BUI 1: CHUYN SNG C

CHUYN SNG C

BUI 1: L THUYT V SNG CI. MC TIU :

1. Kin thc:

- Nu c nh ngha sng. Phn bit c sng dc v sng ngang.

- Gii thch c nguyn nhn to thnh sng.

- Nu c ngha ca cc i lng c trng cho sng c (bin , chu k, tn s, bc sng, vn tc truyn sng)

- Lp c phng trnh sng v nu c ngha ca cc i lng trong phng trnh.

2. Ky nng:

- Quan st v rt ra kt lun.

- Vit c phng trinh sng. Ve thi u theo t v u theo x .

3. Thai :

II.PHNG PHP :

1. Gio vin : Cac tranh ve 14.3; 14.4. Thit bi tao song nc trong hp bng kinh H14.1

2. Hc sinh : III. TIN TRNH:A.L THUYT:

1. Khi nim v sng c, sng ngang, sng dc?a. Sng c: l dao ng dao ng c lan truyn trong mt mi trngkhng truyn c trong chn khngc im:

- Sng c khng truyn c trong chn khng.

- Khi sng c lan truyn, cc phn t vt cht ch dao ng ti ch, pha dao ng v nng lng sng chuyn di theo sng.

- Trong mi trng ng tnh v ng hng, tc khng i.

b. Sng dc: l sng c c phng dao ng trng vi phng truyn sng. Sng dc truyn c trong cht kh, lng, rn. V d: Sng m trong khng kh.c. Sng ngang: l sng c c phng d vung gc vi phng truyn sng. Sng ngang truyn c trong cht rn v trn mt cht lng.

V d: Sng trn mt nc.

2. Cc c trng ca sng c:

a. Chu k (tn s sng): l i lng khng thay i khi sng truyn t mi trng ny sang mi trng khc.

b. Bin sng: l bin dng ca mt phn t c sng truyn qua.c. Tc truyn sng: l tc lan truyn dao ng trong mi trng; ph thuc bn cht mi trng () v nhit (nhit ca mi trng tng th tc lan truyn cng nhanh)

d. Bc sng ((m): : Vi v(m/s); T(s); f(Hz) ( (( m)C1: l khong cch gia hai im gn nhau nht trn phng truyn sng dao ng cng pha vi nhau.

C2: l qung ng sng lan truyn trong mt chu k

Ch : Trn vng trn lng gic:

e. Nng lng sng: Qtrnh truyn sng l qu trnh truyn nng lng.3. Ch :

+ S chu k bng s gn sng tr 1.

+ Khong cch gia hai ngn sng lin tip l .

+ Qung ng truyn sng: S = v.t+ Khong cch gia n ngn sng l (n 1)

4. Phng trnh truyn sng

a. Phng trnh d: um = Acos

vi d = MO th phng trnh sng phn x ti M l:

b. lch pha ca 2 d ti 2 im cch ngun:

+Nu 2 im M v N dao ng cng pha th:

. ( k ( Z )

+Nu 2 im M v N dao ng ngc pha th:

. ( k ( Z )

+Nu 2 im M v N dao ng vung pha th:

. ( k ( Z )

-Nu 2 im M v N nm trn mt phng truyn sng v cch nhau mt khong x th:

(Nu 2 im M v N trn phng truyn sng v cch nhau mt khong d th : (( = eq \f(2(d,() ) - Vy 2 im M v N trn phng truyn sng s:

+ dao ng cng pha khi:d = k( + dao ng ngc pha khi:d = (2k + 1) eq \f((,2)+ dao ng vung pha khi: d = (2k + 1) eq \f((,4)

vi k = 0, 1, 2 ...Ch :

+ Nu ngun kch thch bng dng in c tn s f th sng d vi 2f.

+ Hai im gn nhau nht cng pha cch nhau 1 bc sng

+ Hai im gn nhau nht ngc pha cch nhau na bc sng

+ Hai im gn nhau nht vung pha cch nhau mt phn t bc sng

B.CU HI L THUYTCu 1. Sng c

A. l dao ng lan truyn Trong mt mi trng.

B. l dao ng ca mi im Trong mi trng.

C. l mt dng chuyn ng c bit ca mi trng.

D. l s truyn chuyn ng ca cc phn t Trong mi trng

Chn p n ACu 2. phn loi sng ngang v sng dc ngi ta da vo

A. tc truyn sng v bc sng. B. phng truyn sng v tn s sng.

C. phng dao ng v phng truyn sng. D. phng dao ng v tc truyn sng.Chn p n D

Cu 3. Sng dc l sng c phng dao ng

A. nm ngang.

B. trng vi phng truyn sng.

C. vung gc vi phng truyn sng. D. thng ng.Chn p n BCu 4. Mt sng c hc lan truyn trn mt si dy n hi. Bc sng khng ph thuc vo

A. tc truyn ca sng. B. chu k dao ng ca sng.

C. thi gian truyn i ca sng. D. tn s dao ng ca sng. Chn p n CCu 5. Pht biu no sau y v i lng c trng ca sng c hc l khng ng?

A. Chu k ca sng chnh bng chu k dao ng ca cc phn t dao ng.

B. Tn s ca sng chnh bng tn s dao ng ca cc phn t dao ng.

C. Tc ca sng chnh bng tc dao ng ca cc phn t dao ng.

D. Bc sng l qung ng sng truyn i c Trong mt chu k.Chn p n CCu 6. Chu k sng l

A. chu k ca cc phn t mi trng c sng truyn qua.

B. i lng nghch o ca tn s gc ca sng

C. tc truyn nng lng Trong 1 (s).

D. thi gian sng truyn i c na bc sng. Chn p n ACu 7. Bc sng l

A. qung ng sng truyn Trong 1 (s).

B. khong cch gia hai im c li bng khng.

C. khong cch gia hai bng sng.

D. qung ng sng truyn i Trong mt chu k. Chn p n DCu 8. Sng ngang l sng c phng dao ng

A. nm ngang.

B. trng vi phng truyn sng.

C. vung gc vi phng truyn sng. D. thng ng.Chn p n CCu 9. Khi mt sng c hc truyn t khng kh vo nc th i lng no sau y khng thay i?

A. Tc truyn sng.

B. Tn s dao ng sng.

C. Bc sng.

D. Nng lng sng.Chn p n BCu 10. Tc truyn sng l tc

A. dao ng ca cc phn t vt cht. B. dao ng ca ngun sng.

C. truyn nng lng sng. D. truyn pha ca dao ng. Chn p n DCu 11. Tc truyn sng c hc gim dn Trong cc mi trng

A. rn, kh, lng. B. kh, lng, rn. C. rn, lng, kh. D. lng, kh, rn.Chn p n DCu 12. Tc truyn sng c hc tng dn Trong cc mi trng

A. rn, kh, lng. B. kh, lng, rn. C. rn, lng, kh. D. lng, kh, rn.Chn p nBCu 13. Tc truyn sng c hc ph thuc vo

A. tn s sng.

B. bn cht ca mi trng truyn sng.

C. bin ca sng.

D. bc sng.Chn p n BCu 14. Mt sng c hc lan truyn Trong mt mi trng tc v. Bc sng ca sng ny Trong mi trng l . Chu k dao ng ca sng c biu thc l

A. T = v/ B. T = v. C. T = /v D. T = 2v/Chn p n CCu 15. Mt sng c hc lan truyn Trong mt mi trng tc v. Bc sng ca sng ny Trong mi trng l . Tn s dao ng ca sng tha mn h thc

A. = v/ B. = v. C. = /v D. = 2v/Chn p n ACu 16. Mt sng c hc c tn s lan truyn Trong mt mi trng tc v. Bc sng ca sng ny Trong mi trng c tnh theo cng thc

A. = v/ B. = v. C. = /v D. = 2v/Chn p n ACu 17. Sng c lan truyn Trong mi trng n hi vi tc v khng i, khi tng tn s sng ln 2 ln th bc sng s

A. tng 2 ln. B. tng 1,5 ln. C. khng i.D. gim 2 ln.

Chn p n D v bc song t l nghich vi tn sC. BI TP V NH

Cu 1: Khi ni v sng c hc, nhn xt no sau y cha chnh xc:

A. Hm sng l hm tun hon theo thi gian.

B. Vn tc dao ng ca cc phn t bin thin tun hon.

C. Hm sng l hm tun hon theo khng gian.

D. Tc truyn pha dao ng bin thin tun hon.Chn p n dCu 2: Tc truyn sng trong mt mi trng

A. ph thuc vo bn cht mi trng v tn s sng

B. ph thuc vo bn cht mi trng v bin sng.

C. ch ph thuc vo bn cht mi trng. D. tng theo cng sng.

Chn p n CCu 3: Bc sng l

A. qung ng m mi phn t ca mi trng i c trong 1s.

B. khong cch gia hai phn t ca sng dao ng ngc pha.

C. khong cch gia hai phn t sng gn nht trn phng truyn sng d cng pha.

D. khong cch gia hai v tr xa nhau nht ca mi phn t ca sng.Chn p n CCu4: Chn cu tr li ng. phn loi sng ngang hay sng dc ngi ta da vo:

A. tc truyn sng v bc sng.

B. phng truyn sng v tn s sng.

C. phng truyn sng v tc truyn sng.

D. phng dao ng v phng truyn sng.Chn p n DCu 5: Chn cu tr li ng. Sng dc

A. ch truyn c trong cht rn.

B. truyn c trong cht rn, cht lng v cht kh.

C. truyn c trong cht rn, cht lng, cht kh v c trong chn khng

D. khng truyn c trong cht rn.Chn p n BCu 6: Sng (c hc) ngang

A. truyn c trong cht rn v trong cht lng.

B. khng truyn c trong cht rn.

C. truyn c trong cht rn, lng v kh.

D. truyn c trong cht rn v trn b mt cht lng.Chn p n DCu 7: Chn pht biu sai v qu trnh lan truyn ca sng c hc?

A. L qu trnh truyn nng lng.

B. L qu trnh truyn d trong mi trng vt cht theo thi gian.

C. L qu trnh lan truyn ca pha dao ng.

D. L qu trnh lan truyn ca cc phn t vt cht trong khng gian v theo thi gian.

Chn p n BCu 8: Mt chic phao nh ln cao 10 ln trong 36s, khong cch hai nh sng ln cn l 10m. Vn tc truyn sng l

A. (m/s) B. (m/s) C. 5(m/s)D. 2,5(m/s)Chn p n D CHUYN SNG C

BUI 2: BI TP XC NH CC I LNG C TRNG

I. MC TIU :

1. Kin thc:

-Nm c cc cng thc v cch gii cc bi tp.

2. Ky nng:

- K nng gii c cc bi tp xc nh cc i lng c trng ca song c.

II.PHNG PHP :

1. Gio vin : chun b cc bi tp

2. Hc sinh : Hc bi trc nhIII. TIN TRNH:1. Kin thc c bn:a) nh ngha sng c: Sng c dao ng an truyn trong mt mi trng vt cht.

b) Sng ngang: sng trong cc phn t ca mi trng dao ng theo phng vung gc vi phng truyn sng. Sng ngang (sng c) truyn trong cht rn v mt cht ng.

c) Sng dc: sng c trong cc phn t ca mi trng dao ng theo phng trng vi phng truyn sng. Sng dc truyn c c trong mi trg rn, ng, kh.

d) c trng ca sng hnh sin:

- Bin sng (U0): bin ca sng bin dao ng ca mt phn t mi trng c sng truyn qua.

- Chu k sng (T): thi gian sng an truyn c mt bc sng. Chu k sng bng vi chu k dao ng ca mt phn t ca mi trng c sng truyn qua.

- Tn s ca sng (f): s bc sng m sng an truyn c trong 1s. Tn s sng bng vi tn s dao ng ca phn t mi trng.

- Tc truyn sng (v): Tc truyn sng v tc an truyn dao ng trong mi trng. Vi mi mi trng tc c gi tr nht nh khng ph thuc vo tn s ca ngun sng.

- Bc sng (():

+ ( qung ng m sng truyn trong mt chu k.

+ Hoc khong cch gn nht ca hai im cng pha trn phng truyn sng. -Chu k (T), vn tc (v), tn s (f), bc sng (() lin h vi nhau :

; ; vi (s l qung ng sng truyn trong thi gian (t.

+ Quan st hnh nh sng c n ngn sng lin tip th c n-1 bc sng. Hoc quan st thy t ngn sng th n n ngn sng th m (m > n) c chiu di l th bc sng ; + S ln nh ln trn mt nc l N trong khong thi gian t giy th

- th truyn song

Bc 1: Chn im c bit (im C)

Bc 2: Chn 2 nh sng gn im c bit nht (A; B)

Bc 3: V mi tn t A hoc B song song vi mt phng cn bng, hng v C. Mi tn no chn chiu dao ng ti thi im ca C s l chiu truyn sng.

Nh hnh di l chiu t A n C.

2.Bi tp: A. CC I LNG C TRNG

BI TP C BN

V d 1: Mt ngi ngi b bin trng thy c 10 ngn sng qua mt trong 36 giy, khong cch gia hai ngn sng l 10m.. Tnh tn s sng bin.v vn tc truyn sng bin.

A. 0,25Hz; 2,5m/s B. 4Hz; 25m/s C. 25Hz; 2,5m/s D. 4Hz; 25cm/s

Hng dn gii: Xt ti mt im c 10 ngn sng truyn qua ng vi 9 chu k. T== 4s. Xc nh tn s dao ng. .Vn tc truyn sng: . p n AV d 2. Mt ngi quan st sng trn mt h thy khong cch gia hai ngn sng lin tip bng 2 m v c 6 ngn sng truyn qua trc mt Trong 8 (s). Tc truyn sng nc l

A. v = 3,2 m/s. B. v = 1,25 m/s. C. v = 2,5 m/s. D. v = 3 m/s.

Hng dn gii:Khong cch gia 2 ngn sng lin tip l nn ta c = 2 m.

6 ngn sng truyn qua tc l sng thc hin c 5 chu k dao ng, khi 5T = 8 T = 1,6 (s).

T , tc truyn sng l v = /T = 1,25 m/s chn p n B.V d 3: Hnh bn biu din sng ngang truyn trn mt si dy, theo chiu t tri sang phi. Ti thi im nh biu din trn hnh, im P c i bng 0, cn im Q c i cc i. Vo thi im hng chuyn ng ca P v Q n t s :

A. i xung; ng yn B. ng yn; i xung

C. ng yn; i n D. i n; ng ynHng dn gii

Tng t P i xung ,Q ng yn

BI TP NNG CAOV d 4: Mt sng ngang truyn trn b mt vi tn s f=10Hz. Ti mt thi im no mt phn mt ct ca nc c hnh dng nh hnh v. Trong khong cch t v tr cn bng ca A n v tr cn bng ca D 60cm v im C ang i xung qua v tr cn bng. Chiu truyn sng v tc truyn sng :

A. T A n E vi tc 8m/s. B. T A n E vi tc 6m/s.

C. T E n A vi tc 6m/s. D. T E n A vi tc 8m/s.Hng dn gii

Khong cch A n D l S= 3/4 =60 =80cm

V= .f=8m/s

Ta c C i xung th E i ln vy song truyn t E n A

P N DV d 5: Lc t=0 u O dy cao su cng ngang bt u dao ng vi T=2s ,A=5cm, to thnh song v=2cm/s. im M cch O mt khong 1,4 cm. Thi im u tin M n im thp nht l?

Hng dn gii:Thi gian truyn sng t O ti M : t1=S/v=1,4/2=0,7s

Thi gian M bt u dao ng ti v tr M thp nht : t2=T/2+T/4=1,5s

Vy t=t1+t2 =2,2s

C, BI TP T LUYN NHBI TP C BNBi 1: Ti mt im trn mt cht lng c mt ngun dao ng vi tn s 120Hz, to ra sng n nh trn mt cht lng. Xt 5 gn li lin tip trn mt phng truyn sng, v mt pha so vi ngun, gn th nht cch gn th nm 0,5m. Tc truyn sng l

A. 30 m/s

B. 15 m/s

C. 12 m/s

D. 25 m/s

Gii : 4( = 0,5 m ( ( = 0,125m ( v = 15 m/s (

p n BBi 2. Mt chic phao nh ln cao 10 ln trong 36s, khong cch hai nh sng ln cn l 10m. Vn tc truyn sng l

A. 25/9(m/s)

B. 25/18(m/s)

C. 5(m/s)

D. 2,5(m/s)Gii: Chn D HD: phao nh ln cao 10 ln trong 36s ( 9T = 36(s) ( T = 4(s)

Bi 3: Ngun pht sng S trn mt nc to dao ng vi tn s f = 100Hz gy ra cc sng trn lan rng trn mt nc. Bit khong cch gia 7 gn li lin tip l 3cm. Vn tc truyn sng trn mt nc bng bao nhiu?

A. 25cm/s.

B. 50cm/s. * C. 100cm/s. D. 150cm/s.

Chn B HD:

EMBED Equation.DSMT4 Bi 4: Mt sng ngang truyn trn b mt vi tn s f=10Hz. Ti mt thi im no mt phn mt ct ca nc c hnh dng nh hnh v. Trong khong cch t v tr cn bng ca A n v tr cn bng ca D 30cm v im C ang i xung qua v tr cn bng. Chiu truyn sng v tc truyn sng :

A. T A n E vi tc 1m/s. B. T A n E vi tc 6m/s.

C. T E n A vi tc 6m/s. D. T E n A vi tc 1m/s.Hng dn gii

Khong cch A n D l S= 3/4 =30 =10cm

V= .f=1m/s

Ta c C i xung th E i ln vy song truyn t E n ABi5: Mt sng truyn theo phng AB. Ti mt thi im no , hnh dng sng c dng nh hnh v. Bit rng im M ang i n v tr cn bng. Khi im N ang chuyn ng:

A. i n. B. i xung. C. ng yn. D. chy ngang.

Hng dn gii

p n A

BI NNG CAO

Bi 6: Sng co tn s 20Hz truyn trn cht long vi tc 200cm/s, gy ra cac dao ng theo phng thng ng cua cac phn t cht long. Hai im M va N thuc mt cht long cng phng truyn song cach nhau 22,5cm. Bit im M nm gn ngun song hn. Tai thi im t im N ha xung thp nht. Hi sau thi gian ngn nht la bao nhiu thi im M se ha xung thp nht?

A.

B.

C.

D.

Gii: + Ta c: = v/f = 10 cm . Vy M v N dao ng vung pha.

+ Tai thi im t im N ha xung thp nht th sau thi gian ngn nht la 3T/4 thi im M se ha xung thp nht. . Chn BBi 7: Sng truyn theo phng ngang trn mt si dy di vi tn s 10Hz. im M trn dy ti mt thi im ang v tr cao nht v ti thi im im N cch M 5cm ang i qua v tr c li bng na bin v i ln. Coi bin sng khng i khi truyn. Bit khong cch MN nh hn bc sng ca sng trn dy. Chn p n ng cho tc truyn sng v chiu truyn sng.

A. 60cm/s, truyn t M n N B. 3m/s, truyn t N n M

C. 60cm/s, t N n M D. 30cm/s, t M n N

Hng dn gii

Vi MN =5cm .suy ra ( c 2 trng hp:

(/6 =5 => (=30cm; =>Tc v=(.f =30.10=3m/s

5(/6 =5=> ( =6cm; =>Tc v=(.f =6.10 = 60 cm/s

Vy p n phi l : 3m/s, t M n N; hoc: 60cm/s, truyn t N n M.

Vi cho ta chn .p n CBi 8. Mt song ngang co chu ki T=0,2s truyn trong mt mi trng an hi co tc 1m/s. Xet trn phng truyn song Ox, vao mt thi im nao o mt im M nm tai inh song thi sau M theo chiu truyn song , cach M mt khoang t 42 n 60cm co dim N ang t vi tri cn bng i ln inh song . Khoang cach MN la:

A. 50cm B.55cm C.52cm D.45cm

Gii:Khi im M nh sng, im N v tr cn bng

ang i ln, theo hnh v th khong cch MN

MN = ( + k( vi k = 0; 1; 2; ...

Vi ( = v.T = 0,2m = 20cm

42 < MN = ( + k( < 60 -------> 2,1 0,75 < k < 3 0,75 ------> k = 2 Do MN = 55cm. Chn p n B

CHUYN SNG C

BUI 3,4: BI TP LIN QUAN N PHNG TRNH SNG C.

LCH PHA,LI VN TCI. MC TIU :

1. Kin thc:

-Nm c cc cng thc phng trnh truyn sng. - Nm c cng thc lch pha v li ,vn tc ca 1 im v hai im.

2. Ky nng:

- K nng gii c cc bi tp.

II.PHNG PHP :

1. Gio vin : chun b cc bi tp

2. Hc sinh : Hc bi trc nhIII. TIN TRNH1. Kin thc c bn:a. Phng trnh sng

Xt ti ngun O: c phng trnh sng : u0 = U0cos((t + )

Vit phng trnh dao ng ti M cch O mt on l d, trong mi trng c bc sng , c tc truyn sng l v

Sng truyn t O n M:

uM = U0cos(((t -(t)+) = U0cos((t+ -(t) = U0cos((t + - (d))

,v))

) = U0cos((t + - eq \s\don1(\f()

Nhn xt: Chiu truyn sng l chiu t im nhanh pha ti im tr pha

Phng trnh dao ng ti M: uM = U0cos((t + - ) c gi l phng trnh truyn sng

b. lch pha gia hai im

lch pha dao ng ca hai im trn phng truyn sng: =

Ta c cc trng hp sau:

+ Hai im cng pha (khng trng nhau) = = k2 ( |d| = k; k = 1; 2; 3..

im cng pha gn nht: k = 1 ( d =

im cng pha gn th hai: k = 2 ( d = 2

im cng pha th n: k = n ( d = n

+ Hai im ngc pha = = (2k+1) ( |d| = (k + ); k = 0; 1; 2; 3..

im ngc pha gn nht: k = 0 ( d = 0,5

im ngc pha gn th hai: k = 1 ( d = 1,5

im ngc pha th n: k = n-1 ( d = (k 0,5)

+ Hai im vung pha = = (2k+1) ( |d| = (k + ); k = 0; 1; 2; ..

im vung pha gn nht: k = 0 ( d = 0,25

im vung pha gn th hai: k = 1 ( d = 0,75

im vung pha th 3: k = 2 ( d = 1,25

Ch 1: Nu bi yu cu khong cch ca hai im lch pha gn nht ta c = ( d =

+ Hai im gn nht cng pha: = = 2 ( d =

+ Hai im gn nht ngc pha: = = ( d =

+ Hai im gn nht vun pha: = = ( d =

Ch 2: Cc im cng cch ngun mt on nh nhau th lun dao ng cng pha

c. Bi ton lien quan n li , vn tc:+ li vn tc cng mt im hai thi im-Phng trnh dao ng l uO =Acos(t),phng trnh vn tc v=u,= -Asint

-Da vo vng trn lng gic gii

-Trng hp c bit:

+ Li v vn tc hai thi im2. Bi tp:DNG 1: PHNG TRNH TRUYN SNGBI C BN

V d 1: Ti t = 0, u A ca mt si dy dao ng iu ha vi phng trnh u = 5cos(10t + /2) cm. Dao ng truyn trn dy vi bin khng i v tc truyn sng l v = 80 cm/s.

a) Tnh bc sng.b) Vit phng trnh dao ng ti im M cch A mt khong 24 cm.Hng dn gii:a) T phng trnh ta c = eq \s\don1(\f(,2)) = 5 Hz = eq \s\don1(\f(v,)) = eq \s\don1(\f(80,5)) = 16 cm/s.

b) Sng truyn t A n M nn dao ng ti M chm pha hn dao ng ti A khi A > M ( M = A - eq \s\don1(\f(2d,)) = (10t + eq \l(\f((,2))) - eq \s\don1(\f(2.24,16)) = 10t - eq \s\don1(\f(5,2)) uM = 5cos(10t - eq \s\don1(\f(5,2))) cm

Thi gian sng truyn t A n M l t = eq \s\don1(\f(d,v)) = 0,3(s)

Vy phng trnh dao ng ti M l uM = = 5cos(10t - eq \s\don1(\f(5,2))) cm, vi t 0,3 (s).

V d 2. Sng truyn t im M n im O ri n im N trn cng 1 phng truyn sng vi tc v = 20 m/s. Cho bit ti O dao ng c phng trnh uO = 4cos(2t /6) cm v ti hai im gn nhau nht cch nhau 6 m trn cng phng truyn sng th dao ng lch pha nhau gc 2/3 rad. Cho ON = 0,5 m. Phng trnh sng ti N l

A. uN = 4cos cmB. uN = 4cos cm

C. uN = 4cos cmD. uN = 4cos cm

Hng dn gii: T gi tht ta c = eq \s\don1(\f(2,3)) = eq \s\don1(\f(2d,)) ( eq \s\don1(\f(2,3)) = eq \s\don1(\f(2.6,)) = 18 m = eq \s\don1(\f(v,)) = eq \s\don1(\f(10,9)) Hz.

lch pha ca sng ti O v ti N l O/N = eq \s\don1(\f(2.ON,)) = eq \s\don1(\f(,18)) = eq \s\don1(\f(,18)) rad

Khi phng trnh dao ng ti N l uN = 4cos cm = 4cos cm

chn A.V d 3 Mt si dy n hi nm ngang c im u O dao ng theo phng ng vi bin A=5cm, T=0,5s. Vn tc truyn sng l 40cm/s. Vit phng trnh sng ti M cch O d=50 cm.

A.

B

C.

D

Gii: Phng trnh dao ng ca ngun:

Vi :

EMBED Equation.DSMT4 .Phng trnh dao ng tai M:

Trong : ;d= 50cm . .

BI NNG CAO

V d 4 Mt sng c hc truyn theo phng Ox vi bin coi nh khng i. Ti O, dao ng c dng u = acost (cm). Ti thi im M cch xa tm dao ng O l bc sng thi im bng 0,5 chu k th ly sng c gi tr l 5 cm?. Phng trnh dao ng M tha mn h thc no sau y:

A. B.

C. D.

Gii : Sng truyn t O n M mt mt thi gian l:t = = Phng trnh dao ng M c dng: .Vi v =(/T .Suy ra: Ta c: Vy Hay : V d 5. Mt sng ngang c phng trnh sng u = 6coscm, vi d c n v mt, t n v giy. Tc truyn sng c gi tr l

A. v = 100 cm/s. B. v = 10 m/s. C. v = 10 cm/s. D. v = 100 m/s.Hng dn gii:T phng trnh sng ta c:

u = 6coscm Acos ( ( v = = 100 cm/sDNG 2: LCH PHA GIA HAI IM TRN PHNG TRUYN SNG

BI C BN

V d 1. Mt sng c hc c tn s 45 Hz lan truyn vi tc 360 cm/s. Tnha) khong cch gn nht gia hai im trn phng truyn sng dao ng cng pha.

b) khong cch gn nht gia hai im trn phng truyn sng dao ng ngc pha.

c) khong cch gn nht gia hai im trn phng truyn sng dao ng vung pha.Hng dn gii:T gi tht ta tnh c bc sng = v/ = 360/45 = 8 cm.

a) Khong cch gn nht gia hai im dao ng cng pha l dmin = = 8 cm.

b) Khong cch gn nht gia hai im dao ng ngc pha l dmin = /2 = 4 cm.

c) Khong cch gn nht gia hai im dao ng vung pha l dmin = /4 = 2 cm.

V d 2. Mt sng c lan truyn vi tn s 50 Hz, tc 160 m/s. Hai im gn nhau nht trn cng phng truyn sng dao ng lch pha nhau l /4 th cch nhau mt khong

A. d = 80 cm. B. d = 40 m. C. d = 0,4 cm. D. d = 40 cm.Hng dn gii:

T gi tht ta c bc sng = 160/50 = 3,2 m.

Li c eq \s\don1(\f(,4)) = eq \s\don1(\f(2d,)) d = eq \s\don1(\f(,8)) = eq \s\don1(\f(320,8)) =40 cm. Vy d = 40 cm chn D.V d 3: Cho mt mi nhn S chm nh vo mt nc v dao ng iu ho vi tn s = 20 Hz. Ngi ta thy rng hai im A v B trn mt nc cng nm trn phng truyn sng cch nhau mt khong d = 10 cm lun dao ng ngc pha vi nhau. Tnh vn tc truyn sng, bit rng vn tc ch vo khong t 0,8 m/s n 1 m/s.Hng dn gii:

Hai im A v B dao ng ngc pha nn ta c = (2k + 1) ( eq \s\don1(\f(2d,)) = (2k + 1)

Thc hin php bin i ta c = eq \s\don1(\f(2d,2k+1)) ( eq \s\don1(\f(v,)) = eq \s\don1(\f(2d,2k+1)) ( v = eq \s\don1(\f(2d,2k+1))

Thay gi tr ca d = 10 cm, = 20 Hz vo ta c v = eq \s\don1(\f(400,2k+1)) cm/s = eq \s\don1(\f(4,2k+1)) m/s

Do 0,8 ( v ( 1 ( 0,8 ( eq \s\don1(\f(4,2k+1)) ( 1 ( eq \s\don1(\f(3,2)) ( k ( 2 ( Chn k = 2 ( v = 0,8 m/s = 80 cm/s

Vy tc truyn sng l v = 80 cm/s.

BI NNG CAO

V d 4: Sng ngang truyn trn mt cht lng vi tn s = 100 Hz. Trn cng phng truyn sng ta thy 2 im cch nhau 15 cm dao ng cng pha nhau. Tnh tc truyn sng, bit tc sng ny nm Trong khong t 2,8 m/s n 3,4 m/s.

A. v = 2,8 m/s. B. v = 3 m/s. C. v = 3,1 m/s. D. v = 3,2 m/s.Hng dn gii:

Hai im dao ng cng pha nn eq \s\don1(\f(2d,)) = k2 ( d = k = k.eq \s\don1(\f(v,)) v = eq \s\don1(\f(d,k))

M 2,8 (m/s) ( v ( 3,4 (m/s) ( 2,8 ( eq \s\don1(\f(,k)) = eq \s\don1(\f(15,k)) ( 3,4 ( k = 5 ( v = 3 m/s

Vy chn p n B.V d 5: Mt sng c c bc sng , tn s f v bin a khng i, lan truyn trn mt ng thng t im M n im N cch M 19/12. Ti mt thi im no , tc dao ng ca M bng 2(fa, lc tc dao ng ca im N bng:A. (fa B. fa C. 0 D. (fa

Hng dn gii:Dng trc Ou biu din pha dao ng ca M thi im t (vec t quay ca M)Tai thi im t, im M c tc dao ng M bng 2(fa

M v tr cn bng (hnh v): MN =

thi im t: N tr pha hn M mt gc : =

Quay ngc chiu kim ng h mt gc ta c vc t quay ca N

Chiu ln trc Ou/ ta c u/N = == (fa. Chn BDNG 3: LI V VN TCBi c bn

V d 1: Mt sng c c pht ra t ngun O v truyn dc theo trc Ox vi bin sng khng i khi i qua hai im M v N cch nhau MN = 0,25( (( l bc sng). Vo thi im t1 ngi ta thy li dao ng ca im M v N ln lt l uM = 4cm v uN = (4 cm. Bin ca sng c gi tr l

A. . B. . C. . D. 4cm.

Gii: Bc sng l qung ng vt c trong 1 T

MN = 0,25(, tc t M n c N l T/4 , hay gc MON = /2= 900M Vo thi im t1 ngi ta thy li dao ng ca im M v N ln lt l

uM = 4cm v uN = (4 cm.

Suy ra Ch c th l M, N i xng nhau nh hnh v v gc MOA = 450

Vy bin M: UM = U0 /= 4 . Suy ra UO = . Chn CV d 2: Mt ngun O dao ng vi tn s f = 50Hz to ra sng trn mt nc c bin 3cm(coi nh khng i khi sng truyn i). Bit khong cch gia 7 gn li lin tip l 9cm. im M nm trn mt nc cch ngun O on bng 5cm. Chn t = 0 l lc phn t nc ti O i qua v tr cn bng theo chiu dng. Ti thi im t1 li dao ng ti M bng 2cm. Li dao ng ti M vo thi im t2 = (t1 + 2,01)s bng bao nhiu ?

A. 2cm. B. -2cm.

C. 0cm.

D. -1,5cm.

Phng trnh truyn sng t ngun O n M cch O on x theo chiu dng c dng:

.

Theo gi thit: ,

im M tai thi im .

Vy sng ti hai thi im trn c li ngc pha nhau nn .p n B.V d 3: Sng lan truyn t ngun O dc theo 1 ng thng vi bin khng i. thi im t = 0 , im O i qua v tr cn bng theo chiu (+). thi im bng 1/2 chu k mt im cch ngun 1 khong bng 1/4 bc sng c li 5cm. Bin ca sng l A. 10cm B. 5cm C. 5 cm D. 5cm

Gii: Biu thc ca ngun sng ti O: u0 = acos(t - ) (cm)

Biu thc ca sng ti M cch O d = OM uM = acos(t - ) (cm)

Vi : du (+) ng vi trng hp sng truyn t M ti O;

du (-) ng vi trng hp sng truyn t O ti M

Khi t = T/2; d = (/4 th uM = 5 cm => acos(t - )

=> acos(

EMBED Equation.3 - ) = a cos( ) = a = 5 Do a > 0 nn a = 5 cm. Chn DBi nng caoV d 4: Hai im M, N cng nm trn mt phng truyn sng cch nhau x = /3, sng c bin A, chu k T. Ti thi im t1 = 0, c uM = +3cm v uN = -3cm. thi im t2 lin sau c uM = +A, bit sng truyn t N n M. Bin sng A v thi im t2 l

A. v

B. v

C.v

D. v

Gii: + Ta c lch pha gia M v N l: , + T hnh v, ta c th xc nh bin sng l: A = (cm)

+ thi im t1, li ca im M l uM = +3cm, ang gim. n thi im t2 lin sau , li ti M l uM = +A.

+ Ta c

vi:

Vy: . Chon A.V d 5: Mt sng c lan truyn trn si dy vi chu k T, bin A. thi im t0 , ly cc phn t ti B v C tng ng l -24 mm v +24 mm; cc phn t ti trung im D ca BC ang v tr cn bng. thi im t1, li cc phn t ti B v C cng l +10mm th phn t D cch v tr cn bng ca n

A.26mm B.28mm C.34mm D.17mm

Gii :

* Tai t1 ta co cac vi tri B, D, C nh hinh 1,

nh vy khoang cach BC= 24.2= 48 mm

* Tai t2 ta co cac vi tri B, D, C nh hinh 2. Khong cch BC= 48mm khng i

B va C co cung li 10 mm nn:

OH = 10 mm;BH= 0,5.BC = 24mm

Vy :

3.bi tp v nhDNG 1:Bi c bnBi1: Mt sng c truyn dc theo trc Ox c phng trnh l (cm), vi t o bng s, x o bng m. Tc truyn sng ny l

A. 3 m/s. B. 60 m/s. C. 6 m/s. D. 30 m/s.

Gii: Phng trnh c dng .Suy ra: ;

= (x => v = = 2.3 = 6(m/s) p n C

Bi 2: Sng c truyn trong mt mi trng dc theo trc Ox vi phng trnh u = cos(20t - 4x) (cm) (x tnh bng mt, t tnh bng giy). Vn tc truyn sng ny trong mi trng trn bng

A. 5 m/s.

B. 4 m/s.

C. 40 cm/s.

D. 50 cm/s. Gii:+ Ta c:

Bi 3: Trn mt si dy di v hn c mt sng c lan truyn theo phng Ox vi phng trnh sng u = 2cos(10t - x) (cm) ( trong t tnh bng s; x tnh bng m). M, N l hai im nm cng pha so vi O cch nhau 5 m. Ti cng mt thi im khi phn t M i qua v tr cn bng theo chiu dng th phn t N

A. i qua v tr cn bng theo chiu dng.

B. i qua v tr cn bng theo chiu m.

C. v tr bin dng.

D. v tr bin m.Ta c : = (x ( ( = 2 m. Trong bi MN = 5 m = 2,5( ( M v N dao ng ngc pha nhau.Bi 4: Cho phng trnh sng: (m, s). Phng trnh ny biu din:

A. Sng chy theo chiu m ca trc x vi vn tc (m/s)

B. Sng chy theo chiu dng ca trc x vi vn tc (m/s)

C. Sng chy theo chiu dng ca trc x vi vn tc 17,5 (m/s)

D. Sng chy theo chiu m ca trc x vi vn tc 17,5 (m/s

Gii:* Cng thc vng tnh lch pha ca 2 im cch nhau dc theo 1 phng truyn l:

* Nu ti O l ( PT dao ng ti M :

* p dng: Ta c phng trnh tng qut :

Ta so snh PT ca bi cho: (m, s)

( ( v=17,5 m/s

Ta nhn du ca ko phi l tr m l cng ( sng truyn ngc chiu dng. Chn D

Bi5. Sng truyn t O n M vi vn tc v=40cm/s, phng trnh sng ti O l u= 4sint(cm). Bit lc t th li ca phn t M l 3cm, vy lc t + 6(s) li ca M l

A. -3cm

B. -2cm

C. 2cm

D. 3cm

Gii: Chn A.T= 4s => 3T/2 =6s ( Li ca M lc t + 6 (s) l -3cmBI NNG CAOBi 6: Ngun sng O dao ng vi tn s 10Hz, dao ng truyn i vi vn tc 0,4m/s theo phng Oy; trn phng ny c hai im P v Q vi PQ = 15cm. Bin sng bng a = 1cm v khng thay i khi lan truyn . Nu ti thi im t no P c li 1cm th li ti Q l

A. 1cm B. -1cmC. 0D. 2cm

= 4cm; lc t, uP = 1cm = acost cost =1

uQ = acos(t - ) = acos(t - )= acos(t -7,5) = acos(t + 8 -0,5)

= acos(t - 0,5) = asint = 0

Bi7: Mt sng c lan truyn t ngun O, dc theo trc Ox vi bin sng khng i, chu k sng T v bc sng . Bit rng ti thi im t = 0, phn t ti O qua v tr cn bng theo chiu dng v ti thi im t = phn t ti im M cch O mt on d = c li l -2 cm. Bin sng l

A. 4/ cmB. 2

C. 2 cm D. 4 cm

Bi8: Mt ngun O pht sng c dao ng theo phng trnh: ( trong u(mm),t(s) ) sng truyn theo ng thng Ox vi tc khng i 1(m/s). M l mt im trn ng truyn cch O mt khong 42,5cm. Trong khong t O n M c bao nhiu im dao ng lch pha vi ngun?

A. 9

B. 4

C. 5

D. 8

Gii : Ta c pha ca mt im M bt k trong mi trng c sng truyn qua:

M l im lch pha vi O mt gc nn ta c:

(v M tr pha hn O nn loi trng hp ). Vy c tt c 4 im lch pha i vi O

DNG 2:

BI C BNBi 1: Sng ngang truyn trn mt cht lng vi tn s = 100 Hz. Trn cng phng truyn sng ta thy 2 im cch nhau 15 cm dao ng cng pha nhau. Tnh tc truyn sng, bit tc sng ny nm Trong khong t 2,8 m/s n 3,4 m/s.

A. v = 2,8 m/s. B. v = 3 m/s. C. v = 3,1 m/s. D. v = 3,2 m/s.Hng dn gii:

Hai im dao ng cng pha nn eq \s\don1(\f(2d,)) = k2 ( d = k = k.eq \s\don1(\f(v,)) v = eq \s\don1(\f(d,k))

M 2,8 (m/s) ( v ( 3,4 (m/s) ( 2,8 ( eq \s\don1(\f(,k)) = eq \s\don1(\f(15,k)) ( 3,4 ( k = 5 ( v = 3 m/s

Vy chn p n B.Bi 2: Mt sng ngang truyn trn trc Ox c m t bi phng trnh u = 0,5cos(50x 1000t) cm, Trong x c n v l cm. Tc dao ng cc i ca phn t mi trng ln gp bao nhiu ln tc truyn sng?

A. 20 ln. B. 25 ln. C. 50 ln. D. 100 ln.Hng dn gii:

Tc cc i ca phn t mi trng l vmax = A = 1000.0,5 = 500 cm/s.

Tc truyn sng l = 1000/50 = 20 cm/s ( tc ca phn t mi trng c sng truyn qua gp 25 ln tc truyn sng.

Bi 3: Hai im cng nm trn phng truyn sng cch nhau . Ti thi im c v . Tnh bin sng A?

A:

B:

C:

D:

Gii:

Gc lch pha gia M, N:

= = = >

nhn M hoc P

nhn N hoc P v >

nn nhn 2 im M, N (nh hnh)

Vy cos = cos = =

A = 3 cm, p n ABi 4: Mt sng c hc c c truyn theo phng OX vi tc 20 . Cho rng khi truyn sng bin khng i . Bit phng trnh sng ti O l: , di sng ti M cch O 40 lc di sng ti O t cc i l:

A: 4

B: 0

C: -2

D: 2

Gii:

Bc sng = = 240(cm)

Gc lch pha O, M: =

V tr ca M (hv). Li ca M: uM = Acos = 2 cm

Bi 5: Trn mt nc ti hai im AB c hai ngun sng kt hp dao ng cng pha, lan truyn vi bc sng . Bit AB = 11. Xc nh s im dao ng vi bin cc i v ngc pha vi hai ngun trn on AB( khng tnh hai im A, B)

A. 12B. 23C. 11D. 21

M cc i th

M cc i cng pha ngun th

M cc i ngc pha ngun th

Yu cu bi ton suy ra suy ra c 11 gi tr ca

BI NNG CAO

Bi6: Mt song ngang co chu ki T=0,2s truyn trong mi trng an hi co tc 1m/s. Xet trn phng truyn song Ox, vao mt thi im nao o mt im M nm tai inh song thi sau M theo chiu truyn song, cach M mt khoang t 42cm n 60cm co im N ang t vi tri cn bng i ln inh song . Khoang cach MN la:

A. 50cm B.55cm C.52cm

D.45cm

Gii: Khi im M nh sng, im N v tr cn bng ang i ln, theo hnh v th khong cch MN

MN = ( + k( vi k = 0; 1; 2; ...Vi ( = v.T = 0,2m = 20cm

42 < MN = ( + k( < 60 => 2,1 0,75 < k < 3 0,75 => k = 2. Do MN = 55cm. Chn BBi 7: Mt ngun dao ng iu ho vi chu k 0,04s. Vn tc truyn sng bng 200cm/s. Hai im nm trn cng mt phng truyn sng v cch nhau 6 cm, th c lch pha:

A. 1,5(.

B. 1(.

C.3,5(.

D. 2,5(.

Gii: Chn A HD: . lch ch pha:

Bi8: Mt ngun 0 pht sng c c tn s 10hz truyn theo mt nc theo ng thng vi V = 60 cm/s. Gi M v N l im trn phng truyn sng cch 0 ln lt 20 cm v 45cm. Trn on MN c bao nhiu im dao ng lch pha vi ngun 0 gc / 3.

A. 2 B. 3 C. 4 D. 5

Gii: - lch pha ca ngun 0 v im cch n mt khong d l :

- lch pha /3 th v:c 4 im

Bi 9: AB l mt si dy n hi cng thng nm ngang, M l mt im trn AB vi AM=12,5cm. Cho A dao ng iu ha, bit A bt u i ln t v tr cn bng. Sau khong thi gian bao lu k t khi A bt u dao ng th M ln n im cao nht. Bit bc sng l 25cm v tn s sng l 5Hz.

A. 0,1sB. 0,2s.C. 0,15sD. 0,05s

Gii: C (=25 cm; f=5Hz; v=125 cm/s

DNG 3BI C BN

Bi 1: Hai im M, N cung nm trn mt phng truyn song cach nhau (/3. Tai thi im t, khi li dao ng tai M la uM = +3 cm thi li dao ng tai N la uN = 0 cm. Bin song bng:

A. A = cm.

B. A = 3 cm.

C. A = 2cm.

D. A = 3cm.Ta c th vit: uM = Acos((t) = +3 cm (1), uN = Acos((t - ) = 0 cm (2)

T (2) ( cos((t - ) = 0 ( (t - = , k ( Z ( (t = + k(, k ( Z.

Thay vo (1): Acos(+ k() = 3. Do A > 0 nn Acos(- () = Acos() = = 3 (cm) ( A = 2cm.Bi 2: Mt sng ngang tn s 100 Hz truyn trn mt si dy nm ngang vi vn tc 60 m/s. M v N l hai im trn dy cch nhau 0,15 m v sng truyn theo chiu t M n N. Chn trc biu din li cho cc im c chiu dng hng ln trn. Ti mt thi im no M c li m v ang chuyn ng i xung. Ti thi im N s c li v chiu chuyn ng tng ng lA. m; i xung. B. m; i ln.

C. Dng; i xung.

D. Dng; i ln. ( = = = 0,6 m. Trong bi MN = 0,15 m = , do sng truyn t M n N nn dao ng ti M sm pha hn dao ng ti N mt gc (/2 (vung pha). Dng lin h gia dao ng iu ha v chuyn ng trn u.Chn CBi 3: Ngun sng O dao ng vi tn s 10 Hz , dao ng truyn i vi vn tc 0,4 m/s trn phng Ox . Trn phng ny c 2 im P v Q theo chiu truyn sng vi PQ = 15 cm. Cho bin sng a = 1 cm v bin khng thay i khi sng truyn. Nu ti thi im no P c li 1 cm th li ti Q l:

A. 1 cm B. 1 cm C. 0

D. 0,5 cmTnh c ( = 4 cm ; = 3,75 hay PQ = 3( + 0,75( ; (( = 2(. = 7,5( hay (( = 0,75.2( =

(Nh: ng vi khong cch ( th lch pha l 2( ; ng vi 0,75( th (( = 0,75.2( = ).

( dao ng ti P sm pha hn dao ng ti Q mt gc hay dao ng ti P tr pha hn dao ng ti Q mt gc . ( Lc uP = 1 cm = a th uQ = 0.Bi 4: Mt sng c lan truyn trn si dy vi chu k T, bin A. thi im t0 , ly cc phn t ti B v C tng ng l -24 mm v +24 mm; cc phn t ti trung im D ca BC ang v tr cn bng. thi im t1, li cc phn t ti B v C cng l +10mm th phn t D cch v tr cn bng ca n

A.26mm B.28mm C.34mm D.17mmGii 1: T thi im t0 n t1 :

+ vc t biu din d ca B quay gc B00B1 = ( - (( + () + vc t biu din d ca C quay gc C00C1= (( + ()

=> Ta c : (t = t1 t0 =

=> ( = 2() => = ( /2

+ Ta c : cos( = sin =

=> 24/A = => A = 26 cm

Bi5 Mt sng c hc lan truyn dc theo mt ng thng vi bin sng khng i c phng trnh sng ti ngun O l: u = A.cos(t - /2) cm. Mt im M cch ngun O bng 1/6 bc sng, thi im t = 0,5/ c ly cm. Bin sng A l:

A. 2 (cm) B. 2(cm)

C. 4 (cm)

D. (cm)

Gii:

BI NNG CAOBi 6: Mt sng c hc lan truyn dc theo 1 ng thng c phng truyn sng ti ngun O l :

uo = Acos(t + ) (cm). thi im t = 1/2 chu k mt im M cch ngun bng 1/3 bc sng c dch chuyn uM = 2(cm). Bin sng A l

A. 4cm. B. 2 cm. C. 4/cm. D. 2 cmGii: Biu thc ca ngun sng ti O: uo = Acos(t + ) (cm).

Biu thc ca sng ti M cch O d = OM uM = Acos(t + ) (cm)

Vi : du (+) ng vi trng hp sng truyn t M ti O;

du (-) ng vi trng hp sng truyn t O ti M

Khi t = T/2; d = (/3 th uM = 2 cm

uM = Acos(t + ) = Acos(

EMBED Equation.3 + ) = Acos() = 2 cm

=> Acos() = Acos() = 2 (cm) => A= 4/cm. Chn C => Acos() = 2 (cm) => A < 0

Bi 7: Mt sng c hc lan truyn trn mt phng truyn sng vi vn tc v = 50cm/s. Phng trnh sng ca mt im O trn phng truyn sng l: u0 = acos(t) cm. thi im t = 1/6 chu k mt im M cch O khong (/3 c dch chuyn uM = 2 cm. Bin sng alA. 2 cm. B. 4 cm. C. 4/ cm D. 2 cm. Gii: Biu thc ca ngun sng ti O: uo = acos(t ) (cm).

Biu thc ca sng ti M cch O d = OM uM = acos(t ) (cm)

Vi : du (+) ng vi trng hp sng truyn t M ti O;

du (-) ng vi trng hp sng truyn t O ti M

Khi t = T/6; d = (/3 th uM = 2 cm

uM = acos(t ) = acos(

EMBED Equation.3 )

=> acos( = - a = 2 cm => a < 0 loi => acos(-) = 2 (cm) => a = 4cm. Chn BBi 8: Sng lan truyn t ngun O dc theo 1 ng thng vi bin khng i. thi im t = 0 , im O i qua v tr cn bng theo chiu (+). thi im bng 1/2 chu k mt im cch ngun 1 khong bng 1/4 bc sng c li 5cm. Bin ca sng l A. 10cm B. 5cm C. 5 cm D. 5cmGii: Biu thc ca ngun sng ti O: u0 = acos(t - ) (cm)

Biu thc ca sng ti M cch O d = OM uM = acos(t - ) (cm)

Vi : du (+) ng vi trng hp sng truyn t M ti O;

du (-) ng vi trng hp sng truyn t O ti M

Khi t = T/2; d = (/4 th uM = 5 cm => acos(t - )

=> acos(

EMBED Equation.3 - ) = a cos( ) = a = 5 Do a > 0 nn : a = 5 cm. Chn D

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

EMBED Equation.DSMT4

d1

0

NN

d

d2

M

M

N

t

((

M

M2

M1

u(cm)

N

A

3

-3

(

((

-A

v

M

N

P

Q

EMBED Equation.3

3

-3

A

O

M

EMBED Equation.3

A EMBED Equation.3

ly k=0

- 24

24

10

A

B0

B1

C1

C0

(

(

(

D

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