S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin1 LP 10 CHNG I C HC NG HC Chuyn ng c l s di ch ca vt theo thigian. Cht im:l vt c kch thc rt nh so vi phm vi chuyn ng ca nnn ta b qua kch thc ca n , xem n nh mt im hnh hc c khi lngm. Qu o ca cht im :l con ng cht im vch ra trong khng gian khi n chuyn ng . Chuyn ng tnh tin l chuyn ng ca vt m qu oca mi im trn vtu ging ht nh nhau ,c th chng kht ln nhau c . di 1 2M M, ca mt cht iml mt vc t c : -Phng l ng thng i qua v tr M1 thi im t1 v v tr cui M2 thi im t2. -Chiu t M1 n M2 - ln bng chiu di on thng 1 2M M di trong chuyn ng thng 2 1X X X A = ; Trong X1 ,X2 l ta ca cht im ti cc thi im t1 ,t2 tng ng . x s A= A( s Alun dngl qung ng i c ca cht im). H quy chiu = h ta gn vi vt lm mc +ng h v gc thi gian Vc t vn tc trung bnh 1 2tbM Mvt=A,, Vn tc trung bnh tbVl gi tr i s ca tbv, 2 12 1tbx x xVt t t A= = A

Nu0 x A >th tbVdng :Chiu dng ca trcox cng chiuvi tbv, Nu0 x A 0 ; chuyn ng thng chm dn ua.v < 0. Ri t do :l s ri theo phng thng ngch di tc dng ca trng lc ( ). P m g =,,. .tv g t =21. .2h g t = 22. .tv g h = g : gia tc ri t do (gia tc trng trng )g, c hngthng ng t trn xung, gi tr g ph thuc v a l, cao v cu trc a l ni o.Thng ly g, =9,8 m/s2 , khng yu cu chnh xc cao c th ly 2210mgst = =, Chuyn ng trn u :l chuyn ngc qu o l ng trn v cht im i c nhng trn c di bng nhau trong nhng khong thi gian bng nhau bt k Gc quay : l gcm bn knh ni t tm n vtqut c khi vt i c cung trn c distrong khong thigian tsR =

0s M M =3,14( rad) Tng ng 1800 Chu k quay T : l khong thi gian vt i c mt vng trn . Tn sn : l s vng m vt quay c trong mt n v thi gian (1 giy) 1Tn= 1nT= Vn tc gc e:22. .nt t te t = = ={M rng: vc te, c ln le,nm trn trc quay , c hng ca ngn cikhi chng ta a bn tay phi nm ly trc quay sao cho chiu nm 4 ngn tay l chiu quay ca vt v ngn ci vung gc ngn tr} Vn tc di .sv RteA= =A{M rng:dng vc t| | , v r e =, , , trong r, l vc t gc tm quay ,ngn vt} Gia tc hng tm hta22.h tva rre = =hta, lun nm theo bn knhhng vo tm qu o . n v : T,t:(s)s,R (m)( ) rad n (Hz)( / ) rad s e v(m/s)a(m/s2) Tnh tng i ca chuyn ng i vi cch quy chiu khc nhau th ta (v qu o ) ca vt s khc nhau .Vn tc ca cng mt vt i vi cc h quy chiu khc nhau th khc nhau . Cng thc cng vn tc1,3 1,2 2,3v v v = +, , ,;12v, l vn tc cavt 1 sovi vt 2; 23v, l vn tc cavt 2 sovi vt 3 ; 13v, l vn tc cavt 1 sovi vt 3 .Ch :12 21 23 32 13 31; ; v v v v v v = = = , , , , , , LP 10 CHNG IING LC HC CHT IM CC NH LUT NIU TN e, v, r, S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin3 Lc : l i lng c trng cho tc dng ca vt ny vo vt khcm kt qu l gy ra gia tc cho vthay lm vt bin dng .n v lc l N (Niu tn) Tng hp lc :l thay th nhiu lc tc dng ng thivo cng mt vt bngmt lc c tc dng ging htnh ton b cc lc y .Lc thay th gi l hp lc .cc lc c thay th gi l cc lc thnh phn . Phn tch lc :l thay th mt lcbng hai hay nhiu lc thnh phnc tc dng ng thi v gy hiu qu ging ht nh lcban u .cc lc thay th gi l cc lc thnh phn . (tng hp v phn tch lc phi tun theo quy tc cng vc t) H vt : l mt tp hp hai hay nhiu vt m gia chng c tng tc lc vi nhau .Cc lc tng tc gia cc vt trong trong hgi l ni lc ,lc do cc vt ngoi h tc dng ln cc vt trong h gi l ngoi lc H kn: l h vt ch c cc ni lc khng c ngoi lc hoc c cc ngoi lc song cc ngoi lc trit tiu ln nhau (tng cc vc t ngoi lc bng khng ). nh lut I Niu tn :" Khi khng chu tc dng ca lc no hoc khi chu tc dng ca cc lc cn bng (hp lc =0),mt vt ang ng yn s tip tc ng yn,ang chuyn ng s tip tc chuyn ng thng u ". Qun tnh: l tnh cht ca mi vtc xu hng bo ton vn tcc v hng v ln .(chuyn ng thng u gi l chuyn ng theo qun tnh ) H quy chiu qun tnh :l h quy chiu trong nh lut I Niu tn c nghim ng .H quy chiu gn vi mt t hoc chuyn ng thng uso vi mt t c th coi lh quy chiu qun tnh (gn ng).

10 0niiF a== =,, nh lut II Niu tn: "Gia tc ca mt vt lun cng hng vi lc (hoc hp lc )tc dng ln vt . ln ca gia tct l thun vi lc (hoc hp lc ) tc dng ln vtv t l nghch vi khi lng ca vt . " Khi lng ca vt l i lng c trng cho mc qun tnh ca vt . nh lut III Niu tn:" Trong mi trng hp ,khi vt A tc dng vo vt Bmt lc ( )A BF ,th vt B cng tc dng li vt A mt lc ( )B AF ,.hai lc ny l hai lc trc i. "A B B AF F = , ,

Tng tc gia hai vt ,mt lc gi l lc tc dung ,lc kia gi l phn lc .Chng lun xut hin tng cp ,cng bn cht ,xut hin v mt i ng thi ,c gi l ng thng ni hai vt . Lc qun tnh : Trong h quy chiu phi qun tnhxut hin lc qun tnh qtF, .qtF ma = ,,( a, l gia tch quy chiu phi qun tnh i vi h quy chiu qun tnh ).lc qun tnh khng c phn lc. nh lut vn vt hp dn ; "Hai vt bt k u ht nhau .Lc ht gia hai vt (coi nh cht im )t l thun vi tch ca hai khi lng cachng v t l nghch vi bnh phng khong cch gia chng"1 22..hdm mF Gr=G=6,67.10-11 m2N/kg2 m1m2 :khi lng ca hai vt(hai cht im ) r : khong cch gia hai vt Lc ma st trt: xut hin mt tip xc ca hai vt ang trt trn nhau.c hng ngc vi hng ca vn tc tng ica vt ny so vi vt kia ,c ln t l vi ln ca p lc (lc php tuyn ) ..ms trF N =Vi l h s ma st ph thuc tnh cht mt tip xc v bn cht vt liu .N l p lc Lc ma st ln :Xut hin mt tip xc khi mt vt ln trn mt vt khc .c ln t l vi ln ca p lc .h s ma st ln nh hn h s ma st trt hng chc ln1niiFam==,,S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin4 Lc ma st ngh:Xut hin mt tip xcv gi cho vt ng ynkhi n b mt lctc dng song song mt tip xc .Khng c hng xc nh ,hng ca n ngc vi hng ca lc tc dng .Khng c ln nht nh , ln ca n bng ln ca lc tc dng .C ln cc i ,max.msnF k N = Vi k ~ ( thc chtkhi ln hn;cng c trng hp k vchnh lch ng k ) Lc n hi:L lc xut hin khi mt vt b bin dng ,c xu hng chng li nghuyn nhn gy ra bin dng . Lc n hi ca l xuthin c hai u ca n v tc dng vocc vt tip xc (hay gn) vi n lm n bin dng .Khi b dn lc n hi ca l xo hng vo trong l xo cn khi b nn lc n hi ca l xo hng ra ngoi. nh lut Hc : "Trong gii hn n hi ,lc n hi ca l xo t lvi bin dng ca l xo " .hF k l = A Vi kl cng vt n hi . l Al bin dng vt n hi. (Du tr ch lc n hi lun ngc vi chiu bin dngtc chiu dch chuyn tng i ca mi u l xo so vi u kia) i vi mt si dyb ko cng n s tc dngln hai vt buc hai u dynhng lc (hay gi lc cng)c c im l :im t ca lc l im m u dy tip xc vi vt ,phng ca lc trng vi chnh si dy vchiu ca lc hng t hai u dy vophn gia dy. i vi cc mt tip xc b bin dng khi p vo nhau lc n hi c gi l lc php tuyn . Trng lcP,:hd qtP F F = +, , , hdF, l lc hp dngia vt v qu t. qtF, l lc qun tnhli tm xut hin do s t quay ca tri t quanh trc n.ni chung phng hdF, khc phng qtF,,tr vng xch o . {Lu : hdF, ph thuc hnh dng v s phn b khi lngca qu tcng nh ca vt .Nu xem vt nh l mt cht im khi lng m cch mt t mt caoh. Mt cch gn ng xem qu t hnh cu phn b khi lng ng u quanh tm thhdF, c hng t vt n tm qu tv( )2..hdmMF GR h=+, qtF, c hng ra xa tm quay ca vt quanh trc qu t . ( )2. . cosqtF m R h e = +,vil v a l, ( )57, 27.10 / rad s e=l vn tc gc quay quanh qu t .Theo nh lut 2 Niu tn th. P m g =,, . Nh vyg, ph thucvo cu to a chtca vng o n , cao ni og,,v a l .g, c hng khng i qua tm qu t .Trong trng hp h R 2cth 1 2 bc c c = 1 2 br r r = + dng in i ra t cc dng ca 1c Mc hn hp i xng mhngmi hng cnngun ging nhau . .bn c c =.bnr rm=9. in nng v cng sut ca dng in .nh lut Jun-Len-x Cng ca dng in. . A U I t =(J) Cng sut ca dng in . P U I =(w) nh lut Jun-Len-x2. . Q R I t = (J) S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin15 Cng ca ngun in. .ngA I t c = (J) Cng sutca ngunin.ngP I c = (w) Cng sut ca dng c tiu th in ch ta nhit22. .UP U I R IR= = = (w) Cng sut ca my thu in2. .p pP I r I c = +(w) LP 11 CHNG IIIDNG IN TRONG CC MI TRNG 1.Dng in trong kim loi Dng in trong kim loi l dng dch chuyn c hng cacc electron t do . S va chm ca electron t do vicc ch mt trt t camng tinh th l nguyn nhn gy ra in tr v tc dng nhit ca dy dn kim loi. S khuch tn cc electron t do ga cc kim loi khc nhaukhi chng tip xc nhau l nguyn nhnca hin tng nhit in . Khi nhit gim xung di mt gi tr Tc no ,in tr ca kim loi ( hay hp kim ) gim t ngt xung gi tr bngkhng , l hin tng Siu Dn 2.Dng in trong cht in phn Dng in trong cht in phn l dng dch chuyn c hng cacc Ion dng ( v Ca tt-in cc m) vI on m(V A nt - in cc dng) . cc Ionkhi v in ccs trao i electronvi cc in cc, to thnh nguyn t hay phn t trung hari gii phng ra , hoc tham gia cc phn ng ph .Mt trong cc phn ng ph l phn ng cc dng tan, phn ng ny xy ra trong ccbnh in phn mc A nt l kim loim mi ca n c mt trong dung dch in phn . nh lut Fa- ra - y vin phn 1. . .Am I tF n=trong m l khi lng cht gii phng ra in cc ( tnh bng gam) .A l nguyn t lng ca cht .nl ha tr ca ca cht y.I l cng dng in v t l thi giandng in i qua cht in phn . F=96 500 C/mol. 3.Dng in trong chn khng Dng in trong chn khng l dng chuyn di c hng ca cc electron b bt ra t ca tt b nung nng do tc dng ca in trng . c im l dng in trong chn khng ch chy theo mt chiu nht nh t A nt sang Ca tt4.Dng in trong cht kh Dngin trong cht kh l dng chuyn di c hng ca cc Ion dngv Ca tt ,cc Ion m v cc electron v A nt. Khi in trng cn yu , mun c cc Ion v cc electrondn in trong cht khcn phi c tc nhn ion ha ( ngn la ,nh sng ....)( phng i khng t lc ).Cn khi in trng trong cht kh mnh thc xy ra s ion ha do va chmlm cho s in tch t dotrong cht kh tng vt ln ( phng in t lc ). S ph thuc giacng dng in vo hiu in thgia A nt v ca ttc dng phc tp , khng tun theo nhlut m (Tr hiu in th rt thp ). Tia la in v h quang inl hai dng phng in trong khng kh iu kin thng . S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin16 C ch ca tia la in l s Ion ha do va chmkhi cng in trng trong khng kh ln hn 3.106 v/m. Khi p sut trong cht kh ch cn vo khong 1 n 0,001 mmHg trong ng phng in c s phng in thnh min : Ngay phn mt Ca ttc min ti ca tt , phn cn li ca ng cho n A ntl ct sng A nt . Khi p sut trong ng gimdi 10-3 mmHg th min ti Ca tt chimton b ng , lc ta c tia Ca tt ( dngelectron) ( pht x lnh ). 5.Dng in trong bn dn Dng in trong bn dn l dng dch chuyn c hngca c electron t do v cc l trng . Ty theo loi tp cht pha vo bn d tinh khit ma fta c bn dn loinhay bn d loip.Dng in trong bn dn loinch yu l dng elec tron cn trong bn d loi p ch yu l dng cc l trng . Lp chuyn tip p - n c tnh cht dn in ch yu theo mt chiu nht nh t p sang n. LP 11 CHNG IVT TRNG 1.T trng .Cm ng t. Xung quanh nam chm v xung quanh dng in (ni chung l xung quanhin tch chuyn ng) tn ti t trng.T trng c tnh cht c bn l tc dng lc t ln nam chm hay dng in (in tch chuyn ng) t trong n. Cm ng t( B,) l i lng c trngcho t trng v mt tc dng lc t.n v cm ng t l tla k hiu T Cm ng tc ln ph thuc cng dng in , dng mch in v mi trngbao quanhmch in . .oB B =, , oB, vc t cm ng t trong chn khng . B, vc t cm ng t trong mi trng khc chn khng. l hng s gi l t thm ph thuc bn cht mi trng . >1 cht thun t ; < 1 cht nghch t;1 >cht st t . ng sc tl ng cong c hng c vtrong t trng sao chovc t cm ngt ti bt k im notrn ng cong cng c phng tip tuyn vi ng congv c chiu trng chiu vichiu ca ng cong ti im ta xt. (Vc t cng t trng 01H B =, , vi 704 .10 / H m t= ) 2.Lc t tc dng ln mt on dy dn mang dng in . . .sin F B I l o =( cng thc Ampe); ol gc hp bi hng I vB,. Quy tc bn tay tri : Dui thng bn tay tri sao choB, xuynvo lng bn tay , chiu t c tay n ngn gia ch chiu I .Ngn ci choi 90o so vi ngn tr ch chiuF,. 3.T trng ca dng in trong cc mch c dng khc nhau Dy dn thng di 72.10IBr= ;Ngn ci bn tay phichoi 90o ch chiu I ,chiu nm 4 ngn tay ch chiung sc t(ST) .B,tip tuyn ST ,c chiu ST ti tip im . S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin17 Chm :Song songHi t Phn k Khung dy trn7.2. .10 .N IBRt=N :S vng dy trong khung .Chiu nm 4 ngn tay bn tay phitheochiuI,ngn ci choi 90o ch chiu ,ng sc t(ST) xuyn qua khung dy.B, tip tuyn ST ,c chiu ST ti tip im . ng dy di(xlnit) 74. .10 . . B n I t= n:S vng dy ng vi mi mt chiu di ng . Chiu nm 4 ngn tay bn tay phitheo chiuI, Ngn ci choi 90o ch chiu ,ng sc t(ST)xuyn qua khung dy.B,tip tuyn ST ,c chiu ST ti tip im .4.m men ngu lc t. . .sin M I B S u =S: din tch phn mt phng gii hn bi khung . ul gc hp bivc t php tuyn ca khung vB, 5.Lc Lo-ren x. . .sin f q v B o = ol gc hp biv,vB, 6.Nguyn l chng cht t trng 1 2 3...nB B B B B = + + + +, , , , , LP 11 CHNG VCM NG IN T 1.T thng qua din tch S : . .cos B S | o =n v V-be ( Wb) ; ol gc hp bi vc t php tuyn ca mt phng khung dy v B, 2. t cm ca ng dyLi|= n v Hen ry (H) 3.Sut in ng cm ng trong mch in kn ct|cA= A(v) ln sut in ngcm ng trong mt on dy chuyn ng : . . .since Bl v o = (v) o l gc to biv,ca thanh vB,( v,vB,cng vung gc vi thanh ). Sut in ng t cm .tcie LtA= A(vn) 4.Nng lngt trng trong ng dy21. .2W Li =(J) 5.Mt nng lng t trng 7 21.10 .8.W Bt=6.Dng in Fu C : dng in cm ngc sinh ra trong khi vt dnkhi vt dn chuyn ng trong t trng hayc t trong t trng bin thin theo thi giangi l dng in Fu- c. 7.nh lutLen X : Dng in cm ng c chiu sao chot trng do n sinh rac tc dng chng li nguyn nhn sinh ra n . (du tr cng thc trongmc 3 biu din nh lutLen X) LP 11 CHNG VIPHN X V KHC X NH SNG 1. Cc khi nin Ngun sng l cc vt t pht ra nh sng. S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin18 i i N I S S Gng cu li Vt sng l bao gm cc ngun sng v cc vt c chiu sng. Vt chn sng l vt ngn khng cho nh sng truyn qua. Vt trong sut (mi trng trong sut) l vt (mi trng) cho nh sng truyn qua hu nh hon ton. Tia sngl ng truyn ca nh sng . Chm sng l tp hp ca v s tia sng . 2.nh lut truyn thng nh sng :Trong mt mi trng trong sut v ng tnh , nh sng truyn theo mt ng thng . 3.Nguyn l v tnh thun nghch chiu truyn nh sng : Nu AB l mt ng truyn nh sng ( mt tia sng )th trn ng truyn c th cho nh sngi t A n B , hoc t B v A. 4.nh lut phnx nh sng :Tia phn x trong mt phng ti v bn kia php tuyn so vi tia ti .Gc phn x bng gc ti . ' i i =Mt phng ti l mt phngcha tia ti SI v php tuyn IN 5. Gng a ) Gng phng: l mt phn mt phng phn xnh ng tt . c im gng phng :Vt tht( trc gng) cho nh o(sau gng) .vt o(sau gng) cho nh tht (trc gng).nh i xng vi vt qua mt phng ca gng , nh ln bng vt , khng chng kht c ln vt . ' d d = ' ' '1A B dkd AB= = =Vt tht d > 0 : Vt o d < 0 ; nh tht d > 0 ;nh od0l bin bng li cc i . el mt i lng trung giancho php tnh tn s 2fet= .e gi l tn s gcc n v l (rad/s)..t | e = +gi l pha dao ng thi im t,khi t=0 th | =v c gi l pha ban uca dao ng n v l rad . Mt dao ng iu hac thbiu dinbng mt vc t quay OM, c dibng bin A,vc t ny quay quanh quanhOvi tc gc e , vo thi im ban ut=0 , vc t quay hp vi trc0x mt gc bng pha ban u .Hnh chiu ca vc t quayOM, ln trc ox th bng li ca dao ng . Trong phng php Fresnen mi dao ng iu ha c biu din bng mt vc t c mun bng A , quay quanh gcOca vc tvi vn tcbng tn s gce . Vn tc vt dao ng iu ha ( ) ' . .sin . v x A t e e = = +Vi xc dng nh biu thc () Gia tc ca vt dao ng iu ha( )2' '' . .cos . a v x A t e e = = = +Vi xc dng nh biu thc () Vt VTCB: x = 0;v =,v,Max = eA; a =,a,Min = 0 S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin25 Vt bin: x = A;v = ,v,Min = 0;a =,a,Max = e2A Nu mt vtkhi lngm , mi khi ri khiv tr cn bng 0 mt on x , chu mtlc tc dng . F k x = (lc hi phc ,khng bt buc l lc n hi) th vt y sdao ng iu ha quanh 0 vi tn s gc kme = .bin A v pha ban uph thuc cch kch thch dao ng v chn gc thi gian . Biu thc c lp 2. a x e = 2 22 2 21.v xA A e + = Chiu di qu o: 2A C nng: 2 212tE E E m A e = + =Vi 2 2 2 21sin ( ) sin ( )2E m A t E t e e e = + = + 2 2 2 21cos ( ) cos ( )2tE m A t E t e e e = + = +Dao ng iu ho c tn s gc l e, tn s f, chu k T. Th ng nng v th nng bin thin vi tn s gc 2e, tn s 2f, chu k T/2 ng nng v th nng trung bnh trong thi gian nT/2 ( neN*, T l chu k dao ng) l: 2 212 4Em A e = Khong thi gian ngn nht vt i t v tr c to x1 n x2 2 1t e e AA = =vi 1122coscosxAxA==v (1 2,2 2t t s s ) Qung ng i trong 1 chu k lun l 4A; trong 1/2 chu k lun l 2A Qung ng i trong l/4 chu k l A khi vt xut pht t VTCB hoc v tr bin (tc l = 0; t; t/2) Qung ng vt i c t thi im t1 n t2.Xc nh: 1 1 2 21 1 2 2Acos( ) Acos( )sin( ) sin( )x t x tvv A t v A te e e e e e = + = + = + = + (v1 v v2 ch cn xc nh du) Phn tch: t2 t1 = nT + At (n eN; 0s At < T)Qung ng i c trong thi gian nT l S1 = 4nA, trong thi gian At l S2. Qung ng tng cng l S = S1 + S2 * Nu v1v2> 0 2 2 12 2 1242Tt S x xTt S A x x

A < =

A > =

* Nu v1v2 < 0 1 2 1 21 2 1 20 20 2v S A x xv S A x x> =

< = + + Cc bc lp phng trnh dao ng dao ng iu ho: * Tnh e* Tnh A (thng s dng h thc c lp) S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin26 * Tnh da vo iu kin u: lc t = t0 (thng t0 = 0)00Acos( )sin( )x tv A te e e = + = + Lu : + Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0 + Trc khi tnh cn xc nh r thuc gc phn t th my ca ng trn lng gic(thng lyt < t s ) Cc bc gii bi ton tnh thi im vt i qua v tr bit x (hoc v, a, E, Et, E, F) ln th n * Gii phng trnh lng gic ly cc nghim ca t (Vi t > 0 phm vi gi tr ca k ) * Lit k n nghim u tin (thng n nh) * Thi im th n chnh l gi tr ln th n Lu : ra thng cho gi tr n nh, cn nu n ln th tm quy lut suy ra nghim th n Cc bc gii bi ton tm s ln vt i qua v tr bit x (hoc v, a, E, Et, E, F) t thi im t1 n t2. * Gii phng trnh lng gic c cc nghim * T t1 < ts t2 Phm vi gi tr ca (Vi k e Z) * Tng s gi tr ca k chnh l s ln vt i qua v tr . Cc bc gii bi ton tm li dao ng sau thi im t mt khong thi gian At. Bit ti thi im t vt c li x = x0.* T phng trnh dao ng iu ho: x = Acos(et + ) cho x = x0 Ly nghim et + = o (ng vi x anggim ) hoc et + = t - o(ng vi x ang tng) vi 2 2t to s s* Li sau thi im At giy l: x = Acos(eAt + o) hoc x = Acos(t - o + eAt) = Acos(eAt - o) Dao ng iu ho c phng trnh c bit: Phng trnh dao ng dngx = a Acos(et + ) vi a = const Bin l A, tn s gc l e, pha ban u x l to ,x0 = Acos(et + ) l li .To v tr cn bng x = a, to v tr bin x = a A Vn tc v = x = x0, gia tc a = v = x = x0 H thc c lp: a = -e2x0

2 2 20( )vA xe= + Phng trnh dao ng dngx = a Acos2(et + ) (ta h bc) Bin A/2; tn s gc 2e, pha ban u 2. 2Conlclxo:Lh gmmt vt nng khilngm gnvimt ulxo cngk khilng khng ng k ,u cn li ca l xo c nh .Khi Fms=0th sau khi vt c kch thch th n s dao ng iu haTn s gc: kme = ; chu k: 22mTktte= = ; tn s: 1 12 2kfT met t= = = C nng:2 2 21 12 2tE E E m A kA e = + = = Vi 2 2 2 21 1sin ( ) sin ( )2 2E mv kA t E t e e = = + = +

2 2 2 21 1cos ( ) cos ( )2 2tE kx kA t E t e e = = + = + S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin27 * bin dng ca l xo thng ng khi vt v tr cn bng : mglkA = 2lTgtA=* bin dng calxo nm trnmt phngnghing c gc nghing o v0msF= : sin mglkoA = 2sinlTgtoA=* Trng hp vt di: + Chiu di l xo ti VTCB: lCB = l0 +l A(l0 l chiu di t nhin) + Chiu di cc tiu (khi vt v tr cao nht): lMin = l0 +l A A + Chiu di cc i (khi vt v tr thp nht): lMax = l0 +l A+ A lCB = (lMin + lMax)/2 + Khi A > Al th thi gian l xo nn l t ^^= , vi cos = Al +Thi gian l xo gin l T/2 - At, vi At l thi gian l xo nn (tnh nh trn)* FNmax = FMax = k(Al + A) * Nu A < Al FNmin = FMin = k(Al - A) * Nu A> Al FMin = 0 * Trng hp vt trn: lCB = l0 - l A ;lMin = l0 -l A A; lMax = l0 - l A+ A lCB = (lMin + lMax)/2 Lc hi phc hay lc phc hi (l lc gy dao ng cho vt) l lc a vt v v tr cn bng (l hp lccacclctcdnglnvtxtphngdaong),lunhngvVTCB,clnFhp=k,x,= me2,x,. Lc n hi l lc a vt v v tr l xo khng bin dng. C ln Fh = kx* (x* l bin dng ca l xo) *Viconlclxonmngang thlchi phcvlc nhilmt (v ti VTCBlxo khng bin dng) * Vi con lc l xo thng ng hoc t trn mt phng nghing + ln lc n hi c biu thc: * Fh = k,Al + x, vi chiu dng hng xung * Fh = k,Al - x,vi chiu dng hng ln + Lc n hi cc i : FMax = k( l A+ A) = FKMax + Lc n hi cc tiu: * Nu A < Al FMin = k( l A- A) = FKMin * Nu A>Al FMin = 0 (lc vt i qua v tr l xo khng bin dng) Mt l xo c cng k, chiu di l c ct thnh cc l xo c cng k1, k2, v chiu di tng ng l l1, l2, th ta c: kl = k1l1 = k2l2 = Ghp l xo:* Ni tip 1 21 1 1...k k k= + + cng treo mt vt khi lng nh nhau th: T2 = T12 + T22+... * Song song: k = k1 + k2 + cng treo mt vt khi lng nh nhau th:2 2 21 21 1 1...T T T= + +Gn l xo k vo vt khi lng m1 c chu k T1, vo vt khi lng m2 c T2, vo vt khi lngm1+m2 c chu k T3, vo vt khi lng m1 m2 (m1 > m2)c chu k T4. k m Vt di m k Vt trn S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin28 k m1 m2 Hnh 1 m2 m1 k Hnh 2 Hnh 3 m1 k m2 Q o s s0 O M Th ta c: 2 2 23 1 2T T T = +v 2 2 24 1 2T T T = Vtm1cttrnvtm2daongiuhotheophngthngng. (Hnh 1) m1 lun nm yn trn m2 trong qu trnh dao ng th:

1 2ax 2( )Mm m g gAk e+= =Vtm1vm2cgnvohaiulxot thngng,m1daongiu ho.(Hnh 2) m2 lun nm yn trn mt sn trong qu trnh m1 dao ng th: 1 2ax( )Mm m gAk+= Vt m1 t trn vt m2 dao ng iu ho theo phng ngang. H s ma st gia m1 v m2 l , b qua ma st gia m2 v mt sn. (Hnh 3) m1 khng trt trn m2 trong qu trnh dao ng th: 1 2ax 2( )Mm m g gAk e+= = 3 Con lc n(con lc ton hc ) :l h gmmt si dy khng dn, c khi lng khng ng k, treo 1 vt nng c kch thc rt nh so vi chiu di dy treo. Nu 0010Maxo o = s ;Fms=0 th sau khi kch thch vt s dao ng iu ha (gn ng ) Tnsgc: gle = ;chuk: 22lTgtte= = ;tns: 1 12 2gfT let t= = = Phng trnh dao ng: s = S0cos(et + ) hoco= 0o cos(et + )vi s =o l, S0 = 0o l v 010 o s v = s = - eS0sin(et + ) = 0l e o sin(et + ) a = v = -e2S0cos(et + ) = 20l e o cos(et + ) = -e2s = 2. l eo Lu : S0 ng vai tr nh A cn s ng vai tr nh x H thc c lp: * a = -e2s = -e2o l * 2 2 20( )vS se= +* 22 20vglo o = + C nng:2 2 2 2 2 2 0 0 0 01 1 1 12 2 2 2tmgE E E m S S mgl m lle o e o = + = = = = Vi 2 21sin ( )2E mv E t e = = +

2(1 os ) cos ( )tE mgl c E t o e = = + Ti cng mt ni con lc n chiu di l1 c chu k T1, con lc n chiu di l2 c chu k T2, con lc n chiu di l1 + l2 c chu k T2,con lc n chiu di l1 - l2 (l1>l2) c chu k T4. Th ta c: 2 2 23 1 2T T T = +v 2 2 24 1 2T T T = S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin29 Vn tc v lc cng ca si dy con lc n ( )02. . cos cos v g l o o = ( )0. 3cos 2cos T m g o o = Con lc n c chu k ng T cao h1, nhit t1. Khi a ti cao h2, nhit t2 th ta c: 2T h tT R A A A= +Vi R = 6400km l bn knh Tri t, cn l h s n di ca thanh con lc. Con lc n c chu k ng T su d1, nhit t1. Khi a ti su d2, nhit t2 th ta c: 2 2T d tT R A A A= + Con lc n c chu k ng T cao h, nhit t1. Khi a xung su d, nhit t2 th ta c: 2 2T d h tT R R A A= + Con lc n c chu k ngT sud , nhit t1. Khi a ln caoh , nhit t2th ta c: 2 2T h d tT R R A A= +Lu : * Nu AT > 0 th ng h chy chm (ng h m giy s dng con lc n) * Nu AT < 0 th ng h chy nhanh * Nu AT = 0 th ng h chy ng * Thi gian chy sai mi ngy (24h = 86400s):.86400( )TsTAu = Khi con lc n chu thm tc dng ca lc ph khng i: Lc ph khng i thng l: * Lc qun tnh:F ma = , ,, ln F = ma (F a |+, ,) Lu : + Chuyn ng nhanh dn ua v ||, , ( v, c hng chuyn ng) + Chuyn ng chm dn ua v |+, , * Lc in trng:F qE =, ,, ln F = ,q,E (Nu q > 0 F E ||, ,; cn nu q < 0 F E |+, ,) * Lc y csimt: F = DgV( F,lung thng ng hng ln) Trong : D l khi lng ring ca cht lng hay cht kh. g l gia tc ri t do. V l th tch ca phn vt chm trong cht lng hay cht kh . Khi :' P P F = +, , , gi l trng lc hiu dng hay trong lc biu kin (c vai tr nh trng lcP,) 'Fg gm= + ,, , gi l gia tc trng trng hiu dng hay gia tc trng trng biu kin. Chu k dao ng ca con lc n khi :' 2'lTgt = Cc trng hp c bit: *F, c phng ngang: + Ti VTCB dy treo lch vi phng thng ng mt gc c: FtgPo = + 2 2' ( )Fg gm= +*F,c phng thng ng th'Fg gm= S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin30 + NuF, hng xung th'Fg gm= ++ NuF, hng ln th 'Fg gm= Tng hp hai dao ng iu ha cng phngcng tn sMt vt tham gia ng thi hai daong iu hacng phngc phng trnh 1 1 1.cos( . ) x A t e = + v 2 2 2.cos( . ) x A t e = +.Ta c th dng phng phpgin Fre-nen tm dao ng tht ca vt : v cc vc t quay 1OM, v 2OM,biu din cho 1xv 2xvo thi im t=o. vc t tng 1 2OM OM OM = +, , , chnh l vc t quay biu din dao ng tng hp1 2x x x = +. Bngbin i ton hcta ckt qu dao ng tht ca vt cng l mt dao ng iu ha cng tn s v cng phng vi hai dao ng thnh phnc phng trnh dao ng l .cos( . ) x A t e = + trong : ( )2 2 21 2 1 2 2 12. . .cos A A A A A = + + 1 1 2 21 1 2 2.sin .sintan.cos .cosA AA A +=+ vi 1s s 2( nu 1s 2 ) * Nu A = 2kt(x1, x2 cng pha) AMax = A1 + A2 ;k N e` * Nu A = (2k+1) t(x1, x2 ngc pha) AMin = ,A1 - A2, Khibit mt dao ng thnh phn x1 = A1cos(et + 1) v dao ng tng hpx = Acos(et + ) th dao ng thnh phn cn li l x2 = A2cos(et + 2). Trong : 2 2 22 1 1 12 os( ) A A A AAc = + 1 121 1sin sintanos osA AAc Ac =vi 1s s 2( nu 1s 2 ) Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cng tn s x1 = A1cos(et + 1; x2 = A2cos(et + 2) , th dao ng tng hp cng l dao ng iu ho cng phng cng tn s x = Asin(et + ). Ta c: 1 1 2 2sin sin sin ...xA A A A = = + + 1 1 2 2os os os ... A Ac Ac A c A = = + +2 2xA A AA = +vtanxAAA= vi e[Min;Max] 4 Dao ng t dol dao ng xy ra trong mt hdi tc dng ca ni lc, sau khi h c kch thch ban u :a h ra khi trng thi cn bng ri th ra . H dao ng l h c kh nng thc hindao ng t do .Mi dao ng t do ca mt h dao ngu c cng tn s gc 0egi ltn s gc ring ca h y.002fet=gi l tn s dao ng ring,Tn s ny ch ph thuc cc c tnh ca h dao ng ,khng ph thuc cc yu t bn ngoi h. 5H t dao ngl h dao ngc ma st nhng c thm c cu b lis tiu hao nng lng v ma st .Chu k h t dao ng bngchu k ring ca n . 6Daongcngbcldaongcamthdaongchutcdngcamt ngoi lc bin i iu hatm sO.cim :trong khong thi gian t A ( thng rtb)k t khi bt u chu tc dng ca ngoilch dao ng rt phc tp(chng trnhTPkhngxt)saukhonthigianchuyntipt A hdaongvitnsOca ngoilc .bin ca dao ng cngbc ph thucbin v tn sO ca ngoi lc . Ox y y1 y2 x1 x2 1 2 M1 M2 M A A1 A2 S hc l v b Sing hc l bn y l bn tho chc cn nhiu sai st .Rt mong nhn c kin ng gp hon thin31 7 Dao ng tt dnl dao ngcamt h dao ng trong c ma st:"bin " dao ng gim dn theo thi gian .Khi ma st ln dao ng khng xy ra .Khi ma st rt nhc th coi dao ngtt dn l dao ng iu havi tn s gc bng tn s gc0eca dao ng iu ha. Mt con lcl xo daong tt dn vibin A(bin ln nht ), h s mast .Qungngvticnlcdnglil: 2 2 22 2kA ASmg ge = =Mtvtdaongttdnthgimbinsaumichukl: 24 4 mg gAk eA= = s dao ng thc hin c 24 4A Ak ANA mg ge = = =A 8 Hin tng cng hngl hin tng bin ca dao ng cng bct gi tr cc i khi tn sngoi lcO bng (gn ng ) tn s dao ng ring 0eca h dao ng. f = f0 hay e = e0 hay T = T0 9.cc c trng chnh ca mt s h dao ng Con lc l xoCon lc n v tri t Con lc vt l v tri t Cu trc Hn bi (m) gn vi l xo(k) Hn bi (m) treo u si dy (l) Vt rn (m,I) quay quanh trc nm ngang (Q) V tr cn bng L xo khng dn ( nm ngang )Dy treo thng ngQG thng ng Lc tc dngLc n hi ca l xo c gi tr. F k x = x: li thng Trng lc ca hn bi v lc ca dy treo . .tgF P m sl= = s:li cong Trng lc ca vt rnv lc ca trc quayc mmen . . .sin M m g d o = o: li gc Phng trnh ng lc hcca dao ng2'' . 0 x x e + =2'' . 0 S S e + =2'' . 0 o e o + =Tn s gckme =gle =. . m g dIe =Phng trnh dao ng.cos( . ) x A t e = +xtrong gii hn n hi 0.cos( . ) s s t e = +0s l < hoc0010 o s 0.cos( . ) t o o e = +01 o