Tai Lieu Luyen Thi Tot Nghiep

8/2/2019 Tai Lieu Luyen Thi Tot Nghiep

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Ti liu n thi tt nghip mn Vt L 12 nm hc 2011 - 2012

DAO NG C

I/. Dao ng iu ha1. Chuyn ng ca vt qua li quanh v tr cn bng gi l dao ng c. V tr cn bng l v tr ca

vt khi ng yn.2. Khi vt dao ng, nu sau nhng khong thi gian bng nhau, gi l chu k, vt tr li v tr c theo

hng c th dao ng ca vt gi l dao ng tun hon.3. Dao ng iu ha l dao ng trong li ca vt l mt hm cosin (hay sin) ca thi gian.

4. Phng trnh dao ng iu ha ( )x A cos t= + Trong A, , l nhng hng s.

x l li dao ng, xmax = A A l bin dao ng, A > 0.

( )t + l pha ca dao ng ti thi im t (rad) l pha ban u (rad).

l tn s gc2

2 fT

= = (rad/s). ( )x Asin t A cos t

2

= + = +

5. Chu k l khong thi gian vt thc hin c mt dao ng ton phn.K hiu T, n v giy (s).

6. Tn s l s dao ng ton phn thc hin c trong mt giy. K hiu f, n v hc (Hz).

2 1 t

Tf n

= = =

v

1 nf

2 T t

= = =

Vi n l s dao ng ton phn thc hin c trong khong thi gian t .

7. Vn tc: ( )v x ' Asin t= = + Hay: v A cos t2

= + +

+ Vn tc bin i iu ha v sm pha hn li 1 gc2

.

+ Vn tc li x: 2 2v A x= + Vn tc cc i (tc cc i): maxv A=

+ Vn tc trung bnh: tb xv t=

+ Tc trung bnh: svt

=

+ Tc trung bnh trong mt chu k dao ng:4A

vT

=

+ Cng thc lin h gia bin , li v vn tc:2

2 2

2

vA x= +

8. Gia tc: ( )2a v ' x" A cos t= = = + Hay: ( )2a A cos t= + + .

+ Gia tc bin i iu ho sm pha hn vn tc 1 gc

2

v ngc pha so vi li . Gia tc lun

lun tri du vi li . Vect gia tc lun hng v v tr cn bng.

+ Gia tc li x: 2a x= + Gia tc cc i: 2maxa A=

9. im P dao ng iu ho trn mt on thng c th coi l hnh chiu ca mt im M chuynng trn u ln ng knh l on thng .

10. th ca li theo thi gian l mt ng hnh sin. Dao ng iu ha gi l dao ng hnh sin.II/. Con lc l xo

1. Con lc l xo gm mt vt nh c khi lng m gn vo u ca mt l xo c cng k. Vt m cth trt trn mt phng nm ngang khng ma st. Khi c kch thch, con lc l xo s dao ngiu ha.

2. Tn s gc: km

= Chu k: mT 2k

= Tn s: 1 kf2 m

=

n v: k (N/m) ; m (kg)

3. Lc ko v: F kx ma= = lun hng v v tr cn bng.

Gio vin:Nguyn nh V - Trng Cao ng cng nghip v xy dng - Phone: 0948249333 - Email:[email protected]
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3/95

xxO

VTCB

MM MM II II NN


+ Gc ti gc to ca trc Ox.+ di bng bin dao ng, OM = A.+ Hp vi trc Ox mt gc bng pha ban u . Chiu dng l chiu dng ca ng trnlng gic.

2. lch pha ca hai dao ng ( ) ( ) ( ) ( )1 1 1 2 2 2x A cos t 1 ; x A cos t 2= + = + : 1 2 =

+ Khi 1 2 > dao ng (1) sm pha hn dao ng (2) v ngc li.

+ Khi ( )2n n 0, 1, 2,... = = hai dao ng cng pha.+ Khi ( ) ( )2n 1 n 0, 1, 2,... = + = hai dao ng ngc pha.

+ Khi ( ) ( )2n 1 n 0, 1, 2,...2

= + = hai dao ng vung pha.

3. Dao ng tng hp ca hai dao ng iu ha cng phng, cng tn s: ( )1 1 1x A cos t= + v

( )2 2 2x A cos t= + l mt dao ng iu ha cng phng, cng tn s vi hai dao ng thnh

phn. Phng trnh dao ng tng hp ( )x A cos t= + , trong

+ Bin A ca dao ng tng hp c xc nh bi: ( )2 21 2 1 2 2 1A A A 2A A cos= + +

+ Pha ban u ca dao ng tng hp c xc nh bi: 1 1 2 21 1 2 2

A sin A sintan

A cos A cos

+ =

+

+ Khi 1 2x & x cng pha th 1 2A A A= + v 1 2 = = .

+ Khi 1 2x & x ngc pha th 1 2A A A= v 1 = nu 1 2A A> ; 2 = nu 2 1A A> .

+ Khi 1 2x & x vung pha th2 2

1 2A A A= +

+ Trong mi trng hp th 1 2 1 2A A A A A + .VI/. Cc trng hp thng gp

1. Thi gian trong dao ng iu haXt dao ng vi chu k T, bin A trn trc Ox theo phng trnh ( )x A cos t= +

Thi gian ngn nht, khi vt dao ng: + T M n M hoc ngc li:T

t2

= .

+ T O n M hoc ngc li:T

t4

= . + T O n I hoc ngc li:T

t12

= .

+ T I n M hoc ngc li:T

t6

= . + T O n N hoc ngc li:T

t8

= .

2. Vit phng trnh dao ng l i tm A, v ri th vo phng trnh ( )x A cos t= +

+ Tm t cng thc2

T

= hay 2 f =

Vi con lc l xo:k

m = Vi con lc n:

g =

l

n v ca k (N/m) ; m (kg) ; l (m) v g = 9,8 m/ 2s .

+ Tm A c th da vo cng thc

2

2 22vA x= +

+ Tm da vo gc thi gian (t = 0). Trng hp tng qut:

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Khi t = 0 m0

0

x x Acos

v v Asin

= = = =

Suy ra:

0

0

xcos

A

vsin

A

= =

Cc trng hp thng gp:+ Khi t 0= m x A= + th 0 = . + Khi t 0= m x A= th = .

+ Khi t 0= m x 0= vv 0 th .

2

v 0 th .2

> =

< = +

+ Khi t 0= mA

x2

= + vv 0 th .

3

v 0 th .3

> =

< = +3. Cc cng thc suy ra t cng thc gc

Vi con lc l xo:

+ T 22

k kk m m

m = = =

+ T2 2

2 2

m 4 m T k T 2 k m

k T 4

= = =

+ T 2 22 2

1 k kf k 4 f m m

2 m 4 f = = =

Vi con lc n:

+ T2 2

2 2

4 T gT 2 g

g T 4

= = =

l ll

+ T 2 22 2

1 g gf g 4 f

2 4 f= = =

l l

l4. Xc nh lc n hi ca l xo

a) Vi con lc l xo nm ngang : hF kx= maxF kA=b) Vi con lc l xo treo thng ng

+ Chiu dng hng xung: hF k x= +l

+ Chiu dng hng ln: hF k x= l

c) Lc n hi cc i: ( )maxF k A= +l

d) Lc n hi cc tiu:( )

min

0 khi AF

k A khi A

= >

l

l l

k (N/m) ; m (kg) ; A, x, l (m) ; F (N).

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Bi tp trc nghim

1. Mt vt dao ng iu ha theo phng trnh ( )x 20cos 2 t mm

4

= +

. thi im

( )1

t = s4

, li ca vt l

A. -14,4 mm . B. 5 mm. C. 0 mm . D. 14,4mm.2. Con lc l xo dao ng iu ha vi chu k 0,5 s , khi lng qu nng l 400 gam. Ly2 10. = cng ca l xo l

A. 0,156 N/m. B. 32 N/m. C. 64 N/m. D. 6400 N/m.3. Con lc l xo nm ngang dao ng iu ha. Vn tc ca vt bng khng khi vt chuynng quaA. v tr cn bng. B. v tr vt c li cc i.C. v tr m l xo khng b bin dng. D. v tr m lc n hi ca l xo bng khng.4. Con lc l xo gm vt nng c khi lng m = 100 g v l xo c cng k = 100 N/m, daong iu ha vi chu k lA. 0,1 s. B. 0,2 s. C. 0,3 s. D. 0,4 s.5. Mt vt dao ng iu ha vi bin 4 cm v chu k 2 s. Chn gc thi gian l lc vt iqua v tr cn bng theo chiu dng. Phng trnh dao ng ca vt l

A. ( )

x = 4cos 2t - cm .2

B. ( )

x = 4cos 2t + cm .2

C. ( ) ( )x = 4cost cm . D. ( )

x = 4cost - cm2

.

6. Trong dao ng iu haA. vn tc bin i iu ha cng pha vi li .B. vn tc bin i iu ha ngc pha vi li .

C. vn tc bin i iu ha sm pha2 so vi li .

D. vn tc bin i iu ha tr pha2

so vi li .

7. Nu chn gc ta trng vi v tr cn bng th thi im t, biu thc quan h gia bin A, li x, vn tc v v tn s gc ca cht im dao ng iu ha lA. 2 2 2 2A v x= + . B. 2 2 2 2A x A= + .

C.2

2 2

2

xA v= +

. D.

22 2

2

vA x= +

.

8. Phng trnh dao ng ca mt cht im c dng ( )x 6cos t cm2

= + . Gc thi gian

c chn vo lcA. cht im i qua v tr x = 6 cm.B. cht im i qua v tr cn bng theo chiu dng.C. cht im i qua v tr x = - 6 cm.D. cht im i qua v tr cn bng theo chiu m.9. Trong dao ng iu ha li , vn tc, gia tc l ba i lng bin i iu ha theo thigian v cA. cng bin . B. cng pha.C. cng tn s gc. D. cng pha ban u.

10. Gia tc ca vt dao ng iu ha bng 0 khiA. vt v tr c li cc i. B. vt v tr bin m.C. vt v tr c li bng khng. D. vt v tr c pha dao ng cc i.11. Hai dao ng iu ha cng phng, cng tn s, c lch pha . Bin ca hai daong ln lt l A1 v A2. Bin A ca dao ng tng hp c gi tr

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A. ln hn A1 + A2. B. nh hn 1 2A A .

C. lun lun bng ( )1 21

A A2

+ . D. nm trong khong t 1 2A A n A1 + A2.

12. Xt dao ng tng hp ca hai dao ng c cng tn s v cng phng dao ng. Bin ca dao ng tng hp khng ph thucA. bin ca dao ng th nht. B. bin ca dao ng th hai.C. tn s chung ca hai dao ng. D. lch pha ca hai dao ng.13. Mt con lc l xo gm vt nng 400 g gn vo u l xo c cng 40 N/m. Ko qu nngra khi v tr cn bng mt on 4 cm ri th nh cho vt dao ng. Chn gc ta ti v tr cn

bng, chiu dng theo chiu ko vt, gc thi gian l lc th cho vt dao ng. Phng trnh daong ca vt l

A. ( )

x = 4cos 10t + cm2

. B. ( ) ( )x = 4cos 10t cm .

C. ( )

x = 4cos 10t + cm2

. D. ( ) ( )x = 4cos 10t cm .

14. Mt con lc n c chu k dao ng l 4 s, thi gian con lc i t v tr cn bng n v

tr c li cc i lA. 0,5 s. B. 1,0 s. C. 1,5 s. D. 2,0 s.15. Dao ng cng bc cA. chu k dao ng bng chu k bin thin ca ngoi lc.B. tn s dao ng khng ph thuc tn s ca ngoi lc.C. bin dao ng ch ph thuc tn s ca ngoi lc.D. nng lng dao ng khng ph thuc ngoi lc.16. Mt im M chuyn ng u vi tc 0,60 m/s trn mt ng trn c ng knh 0,40m. Hnh chiu P ca im M ln mt ng knh ca ng trn dao ng iu ha vi bin ,tn s gc v chu k ln lt lA. 0,40 m ; 3,0 rad/s ; 2,1 s. B. 0,20 m ; 3,0 rad/s ; 2,48 s.

C. 0,20 m ; 1,5 rad/s ; 4,2 s. D. 0,20 m ; 3,0 rad/s ; 2,1 s.17. Con lc l xo gm vt nh khi lng m dao ng iu ha dc trn trc Ox vi phng

trnh dao ng ( ) ( )x 5cos t cm= + . ng nng ca vtA. bo ton trong sut qu trnh dao ng. B. t l vi tn s gc .C. bin i iu ha vi tn s gc . D. bin i tun hon vi tn s gc 2 .18. Mt con lc n c khi lng vt nng m dao ng iu ha vi tn s f. Nu khi lngvt nng l 2m th tn s dao ng ca vt l:A. 2f . B. 2f . C. f / 2 . D. f .19. Gi tr cc i ca lc n hi tc dng vo qu nng (khi lng m) ca con lc l xo daong iu ha theo phng thng ng vi bin A l

A. max mgF k 2A .k = +

B. max mgF k A .k =

C. maxmg

F k A .k

= +

D. max2mg

F k A .k

= +

20. Vn tc ca cht im dao ng iu ha c ln cc i khiA. li bng khng. B. pha dao ng cc i.C. gia tc c ln cc i. D. li c ln cc i.

21. Mt vt dao ng iu ha theo phng trnh ( )x = 8cos3,14t cm , ly 3,14 = . lnvn tc ca vt ti v tr cn bng l

A. 25,12 cm/s. B. 0 cm/s. C. 78,88 cm/s. D. 52,12 cm/s.22. Mt vt dao ng iu ha vi phng trnh ( )x 4sin t cm= . Thi gian ngn nht vti t v tr cn bng n v tr c li x = 2 cm l

A.1

s6

. B. 0,7 s. C. 0,06 s. D.1

s12

.

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23. Dao ng tng hp ca hai dao ng iu ha cng phng ( )1x 4cos10 t cm= v

( )2x 4cos 10 t+ cm2

=

c bin v pha ban u l

A. ( )3

4 2 cm &4

. B. ( )4 2 cm &

4

.

C. ( )4 2 cm &2

. D. ( )8 2 cm &

2

.

24. Mt vt dao ng iu ha theo phng trnh x 4sin 5t3

= +

(x tnh bng cm, t tnh bng

s). Vn tc v gia tc ca vt c gi tr cc i bngA. 0,2 m/s v 21 m / s . B. 0,4 m/s v 21,5m/s .

C. 0,2 m/s v 22 m / s . D. 0,6 m/s v 22 m / s .25. Mt con lc l xo gm vt nng 400 g gn vo u l xo c cng 40 N/m. Ko qu nngra khi v tr cn bng mt on 4 cm ri th nh cho vt dao ng. Vn tc cc i ca vt lA. 160 cm/s. B. 80 cm/s. C. 40 cm/s. D. 20 cm/s.26. Mt con lc l xo gm vt nng 400 g gn vo u l xo c cng 40 N/m. Ko qu nngra khi v tr cn bng mt on 5 cm ri th nh cho vt dao ng. Vn tc ca vt khi v trcch v tr cn bng 3 cm lA. 20 cm/s. B. 30 cm/s. C. 40 cm/s. D. 10 cm/s.

27. Mt vt dao ng iu ha theo phng trnh ( )x 10cos 4 t cm6

= +

. Tc trung bnh

ca vt trong mt chu k dao ng l

A. ( )80 cm/s . B. ( )40 cm / s . C. ( )40 cm/s . D. ( )20 cm/s .28. Pht biu no sau y l khng ng i vi con lc l xo t nm ngang, chuyn ngkhng ma st?

A. Chuyn ng ca vt l chuyn ng thng.B. Chuyn ng ca vt l mt dao ng iu ha.C. Chuyn ng ca vt l chuyn ng bin i u.D. Chuyn ng ca vt l chuyn ng tun hon.29. Mt con lc l xo c khi lng m v l xo c cng k. Nu tng khi lng ln 2 ln vgim cng i 2 ln th chu k sA. tng 4 ln. B. khng i. C. gim 2 ln. D. tng 2 ln.30. Chu k dao ng iu ha ca con lc l xo khng ph thuc voA. khi lng ca con lc. B. bin dao ng.C. cng ca l xo. D. tn s dao ng.31. Trong dao ng iu ha

A. gia tc bin i iu ha cng pha so vi vn tc.B. gia tc bin i iu ha ngc pha so vi vn tc.

C. gia tc bin i iu ha sm pha2

so vi vn tc.

D. gia tc bin i iu ha tr pha2

so vi vn tc.

32. Mt con lc l xo treo thng ng. L xo c cng 20 N/m, vt treo c khi lng m. Khivt v tr cn bng l xo dn ra mt on 4 cm. Con lc dao ng iu ha vi bin 3 cm.Lc n hi ca l xo c gi tr nh nht trong qu trnh vt dao ng lA.2 N. B.14 N. C.0,2 N. D.1,4 N.

33. Mt con lc l xo treo thng ng. L xo c cng 20 N/m, vt treo c khi lng m. Khivt v tr cn bng l xo dn ra mt on 4 cm. Con lc dao ng iu ha vi bin 3 cm.Lc n hi ca l xo c gi tr ln nht trong qu trnh vt dao ng lA.2 N. B.14 N. C.0,2 N. D.1,4 N.34. Mt con lc l xo gm vt nng c khi lng m = 400g v l xo c cng k = 160 N/m.Vt dao ng iu ha theo phng thng ng vi bin 10 cm. Vn tc ca vt khi qua v tr

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cn bng c ln lA. 4 m/s . B. 0 m/s. C. 2 m/s . D. 6,28 m/s.

35. Con lc l xo dao ng iu ha theo phng ngang vi bin A. Li ca vt khi ngnng bng th nng ca l xo l

A.A 2

x2

= . B.A

x2

= .

C.A 3

x 2= . D.A

x 4= .36. Ti cng mt v tr a l, hai con lc n c chu k dao ng ring ln lt l 1,5 s v 2 s.Chu k dao ng ring ca con lc th ba c chiu di bng tng chiu di ca hai con lc ni trnl

A. ( )0,5 s . B. ( )1, 75 s . C. ( )2,5 s . D. ( )3,5 s .37. Con lc l xo gm vt nng c khi lng m v l xo c cng k = 80 N/m, dao ng iuha vi bin 5 cm. ng nng ca con lc khi n qua v tr c li x = - 3 cm lA. 0,032 J. B. 0,064 J. C. 0,096 J. D. 0,128 J.38. iu no sau y l sai khi ni v nng lng trong dao ng iu ha ca con lc l xo?A. C nng t l vi bnh phng bin dao ng.B. C s chuyn ha qua li gia ng nng v th nng nhng c nng c bo ton.C. C nng t l vi cng ca l xo.D. C nng bin thin theo thi gian vi chu k bng na chu k bin thin ca vn tc.39. Mt con lc l xo gm vt nng 400 g gn vo u l xo c cng 40 N/m. Ko qu nngra khi v tr cn bng mt on 4 cm ri th nh cho vt dao ng. C nng dao ng ca con lclA. 320 J. B. 6,4 . 210 J. C. 3,2 . 210 J D. 3,2 J.

40. Mt vt dao ng iu ha theo phng trnh ( )x 10cos4 t cm= . Tc trung bnh ca vt

trong1

4

chu k dao ng, k t lc t = 0 l

A. ( )80 cm/s . B. ( )40 cm/s . C. ( )40 cm/ s . D. ( )20 cm/s .41. Con lc n dao ng iu ha vi chu k 1s ti ni c gia tc trng trng 9,8 m/s2. Chiudi ca con lc lA. 12,4 cm. B. 24,8 cm. C. 1,56 m. D. 2,45 m.42. Mt vt dao ng iu ha vi chu k T. ng nng ca vt sA. bin i theo thi gian di dng hm s sin.B. bin i tun hon theo thi gian vi chu k T/2.C. bin i tun hon vi chu k T.D. khng thay i theo thi gian.43. Khi gn qu nng m1 vo mt l xo th n dao ng vi chu k 1,2 s. Khi gn qu nng m2vo l xo th n dao ng vi chu k 1,6 s. Khi gn ng thi m1 v m2 vo l xo th n daong vi chu kA. 1,4 s. B. 2,8 s. C. 2,0 s. D. 4,0 s.44. Pht biu no sau y v ng nng v th nng trong dao ng iu ha l khng ng?A. ng nng v th nng bin i iu ha cng chu k.B. ng nng bin i iu ha cng chu k vi vn tc.C. Th nng bin i iu ha vi tn s gp 2 ln tn s ca li .D. Tng ng nng v th nng khng ph thuc vo thi gian.45. Pht biu no sau y l sai?A. Dao ng tt dn l dao ng c bin gim dn theo thi gian.

B. Dao ng cng bc l dao ng chu tc dng ca mt ngoi lc bin thin tun hon.C. Khi c cng hng dao ng, tn s ca dao ng cng bc bng tn s dao ng ring cah dao ng .D. Tn s ca dao ng cng bc lun bng tn s ring ca h dao ng.

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46. Mt vt chu tc ng ng thi hai dao ng iu ha cng phng vi cc phng trnh:

( ) ( )1 2

x = 5cost - cm ; x = 5cost cm2

. Phng trnh dao ng ca vt s l

A. ( )

x = 5 2cost - cm .4

B. ( )x = 5 2sin t - cm .4

C. ( )

x = 5 3cost + cm .4

D. ( )

x = 5cost + cm .6

47. Hai dao ng iu ha cng phng cng tn s v cng pha c bin A1 & A2 vi

2 1A 3A= . Dao ng tng hp c bin bngA. A1. B. 2 A1. C. 3 A1. D. 4 A1.48. Con lc n c chiu di khng i, dao ng iu ha vi chu k T. Khi a con lc lncao th chu k dao ng ca nA. tng ln. B. gim xung .C. khng thay i. D. khng xc dnh c tng hay gim.49. Ti mt ni xc nh, chu k dao ng iu ha ca con lc n t l thun viA. chiu di con lc. B. cn bc hai chiu di con lc.

C. cn bc hai gia tc trng trng. D. gia tc trng trng.

50. Mt vt dao ng iu ha theo phng trnh x 4sin 8 t6

= +

, vi x tnh bng cm, t tnh

bng s. Chu k dao ng ca vt l

A.1

s8

. B. 4 s. C.1

s4

. D.1

s2

.

51. Mt con lc l xo dao ng iu ha vi phng trnh x Acos t= v c c nng l W.ng nng ca vt ti thi im t l

A.

WW = cost

2

. B.

WW = sint

4

.

C.

W = cost2W . D.

W =t 2Wsin .

52. Li v gia tc ca mt vt dao ng iu ha lun bin thin iu ha cng tn s v

A. lch pha vi nhau4

. B. lch pha vi nhau

2

.

C. ngc pha vi nhau. D. cng pha vi nhau.53. Mt cht im dao ng iu ha trn trc Ox vi chu k T. V tr cn bng ca cht imtrng vi gc ta , khong thi gian ngn nht n i t v tr c li x A= n v tr c li

Ax

2= l

A.T

6. B.

T

4. C.

T

2. D.

T

3.

54. Trong dao ng iu ha, vn tc tc thi ca vt dao ng ti mt thi im t lun

A. sm pha4

so vi li dao ng. B. cng pha vi li dao ng.

C. lch pha2

so vi li dao ng. D. ngc pha vi li dao ng.

55. Hai dao ng iu ha cng phng, cng tn s, c cc phng trnh dao ng l:

( )1

x = 3cost - cm

4

v ( )

2

x = 4cost + cm

4

. Bin ca dao ng tng hp hai dao ng

trn lA. 1 cm. B. 7 cm. C. 5 cm. D. 12 cm.56. Mt con lc l xo gm mt l xo khi lng khng ng k, mt u c nh v mt ugn vi mt vin bi nh. Con lc ny ang dao ng iu ha theo phng nm ngang. Lc n

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10/95


hi ca l xo tc dng ln vin bi lun hngA. theo chiu chuyn ng ca vin bi. B. theo chiu dng quy c.C. v v tr cn bng ca vin bi. D. theo chiu m quy c.57. Mt con lc n gm mt hn bi nh khi lng m, treo vo mt si dy khng dn, khilng si dy khng ng k. Khi con lc n ny dao ng iu ha vi chu k 3 s th hn bichuyn ng trn mt cung trn di 4 cm. Thi gian hn bi i c 2 cm k t v tr cn bng lA. 0,5 s. B. 1,5 s. C. 0,25 s. D. 0,75 s.

58. Mt con lc l xo gm mt l xo khi lng khng ng k, cng k, mt u c nh vmt u gn vi mt vin bi nh khi lng m. Con lc ny ang dao ng iu ha c c nngA. t l vi bnh phng bin dao ng. B. t l nghch vi khi lng m ca vin bi.C. t l vi bnh phng chu k dao ng. D. t l nghch vi cng k ca l xo.59. Mt h dao ng chu tc dng ca ngoi lc tun hon n 0F F sin10 t= th xy ra hintng cng hng. Tn s dao ng ring ca h phi lA. 5 Hz . B. 10 Hz . C. 5 Hz . D. 10 Hz .60. Mt con lc l xo gm vt nng c khi lng 400 gam v l xo c cng 40 N/m. Conlc ny dao ng iu ha vi chu k bng

A. s

5

. B.

1s

5

. C. s5 . D. s

5.

61. Hai dao dng iu ha cng phng, cng tn s, c phng trnh ( )1x 6sin t cm3

= +

v ( )2x 8sin t cm6

=

. Dao ng tng hp ca hai dao ng ny c bin

A. 10 cm. B. 2 cm. C. 7 cm. D. 14 cm.62. Trong dao ng c hc, khi ni v vt dao ng cng bc (giai on n nh), pht biuno sau y l ng?A. Bin ca dao ng cng bc lun bng bin ca ngoi lc tun hon tc dng ln vt.B. Chu k ca dao ng cng bc lun bng chu k dao ng ring ca vt.

C. Bin ca dao ng cng bc ch ph thuc vo tn s ca ngoi lc tun hon tc dng lnvt.D. Chu k ca dao ng cng bc bng chu k ca ngoi lc tun hon tc dng ln vt.

63. Hai dao ng iu ha cng phng, c phng trnh l ( )1x 6cos 10 t cm4

=

v

( )2x 8cos 10 t cm4

= +

. Bin ca dao ng tng hp hai dao ng trn bng

A. 10 cm. B. 2 cm. C. 14 cm. D. 12 cm.64. Mt cht im dao ng iu ha trn on thng AB. Khi qua v tr cn bng, vect vn

tc ca cht imA. lun c chiu hng n A. B. c ln cc i.C. bng khng. D. lun c chiu hng n B.

65. Hai dao ng iu ha cng phng, c phng trnh l ( )1x 3cos t cm3

= +

v

( )2x 4cos t cm3

=

. Hai dao ng ny

A. lch pha nhau gc2

3

. B. ngc pha nhau.

C. cng pha nhau. D. lch pha nhau gc 3 .66. Cho hai dao ng iu ha cng phng c cc phng trnh ln lt l

1x 4cos( t )(cm)

6

=

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v 2x 4cos( t )(cm)2

= . Dao ng tng hp ca hai dao ng ny c bin l

A. 4 2 cm . B. 2 cm. C. 4 3 cm . D. 8 cm.67. Dao ng tt dnA. lun c hi. B. c bin gim dn theo thi gian.C. lun c li. D. c bin khng i theo thi gian.68. Mt cht im dao ng iu ha trn trc Ox theo phng trnh x 5 cos 4 t= (x tnh bngcm, t tnh bng s). Ti thi im t = 5 s, vn tc ca cht im ny c gi tr bngA. 20 cm /s . B. 0 cm/s . C. 5cm/s . D. 20 cm /s .69. Mt vt nh dao ng iu ha theo mt trc c nh. Pht biu no sau y ng?A. Lc ko v tc dng vo vt khng i.B. Li ca vt t l vi thi gian dao ng.C. Qu o chuyn ng ca vt l mt ng hnh sin.D. Qu o chuyn ng ca vt l mt on thng.70. Mt con lc l xo gm vt nh khi lng 400 g, l xo khi lng khng ng k v c cng 100N/m. Con lc dao ng iu ha theo phng ngang. Ly 2 10 = . Dao ng ca con lcc chu k l

A. 0,6 s. B. 0,4 s. C. 0,2 s. D. 0,8 s.71. Mt con lc n gm qu cu nh khi lng m c treo vo u mt si dy mm, nh,khng dn, di 64 cm. Con lc dao ng iu ha ti ni c gia tc trng trng g. Ly

2 2g (m / s )= . Chu k dao ng ca con lc lA. 2 s. B. 1,6 s. C. 1 s. D. 0,5 s.

72. Mt cht im dao ng iu ha vi chu k ( )0,5 s v bin 2 cm. Vn tc ca chtim ti v tr cn bng c ln bngA. 4 cm/s. B. 8 cm/s. C. 3 cm/s. D. 0,5 cm/s.73. Mt con lc n gm si dy c khi lng khng ng k, khng dn, c chiu di l vvin bi nh c khi lng m. Kch thch cho con lc dao ng iu ha ni c gia tc trng

trng g. Nu chn mc th nng ti v tr cn bng ca vin bi th th nng ca con lc ny li gc c biu thc lA. ( )mg 3 2cos . l B. ( )mg 1 sin . l

C. ( )mg 1 cos .+ l D. ( )mg 1 cos . l74. Khi a mt con lc n ln cao theo phng thng ng (coi chiu di ca con lc khngi) th tn s dao ng iu ha ca n sA. tng v tn s dao ng iu ha ca n t l nghch vi gia tc trng trng.B. gim v gia tc trng trng gim theo cao.C. khng i v chu k dao ng iu ha ca n khng ph thuc vo gia tc trng trng.D. tng v chu k dao ng iu ha ca n gim.

75. Mt con lc l xo gm vt c khi lng m v l xo c cng k khng i, dao ng iuha. Nu khi lng m = 200 g th chu k dao ng ca con lc l 2 s. chu k con lc l 1 s thkhi lng m bngA. 800 g. B. 200 g. C. 50 g. D. 100 g.76. Mt vt nh dao ng iu ha c bin A, chu k T, thi im ban u to = 0 vt ang

v tr bin. Qung ng m vt i c t thi im ban u n thi imT

t4

= l

A.A

.2

B. 2A. C. A. D.A

.4

77. Ti mt ni, chu k dao ng iu ha ca mt con lc n l 2,0 s. sau khi tng chiu di

ca con lc thm 21 cm th chu k dao ng iu ha ca n l 2,2 s. Chiu di ban u ca conlc ny lA. 101 cm. B. 99 cm. C. 98 cm. D. 100 cm.78. Pht biu no sau y l sai khi ni v dao ng c hc?A. Bin dao ng cng bc ca mt h c hc khi xy ra hin tng cng hng khng phthuc vo lc cn ca mi trng.

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B. Tn s dao ng cng bc ca mt h c hc bng tn s ca ngoi lc iu ha tc dng lnh y.C. Hin tng cng hng xy ra khi tn s ca ngoi lc iu ha bng tn s dao ng ring cah.D. Tn s dao ng t do ca mt h c hc l tn s dao ng ring ca h y.79. Mt vt dao ng iu ha dc theo trc Ox, quanh v tr cn bng O vi bin A v chu

k T. Trong khong thi gianT

4

, qung ng ln nht m vt c th i c l

A. A. B. A 2. C.3A

.2

D. A 3.

80. Cht im c khi lng 1m 50 g= dao ng iu ha quanh v tr cn bng ca n vi

phng trnh dao ng ( )1x sin 5 t cm6

= +

. Cht im c khi lng 2m 100 g= dao ng

iu ha quanh v tr cn bng ca n vi phng trnh dao ng ( )2x 5sin t cm6

=

. T s

c nng trong qu trnh dao ng iu ha ca cht im 1m so vi cht im 2m bng

A. 2. B. 1. C. 15

. D. 12

.

81. Mt con lc l xo gm vin bi nh khi lng m v l xo khi lng khng ng k c cng 10 N/m. Con lc dao ng cng bc di tc dng ca ngoi lc tun hon c tn s gc

F . Bit bin ca ngoi lc tun hon khng thay i. Khi thay i F th bin dao ngca vin bi thay i v khi F 10 rad / s = th bin dao ng ca vin bi t gi tr cc i. Khilng m ca vin bi bngA. 10 gam. B. 40 gam. C. 100 gam. D. 120 gam.82. Cho hai dao ng iu ha cng phng c phng trnh dao ng ln lt l

( )1x 3 3sin 5 t cm2

= + v ( )2x 3 3sin 5 t cm2

= . Bin dao ng tng hp ca haidao ng trn bng

A. 3 3cm. B. 6 3cm. C. 3cm. D. 0 cm.

83. Mt con lc l xo gm vin bi nh c khi lng m v l xo khi lng khng ng k c cng k, dao ng iu ha theo phng thng ng ti ni c gia tc ri t do l g. Khi vin bi v tr cn bng, l xo dn mt on l . Chu k dao ng iu ha ca con lc ny l

A. g2l

. B. 2g

l. C.

1 m

2 k. D.

1 k

2 m.

84. Mt vt dao ng iu ha dc theo trc Ox vi phng trnh = x Acos t. Nu chn gc

ta O ti v tr cn bng ca vt th gc thi gian t = 0 l lc vtA. qua v tr cn bng theo chiu dng ca trc Ox.B. v tr li cc i thuc phn dng ca trc Ox.C. qua v tr cn bng O ngc chiu dng ca trc Ox.D. v tr li cc i thuc phn m ca trc Ox.85. Khi ni v mt h dao ng cng bc giai on n nh, pht biu no di y l sai?A. Tn s ca h dao ng cng bc bng tn s ca ngoi lc cng bc.B. Bin ca h dao ng cng bc ph thuc vo tn s ca ngoi lc cng bc.C. Tn s ca h dao ng cng bc lun bng tn s dao ng ring ca h.D. Bin ca h dao ng cng bc ph thuc bin ca ngoi lc cng bc.86. Dao ng c hc ca con lc vt l trong ng h qu lc khi ng h chy ng l daongA. duy tr. B. t do. C. cng bc. D. tt dn.87. Pht biu no sau y l ng khi ni v dao ng tt dn?A. Dao ng tt dn c bin gim dn theo thi gian.B. C nng ca vt dao ng tt dn khng i theo thi gian.

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C. Lc cn mi trng tc dng ln vt lun sinh cng dng.D. Dao ng tt dn l dao ng ch chu tc dng ca ni lc.88. Khi ni v mt vt dao ng iu ha c bin A v chu k T, vi mc thi gian (t = 0) llc vt v tr bin, pht biu no sau y l sai?

A. Sau thi gianT

8, vt i c qung ng bng 0,5A.

B. Sau thi gian

T

2 , vt i c qung ng bng 2A.

C. Sau thi gianT

4, vt i c qung ng bng A.

D. Sau thi gian T, vt i c qung ng bng 4A.89. Ti ni c gia tc trng trng l 9,8 m/ 2s , mt con lc n dao ng iu ha vi bin gc o6 . Bit khi lng vt nh ca con lc l 90 g v chiu di dy treo l 1 m. Chn mc thnng ti v tr cn bng, c nng ca con lc xp x bngA. 6,8. 310 J. B. 3,8. 310 J. C. 5,8. 310 J. D. 4,8. 310 J.90. Mt cht im dao ng iu ha c phng trnh vn tc l v = 4cos2t (cm/s). Gc ta v tr cn bng. Mc thi gian c chn vo lc cht im c li v vn tc l:

A. x = 2 cm, v = 0. B. x = 0, v = 4 cm/s.C. x = 2 cm, v = 0. D. x = 0, v = 4 cm/s.91. Mt vt dao ng iu ha dc theo trc ta nm ngang Ox vi chu k T, v tr cn bngv mc th nng gc ta . Tnh t lc vt c li dng ln nht, thi im u tin m ngnng v th nng ca vt bng nhau l

A.T

4. B.

T

8. C.

T

12. D.

T

6.

92. Ti ni c gia tc trng trng g, mt con lc n dao ng iu ha vi bin gc 0 .Bit khi lng vt nh ca con l

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