cting bien dp y/l cm. 13.4. Ne'u hai nguon ket hdp dao dong cijng pha thi bien dp dao dong tong hdp Birdc song la 0,3m. M la mot diem tren mat nu'dc each hai nguon Ian lu'dt cua hai song truyen tdi diem M cho bdi: 7cm va 12cm. Bien dp dao dong tong hdp tai M la 2n{d2-d,) 27r(d2-d,) A. A M = lyfZcm B. A M = 2cm C . A M = V2 cm D. A M = 1 cm A. AM = A cos B. A M = 2 A cos X 13.11. Hai nguon ke't hdp tren mat nu'dc cd phu'dng trlnh U | = U 2 = \fi cos(87tt)(cm) Tt(d2-d,) 7c(d2-d,) To'c dp truyen song tren mat nu'dc la 48cm/s. M la mot diem tren mat nu'dc C. A M = 2 A cos D. A M = A cos X X each hai nguon Ian lu'dt 20cm va 18cm. Phu'dng trlnh dao dong tong hdp tai M la f 13.5. Hai nguon ket hdp dao dong ngu'dc pha, nhilng diem tai do dao dong c6 A. U M = 3cos(87it - — )(cm) B. U M - 2cos(8Ttt - — )(cm) bien dp CLTC dai la nhiJng diem ma hieu du'dng di cua hai song tif nguon 6 2 truyen tdi la: ,' C. U M = 3cos(87it - — )(cm) D. U M = 2cos(87tt - — )(cm) 4 6 A. d2-d, = k - B. d 2 - d i =k?u 13.12. Hai nguon ke't hdp S,, S2 tren mat nifdc dao dong gio'ng nhau: U | = U 2 = Acos(cot) C.d2-d, = k + - D. d . - d , = k +- X J Mot diem nSm tren du'dng trung trifc cua S1S2 dao dong vdi bien dp 2 13.6. Tai hai diem S|, S2 tren mat nu'dc c6 hai nguon ke't hdp cijng pha, cung tan A. AM =0 j B.AM = A C. A M = Y D. AM=2A so" f - 20Hz. To'c d p truyen song tren mat nu'dc la v = 30cm/s. D p lech ph'' |i.l3. Hai nguon ket hdp Sj, S2 tren mat nu'dc dao dong ngu'dc pha: ciia hai song tai diem M tren mat nu'dc each hai nguon Ian lu'dt d | = 12cm. U | = Acos(cot) U 2 = Acos(cot - 71) dn = 13cm la: Biem nhm tren du'dng trung triTc cua S1S2 dao dong vdi bien dp A. Acp = — R a d B. A(p = Y ^ a d C. A(p = ^ R a d D. Acp = —Rad A.AM =0 , B . A M =A C.AM=2A D . A M = 3A ^ • , 227
Phuong phap giai bai t$p trSc nghijm Co hgc 12 - Trjn Trpng Hung 13.19. Hai nguon ket hdp S,, S2 tren mat nirdc each nhau khoang / c6 cung 13.14. Hai nguon ke't hdp S i , S2 tren mat nu'dc dao dong vdi phu'dng trinh: phu'dng trinh dao dong: u, = U2 = Acos(o)t) U i = Acos(cot) ' • U2 = Acos(cot - ~ ) '• " Bifdc song truyen tren mat nu'dc l i i X. De trung diem I cua S1S2 dao dong (tdng hdp) Cling pha vdi hai nguon thi khoang each hai nguon la Trung d i e m O cua S1S2 dao dong vdi bien do tdng hcfp A.I:=kX B . / = k-- C . / = 2k^ D./ = k - A. Ao = 0 B. Ao = A C . AO = AV2 D . AO = 2A 2 4 13.15. Trong thi nghiem giao thoa tren mat mldc, hai nguon ket hcfp ngUctc pha ( v d i k = 1,2, 3,...) CO phu'dng trinh: 13.20. Hai nguon ke't hdp S|, S2 tren mat nu'dc each nhau 20cm cung dao dong U | = 2cos(20nt) (cm) " U j = 2cos(207it + 7t) (cm) • vdi phu'dng trinh: U i = U2 = Acos(lOTtt). V a n to'c truyen song la 0,23m/s. So To'c do truyen song tren mat nirctc la 0,4 m/s. M la mot d i e m tren mat nirdc diem dao dong cifc dai nam tren doan S1S2 la each hai nguon Ian lu'dt 6cm va 7cm. Phu'dng trinh dao dong tong hdp tai M la A. 5 B.9 C. 12 D. 7 A. UM = 4cos(207it - - ) (cm) , B. UM = 2^2 cos(207it - — ) (cm) 13.21. Tre n mat nu'dc cd hai nguon ke't hdp cung pha Si, S2 each nhau l0,75cm phat ra hai song cung bien dp, cung tan so' gdc 20rad/s. To'c dp truyen song la C. U M = 4cos(20TCt + — ) (cm) D. U M = 2V2 c o s ( 2 0 7 r t + -—) (cm) 3,18cm/s. Lay — ~ 0,318. So' diem dao dong cifc tieu tren ,|, L2 la 4 4 71 13.16. Mai nguon ket hdp S|, S2 tren mat niidc giong het nhau. Bu'dc song truyen A. 18 B. 20 . C. 22 D . 16 tren mat nu'dc la 4cm. M la mot diem nam tren mat nufdc each S|, S2 Ian krdt 13.22. Tre n mat nifdc c6 cac van giao thoa cua hai nguon ke't h d p i ' i , "2 cung pha. 16cm va 10cm. Ke tu" du'dng trung triTc cua S1S2, d i e m M nam tren van giao Bu'dc song la 8cm. Khoang each tiT dirdng trung triTc cua : ii2 den van giao thoa : •'. .\: thoa ci/c tieu dau tien (ke tij' du'dng trung trifc) doc theo S1S2 la A. ciTc dai thti" nha't , B. cifc dai thu" hai A . 4cm B. 8cm C. 6cni D . 2cm C. ci/c tieu thu-nhat ,,,, , 4 , , ^,1,, , D . ci/c tieu thi?hai 13.23. Khoang each giffa 4 van giao thoa circ dai lien tiep dpc theo du'dng no'i hai ngupn ke't hdp la 4,5cm. Bie't tpc dp truyen song la 24cm/s. Tan so dao dong 13.17. Hai nguon ket hdp 0 | , O2 tren mat nUdc c6 phu'dng trinh dao dong: cua nguon la i ^ • U | = U2 = Acos(57tt) A. f = 8 H z B. f = 1 6 H z C. f = 72Hz D. f=10,67Hz V a n toe truyen song la 15cm/s. M la mot diem tren mat nUdc each O i , O2 Ian HLfdNG DAN GIAI lu'dt 0 | M = 16cm va O j M = 25cm. Ke tiT du'dng trung triTc cua O1O2 d i e m M 13.1. Dap an B 13.2. Dap an C 13.3.DapanA 13.4. Dap an C nam tren van giao thoa 13.5. Dap an D A. ciTc tieu thu'nhat ' ' ' B . cifc tieu thii'hai C. ci/c dai thu' nha't D . ciTc dai thiJ hai 13.6. Dap an B =^ ' 13.18. T re n mat nu'dc c6 hai nguon ket hdp A va B each nhau 20cm vdi phu'dng D o lech pha: z\cp - 2 7 i ^ ^ , vdi X = - = — = 1,5 (cm) trinh dao dong: UA = UB = coscot (cm) , 13-12 471 , , nen A(p = 27t = — (rad) Bifck- song A. = 8cm. B i e n dp song khong d o i . Gpi I la mot d i e m tren du'dng 1^5 3 ; ; :, .. J ., trung triTc cua A B dao dong cDng pha vdi cac nguon A, B va gan trung diem 13.7. Dap an A .' U O cua A B nha't (hinh ve). Hai phu'dng trinh song tdi tai M : U I M = Acos(207tt - 27i—^) Khoang each 01 do dUdc la j A . d, A. 0 1 = 0 B. O I = > / l 5 6 c m A- - U2M = AcOS(207lt + 71 - 27t ) • A, C. 01 = x/l25 cm D. 01 = 15cm i; • 228 m' 229
iidi^ iiyiMCHi riyi; I - iidii ! rpng Mung . .12. Dap a n D D o l e c h pha c u a h a i song tai M : Acp = (p, - 92 = 2n~—^-TT , vdi D i e m nhm t r e n d i T d n g t r u n g triTc S1S2 c6 d o l e c h p h a c u a h a i d a o d o n g t a i d o l a T = - ^ = 0,l(s) . , ^18-15 n Acp = 271 =0 (vid|=d;) . - 2071 V, n e n : Acp = 2n— n = — (rad) >. = v.T = 5 0 . 0 , l = 5 ( c m ) , ^ , 5 5 Acp AM = 2 A cos = 2A 13.8. Dap an D Acp 71 13.13. Dap an A B i e n do song tdng hdp tai M : A M = 2 A c o s — ^ = 2.1 c o s — = 1 ( c m ) 2 3 Dp lech pha cua hai dao dong lai M : 13.9. D a p an B d. - d , lirZl Acp = 271 n + 71 = 5X ( V l d 2 = di) 2K_ T = = 0.1 (s) 2071 Acp n -> AM = 2A c o s - = 2A cos — = 0 /• ^ = v.T=:40.0J=4(cm) ' " ' ' ' ' ' 2 13.14. Dap an C • Ais) = 2n~ ^ = 271 = Hai bieu thiJc song tdi tai O tif hai nguon la Acp 37r AM = 2 A cos = 2.2 c o s = 2V2 ( c m ) U : o = Acos(cot - 2 7 t — ) U 2 0 = Acos(cot - — - 2 7 t — ) A, 2 A, 13.10. D a p an C Dp lech pha ciia hai dao dong tai O: 12-7 47r dj - d j 71 71 D o lech pha cua hai dao dong tai M : Acp = 2 7 t ^ ^ . ^ ' + n = 2n 7l = - Acp = cp[ - c p 2 =27t ' • - = - (vi d| =d2) 30 X 2 2 B i e n do dao dong tong hctp tai M : Acp B i e n do dao dong tong hdp tai O: A o = 2 A c o s = 2A cos- = AV2 Acp 27r A M = 2 A cos = 2 ^ cos- = 2N/2 = V2 ( c m ) 13.15. Dap an B 1 271 Phu'dng trinh song tai M do hai nguon truyen tdi Ian lifdt la: • . j,^ • ChiJ y : CO the vie't: Acp = 2 7 t ^ ^ ^ — i - - 7 i = - T = ^ = 0,l(s) U | M =2cos(207ct - 271 ), v d i 2071 A M = 2sy2 COS = 27^ = ^ (cm) if. X A = v.T = 4 0 . 0 , l = 4 ( c m ) 13. I I . Dap an A nen: U,M = 2 c o s ( 2 0 7 t t - 2 7 t - ) = 2cos(207it-37i) (cm) 4 di+d2 T =^ = 0,25(s) U M = 2 A c o s 7 i ^ ^ — ^ . c o s STlt-Tl , vdi 07t va UjM = 2cos(207tt -f 7t - 27i ^ ) = 2cos(207it -f 7t - 27t - ) = 2cos X = v.T = 48.0,25 = 12 (cm) A, 4 18_-20 18 + 20 571^ nen UM = 2 7 ^ c o s 71- - c o s 87rt - 7 : Dao dong tong hdp tai M : U M = 2 cos(207tt-37r) + cos 2 0 T t t - 12 ( 197t^ 19TC A' , u -) ^+b a-b = 2 ^ 3 cos cos 87rt - = 2^3. -.cos 87lt + 2 7 l - A p dung: cosa -f cosb = 2cos ^ cos ^ [ 6 J f = 3cos 8 7 r t - 771 ~ (cm) U M = 2.2.COS 2 0 7 tt-- 1 l7I .COS = 2 ^ COS 2 0 7 i t - l5l (cm) 4 . 231 230
uiy iiuiMi ivi I V L r v v i i iMiuiiy uii^i 13.16. Dap an D d, ^ d,^d, d, - d , 10-16 Ui = 2 A c O S 7 I - ^ - - . C O S 03t - 7t Ta c6; = -1,3 A. X V i I la trung d i e m cua S1S2 nen: di = d2 = — hay d2 - d| = • - 2 . 1 2 / V a y U i = 2Acos0.co.s (Ot - 71 — V a y : M nhn tren van giao thoa ciTc.tieu thuf X. V hai (ve phia gan S, hdn S,) nhir hinh ve. Do lech pha cila dao dong tai I vdi S, (hoac S,) la:, ^; 13.17. Dap an B I Acp = (Ps, -cp, =0- = n— T = ^ = 0,4(s) X 571 I ' X = vT = 15.(),4 = 6 ( c m ) ;.. V\ dao dong cung pha vdi hai nguon nen: Acp = 7t— = 2k7r -> / = 2 k ^ X dj-d, 2 5 - 1 6 = 1,5 13.20. Dap an B " ' " ' 6 \d, - d , - k ^ d:-d| = +— •ii v f ' =; 'n'i Co I" ' ^2d, = kX + l d,+d,=/ M nam tren van cifc tieu thi'r hai (ve phia gan O, hdn O 2 ) nhiThlnh ve. 0 1,25 / 1 I ^ d2-d,= k + - ^ = k + -(cm) k V 2y 2 do = - + 5,625 V a y : d| (,„)„) = 8 . 2 = 16 (cm), d2+di=S|S2 =10,75(cm) CO 0 1 = 7 d f - 0 A - nenOI,„i„= N / I 6 ' - 1 0 - =VF56(cm) ?.^> > ias v d i O < d . = - + 5,625 < 10,75 13.19. Dap an C ' 2 PhiWng trinh dao dong tdng hdp tai trung diem I cua SiS. la: - -ll,25