Hsg Con Lac Lo Xo.thuvienvatly.com.0dee7.29505

Cu 1: (1,5) Mt khi g khi lng M=400g c treo vo l xo c cng k=100N/m

Cu 1: (1,5) Mt khi g khi lng M=400g c treo vo l xo c cng k=100N/m. Mt vin bi khi lng m=100g c bn n vi vn tc v0= 50cm/s va chm vo khi g. Sau va chm h dao ng iu ha.

Xc nh chu k v bin dao ng.

Bit va chm tuyt i n hi.Va chm tuyt i n hi

(1)

inh lut bo ton nng lng

(2)

T (1), (2) suy ra:

Chu k:

nh lut bo ton c nng

Cu 2: (2) Mt qu cu c khi lng

m= 2kg treo mt u mt si dy c khi lng khng ng k v khng co dn. B qua ma st v sc cn. Ly g= 10m/s2.

a) Ko qu cu khi v tr cn bng mt gc ri th ra ( vn tc ban u bng khng). Thit lp biu thc lc cng dy ca dy treo khi qu cu v tr lch mt gc so vi v tr cn bng. Tm v tr ca qu cu trn qu o lc cng t cc i. Tinh ln ca lc cng cc i nu gc =600.

b) Phi ko qu cu khi v tr cn bng mt gc bng bao nhiu khi th cho dao ng, lc cng cc i gp 3 ln trng lng ca qu cu.

c) Thay si dy treo qu cu bng mt l xo c trng lng khng ng k. cng ca l xo l k= 500N/m, chiu di ban u l0=0,6m. L xo c th dao ng trong mt phng thng ng xung quanh im treo O. Ko qu cu khi v tr cn bng mt gc ri th ra. Lc bt u th, l xo trng thi khng b nn dn. Xc nh dn ca l xo khi qu cu n v tr cn bng.

2a

bTmax= 3mg. T h thc trn suy ra:

cChn mc th nng ti VT thp nht.

C nng ti A(ngang):

C nng ti B(thp nht):

Lc n hi ti VT B:

T (1),(2)

EMBED Equation.DSMT4 Thay vo (3):

Gii ra: =0,104(m)

Cu 3(2 im)

1) Mt vt c khi lng , dao ng iu ho theo phng trnh c dng . Bit th lc ko v theo thi gian F(t) nh hnh v. Ly . Vit phng trnh dao ng ca vt.

2) Mt cht im dao ng iu ha vi chu k T v bin . Bit trong mt chu k, khong thi gian vn tc c ln khng vt qu (cm/s) l . Xc nh chu k dao ng ca cht im.3) Mt con lc l xo t trn mt phng nm ngang c (N/m), . a qu cu n v tr m l xo b nn 10cm, ri th nh. Bit h s ma st gia vt v mt phng nm ngang l ( = 0,2. Ly g = 10(m/s2). Tnh vn tc cc i m vt t c trong qu trnh dao ng.1) (1 im)

T th, ta c: = 1(s) ( T = 2s ( ( = ((rad/s).0,25

( k = m.(2 = 1(N/m).

+) Ta c: = kA ( A = 0,04m = 4cm.0,25

+) Lc t = 0(s) t th, ta c: Fk = - kx = - 2.10-2 m ( x = 2cm v Fk ang tng dn (vt ang chuyn ng v VTCB) ( v < 0.

0,25

Vy, phng trnh dao ng ca vt l: x= 4cos((t + (/3) cm.0,25

2) (0,5im)

T gi thuyt, ( 24((cm/s).

Gi x1 l v tr m v = 24((cm/s) v t1 l thi gian vt i t v tr x1 n A.

( Thi gian vn tc c ln khng vt qu 24((cm/s) l: t = 4t1 = ( t1 = ( x1 = A/2.0,25

p dng cng thc:

0,25

3) (0,5im)

Gi x0 l ta ca VTCB, ta c: Fdh = Fms ( k.x0 = (mg

(

0,25

Bin dao ng ca con lc l: A = (l x0 = 9cm.

Vn tc cc i l: vmax = A( = 90(cm/s).0,25

Bi 4 Con lc l xo nh hnh v. Vt nh khi lng m = 200g, l xo l tng c cng k = 1N/cm, gc = 300. Ly g = 10m/s2.

a/ Chn trc ta nh hnh v, gc ta trng vi v tr cn bng. Vit phng trnh dao ng. Bit ti thi im ban u l xo b dn 2cm v vt c vn tc v0 = 10cm/s hng theo chiu dng.

b/ Ti thi im t1 l xo khng bin dng. Hi ti t2 = t1 + s, vt c ta bao nhiu?

c/ Tnh tc trung bnh ca m trong khong thi gian t = t2 - t1.Bi 4 (2,5)

a/ Ti VTCB

=> l = 1cm, = rad/s, T = .

Bin : A = => A = 2cm v .

Vy: x = 2cos()cm.

b/ Ti t1 vt M c vn tc v1, sau t = = 1,25T.

- vt K (nu v1 > 0) => ta x2 = cm.

- vt N (nu v1 < 0) => ta x2 = -cm.

c/ Qung ng m i c: - Nu v1 s1 = => vtb = 26,4m/s.

- Nu v1>0 => s2 = => vtb = 30,6m/s.0,25

0,25

0,25

0,25

0,25

0,25

0,25

0,25

0,25

0,25

5. Mt l xo nh c cng , u trn c gn vo gi c nh trn mt nm nghing mt gc so vi phng ngang, u di gn vo vt nh c khi lng (hnh v 1). B qua ma st mt nm v ma st gia nm vi sn ngang. Nm c khi lng M. Ban u nm c gi cht, ko m lch khi v tr cn bng mt on nh ri th nh vt v ng thi bung nm. Tnh chu k dao ng ca vt m so vi nm.

Tnh chu k dao ng ca vt so vi nm (1im):

+ Trong h quy chiu gn vi nm:

- Ti VTCB ca m trn nm (khi m cn bng trn nm th nm cng cn bng

trn bn): l xo gin mt on: (1)

- Chn trc Ox gn vi nm v trng mt nm hng xung, O l VTCB ca m

trn nm.

- Ti v tr vt c li x: theo nh lut II Niu Tn:

............................................................

vi a l gia tc ca nm so vi sn.

+ Trong hqc gn vi bn, vi nm ta c:

.....................................................

thay (1) vo biu thc va tm ta c:

+ Thay (3) vo (2) cho ta:

chng t m dao ng iu ho so vi nm vi chu k:

Bi 6 (6 im). Cho con lc l xo l tng K = 100N/m,

m1 = 200gam, m2 = 50gam, m0 = kg. B qua

lc cn khng kh, lc ma st gia vt m1 v mt sn.

H s ma st gia vt m1 v m2 l . Cho g = 10m/s2.

1) Gi s m2 bm m1, m0 c vn tc ban u v0 n va chm n hi xuyn tm vi m1, sau va chm h (m1 + m2) dao ng iu ho vi bin A = 1 cm .

a. Tnh v0.

b. Chn gc thi gian ngay sau va chm, gc to ti v tr va chm, chiu dng ca trc to hng t tri sang phi (hnh v). Vit phng trnh dao ng ca h (m1 + m2). Tnh thi im h vt i qua v tr x = + 0,5 cm ln th 2011 k t thi im t = 0.

2) Vn tc v0 phi trong gii hn no vt m1 v m2 khng trt trn nhau (bm nhau) trong qu trnh dao ng ?

1) a. t m1 + m2 = 250 g = 0,25 kg, p dng hai LBT ta tnh c vn tc hai vt sau va chm: (1)

Hai vt dao ng iu ho vi tn s: (2)

Vn tc ca hai vt ngay sau va chm chnh l vn tc cc i ca dao ng. T cng thc (1), vi A = 1 cm, ta c: (3)

b. Lc t = 0, ta c:

Phng trnh dao ng ca h (m1 + m2) l: .

+ Dng PP vc t quay, ta tm c thi im vt i qua v tr c li x = + 0,5 cm ln th 2011 l: t = t1 + t2 =

2) Khi hai vt ng yn vi nhau th lc lm cho vt m2 chuyn ng chnh l lc ma st ngh gia hai vt, lc ny gy ra gia tp cho vt m2 :

(5)

M: (6)

T (5) v (6) ta c:

Cu 7 (5,0 im): Mt si dy cao su nh n hi c cng k = 25N/m u trn c gi c nh, u di treo vt m = 625g. Cho g = 10m/s2, .

1) Ko vt ri khi v tr cn bng theo phng thng ng hng xung di mt on bng 5cm ri th nh cho vt dao ng iu ha. Chn gc thi gian l lc th vt, gc ta ti v tr cn bng, chiu dng hng xung.

a) Vit phng trnh dao ng ca vt.

b) Tnh tc trung bnh ca vt k t lc bt u chuyn ng n lc vt qua v tr c x = -2,5cm ln th 2.

2) Vt ang v tr cn bng, truyn cho vt vn tc 2m/s hng thng ng xung di. Xc nh cao cc i ca vt so vi v tr cn bng.

- Phng trnh dao ng ca vt c dng: .

- Tn s gc: ..

- Ti thi im t = 0: .

- Phng trnh dao ng l: ..

- T mi quan h gia dao ng iu ha v chuyn ng trn u ta xc nh c thi

gian k t lc vt bt u chuyn ng n lc vt qua v tr

x = -2,5cm l:

EMBED Equation.3

- Tc trung bnh: ttb

- Ti v tr cn bng gin ca dy l V vy vt ch dao ng iu ha khi A < 25cm..

- Nu ti VTCB truyn vn tc v = 2m/s th bin c th t l , nn khi i ln qua v tr 25cm th dy b chng do vy vt khng dao ng iu ha..

- p dng nh lut BTNL, chn gc th nng hp dn ti VTCB th :

Ti VTCB: W1 = Ti v tr cao nht: W2 = mghmax..

W1 = W2 => hmax = 32,5cm.

Bi 7(5,0 im)Vt nng c khi lng m nm trn mt mt phng nhn nm ngang, c ni vi mt l xo c cng k, l xo c gn vo bc tng ng ti im A nh hnh 2a. T mt thi im no , vt nng bt u chu tc dng ca mt lc khng i F hng theo trc l xo nh hnh v.

a) Hy tm qung ng m vt nng i c v thi gian vt i ht qung ng y k t khi bt u tc dng lc cho n khi vt dng li ln th nht.

b) Nu l xo khng khng gn vo im A m c ni vi mt vt khi lng M nh hnh 2b, h s ma st gia M v mt ngang l (. Hy xc nh ln ca lc F sau vt m dao ng iu ha.

Bi 7(5)a) Chn trc ta hng dc theo trc l xo, gc ta trng vo v tr cn bng ca vt sau khi c lc F tc dng nh hnh 1. Khi , v tr ban u ca vt c ta l x0. Ti v tr cn bng, l xo b bin dng mt lng x0 v:

0.5

Ti ta x bt k th bin dng ca l xo l (xx0), nn hp lc tc dng ln vt l:

0.5

Thay biu thc ca x0 vo, ta nhn c:

0.5

Trong . Nghim ca phng trnh ny l:

0.25

Nh vy vt dao ng iu ha vi chu k . Thi gian k t khi tc dng lc F ln vt n khi vt dng li ln th nht (ti ly cc i pha bn phi) r rng l bng 1/2 chu k dao ng, vt thi gian l:

0.5

Khi t=0 th:

0.5

Vy vt dao ng vi bin F/k, thi gian t khi vt chu tc dng ca lc F n khi vt dng li ln th nht l T/2 v n i c qung ng bng 2 ln bin dao ng. Do , qung ng vt i c trong thi gian ny l:

0.5

b) Theo cu a) th bin dao ng l

sau khi tc dng lc, vt m dao ng iu ha th trong qu trnh chuyn ng ca m, M phi nm yn.

0.5

Lc n hi tc dng ln M t ln cc i khi bin dng ca l xo t cc i khi vt m xa M nht (khi l xo gin nhiu nht v bng: ).

0.5

vt M khng b trt th lc n hi cc i khng c vt qu ln ca ma st ngh cc i:

0.5

T suy ra iu kin ca ln lc F:

0.25

Bi 8 (4 im) Hai qu cu nh m1 v m2 c tch in q v -q, chng c ni vi nhau bi mt l xo rt nh c cng K (hnh 1). H nm yn trn mt sn nm ngang trn nhn, l xo khng bin dng. Ngi ta t t ngt mt in trng u cng , hng theo phng ngang, sang phi. Tm vn tc cc i ca cc qu cu trong chuyn ng sau . B qua tng tc in gia hai qu cu, l xo v mt sn u cch in.

Bi8.Do tng ngoi lc tc dng h kn theo phng ngang nn khi tm ca h ng yn v tng ng lng ca h c bo ton. Chn trc Ox c phng ngang hng sang phi, gc O khi tm ca h. Ta c:

m1v1 + m2v2 = o v2 = -(1)

.Vt m1 v m2 s dao ng iu ha xung quanh v tr cn bng ca chng, ti hp lc tc dng ln mi vt bng 0 v vn tc ca chng t cc i. Ta c:

qE = k(x1-x2) (2)

++= qE(x1-x2) (3)

.T (1) v (2) v (3) ta c:

V1=, V2=

Cu 9(4): Mt con lc l xo treo thng ng gm vt nh khi lng m = 250g v mt l xo nh c cng k = 100 N/m. Ko vt m xung di theo phng thng ng n v tr l xo gin 7,5 cm ri th nh. Chn gc ta v tr cn bng ca vt, trc ta thng ng, chiu dng hng ln trn, gc thi gian l lc th vt. Cho g = 10m/s2. Coi vt dao ng iu ha

a. Vit phng trnh dao ng

b. Tnh thi gian t lc th vt n thi im vt i qua v tr l xo khng bin dng ln th nht.

c. Thc t trong qu trnh dao ng vt lun chu tc dng ca lc cn c ln bng trng lc tc dng ln vt, coi bin dao ng ca vt gim u trong tng chu k tnh s ln vt i qua v tr cn bng k t khi th.

a. Vt chu tc dng ca 2 lc: trng lc

v lc n hi ca l xo:

- Ti VTCB c:

- Phng trnh dao ng ca vt c dng:

Vi

-Ti lc t = 0

Vy pt:

b. Vt bt u chuyn ng n lc x = 2,5 cm th l xo ko gin l th nht. khi ta c bn knh vc t ca chuyn ng trn u qut c mt gc

c.Gi A1, A2, .., An l bin dao ng ca vt trong nhng ln k tip. Mi ln vt i qua v tr cn bng nng lng gim:

Vy s ln vt i qua v tr cn bng l: ln

Cu 10Mt con lc l xo treo thng ng gm vt nng c khi lng m = 100(g) v l xo nh c cng k = 100(N/m). Nng vt nng ln theo phng thng ng n v tr l xo khng b bin dng, ri truyn cho n vn tc (cm/s) thng ng hng ln. Chn gc thi gian l lc truyn vn tc cho vt nng. Chn trc ta Ox thng ng, chiu dng hng xung, gc ta O v tr cn bng.

Ly g = 10(m/s2); .

a) Nu sc cn ca mi trng khng ng k, con lc l xo dao ng iu ha. Tnh:

- ln ca lc n hi m l xo tc dng vo vt lc t = 1/3(s).

- Tc trung bnh ca vt trong khong thi gian 1/6(s) u tin.

b) Nu lc cn ca mi trng tc dng ln vt nng c ln khng i v bng FC=0,1(N). Hy tm tc ln nht ca vt sau khi truyn vn tc.

+ Khi vt VTCB (rad/s)

+ Phng trnh dao ng ca vt: (cm)

+ t =1/3(s) => x = 2(cm). ln lc n hi: Fh=k= 3(N)

+ Biu din bng vc t quay .

Sau t =1/6s quay

Qung ng vt dao ng iu ha

i c sau 1/6s l:

S= 2A+ 2HM = 2A + A=3A=6cm

+ Tc trng bnh:

Vtb=

Chn mc tnh th nng l VTCB

+ C nng ban u W0 =

+ Vt chuyn ng chm dn n v tr cao nht cch VTCB A:

Cu 11. (2,5 im) Mt con lc l xo c treo thng ng gm vt nng khi lng m = 1kg, l xo nh c cng k = 100N/m. t gi B nm ngang vt m l xo c chiu di t nhin. Cho gi B chuyn ng i xung vi gia tc a = 2m/s2 khng vn tc ban u.

a. Tnh thi gian t khi gi B bt u chuyn ng cho n khi vt ri gi B.

b. Chn trc ta c phng thng ng, chiu dng hng xung, gc ta ti v tr cn bng ca vt, gc thi gian l lc vt ri gi B. Vit phng trnh dao ng iu ha ca vt.

Cu 11(2,5 )a. Tm thi gian

( Khi vt VTCB l xo gin:

Tn s ca dao ng:

( Vt m: .

Chiu ln Ox: mg - N - k = ma

Khi vt ri gi th N = 0, gia tc ca vt a = 2 m/s2( Suy ra:

b. Vit phng trnh

( Qung ng vt i c cho n khi ri gi l

Ta ban u ca vt l: x0 = 0,08 - 0,1 = - 0,02 m = -2 cm

Vn tc ca vt khi ri gi l: v0 = at = cm/s

( Bin ca dao ng: = 6 cm

Ti t = 0 th 6cos = -2 v v ( 0 suy ra = -1,91 rad

Phng trnh dao ng: x = 6cos(10t - 1,91) (cm)

Cu 12 (2 im).Mt con lc l xo gm vt nng c khi lng , l xo nh c cng . Khi M ang v tr cn bng th th vt ri t cao so vi M (Hnh 1). Coi va chm gia m v M l hon ton mm. Sau va chm, h M v m bt u dao ng iu ha. Ly . a) Tnh vn tc ca m ngay trc va chm v vn tc ca hai vt ngay sau va chm.

b) Vit phng trnh dao ng ca h (M+m). Chn gc thi gian l lc va chm, trc ta Ox thng ng hng ln, gc O l v tr cn bng ca h sau va chm.

c) Tnh bin dao ng cc i ca h vt trong qu trnh dao ng vt m khng ri khi M

aVn tc ca m ngay trc va chm:

Do va chm hon ton khng n hi nn sau va chm hai vt c cng vn tc V

bTn s dao ng ca h:. Khi c thm m th l xo b nn thm mt on:. Vy VTCB mi ca h nm di VTCB ban u mt on 1cm

Tnh A: (cm)

Ti t=0 ta c:

Vy:

cPhn lc ca M ln m l N tha mn:

(

m khng ri khi M th

EMBED Equation.DSMT4 Vy

Cu 13 (2,5 im).

Cho con lc l xo gm l xo nh c cng , vt nng kch thc nh c khi lng (Hnh 2). Kch thch cho vt dao ng iu ha theo phng thng ng. Chn gc thi gian l lc vt qua v tr c li vi tc theo phng thng ng hng xung di. Chn trc ta Ox theo phng thng ng, chiu dng hng ln trn, gc O trng vi v tr cn bng ca vt. Ly .

a) Vit phng trnh dao ng ca vt.

b) Tnh khong thi gian ngn nht vt i t v tr c li n v tr c li .

c) Tnh qung ng i c ca vt k t lc bt u dao ng n khi ti v tr c ng nng bng th nng ln th hai.

aTn s gc

Ti t = 0, ta c:

( Phng trnh dao ng

bKhong thi gian ngn nht vt i t v tr c li x1 = -2,5cm n v tr c li x2 = 2,5cm

cQung ng vt i t v tr ban u ti v tr c ng nng bng th nng ln th 2

Bai 14: Mot con lac gom mot vat nang co khoi lng m=100g c treo vao

au di cua mot lo xo thang ng au tren co nh. Lo xo co o

cng K=20N/m, vat m c at tren mot gia nam ngang(hnh ve).

Ban au gi gia e lo xo khong b bien dang, roi cho gia chuyen

ong thang xuong nhanh dan eu vi gia toc a=2m/s2. Lay g=10m/s2.

1- Hoi sau bao lau th vat ri khoi gia ?

2- Cho rang sau khi ri gia vat dao ong ieu hoa.Viet phng trnh dao ong cua vat. Chon goc thi gian luc vat va ri gia , goc toa o v tr can bang, truc toa o thang ng, chieu dng hng xuong

* Chon truc toa o Ox thang ng, chieu dng hng xuong, goc O la v tr can bang cua m. Ban au lo xo khong bien dang vat v tr B. Goc thi gian luc cho gia chuyen ong.

*Khi cha ri gia , m chu tac dung cua:trong lc, lc an hoi, phan lc

Theo nh luat II Newton:

*Gia s en C vat ri gia , khi o N= 0, vat van co gia toc a=2m/s2:

. Chieu len Ox: P F = ma hay mg k.BC = ma. B

Suy ra: BC = m C

*Mat khac : goi t la thi gian t luc bat au chuyen ong en luc ri gia , ta co O

x

.

*Tan so goc:

*-o gian cua lo xo v tr can bang:

-Van toc vat tai C :VC = at = 2.0,2 = 0,4 m/s.ieu kien au: t=0

*Giai

* Phng trnh

Phng trnh

Cu 15: Mt con lc lo xo gm vt nng M=300g, cng k=200N/m nh (hnh v 3). Khi M ang

v tr cn bng th vt m=200g t cao h=3,75cm so vi M.Sau va chm h M v m

bt u dao ng iu ha . Bqua ma st,ly g=10m/s2 .Coi va chm gia m v M

l hon ton khng n hi.a.Tnh vn tc ca m ngay trc va chm,v vn tc ca hai vt ngay sau va chm

b.Vit phng trnh dao ng ca h (M+m) chn gc thi gian l lc va chm ,

trc ta 0x thng ng hng ln gc 0 l v tr cn bng ca h sau va chm.

c. Tnh bin dao ng cc i ca hai vt trong qu trnh dao ng

vt m khng ri khi M

3

(4,5)a Vn tc ca m ngay trc va chm: (m/s)=(cm/s)0,5

Do va chm hon ton khng n hi nn sau va chm vng v a c cng vn tc V

(m/s)=(cm/s)

0,5

bVit PT dao ng:(rad/s). Khi c thm m th l xo b nn thm mt on:(cm) vy VTCB mi ca h nm di VTCB ban u mt on 1cm0,75

Tnh A: (cm)0,5

Ti t=0 ta c:(rad/s)0,5

Vy: x=2cos(20t+) (cm)0,5

cLc tc dng ln m l:

Hay N=

0,75

m khng ri khi M th

EMBED Equation.DSMT4 Vy (cm)0,5

Bi 16: (4,0 im)

C mt s dng c gm mt qu cu nh c khi lng m, mt l xo nh c cng k v mt thanh cng nh OB c chiu di l.

1) Ghp l xo vi qu cu to thnh mt con lc l xo v treo thng ng nh hnh v (H.1). Kch thch cho con lc dao ng iu ho vi bin A = 2cm. Ti thi im ban u qu cu c vn tc v gia tc a = - 4m/s2. Hy tnh chu k v pha ban u ca dao ng.

2) Qu cu, l xo v thanh OB ghp vi nhau to thnh c h nh hnh v (H.2). Thanh nh OB treo thng ng. Con lc l xo nm ngang c qu cu ni vi thanh. v tr cn bng ca qu cu l xo khng b bin dng. T v tr cn bng ko qu cu trong mt phng cha thanh v l xo thanh OB nghing vi phng thng ng gc 0 < 100 ri bung khng vn tc u.

B qua mi ma st v lc cn.

Chng minh qu cu dao ng iu ho. Cho bit: l = 25cm,

m = 100g, g = 10m/s2 . Tnh chu ky dao ng ca qu cuBi 17: (4,00 im)

1) Chu k v pha ban u ca dao ng (2,00 im):

- Chu ky: Ta c h thc: ( (1)

0,25 t X = 2, thay cc gi tr ca v0 v a0 ta i n phng trnh bc hai:

4X2 1200X 160000 = 0

(2)

0,25 X2 300X 40000 = 0

Phng trnh cho nghim:

(3)

0,25 Chn nghim thch hp: X = 400 2 = 400 = 20(rad/s)

Vy chu k dao ng:

(4)

0,25 - Pha ban u:

Ti t = 0, ta c: v0 = -Asin = (2)

a0 = -A2co = - 4m/s2 = -400cm/s2. (5)

0,50

T (3): ;

T (2): chn

(6)

0,50 2) H dao ng iu hoa - Chu ky: (2,00 im)

Ti thi im t, qu cu c to x v vn tc v, thanh treo OB c gc lch so vi phng thng ng. Biu thc c nng c nng ton phn ca h:

(7)

Chon gc th nng ti VTCB:

. (8)

0,50 Do nn .

C nng ton phn ca h:

(9)

0,50 Ly o hm bc nht ca c nng E theo thi gian:

V v = x, v = x nn : (10)

Vy qu cu dao ng iu ho vi tn s gc: (11)0,50 - Ta lai co: k = m2 = 0,1.400 = 40N/m.

Vy:

Chu k dao ng:

(12)0,50 Cu 18: Mt con lc l xo nm ngang c cng , vt nh khi lng . Ban u gi vt sao cho l xo b nn 10(cm) ri th nh.

1. B qua mi ma st, vt dao ng iu ho.

a) Vit phng trnh dao ng ca vt, chn gc O l v tr cn bng ca vt, chiu dng l chiu chuyn ng ca vt lc th, gc thi gian lc th vt.

b) Xc nh thi im l xo nn 5cm ln th 2010 k t lc th.

2. Thc t c ma st gia vt v mt bn vi h s ma st trt gia vt v mt bn l . Ly . Tnh tc ca vt lc gia tc ca n i chiu ln th 4.

Phng trnh dao ng:

trong :

Vy:

+ Ta thy l xo nn 5cm cc ln chn lin tip cch nhau mt chu k, do l xo nn

ln th 2010 ti thi im: vi t2 l thi im l xo nn 5cm

ln th 2.

+ Ta xc nh thi im l xo nn 5cm ln

th hai, s dng pp vec t quay ta c: k t

thi im ban u n lc l xo nn 5cm ln

th 2 th vect quay mt gc:

+ Do thi im l xo nn 5cm ln th 2010 l:

+ Lc c ma st, ti VTCB ca vt l

xo bin dng mt on:

+ Ta thy c hai VTCB ca vt ph thuc vo chiu chuyn ng ca vt, nu vt isang phi lc l xo nn 2,5mm th VTCB l bn tri O(v tr C1), lc vt i sang tri m l xo gin 2,5mm th VTCB l bn phi O( v tr C2)

+ p dng inh lut bo ton nng lng, ta tnh c gim to cc i sau mi ln qua O l hng s v bng:

+ Gia tc ca vt i chiu ln th 4 ng vi vt i qua VTCB C2 theo chiu sang tri ln th 2, p dng nh lut bo ton nng lng ta c:

Cu 19: Cho c h gm c mt vt nng c khi lng m c buc vo si

dy khng dn vt qua rng rc C, mt u dy buc c nh vo im A.

Rng rc C c treo vo mt l xo c cng k. B qua hi lng ca l xo, rng rc v ca dy ni. T mt thi im no vt nng bt u chu tc dng ca mt lc khng i nh hnh v

a. Tm qung ng m vt m i c v khong thi gian k t lcvt bt u chu tc dng ca lc n lc vt dng li ln th nht

b. Nu dy khng c nh A m ni vi mt vt khi lng M (M>m).Hy xc nh ln ca lc F sau vt dao ng iu ha

Vt cn bng khi cha tc dng lc F: mg = k

Chn trc Ox thng ng t trn xung. O trng vi VTCB mi khi c lc F tc dng.

Ti VTCB mi: F + P - = 0 (vi xo l khong cch gia VTCB mi so vi VTCB c)

Khi vt c li x l xo gin: + x

F + P - = mx x + x = 0

Vy vt DH vi phng trnh: x = Acos()

Trong

Nh vy chu k dao ng ca vt T = . Thi gian t lc tc dng lc n khi vt dng li ln th nht l

Khi t = 0: x = Acos() = - xo = -

V = -A = 0

A = ,

S = 2A =

Lc tc dng ln M nh hnh v

m dao ng iu ho sau khi tc dng lc F th M phi ng yn N trong qu trnh m chuyn ng

N = P - 0 Mg - = Mg -k 0

F Mg

EMBED Equation.DSMT4

m

M


O

(

(

(

(

(

- A

O

A

- x1

x1

x

m

x

O

-1

x

M

N

K

K'

O

m

K

M

300

Hnh 1

m

N

Fq

P

Fd

N

P/

Q

O

X

m2

m1

m0


K

O

x

-5 -2,5 O 5

F

m

k

Hnh 2a

A

F

m

k

Hnh 2b

M

F

m

k

Hnh 1

O

x0

EMBED Equation.3 K

(Hnh 1)

m1,q K m2, - q

EMBED Equation.3 K

o

m1,q K m2, - q

.

x

EMBED CDraw5

x

0

EMBED Equation.3

2,5

H

M



x

o

-A

A

m

k

EMBED Equation.3

EMBED Equation.3

B

O

x

0,5

M

Hnh 1

m

k

h

M

N

O

x

5

- 5

2,5

- 2,5

(

(

M

N

O

5

- 5


2,5

P

Q

(Ln 1)

(Ln 2)

m

Hnh 2

k

Hnh 2

l

O

(H.2)

(H.1)

B

C2

C1

O

10

-10

-5

M2

M1

x

m

k

m

k



M

A

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