một số bài sóng cơ nâng cao.1

7/31/2019 Mt s bi sng c nng cao.1

1/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Cu 1: mt thong ca mt cht lng c hai ngun kt hp A, B cch nhau 10 cm, dao ng thphng thng ng vi phng trnh ln lt l uA = 3cos(40t + /6) (cm); uB = 4cos(40t2/3) (cm). Cho bit tc truyn sng l 40 cm/s. Mt ng trn c tm l trung im ca Anm trn mt nc, c bn knh R = 4cm. S im dao ng vi bin 5 cm c trn ng trn l

A. 30 B. 32 C. 34 D. 36Hng dn

Phng trnh sng ti M do sng ti A truyn n l:uAM = 3cos(40t +

6

- 1

2 d

)

Phng trnh sng ti M do sng ti B truyn n l:

uBM = 4cos(40t +2

3

- 2

2 d

)

Phng trnh sng tng qut tng hp ti M l:

uM = uAM + uBM = 3cos(40t +6

- 1

2 d

) + 4cos(40t +

2

3

- 2

2 d

)

Bin sng tng hp ti M l: (p dng cng thc dao ng iu ha) A = 2 2 2 1

2 2 23 4 2.3.4. os( ( ))

3 6

d dc

+ +

= 2 22 1

23 4 2.3.4. os( ( ))

2c d d

+ +

Bin sng tng hp ti M bng 5 khi: 2 12

os( ( ))2

c d d

= 0

Khi : 2 12

( )

2

d d

2 12 (

2

d d

) =

2

k

Do : d2 d1 = k2

;

M - 8 d2 d1 8 - 8 k2

8 - 8 k 8

Tng t ti hai im M v N hai u bn knh l im dao ng vi bin bng 5cmNn s im dao ng vi bin 5cm l: n = 17x2 2 = 32

Chn p n B

Cu 2 : mt nc c hai ngun sng c A v B cch nhau 15 cm, dao ng iu ha cng tn scng pha theo phng vung gc vi mt nc. im M nm trn AB, cch trung im O l 1cm, l im gn O nht lun dao ng vi bin cc i. Trn ng trn tm O, ng kn20cm, nm mt nc c s im lun dao ng vi bin cc i l

A. 18. B. 16. C. 32. D. 17.

Trang: 1/29 Email: [email protected]

A R = 4cm O B


2/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Hng dn

Sng ti M c bin cc i khi d2 d1 = kTa c d1 = 15/2 + 1,5 = 9cm; d2 = 15/2 1,5 = 6cmKhi d2 d1 = 3.

Vi im M gn O nht chn k = 1. Khi ta c: = 3S im dao ng vi bin cc i trn on AB l:

- S1S2 d2 d1 S1S2Hay -15 k 15 -5 k 5Vy s im dao ng vi bin cc i trn ng trn tm O bn knh 20cm l

n = 10x2 2 = 18 cc i ( y t A v B l hai cc i do ch c 8 ng cc i ct ngtrn ti 2 im, 2 cc i ti A v B tip xc vi ng trn)

Cu 3: Hai mi nhn S1, S2 cch nhau 9cm, gn u mt cu rung c tn s f = 100Hz c

cho chm nh vo mt mt cht lng. Vn tc truyn sng trn mt cht lng l v = 0,8 m/s. G nhcho cn rung th 2 im S1,S2 dao ng theo phng thng ng vi phng trnh dng: uacos2ft. im M trn mt cht lng cch u v dao ng cng pha S 1 , S2 gn S1S2 nht phng trnh dao ng.

Hng dn

Phng trnh sng tng qut tng hp ti M l:

uM = 2acos( 2 1d d

)cos(20t - 2 1

d d

+)

Vi M cch u S1, S2 nn d1 = d2. Khi d2 d1 = 0 cos( 2 1d d

) = 1 A = 2a

M dao ng cng pha vi S1, S2 th: 2 1d d

+= 2k

suy ra: 2 1 2d d k+ = 1 2 2d d

k

+ = v d1 = d2 = k


d1

d2

A S1

O

S2

B

S1

O S2

x

d1


3/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Gi x l khong cch t M n AB: d1 = d2 =2

2

2

ABx

+

= k

Suy ra ( )2

2

2

ABx k

=

= 20,64 9k ; ( = v/f = 0,8 cm)

Biu thc trong cn c ngha khi 20,64 9k 0 k 3,75Vi x 0 v khong cch l nh nht nn ta chn k = 4;

Khi 1 2 2 8d d

k

+= =

Vy phng trnh sng ti M l:uM = 2acos(200t - 8) = uM = 2acos(200t)

Cu 4: Hai ngun sng kt hp trn mt nc cch nhau mt on S 1S2 = 9 pht ra dao nu=cos(t). Trn on S1S2, s im c bin cc i cng pha vi nhau v ngc pha vi ngu(khng k hai ngun) l:

A. 8. B. 9 C. 17. D. 16.

Hng dnPhng trnh sng tng qut tng hp ti M l:

uM = 2cos( 2 1d d

)cos(20t - 2 1

d d

+)

Vi d1 + d2 = S1S2 = 9Khi : Phng trnh sng tng qut tng hp ti M l:

uM = 2cos(2 1d d

)cos(20t - 9) = 2cos(

2 1d d

)cos(20t - ) = - 2cos(2 1d d

)cos(20t)

Vy sng ti M ngc pha vi ngun khi cos( 2 1d d

) = 1

2 1d d

= k2 d1 - d2 = 2k

Vi - S1S2 d1 - d2 S1S2 -9 2k 94,5 k 4,5Suy ra k = 0; 1, 2; 3; 4. C 9 gi tr (c 9 cc i) Chn p n B

TUYN CHN CC CU HI HAY PHN SNG CCu 1 Khi xy ra hin tng giao thoa sng nc vi hai ngun kt hp ngc pha A, B. Nhng im trn mt n

nm trn ng trung trc ca AB s:A. dao ng vi bin ln nht B. Dao ng vi bin b nhtC. ng yn khng dao ng D. dao ng vi bn c gi tr trung bnhCu 2 Khong cch gia hai ngn sng lin tip l 5 (m). Mt thuyn my i ngc chiu sng th tn s va chca sng vo thuyn l 4 Hz. Nu i xui chiu th tn s va chm l 2 Hz. Tnh Tc truyn sng. Bit tc csng ln hn Tc ca thuyn.A. 5 m/s B. 14 m/s C. 13 m/s D. 15 m/sCu 3 Mt dy n hi rt di c u A dao ng theo phng vung gc vi si dy. Tc truyn sng trn dy4 (m/s). Xt mt im M trn dy v cch A mt on 40 (cm), ngi ta thy M lun lun dao ng lch pha so vi mt gc = (n + 0,5) vi n l s nguyn. Tnh tn s. Bit tn s f c gi tr trong khong t 8 Hz n 13 Hz.



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LTH 2011 GV: L Duy Khnh T: 0914683351A. 8,5 Hz B. 10 Hz C. 12,5Hz D. 12 HzCu 4 Sng truyn vi Tc 5 (m/s) gia hai im O v M nm trn cng mt phng truyn sng. Bit phtrnh sng ti O l u = 5.cos(5t - /6) (cm) v phng trnh sng ti im M l uM = 5.cos(5t + /3) (cm). Xc khong cch OM v cho bit chiu truyn sng.A. truyn t O n M, OM = 0,5 m B. truyn t M n O, OM = 0,5 mC. truyn t O n M, OM = 0,25 m D. truyn t M n O, OM = 0,25 mCu 5 Mt sng c hc c truyn theo phng Ox vi bin khng i. Phng trnh sng ti M c dng u2.sin(t + ) (cm). Ti thi im t1 li ca im M l 3 cm v ang tng th li ti im M sau thi im t 1 mkhong 1/6 (s) ch c th l gi tr no trong cc gi tr sauA. -2,5 cm B. -3 cm C. 2 cm D. 3 cmCu 6 Sng dng trn mt si dy n hi rt di c bc sng . Quan st ti 2 im A v B trn dy, ngi ta thA l nt v B cng l nt. Xc nh s nt v s bng trn on AB (k c A v B).A. s nt = s bng = 2.(AB/) + 0,5 C. s nt + 1 = s bng = 2.(AB/) + 1B. s nt = s bng + 1= 2.(AB/) + 1 D. s nt = s bng = 2.(AB/) + 1Cu 7 Mt si dy OM n hi di 90 cm c hai u c nh. Khi c kch thch trn dy hnh thnh 3 bng s(vi O v M l hai nt), bin ti bng l 3 cm. Ti N gn O nht c bin dao ng l 1,5 cm. Khong cch Onhn gi tr ng no sau y?A.10 cm B. 7,5 cm C. 5,2 cm D. 5 cm

Cu 8 Ti hai im A v B trn mt cht lng c hai ngun pht sng dao ng theo phng thng ng vi cphng trnh ln lt l u1 = a1cos(50t + /2) v u2 = a2cos(50t + ). Tc truyn sng trn mt cht lng l(m/s). Mt im M trn mt cht lng cch cc ngun ln lt l d1 v d2. Xc nh iu kin M nm trn cc (vi m l s nguyn)A. d1 - d2 = 4m + 2 cm B. d1 - d2 = 4m + 1 cm C. d1 - d2 = 4m - 1 cm D. d1 - d2 = 2m - 1 cmCu 9 Trn mt nc c hai ngun pht sng kt hp S1 v S2, dao ng theo cc phng trnh ln lt l: ua1cos(50t + /2) v u2 = a2cos(50t). Tc truyn sng ca cc ngun trn mt nc l 1 (m/s). Hai im P,thuc h vn giao thoa c hiu khong cch n hai ngun l PS1 - PS2 = 5 cm, QS1-QS2 = 7 cm. Hi cc im P,nm trn ng dao ng cc i hay cc tiu?A. P, Q thuc cc i B. P, Q thuc cc tiu C. P cc i, Q cc tiu D. P cc tiu, Q cc iCu 10 Trn mt nc hai ngun sng A v B dao ng iu ho theo phng vung gc vi mt nc vphng trnh: u1 = u2 = acos(10t). Bit Tc truyn sng 20 (cm/s); bin sng khng i khi truyn i. Mim N trn mt nc c hiu khong cch n hai ngun A v B tho mn AN - BN = 10 cm. im N nm trng ng ynA. th 3 k t trung trc ca AB v v pha A B. th 2 k t trung trc ca AB v v pha AB. th 3 k t trung trc ca AB v v pha B D. th 2 k t trung trc ca AB v v pha BCu 11 Trong mt th nghim v giao thoa sng trn mt nc, hai ngun kt hp cng pha A, B dao ng vi tn s= 20 Hz. Ti mt im M cch cc ngun A, B nhng khong 25 cm v 20 cm, sng c bin cc i. Gia M v trung trc ca AB c bn dy cc tiu. Tnh Tc truyn sng trn mt nc.A. 30 cm/s B. 40 cm/s C. 25 cm/s D. 60 cm/sCu 12 Trong mt mi trng vt cht n hi c hai ngun kt hp A v B cch nhau 10 cm, cng tn s. K

ti vng gia hai ngun ngi ta quan st thy xut hin 10 dy dao ng cc i v ct on S 1S2 thnh 11 om hai on gn cc ngun ch di bng mt na cc on cn li. Bit Tc truyn sng trong mi trng l (cm/s). Tnh tn s.A. 25 Hz B. 30 Hz C. 35 Hz D. 40 HzCu 13 Ti hai im A v B trn mt cht lng cch nhau 10 cm c hai ngun pht sng kt hp dao ng thphng trnh: u1 = acos(10t); u2 = bcos(10t + ). Tc truyn sng trn mt cht lng 20 (cm/s). Tm s cc titrn on AB.A.5 B. 6 C. 4 D. 3Cu 14 Hai ngun sng c AB cch nhau 21 cm dao ng theo cc phng trnh u1 = acos(4t); u2 = bcos(4+ ), lan truyn trong mi trng vi Tc 12 (cm/s). Tm s im dao ng cc i trn on thng ABA. 7 B. 8 C. 6 D. 5



5/29

LTH 2011 GV: L Duy Khnh T: 0914683351Cu 15 Ti hai im A v B trn mt cht lng cch nhau 15 cm c hai ngun pht sng kt hp dao ng thphng trnh: u1 = acos(40t); u2 = bcos(40t + ). Tc truyn sng trn mt cht lng 40 (cm/s). Gi E, F l him trn on AB sao cho AE = EF = FB. Tm s cc i trn on EF.A. 7 B. 6 C. 5 D. 4Cu 16 Ti hai im A v B trn mt nc cch nhau 16 cm c hai ngun pht sng kt hp dao ng thphng trnh: u1 = acos(30t); u2 = bcos(30t + /2). Tc truyn sng trn mt nc 30 (cm/s). Gi E, F l him trn on AB sao cho AE = FB = 2 cm. Tm s cc tiu trn on EF.A. 10 B. 11 C. 12 D. 13Cu 17 Trn mt nc c hai ngun kt hp A v B cch nhau 8 cm, dao ng theo phng trnh ln lt l= acos(8t); u2 = bcos(8t). Bit Tc truyn sng 4 cm/s. Gi C v D l hai im trn mt nc sao cho ABCDhnh ch nht c cnh BC = 6 cm. Tnh s im dao ng vi bin cc tiu trn on CD.A. 8 B. 9 C. 10 D. 11Cu 18 Trn mt cht lng c hai ngun kt hp A v B cch nhau 8 cm, dao ng theo phng trnh ln ll u1 = acos(8t); u2 = bcos(8t + ). Bit Tc truyn sng 4 cm/s. Gi C v D l hai im trn mt cht lng mABCD l hnh ch nht c cnh BC = 6 cm. Tnh s im dao ng vi bin cc i trn on CD.A. 8 B. 9 C. 10 D. 11Cu 19 Hai ngun kt hp S1 v S2 cch nhau mt khong l 11 cm u dao ng theo phng trnh uacos(20t) mm trn mt nc. Bit Tc truyn sng trn mt nc 0,4 (m/s) v bin sng khng i khi truyi. Hi im gn nht dao ng ngc pha vi cc ngun nm trn ng trung trc ca S1S2 cch ngun S1 bnhiu?A. 32 cm B. 18 cm C. 24 cm D. 6 cmCu 20 Hai ngun S1 v S2 dao ng theo cc phng trnh u1 = a1cos(90t) cm; u2 = a2cos(90t + /4) cm trmt nc. Xt v mt pha ng trung trc ca S1S2 ta thy vn bc k i qua im M c hiu s MS1-MS2 = 13cm v vn bc k + 2 (cng loi vi vn k) i qua im M' c MS1-MS2 = 21,5 cm. Tm Tc truyn sng trn mnc, cc vn l cc i hay cc tiu?A. 25cm/s, cc tiu B. 180cm/s, cc tiu C. 25cm/s, cc i D. 180cm/s, cc iCu 21 Hai ngun kt hp A v B dao ng trn mt nc theo cphng trnh: u1 = 2cos(100t + /2) cm; u2 = 2cos(100t) cm. Khi trn mt nc, to ra mt h thng vn githoa. Quan st cho thy, vn bc k i qua im P c hiu s PA-PB = 5 cm v vn bc k + 1 (cng loi vi vn k)qua im P c hiu s PA-PB = 9 cm. Tm Tc truyn sng trn mt nc. Cc vn ni trn l vn cc i hcc tiu.A. 150cm/s, cc i B. 180 cm/s, cc i C. 250cm/s, cc tiu D. 200cm/s, cc tiuCu 22 Mt ngun m O (coi nh ngun im) cng sut 4 (mW). Gi s ngun m v mi trng u hng, b qua s hp th m v phn x m ca mi trng. Cho bit ngng nghe v ngng au i vi m llt 10-11 (W/m2) v 10-3 (W/m2). nghe c m m khng c cm gic au th phi ng trong phm vi no trO?A. 1 m - 10000 m B. 1 m - 1000 m C. 10 m - 1000 m D. 10 m - 10000 mCu 23 Mt ci ci c coi nh ngun m im pht ra m phn b u theo mi hng. Cch ngun m

km mt ngi va nghe thy m. Bit ngng nghe v ngng au i vi m ln lt l 10-10

(W/m2

) v(W/m2). Hi cch ci bao nhiu th ting ci bt u gy cm gic au?A. 0,1 m B. 0,2 m C. 0,3 m D. 0,4 m

Bi.Ngun m ti O c cng sut khng i. Trn cng ng thng qua O c ba im A, B, C cng nm v mt phca O v theo th t xa c khong cch ti ngun tng dn. Mc cng m ti B km mc cng m ti A l

(dB), mc cng m ti B hn mc cng m ti C l 3a (dB). Bit OA =2

3OB. Tnh t s

OC

OA


L Quc Thnh


6/29

LTH 2011 GV: L Duy Khnh T: 0914683351

A.81

16B.

9

4C.

27

8D.

32

27BI GII :

Cng thc lin h cng m v cng sut ngun pht :2

PI

4d

=

Ta cn tnh :C

A

dOC

OA d=

- Mc cng m ti B km mc cng m ti A l a (dB)

a

A B A A 10A B

0 0 B B

I I I IaL L a 10lg 10lg a lg 10

I I I 10 I = = = = . (1)

- Mc cng m ti B hn mc cng m ti C l 3a (dB)

3a

CB B B 10B C

0 0 C C

II I I3aL L 3a 10lg 10lg 3a lg 10I I I 10 I = = = = . (2)

- Theo gi thit :B

A

d2 3OA OB

3 d 2= = .

- T (1)

2a a a

A B10 10 10

B A

I d 9: 10 10 10

I d 4

= = =

.

- T (1) v (2) suy ra :

2a 3a 2a 2a

CA B A10 10 5 5

B C C A

dI I I. 10 .10 10 10

I I I d

= = =

2 2a a

C 5 10

A

d 9 8110 10

d 4 16

= = = =

.

Cu 1 Trn mt nc c hai ngun kt hp S1, S2 cch nhau 6 2 cm dao ng theo phng trnh au 20cos=(mm). Bit tc truyn sng trn mt nc l 0,4 m/s v bin sng khng i trong qu trnh truyn. im gnht ngc pha vi cc ngun nm trn ng trung trc ca S1S2 cch S1S2 mt on: A. 6 cm. Bcm. C. 3 2 cm D. 18 cm.

Hng dn

Gi M l im dao ng ngc pha vi ngun

Phng trnh sng tng hp ti M l: uM = 2acos( 2 1d d

)cos(20t - 2 1

d d

+)

M dao ng ngc pha vi S1, S2 th: 2 1d d

+= (2k + 1) suy ra: ( )2 1 2 1d d k + = +

Vi d1 = d2 ta c: ( )2 1 2 1 2d d k

= = +



7/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Gi x l khong cch t M n AB: d1 = d2 =2

2 1 2

2

S Sx

+

= ( )2 12

k

+

Suy ra2 2

1 2(2 1)2 2

S Sx k

= +

= 24(2 1) 18k+ ; Vi = v/f = 4cm

Biu thc trong cn c ngha khi 24(2 1) 18k+ 0 k 0,56Vi x 0 v khong cch l nh nht nn ta chn k = 1 suy ra x = 3 2 cm; Chn p n C

Bi 1:Mt ngi quan st mt chic phao ni ln trn mt bin v thy n nh ln cao 6 ln trong 15 giy, coi sbin l sng ngang. Tnh chu k dao ng ca sng bin?A. 3(s) B.43(s) C. 53(s) D. 63(s)

Bi gii: Ch vi dng bi ny ta nn dng cng thc trc nghim:1n

ft

-= , trong t l thi gian dao n

Phao nh ln 6 ln trong 15 giy ngha l phao thc hin c 5 dao ng trong 15 giy.

Vy ta c1 6 1 1

( )

15 3

nf Hz

t

- -= = = suy ra

13( )T s

f= =

Bi 2 :Mt ngi quan st mt bin thy c 5 ngn sng i qua trc mt mnh trong khong thi gian 10(s) v c khong cch gia hai ngn sng lin tip l 5(m). Tnh vn tc sng bin ?A. 1(m) B. 2m C. 3m D.4m

Bi gii: Tng t nh trn ta c :1 5 1 2

( )10 5

nf Hz

t

- -= = = suy ra

2. .5 2( )

5v f ml= = = Ch khong c

gia hai ngn sng lin tip chnh l l

Cu 3:(H 2007). Mt ngun pht sng dao ng theo phng trnh u = acos20t (cm). Trong khong thi gi2(s) sng truyn I c qung ng bng bao nhiu ln bc sng?

A. 10. B. 20. C. 30. D. 40.

Bi gii: theo phng trnh trn ta thy 20w p= nn suy ra2 2

0,1( )20

T sp p

w p= = =

Do c 1 chu k th tng ng 1 bc sng, nn trong khong thi gian t=2(s) sng truyn c qung ng S. ta c t l0,1(s) l

Vy2(s) S

Hay0,1

2 S

l= suy ra S=20l

Cu 4:Mot song co tan so 500Hz, co toc o lan truyen 350m/s. Hai iem ganhau nhat tren phng truyen song phai cach nhau gan nhat mot khoang la b

nhieu e gia chung co o lech pha bang3

rad ?

A. 0,116m. B. 0,476m. C. 0,233m. D. 4,285m.

Bi gii: Ta bit : trong sng c th lch pha l2 .

3

dp pj

lD = = Suy ra

6d

l=

Trong :350

0,7( )500

vm

fl = = = vy khang cch cn tm l

0, 70,116( )

6 6d m

l= = =



8/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Cu 5: Mt sng m c tn s 450(Hz) lan truyn vi vn tc 360(m/s) trong khng kh. lch pha gia him cch nhau d=1(m) trn mt phng truyn sng l :A. 0, 5 ( )radj pD = B. 1, 5 ( )radj pD =

C. 2,5 ( )radj pD = D. 3, 5 ( )radj pD =Bi gii:

2 . 2. .12,5

0,8

dp pj p

l

D = = = ( trong 360

0,8( )

450

vm

f

l = = = )

Cu6: Vn tc truyn m trong khng kh l 340(m/s) , khong cchgia hai im gn nhau nht trn cng mphng truyn sng dao ng ngc pha nhau l 0,8(m). Tn s m l:A. f=85(Hz) B. f=170(Hz) C. f=200(Hz) D. f=225(Hz)

Bi gii: Ta bit 2 sng dao ng ngc pha khi lch pha2 .

(2. 1)d

kp

j pl

D = = +

Gn nhau nht th ly k=0 vy 2. 2.0, 85 1, 7( )d ml = = = hay340

200( )1,7

vf Hz

l= = =

Cu 7: Khi bin ca sng tng gp i, nng lng do sng truyn tng bao nhiu lA. Gim 1/4 B. Gim 1/2 C. Tng 2 ln D. Tng 4 ln

Bi gii: nng lng2.

2

k AE: Vy khi bin tng gp i th nng lng

2 2 2. ' .4' 4 4

2 2 2

k A k A KAE E: = = = Tng 4 ln

Cu 8: Hiu pha ca 2 sng ging nhau phi bng bao nhiu khi giao thoa sng hon ton trit tiu.A. 0 B. /4 C. /2 D

Bi gii: lch pha ca 2 sng ging nhau l : (2 1)kj pD = + th khi giao thoa chng mi trit tiu . Ly k=0

c j pD =

Cu 9: Tm vn tc sng m biu th bi phng trnh: u = 28cos(20x - 200A. 334m/s B. 331m/s C. 314m/s D. 100m/s

Bi gii: p dng phng trnh sng :2

. . ( )x

U A co s t p

wl

= - i chiu ln phng trnh trn ta thy2

20x

xp

l= s

ra2

20 10

p pl = = m

2000. ( ) .( ) 100

2 10 2v f

w pl l

p p= = = = ( m/s) ( Do 2000w= )

Cu 10: Mt mi nhn S c gn vo u ca mt l thp nm ngang v chm vo mt nc. Khi u

thp dao ng theo phng thng ng vi tn s f = 100 (Hz), S to trn mt nc mt sng c bin = 0,5 (cm). Bit khong cch gia 9 gn li lin tip l 4 (cm). Tnh vn tc truyn sng trn mt ncA. 100 cm/s B. 50 cm/s C. 100cm/s D. 150cm/s

Bi gii: p dng cng thc trc nghim khong cch gia n ngn sng lin tip l :

( 1)l n l= - Trong n l s ngn sng : ta c4

4 (9 1) 0, 58

l l= - = = (cm) Vy . 100.0, 5 50( / )v f cm sl= = =


1 9

l l l l l l l l


9/29

LTH 2011 GV: L Duy Khnh T: 0914683351Nhn vo hnh v ta thy t ngn sng th 1 n ngn sng th 9 cch nhau 8l

Cu11: (Bi tp tng t) : Ngun pht sng trn mt nc to dao ng vi tn s f=100(Hz) gy ra strn mt nc . Bit khong cch gia 7 gn li (bng sng lin tip) l 3(cm) . Tnh vn tc truyn s

trn mt nc ?A. 50(cm/s) B. 25(cm/s) C.100(cm/s) D.150(cm/s)

Bi gii: p dng cng thc trc nghim khong cch gia n ngn sng lin tip l :

( 1)l n l= - Trong n l s ngn sng : ta c3

3 (7 1) 0, 56

l l= - = = (cm) Vy . 100.0, 5 50( / )v f cm sl= = =

Cu12: Mt ngun sng c dao ng iu ho theo phng trnh

+=

210cos

tAx . Khong cch gi

hai im gn nhau nht trn phng truyn sng m ti dao ng ca cc phn t mi trng lch p

nhau2

l 5 (m). Hy tnh vn tc truyn sng.

A. 150m/s B. 120m/s C. 100m/s D. 200m/s

Bi gii: lch pha gia hai phn t trn phng truyn sng l:

2 2 .5

2 2

dp p p pj

l lD = = = Vy bc sng l: 20( )ml = suy ra vn tc truyn sng :

10

. .( ) 20.( ) 200( )2 2

m

v f s

w p

l l p p= = = =

Cu 13: Cho mt mi nhn S chm nh vo mt nc v dao ng iu ho vi tn s f = 20 (Hz). Ngta thy rng hai im A v B trn mt nc cng nm trn phng truyn sng cch nhau mt khong d10 (cm) lun dao ng ngc pha vi nhau. Tnh vn tc truyn sng, bit rng vn tc ch vkhong t 0,8 (m/s) n 1 (m/s).A. 100 m/s B. 90m/s C. 80m/s D. 85m/s

Bi gii: lch pha gia hai phn t theo phng truyn sng l:

2 (2 1)d kpj pl

D = = + (Do hai im dao ng ngc pha) vy ta c :

(2 1) (2 1)

2 2

k k vd

f

l+ += = Suy ra :

2 2.0,1.20 4

(2 1) 2 1 2 1

dfv

k k k= = =

+ + +Do gi thit c

vn tc thuc khong 0,8 1( )v m nn ta thay biu thc ca V vo :



10/29

LTH 2011 GV: L Duy Khnh T: 0914683351

40,8 1

(2 1)v

k =

+gii ra : 2 1 4k+ Suy ra : 1,5k

4

2 10,8

k+ Suy ra 2k

hay: 1, 5 2k do k thuc Z nn ly k=2 v thay vo biu thc4 4

0,8( )2 1 2.2 1v mk= = =+ +

Cu 14:. Mt si dy n hi rt di c u A dao ng vi tn s f v theo phng vung gc vi dy. Bin dao ng l 4 (cm), vn tc truyn sng trn y l 4 (m/s). Xt mt im M trn d

v cch A mt on 28 (cm), ngi ta thy M lun lun dao ng lch pha vi A mt gc (2 1)kjD = +

vi k = 0, 1, 2, Tnh bc sng . Bit tn s f c gi tr trong khong t 22 (Hz) n 26 (Hz).A. 8 cm B. 12 cm C. 14 cm D.16 cm


2(2 1)

2

dk

p pj

lD = = + (ch : bi ny ngi ta cho sn lch pha)

Tng t nh bi trn ta c :(2 1) (2 1)

4 4

k k vd

f

l+ += =

Suy ra : (2 1)4

vf k

d= + thay s vo ta c :

4 2 1(2 1)

4.0, 28 0, 28

kf k

+= + =

Do 22 26( )f Hz nn ta c : 2 122 26( )0,8

k Hz+

Gii ra ta c : 2, 58 3,14 3k k = vy2 1 2.3 1

25( )0, 28 0, 28

kf Hz

+ += = = v

40,16( ) 16

25

vm cm

fl = = = =

Cu15: Mt sng c hc truyn trong mt trng n hi.Phng trnh dao ng ca ngun c dn

4 ( )3

x cos t cm =

.Tnh bc sng . Cho bit vn tc truyn sng v = 40 (cm/s) Tnh lch pha c

hai im cch nhau mt khong 40 (cm) trn cng phng truyn sng v ti cng thi im.A. /12 B. /2 C. /3 D. /6


1( )

2 3.2 6f Hz

w p

p p= = = Suy ra

2 2 2 .40

40.6 3

d df

v

p p p pj

lD = = = =



11/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Cu 16: Mt sng c hc truyn trong mt trng n hi.Phng trnh dao ng ca ngun c dn

4 cos ( )3

x t cmp =

. Tnh lch pha ca dao ng ti cng mt im bt k sau khong thi gi

0,5 (s).

A.6

pB. /12 C. /3 D. /8

Bi gii: sau khong thi giant=0,5 giy sng truyn c qung ng d:

Phng trnh dao ng ti M cch ngun mt khong d l :

24 cos ( )

3M

dx t cm

p p

l

= - Trong

thi im (t) pha dao ng ca M l : 12

3

dt

p p

l

= - . Sau thi im t=0,5(s) th pha dao ng ti

lc ny l: 22

( 0, 5)3

dt

p pj

l

= + - Vy lch p

2 1

2 2( ( 0, 5) ) ( . )

3 3 6

d dt t

p p p p pj j j

l lD = - = + - - - =

Cu 17: Trong th nghim v hin tng giao thoa sng trn mt nc hai ngun kt hp Av B dao ng vi tn f=13(Hz) . Ti 1 im M cch ngun AB nhng khong d1=19(cm) v d2=21(cm) , sng c bin cc i. Giav ng trung trc ca AB khng c cc i no khc. Tnh vn tc truyn sng trn mt nc?A. 10(cm/s) B. 20(cm/s) C. 26(cm/s) D. 30(cm/s)

Bi gii: nhn xt do d1


12/29

LTH 2011 GV: L Duy Khnh T: 0914683351Cu 18: Trong th nghim v hin tng giao thoa sng trn mt nc hai ngun kt hp Av B dao ng vi tn f=13(Hz) . Ti 1 im M cch ngun AB nhng khong d1=16(cm) v d2=20(cm) , sng c bin cc i. Giav ng trung trc ca AB c 3 dy cc i khc. Tnh vn tc truyn sng trn mt nc?A. 26,7(cm/s) B. 20(cm/s) C. 40(cm/s) D. 53,4(cm/s)

Bi gii: Tng t M l mt cc i giao thoa v gia M vi ng trung trc ca AB c thm ba cc i khc tn

cng c 4 cc i, v d1


13/29

LTH 2011 GV: L Duy Khnh T: 0914683351

Bi 22:Sng truyn t A n M vi bc sng 60( )cml = M cch A mt on d=3(cm) . So vi sng ti A tsng ti M c tnh cht no sau y ?

A. ng pha vi nhau B. Sm pha hn mt lng3

2

p

C. Tr pha hn mt lng l p D. Mt tnh cht khc

Bi gii: Ta bit phng trnh sng cch ngun mt on l d l :

2cos( )M

dU a t

pw

l= - nu im M nm sau ngun A

(M chm pha hn A)

2cos( )M

dU a t

pw

l= + Nu im M nm trc ngun A

Theo gi thit ta c lch pha2 2 .30

60

dp pj p

lD = = =

Vy sng ti M tr pha hn sng ti A mt lng l

DNG BI TP XC NH S IM CC I, CC TIU TRN ON THNG AB

TH1: Nu 2 ngun AB dao ng cng pha 2 1 2k = = hoc hiu l: 1 2 =

Theo l thuyt giao thoa s gn sng quan st c trn on AB tng ng s im dao ng vi bin cc i tron AB . V vy hiu khong cch gia chng phi l 2 1d d kl- = Mt khc c bao nhiu ng hypepth tng ng trn on AB c by nhiu gn sng. Hay ta c th a im M xung nm trn on AB v lc nyc 1 2d d AB+ = Vy ta c h :

2 1d d kl- = (1) ly (1) +(2) v theo v ta c 22 2

k ABd

l= +

1 2d d AB+ = (2) do M thuc on AB nn 20 d AB< < Thay vo ta c


AMd

M

A BA BM

1d d


14/29

LTH 2011 GV: L Duy Khnh T: 0914683351

202 2

k ABd AB

l< = + < V rt ra

AB ABK

l l

-< < y chnh l cng thc trc nghim tm

im dao ng vi bin cc i trong giao thoa sng

T ng t s im dao ng vi bin cc tiu trn on AB tho mn:

2 1

2 1

(2 1)(3)2

(4)

d d k

d d AB

= + + =

lm tng t nh trn ta c :1 1

2 2

AB ABK

< < . y chnh l cng thc trc nghi

tnh s im dao ng cc tiu (ng yn) trn on AB.

TH2: Nu hai ngun AB dao ng ng c pha: 2 1 (2 1)k = = + hoc hiu l: 2 1 = th c

thc s im cc i l:1 1

2 2

AB ABK

< <

V cng thc s im cc tiu l:AB AB

Kl l

-< < ( Ngc vi cng pha)

Ch nu cc t s trn nguyn th ta ly du = . VD : 2 2K- cn khng nguyn th khng ly du =.

TH3: Nu hai ngun AB dao ng vung pha: 2 1 (2 1)2

k = = + th s im cc i v cc tiu t

on AB l bng nhau v bng:1 1

4 4

AB ABK

< <

Bi 23: Trn mt nc c hai ngun sng nc ging nhau cch nhau AB=8(cm). Sng truyn trn mt nc c bc s1,2(cm). S ng cc i i qua on thng ni hai ngun l:A. 11 B. 12 C. 13 D. 14

Bi gii:

Do A, B dao ng cng pha nn s ng cc i trn AB tho mn:

AB AB

Kl l

-

< I =

Vy chn C.

Cu11: : to hay nh ca mt m m tai cm nhn c s ph thuc vo

A. cng v bin ca m B. cng ca m v vn tc m

C. cng v tn s ca m. D. tn s ca m v vn tc m

Vy chn C.

C u 12: Pht biu no sau y khngng ?

A. Dao ng m c tn s trong min t 16Hz n 20kHz

B. V bn cht th sng m , sng siu m , sng h m u l sng c

C. Sng siu m l nhng sng m tai ngi khng th nghe thy c

D. Sng m l sng dc

Bi gii: Sng m l sng dc c tn s t 16Hz n 20KHz.

Nhng sng c tn s di 16Hz gi l sng h m v trn 20KHz gi l sng siu m.

Tai ngi khng th nghe c h m v siu m.

Chn p n C.

C u 13 : Pht biu no sau y l sai:

http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#12http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#34http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#40http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#40http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#32http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#16http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=193#42http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#3http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#12http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#34http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#40http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#40http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#32http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#16http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=193#42http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#3http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=194#2


26/29

LTH 2011 GV: L Duy Khnh T: 0914683351

A. Sng c hc l s lan truyn ca trng thi dao ng trong mi trng vt cht.

B. Sng ngang c phng dao ng vung gc vi phng truyn sng.

C. Qu trnh truyn sng l qu trnh truyn nng lng.

D. Sng c hc dc khng truyn c trong chn khng nhng sng c dc ngang truyn c trong chn khn

Bi gii: Sng c truyn c trong khng gian l do s n hi ca mi trng vt cht. Trong khng gian ko vt cht nh trong kh quyn Tri t nn sng c hc khng th truyn c vy D sai. Chn D.

C u 14: Mt sng c hc c phng trnh sng:

Bit khong cch gn nht gia 2 im c lch pha i vi nhau l 1m. Vn tc truyn sng l :

A.2,5(m/s) B. 5(m/s) C.10(m/s) D.20(m/s)

Bi gii: lch pha ca 2 im trn phng truyn sng l :

lch pha ca hai im cch nhau 1m l , ta c: Chn D

A.2,5(m/s) B. 5(m/s) C.10(m/s) D.20(m/s)

Cu 15:Hiu pha ca 2 sng ging nhau phi bng bao nhiu khi giao thoa sng hon ton trit tiu.

A. 0 B.4

pC.

2

pD.

Bi gii: Trong sng giao thoa 2 sng trit tiu nhau th

vi k = 0, 1 ,2 ,................ n nh vy vi k = 0 th

chn cu D l ng

Cu 16: Mt sng ngangtn s 100Hz truyn trn mt si dy nm ngang vi vn tc 60m/s. M v N l hai im trdy cch nhau 0,75m v sng truyn theo chiu t M ti N. Chn trc biu din li cho cc im c chiu dhng ln trn . Ti mt thi im no M c li m v ang chuyn ng i xung. Ti thi im N s c

v chiu chuyn ng tng ng l :A. m, i xung B. m, i ln C. d ng, i xung D. d ng, i ln

Bi gii: Bc sng : lchpha gia M v N :

dao ng ti M v N vungpha .

Do ti thi im N ang c li m v chuyn ng i ln.


M N

O X

+
http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=456http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=190#32http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=419#70http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=188#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=195http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=193#43http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=419#70http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#24http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#24http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=193#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=188#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#1http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=188#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#24http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=456http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=190#32http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=419#70http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=188#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=195http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=193#43http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=422#8http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=419#70http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#24http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#24http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=193#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=188#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#1http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=188#2http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=187#24http://www.onthi.com/?a=OT&ot=LT&hdn_lt_id=418#56


27/29

LTH 2011 GV: L Duy Khnh T: 0914683351Chn p n B. Nhn ln hnh v ta thy

M v N dao ng vung pha th khi M

i xung th im N phi i ln v

v c hai u ang nm di trc OX nn

lc ny c hai u ang c li

Hoc ta c th biu din qua chuyn ng trn u :

Khi M i xung N i ln trn ng trn th tng ng

lch pha ca M v N l gc MON gc ny vung

C u 17: Mt sng c hc truyn t im O ti M . O v M cch nhau mt on bng 5 lnbc sng . Dao ngO v M :

A. Cng pha B. Vung pha C. Ng c pha D. l ch pha3

4

p

Bi gii: M v O cch nhau mt s nguyn ln bc sng nn dao ng cng pha .

Chn p n A.

Cu 18: Thc hin giao thoa trn mt mt cht lng vi hai ngun ging nhau, cch nhau 13cm cng phng trnh dao ng l U = 2sin t. Vn tc truyn sng trn mt cht lng l 80cm/s. Xembin sng khgim khi truyn i t ngun. S im ng yn trn on l:

A. 4 B. 8 C. 5 D. 7

Bi gii: Bin dao ng tng hp ca im M bt k thuc l:

trong ln lt l di v

Gi s im M ng yn, ta c A=0, suy ra (1)

Li c cm

Vy (1) tng ng vi (k thuc Z)

Hay

M M thuc nn cm

T rt ra v 0<


28/29

LTH 2011 GV: L Duy Khnh T: 0914683351Chn C

C u 19: Hai ngun sng c AB cch nhau dao ng chm nh trn mt cht lng, cng tn s 100Hz, cng pha thphng vung vung gc vi mt cht lng. Vn tc truyn sng 20m/s.S im khng dao ng trn on AB=l :

A.11 im B. 20 im C.10 im D. 15 im

Bi gii: Bc sng20

0,2100

vm

fl = = = : Gi s im khng dao ng trn on AB l k , ta c

Suy ra 5,5 4,5k- < < vy k = -5;-4;-3;-2;-1;0;1;2;3;4 .

Hay C 10 im . Chn p n C.

C u 20: Hai ngun sng cdao ng cng tn s, cng pha .Quan st hin tng giao thoa thy trn on AB cim dao ng vibin cc i (k c A v B). S im khng dao ng trn on AB l:

A. 6 B. 4 C. 5 D. 2

Bi gii: Trong hin tng giao thoa sng trn mt cht lng , hai ngun dao ng cng pha th trn on AB , im dao ng vibin cc i s hn s im khng dao ng l 1.

Do s im khng dao ng l 4 im.Chn p n B.

Cu 20: Ti im M cch ngun sng v , sng c bin cc i. Bit rgia M v ng trung trc ca c mt ng dao ng mnh, tn s ca sng l f=15Hz. Tvn tc truyn sng trn mt nc.

A. 18 (cm/s) B. 24(cm/s) C. 36(cm/s) D. 30(cm/s)Bi gii:

Chn B.

C u 21: iu no ng khi ni v nng lng sng?

A. Trong qu trnh truyn sng th nng lng sng khng truyn i v n li lng bo ton

B. Qu trnh truyn sng l qu trnh truyn nng lng.

C. Khi sng truyn t mt ngun im trn mt phng, nng lng sng gim tl bnh phng vi qung truyn sng.

D. Khi sng truyn t mt ngun im trong khng gian, nng lng sng gim tl vi qung ng truysng. Bi gii: (B)

C u 22: iu no sau y l sai khi ni v nng lng ca sngc hc?

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LTH 2011 GV: L Duy Khnh T: 0914683351A. Khi sng truyn t mt ngun im trong khng gian, nng lng sng gim t l vi bnh phng qung truyn sng.

B. Nng lng sng lun lun khng i trong qu trnh truyn sng.

C. Khi sng truyn t mt ngun im trn mt phng, nng lng sng gim t l vi qung ng truyn sn

D. Qu trnh truyn sng l qu trnh truyn nng lng.

Bi gii: trong thc t, nng lng sng c hc s gim trong qu trnh truyn sng do mt mt nng lng ra bngoi mi trng (ma st) nn nng lng lun lun ko i trong qu trnh truyn sng l sai . chn B

C u 23: Khi bin ca sng tng gp i, nng lng do sng truyn thay i bao nhiu ln?

A. Gi m B. Gi m C. Khng thay i D. Tng 4 ln

Bi gii:Nng lng sng:

Cu 23: Sng m truyn trong thp vi vn tc 5000(m/s) . Hai im trong thp dao ng lch pha nhau 900 mgn nhau nht th cch nhau mt on 1,5(m). Tn s dao ng ca m l :

A. 833(Hz) B. 1666(Hz) C. 3,333(Hz) D. 416,5(Hz)Bi gii: lch pha2

2

dp pj

lD = = S

ra bc sng2 .1,5

6

2

mp

lp

= =m

5000833( )

6

vf Hz

l= = =
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