Vong Tron Luong Giac Hay

7/30/2019 Vong Tron Luong Giac Hay

1/12

Vng trn lng gic

VNG TRN LNG GICI.t vn .- Gii bi tp v dao ng iu ha p dng vng trn lng gic (VTLG) chnh l s dng

mi quan h gia chuyn ng thng v chuyn ng trn.- Mt im d...h trn mt on thng lun lun c th c coi l hnh chiu ca mt im

M chuyn ng trn u ln ng knh ca on thng .II.Vng trn lng gic.- Mt vt dao ng iu ha theo phng trnh : x = Acos(t + )cm ; (t o bng s) , c biudin bng vct quay trn VTLG nh sau:B1: V mt vng trn c bn knh bng bin R = AB2: Trc Ox nm ngang lm gc.B3: Xc nh pha ban u trn vng trn (v tr xut pht).Quy c :Chiu dng t tri sang phi.- Chiu quay l chiu ngcchiu kim ng h.- Khi vt chuyn ng trn trc Ox : theo chiu m.- Khi vt chuyn ng ditrc Ox : theo chiu dng.

- C bn v tr c bit trn vng trn:M : v tr bin dng xmax = +A y = 0 ; (y l v tr mc ly gc )N : v tr cn bng theo chiu m y = + /2 hoc = 3/2P : v tr bin m xmax = - A y =

Q : v tr cn bng theo chiu dng y = /2 hoc = +3/2

V d :Biu din phng trnh sau bng vct quay :a. x = 6cos(t + /3)cm b.x = 6cos(t /4)cmGii:

III.Dng bi tp1.Dng mt : Xc nh trong khong thi gian t vt qua mt v tr cho trc my ln.Phng php :

+ Biu din trn vng trn , xc nh v tr xut pht.

+ Xc nh gc qut = t.+ Phn tch gc qut = n1.2 + n2. + ;n1 v n2 : s nguyn ; v d : = 9 = 4.2 +

+ Biu din v m trn vng trn.

1

Mc ly gc > 0

< 0

O xA

-A VTCB +AO +

P M

N

Q

-6 0 +6600

a

.M(t = 0)

-6 0 +6

450

N(t = 0)

b.


2/12

Vng trn lng gic

- Khi vt qut mt gc = 2 (mt chu k th qua mt v tr bt k 2 ln , mt ln theo chiudng , mt ln theo chiu m )V d : Vt d...d vi phng trnh : x = 6cos(5t + /6)cm (1)a.Trong khong thi gian 2,5s vt qua v tr x = 3cm my ln.b.Trong khong thi gian 2s vt qua v tr x = 4cm theo chiu dng my ln.

c.Trong khong thi gian 2,5s vt qua v tr cn bng theo chiu dng my ln.d.Trong khong thi gian 2s vt qua v tr cn bng my ln. Gii:Trc tin ta biu din pt (1) trn vng trn, vi = /6(rad)-Vt xut pht t M , theo chiu m. (Hnh 1 )a.Trong khong thi gian t = 2,5s=> gc qut = t. = 2,5.5 = 12,5 =6.2 + /2T vng trn ta thy: (Hnh 2)- trong mt chu k vt qua x = 3cm c 2 ln ti P(chiu m ) v Q(chiu dng )- trong 1 = 6.2 ; 6 chu k vt qua x = 3cm c 6.2 = 12 ln- cn li 2 = /2 t M N vt qua x = 3cm mt ln ti P(chiu m )Vy: Trong khong thi gian t = 2,5s vt qua x = 3cm c 13 lnb.Trong khong thi gian t = 2 s=> gc qut = t. = 2.5 = 10 = 5.2Vt thc hin c 5 chu k (quay c 5 vng)T vng trn ta thy: (Hnh 3)- trong mt chu k vt quav tr x = +4cm theo chiu dng c mt ln , ti NVy : trong 5 chu k th vt quav tr x = 4cm theo chiu dng c 5 lnc.Trong khong thi gian t = 2,5s

=> gc qut = t. = 2,5.5 = 12,5 = 6.2 + /2T vng trn ta thy: (Hnh 4)- Trong mt chu k vt qua v tr cn bng theo chiu dng 1 ln ti N.-Trong 1 = 6.2 ; 6 chu k vt qua v tr cn bng theo chiudng 6 ln ti N.- Cn li 2 = /2 t M P vt qua khng qua v tr cn bng theo chiu dng ln no.Vy trong khong thi gian t = 2,5s vt qua v tr cn bng theo chiu dng 6 ln.d.Trong khong thi gian t = 2s=> gc qut = t. = 2.5 = 10 = 5.2

Vt thc hin c 5 chu k (quay c 5 vng)T vng trn ta thy: (Hnh 5)- Trong mt chu k vt qua v tr v tr cn bng 2 ln ti P(chiu m ) v Q(chiu dng ) .- Vy trong khong thi gian t = 2s vt qua v tr v tr cn bng 10 ln .2. Dng hai: Xc nh thi im vt qua mt v tr c li bt k cho trc.Phng php :

+ Biu din trn vng trn , xc nh v tr xut pht.+ Xc nh gc qut

+ Thi im c xc nh : t =

(s)

VD1 : Vt d...d vi phng trnh : x = 8cos(5t /6)cm (1)Xc nh thi im u tin :

2

-6 0 3 +6

M

P

Q

N

300

-6 0 +4 +6

M

N

-6 0 +6

M

N

P

Hnh 2

Hnh 3

Hnh 4

-6 0 +6

M

Hnh5

P

Q

-8 0 +8-300

M

Hnh 1

-6 0 +6

M300

Hnh 1


3/12

Vng trn lng gic

a.vt qua v tr bin dng.b.vt qua v tr cn bng theo chiu m.c. vt qua v tr bin m.d. vt qua v tr cn bng theo chiu dng.Gii:

Trc tin ta biu din pt (1) trn vng trn, vi = /6(rad) = 300

-Vt xut pht t M , theo chiu dng. (Hnh 1 )a. Khi vt qua v tr bin dng ln mt: ti v tr N

=> gc qut : =300 = /6(rad) => t =

= 16 ( )

5 30s

=

b.Khi vt qua v tr cn bng theo chiu m ln mt :ti v tr P => gc qut : =300 + 900 = 1200 = 2/3(rad)

=> t =

= 2 23 ( )

5 15s

=

c. Khi vt qua v tr bin m ln mt : ti v tr Q=> gc qut :

=300 + 900 +900 = 2100 = 7/6(rad) => t =

=

776 ( )

5 30s

=

d.Khi vt qua v tr cn bng theo chiu dng ln mt : ti v tr K => gc qut :

= 300

+ 900

+ 900

+900

= 3000

= 5/3(rad) => t =

=

5

13 ( )5 3 s

=

VD2 : Vt d...d vi phng trnh :x = 5cos(5t 2/3)cm. Xc nh thi im th 5 vt qua v tr cli x = 2,5cm theo chiu m.Gii :Trc tin ta biu din pt trn vng trn,

vi = 2/3(rad) = -1200

-Vt xut pht t M , theo chiu dng. (Hnh 1 )Thi im u tin vt qua v tr c li x = 2,5cm theo chiu m : ti v tr N : 1 = 2/3 +/2 + /6 = 4/3(rad)Thi im th hai : 2 = 2(rad), (v quay thm mt vng)Thi im th ba: 3 = 2(rad)Thi im th t : 4 = 2(rad)Thi im th nm :5 = 2(rad)- Gc qut tng cng :

= 4/3 + 4.2 = 1 + 2 + 3 + 4 + 5 = 28/3(rad) => t =

=

28( )

15s

VD3 :Mt vt dao ng iu ha c phng trnh x 8cos10t. Thi im vt i qua v tr

x 4 ln th 2009 k t thi im bt u dao ng l :

A.6025

30(s). B.

6205

30(s) C.

6250

30(s) D.

6,025

30(s)

Gii:

3

-8 0 4 +8

M

N

600

-8 0 +8

M

N

P

Q

K

300

-5 -2,5 0 +5

Hnh 1M

-1200

N

/6


4/12

Vng trn lng gic

Vt xut pht t bin dng (xmax = +8).Trong mt chu k th vt qua v tr x 4 c 2 ln ti M(chiu m) v N(chiu dng) ng thi gc qutl : = 2(rad)Vy khi quay c 1004 vng (quanh +8) th qua x 4 c 1004.2 = 2008 ln, gc qut :1 = 1004.2 = 2008(rad)

Cn li mt ln : t +8 n M : gc qut : 2 = /3(rad)Vy gc qut tng cng l: = 1 + 2 = 2008 + /3 = 6025/3(rad)

Thi im : t =

=

6025

30s => A

BI TP VN DNG DNG 2:1. Mt vt dao ng iu ho vi phng trnh x 4cos(4t + /6) cm. Thi im th 3 vt quav tr x 2cm theo chiu dng.A. 9/8 s B. 11/8 s C. 5/8 s D.1,5 s2.Vt dao ng iu ha c ptrnh : x 5cost (cm).Vt qua VTCB ln th 3 vo thi im :

A. 2,5s. B. 2s. C. 6s. D. 2,4s3. Vt dao ng iu ha c phng trnh : x 4cos(2t - ) (cm, s). Vt n im bin dngB(+4) ln th 5 vo thi im :A. 4,5s. B. 2,5s. C. 2s. D. 0,5s.3. Mt vt dao ng iu ha c phng trnh : x 6cos(t /2) (cm, s). Thi gian vt i tVTCB n lc qua im c x 3cm ln th 5 l :A. 61/6s. B. 9/5s. C. 25/6s. D. 37/6s.4. Mt vt DH vi phng trnh x 4cos(4t + /6)cm. Thi im th 2009 vt qua v trx 2cm, k t t 0, l

A.1204924

s. B.12061s24 C.12025s

24 D. p n khc

5. Mt vt dao ng iu ha c phng trnh x 8cos10t. Thi im vt i qua v tr x 4 lnth 2008 theo chiu m k t thi im bt u dao ng l :

A.12043

30(s). B.

10243

30(s) C.

12403

30(s) D.

12430

30(s)

6. Con lc l xo dao ng iu ho trn mt phng ngang vi chu k T 1,5s, bin A 4cm,pha ban u l 5/6. Tnh t lc t 0, vt c to x 2 cm ln th 2005 vo thi im no:A. 1503s B. 1503,25s C. 1502,25s D. 1503,375s

Bi 101 : Mt vt dao ng iu ha c phng trnh : x 4cos(t /2) (cm, s). Thi im thhai vt qua v tr c li x = 2cm theo chiu dng l:A. 13/6 s. B. 9/5 s. C. 4,5s. D. 3,76s.Bi 102 : Mt vt dao ng iu ha c phng trnh : x 5cos(t /2) (cm, s). Thi im thhai vt qua v tr c li x = 2cm theo chiu m l:A. 13/6 s. B. 17/6 s. C. 4,5s. D. 3,76s.Bi 103 : Mt vt dao ng iu ha c phng trnh : x 4cos(0,5t ) (cm, s). Thi imu tin vt qua v tr c li x = 12cm theo chiu m l:A. 13/6 s. B. 5/3 s. C. 10/3s. D. 3,76s.

Bi 104 : Mt vt dao ng iu ha c phng trnh : x Acos(t 2/3) (cm, s). Thi imvt qua v tr c li x = A/2cm ln th hai l:A. 6 s. B. 1 s. C. 4,5s. D. 3,76s.

4


5/12

Vng trn lng gic

Bi 105 : Mt vt dao ng iu ha vi phng trnh : x 5cos(2t /6)cm. Thi im thhai vt qua v tr x = 2,5cm theo chiu m :A. 5/4s B. 1/6s C. 3/2s D. 1s3. Dng ba: Xc nh qung ng vt i c t thi im t1 n t2.Vn tc ca vt.a.Qung ng:

Phng php :+ Biu din trn vng trn , xc nh v tr xut pht.+ Xc nh gc qut = t. ; vi t = t2 t1+ Phn tch gc qut : (Phn tch thnh cc tch s nguyn ca 2 hoc )

= n1.2 + n2. + ; n1 v n2 : s nguyn ; v d : = 9 = 4.2 + + Biu din v m trn vng trn v tnh trc tip t vng trn.+ Tnh qung ng:

- Khi qut 1 = n1.2 th s1 = n1.4.A- Khi qut 2 th s2 tnh trc tip t vng trn.

- Qung ng tng cng l : s = s1+ s2Khi vt quay mt gc : = n.2 (tc l thc hin n chu k) th qung ng l : s = n.4.AKhi vt quay mt gc : = th qung ng l : s = 2ACc gc c bit :

cos300 = 32

; cos600 = 0,5 ; cos450 = 22

*Tnh qung ng ln nht v nh nht vt i c trong khong thi gian 0 < t < T/2.Gc qut = .t v: rad

Qung ng ln nht : ax 2Asin2

MS

=

Qung ng nh nht : 2 (1 os )2

MinS A c

=

b.Vn tc:

5

Vn tc trung bnh v tc trung bnha. Vn tc trung bnh :

2 1

2 1

tb

x xv

t t

=

trong : 2 1x x x = l di.

-Vn tc trung bnh trong mt chu k lun bng khng

b. Tc trung bnh : lun khc 0 ;

2 1

tb

Sv

t t=

trong S l qung ng vt i c t t1 n t2.

Lu : + Trong trng hp t > T/2 ;

Tch '2

Tt n t = + trong *; 0 '

2

Tn N t < < ;

Trong thi gian2

Tn qung ng lun l 2nA ;

Trong thi gian t th qung ng ln nht, nh nht tnh nh trn.+ Tc trung bnh ln nht v nh nht ca trong khong thi gian t:

ax

axM

tbM

Svt

=

v MintbMin

Svt

=

vi SMax; SMin tnh nh trn.


6/12

M

6-63-3

N

600

600

Vng trn lng gic

V d 1: Mt con lc l xo dao ng iu ha vi phng

trnh : x

12cos(50t

/2)cm. Qung ng vt i ctrong khong thi gian t /12(s), k t thi im gc l :A. 6cm. B. 90cm. C. 102cm. D. 54cm.Trc tin ta biu din pt trn vng trn,

vi = /2(rad) = 900

Vt xut pht t M (v tr cn bng theo chiu dng).

t = t2 t1 = /12(s) ; Gc qut : = t. =25

.5012 6

=

Phn tch gc qut =25 (24 1)

2.26 6 6

+= = = + ; Vy 1 = 2.2 v 2 = 6

Khi qut gc : 1 = 2.2 th s1 = 2.4.A = 2.4.12 = 96cm , (quay 2 vng quanh M)

Khi qut gc : 2 = 6

vt i t M N th s2 = 12cos600 = 6cm

- Qung ng tng cng l : s = s1+ s2 = 96 + 6 = 102cm => CV d 2: Mt con lc l xo dao ng iu ha vi phng trnh : x 6cos(20t /3)cm. Qungng vt i c trong khong thi gian t 13/60(s), k t khi bt u dao ng l :A. 6cm. B. 90cm. C.102cm. D. 54cm.Gii:Vt xut pht t M (theo chiu m)

Gc qut = t. = 13/3 =13/60.20 = 2.2 + /3Trong 1 = 2.2 th s1 = 2.4A = 48cm, (quay 2 vng quanh M)Trong 2 = /3 vt i t M N th s2 = 3 + 3 = 6 cmVy s = s1 + s2 = 48 + 6 = 54cm => p n D

V d 3: Mt con lc l xo dao ng iu ha vi bin 6cm v chu k 1s. Ti t = 0, vt iqua VTCB theo chiu m ca trc to .

a.Tng qung ng i c ca vt trong khong thi gian 2,375s k t thi im c chnlm gc l :A. 56,53cm B. 50cm C. 55,75cm D. 42cmb.Tnh tc trung bnh trong khong thi gian trn.Gii:a. Ban u vt qua VTCB theo chiu m: M ;Tn s gc: = 2 rad/s ; Sau t = 2,375s=> Gc qut = t. = 4,75 = 19/4 = 2.2 + 3/4Trong 1 = 2.2 th s1 = 2.4A = 2.4.6 = 48cm

Trong 2 = 3/4 vt i t M n Ns2 = A(t M - 6) + (A Acos45o)(t -6N )Vy s = s1 + s2 = 48 + A + (A Acos45o) = 55,75cm Cb.ADCT:

6

-12 0 +12

M

N

s2= 12cos600

600

300

M

-6 O +6

N

Acos45o

450


7/12

Vng trn lng gic

2 1

tb

Sv

t t=

=

55,75 55,7523, 47 /

2,375 0 2,375cm s= =

V d 4:Mt cht im M dao ng iu ha theo phng trnh: x 2,5cos 10 t2

= +

cm. Tm

tc trung bnh ca M trong 1 chu k dao ngA. 50m/s B. 50cm/s C. 5m/s D. 5cm/sGii:

Trong mt chu k : s = 4A = 10cm => vtb =10

50 /0,2

s scm s

t T= = = B

BI TP VN DNG DNG 3:a.Qung ng:1. Mt vt dao ng iu ho vi bin 4cm, c sau mt khong thi gian 1/4 giy th ngnng li bng th nng. Qung ng ln nht m vt i c trong khong thi gian 1/6 giy

lA. 8 cm. B. 6 cm C. 2 cm. D.4 cm.2.Mt vt dao ng iu ha dc theo trc Ox, quanh v tr cn bng O vi bin A v chu kT. Trong khong thi gian T/4, qung ng nh nht m vt c th i c l

A. A(2- 2) B. A C. 3A D. 1,5A.3. Mt con lc l xo dao ng iu ha vi bin 6cm v chu k 1s. Ti t = 0, vt i quaVTCB theo chiu m ca trc to . Tng qung ng i c ca vt trong khong thigian 2,375s k t thi im c chn lm gc l :A. 56,53cm B. 50cm C. 55,77cm D. 42cm

4. Mt vt dao ng vi phng trnh x 4 2 cos(5t 3/4)cm. Qung ng vt i t thiim t1 1/10(s) n t2 = 6s l :A. 84,4cm B. 333,8cm C. 331,4cm D. 337,5cm5. Mt cht im dao ng iu ho doc theo trc Ox. Phng trnh dao ng l:

x = 10cos (5

26

t

+ ) cm . Qung ng vt i trong khong thi gian t t1 = 1s n t2 =

2,5s l:A. 60 cm. B. 40cm. C. 30 cm. D. 50 cm.

6.Chn gc to ta VTCB ca vt dao ng iu ho theo phng trnh:3

20 os( t- )

4

x c

= (cm;

s). Qung ng vt i c t thi im t1 = 0,5 s n thi im t2 = 6 s lA. 211,72 cm. B. 201,2 cm. C. 101,2 cm. D. 202,2cm.7.Vt dao ng iu ha theo phng trnh : x = 5 cos (10 t + )(cm). Thi gian vt i qungng S = 12,5cm (k t t = 0 ) lA. 1/15 s B. 2/15 s C. 1/30 s D.1/12 s8. Mt vt dao ng iu ho vi phng trnh x = 6cos (2t /3)cm.cm. Tnh di qung

ng m vt i c trong khong thi gian t1 = 1,5 s n t2 =13/3 sA. (50 + 5 3 )cm B.53cm C.46cm D. 66cm

9. Mt vt dao ng iu ho theo phng trnh: x = 5cos( 22 3t ) cm

1. Tnh qung ng vt i c sau khong thi gian t = 0,5s k t lc bt u dao ngA. 12cm B. 14cm C.10cm D.8cm

7


8/12

Vng trn lng gic

2.Tnh qung ng vt i c sau khong thi gian t = 2,4s k t lc bt u dao ngA. 47,9 cm B.49,7cm C.48,7cm D.47,8cm10.Vt dao ng iu ho vi chu k T = 2s, bin A = 2cm. Lc t = 0n bt u chuynng t bin. Sau thi gian t = 2,25s k t lc t= 0 n i c qung ng

l bao nhiuA.10 - 2cm B.53cm C.46cm D. 67cm11.Mt vt dao ng iu ho vi phng trnh: x = 6cos(4t + /3)cm. ttnh bng giy. Tnh qung ng vt i c t lc t = 1/24s n thiim 77/48sA.72cm B. 76,2cm B. 18cm D.22,2cm

12. Mt vt dao ng vi bin 4cm v chu k 2s. mc thi gian khi vtc ng nng cc i v vt ang i theo chiu dng. Tm qung ng

vt i oc trong 3,25s uA. 8,9cm B. 26,9cm C. 28cmD. 27,14cm

13. Mt vt dao ng theo phng trnh x = 4cos(10t + /4) cm. t tnh bng giy. Tm qungng vt i c k t khi vt c tc 0,23m/s ln th nht n khi ng nng bng 3 lnth nng ln th t:

A.12cm B. 8+ 43cm C. 10+ 23cm D. 16cm14. Con lc l xo treo thng ng, gm l xo cng k=100(N/m) v vt nng khi lngm=100(g). Ko vt theo phng thng ng xung di lm l xo gin 3(cm), ri truyn cho

n vn tc 20 3 (cm / s)hng ln. Ly g= 2

=10(m/s2

). Trong khong thi gian 1/4 chu kqung ng vt i c k t lc bt u chuyn ng lA. 5,46(cm). B. 2,54(cm). C. 4,00(cm). D. 8,00(cm).15. Mt con lc l xo gm mt l xo c cng k = 100N/m v vt c khi lng m = 250g,dao ng iu ho vi bin A = 6cm. Chn gc thi gian lc vt i qua v tr cn bng.Qung ng vt i c trong /10s u tin l:A. 6cm. B. 24cm. C. 9cm. D. 12cm.16. Mt cht im dao ng iu ho quanh v tr cn bng O, trn qu o MN = 20cm. Thigian cht im i t M n N l 1s. Chn trc to chiu dng t M n N, gc thi gian lc

vt i qua v tr cn bng theo chiu dng. Qung ng m cht im i qua sau 9,5s k tlc t = 0: A. 190 cm B. 150 cm C. 180 cm D. 160 cm17.Mt con lc gm mt l x c K= 100 N/m, khi lng khng ng k vmt vt nh khi lng 250g, dao ng iu ho vi bin bng 10 cm.Ly gc thi gian t=0 l lc vt qua v tr cn bng. Qung ng vt ic trong t = /24s u tin l:A. 7,5 cm B. 12,5 cm C. 5cm. D. 15 cm18. Vt dao ng iu ha theo phng trnh : x = 4 cos (20t-/2) (cm). Qung ng vt i

trong 0,05s l?A. 8cm B. 16cm C. 4cm D.2cm19. Vt dao ng iu ha theo phng trnh : x = 2 cos (4t - )(cm). Qung ng vt itrong 0,125s l?

8


9/12

Vng trn lng gic

A. 1cm B.2cm C. 4cm D.2cm20. Vt dao ng iu ha theo phng trnh : x = 4 cos (20 t -2 /3)(cm). Tc ca vt saukhi i qung ng S = 2cm (k t t = 0) lA. 40cm/s B. 60cm/s C. 80cm/s d. Gi tr khc21. Vt dao ng iu ha theo phng trnh : x = cos ( t - 2 /3)(dm). Thi gian vt i qung

ng S = 5cm ( k t t = 0) l :A. 1/4 s B. 1/2 s C. 1/6 s D.1/12 sb.Vn tc:1. Mt cht im d. dc theo trc Ox. P.t dao ng l x = 6 cos (20 t- /2) (cm). Vn tc trungbnh ca cht im trn on t VTCB ti im c li 3cm l :A. 360cm/s B.120cm/s C. 60cm/s D.40cm/s2.Mt cht im dao ng dc theo trc Ox. Phng trnh dao ng l x = 4 cos (4 t- /2)(cm). Vn tc trung bnh ca cht im trong chu k t li cc tiu n li cc i l :A. 32cm/s B.8cm/s C. 16cm/s D.64cm/s

3.Chn gc to ta VTCB ca vt dao ng iu ho theo phng trnh: 320 os( t- )4

x c

= cm.

Tc trung bnh t thi im t1 = 0,5 s n thi im t2 = 6 s lA. 34,8 cm/s. B.38,4 m/s. C. 33,8 cm/s. D. 38,8 cm/s.

4.Dng 4 : p dng vng trn cho phng trnh ca vn tc v gia tc.Phng php :Mt vt dao ng iu ha vi phng trnh li : x = Acos(t + )cm

Th phng trnh ca vn tc ( sm pha hn li l /2) => v = Acos(t + +/2)cm/sphng trnh ca gia tc (ngc pha vi li ) => a = A2cos(t + + ) cm/s2

Nh vy bin ca vn tc l : vmax = Abin ca gia tc l : amax = A2

Biu din bng vct quay :

9

xv

a

-A 0 +A

x,a,v trn cng h trc

-A.2

0 +A.2

a

Biu din gia tc a

-A.

0 +A.

v

Biu din vn tc v


10/12

Vng trn lng gic

VD : Mt con lc l xo dao ng iu ha vi chu k T vbin 5 cm. Bit trong mt chu k, khong thi gian vt nh ca con lc c ln gia tc khng vt qu 100cm/s2 l T/3 Ly 2 = 10. Tn s dao ng ca vt l :A.4 Hz. B. 3 Hz. C. 1

Hz. D. 2 Hz.GiiTa thy t = T/3 l khong thi gian gia tc khng vtqu 100cm/s2.Xt trong na chu k: Vt i t M N c gia tc khngvt qu 100 cm/s2; gc qut 600 => t = T/6.Khi ta c = 600.

M cos = 2100

.A

Suy ra 2 = 0100

. os60A c = 40Khi = 40 2 10= = 2 rad/s. Vy f = 1HzVD : Vt dao ng iu ha c vmax = 3m/s v gia tc cc i bng 30 (m/s2). Thi im banu vt c vn tc 1,5m/s v th nng ang tng. Hi vo thi im no sau y vt c gia tcbng 15 (m/s2):A. 0,10s; B. 0,15s; C. 0,20s D. 0,05s;Gii:Ta c: A. = 3 v A.2 = 30m/s2 => = 10 rad/s

Thi im t = 0, = - /6, do x c biu din nh hnh vV a v x ngc pha nhau nn t = 0 pha ca a c biu din trnhnh vNh vy c hai thi im t tha mn bi ton (a = amax/2)

t1 =5

6

= 0,08s v t2 =3

2

= 0,15s

VD: Mt con lc l xo nm ngang ang dao ng t do.Ban u vt i qua v tr cn bng, sau 0,05s n cha ichiu chuyn ng v vn tc cn li mt na. Khong thi gian gia hai ln lin tip c ngnng bng th nng l:A. 0,05s B. 0,04s C. 0,075s D. 0,15sGii: Hai ln lin tip c ng nng bng th nng l T/4

- gi s vt qua VTCB theo chiu dng:x = Acos(t /2)cm v v sm hn x l /2=> v = Acos(t )cm/s ( tnh t v = +A.); v vt cha i chiu nn vn theo chiu m =>n lc vn tc cn li mt na th vt M

10

-A2 100 +A2

300

= 600

MN

-A O A

t = 0

A2/2

-A2 A2

t = 0

-A. 0 A/2 +A.

M

600


11/12

Vng trn lng gic

v = vmax/2 =>cos = v/vmax = 0,5 gc qut = /3=> = /t = 20/3 rad/s=> t = T/4 = (2/)/4 = 0,075s => C

VD: Mt con lc l xo ,vt nng khi lng m=100g v l xo c cng k =10N/m dao ngvi bin 2cm. Thi gian m vt c vn tc nh hn 103 cm/s trong mi chu k l bao

nhiu?A. 0,628s B. 0,417s C. 0,742s D. 0,219sGii:Tn s gc: = 10rad/s => vmax = A. = 20 cm/s- ta xt v tr c vn tc v = 103 cm/s ti M=> cos = v/vmax = 3/2 => = /6- xt trong na chu k: ti M c v = 103 cm/s=> ti N i xng vi M cng c v = 103 cm/s=> t M n N ( vn tc nh hn 103 cm/s )gc qut = /3 + /3 = 2/3 (rad) => t = 2/30 = /15 (s)

trong mt chu k th khong thi gian : t = (/15).2 = 2/15 = 0,4188(s)

Bi tp: Vt nh c khi lng 200 g trong mt con lc l xo dao ng iu ha vi chu k T v bin 4 cm.Bit trong mt chu k, khong thi gian vt nh c ln gia tc khng nh hn 5002 cm/s 2 l T/2. cng ca l xo l:A. 20 N/m. B. 50 N/m. C. 40 N/m. D. 30 N/m.Chia s bi : [email protected]

11

M

300

N

103 103

-20 0 20
mailto:[email protected]:[email protected]


12/12

Vng trn lng gic 12

Vong Tron Luong Giac Hay

Documents

Transcript of Vong Tron Luong Giac Hay