1. Mc lc Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Phn1 . PHNG PHP GII TON V DAO NG IU HA CA CON LC L XO 15 Ch 1. Lin h gia lc tc dng, gin v cng ca l xo . . . . . . . . . . 15 1.Cho bit lc ko F, cng k: tm gin l0, tm l . . . . . . . . . . . . . 15 2.Ct l xo thnh n phn bng nhau ( hoc hai phn khng bng nhau): tm cng ca mi phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Ch 2. Vit phng trnh dao ng iu ha ca con lc l xo . . . . . . . . . . 15 Ch 3. Chng minh mt h c hc dao ng iu ha . . . . . . . . . . . . . . . 16 1.Phng php ng lc hc . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.Phng php nh lut bo ton nng lng . . . . . . . . . . . . . . . . . . 16 Ch 4. Vn dng nh lut bo ton c nng tm vn tc . . . . . . . . . . . . 16 Ch 5. Tm biu thc ng nng v th nng theo thi gian . . . . . . . . . . . . 17 Ch 6. Tm lc tc dng cc i v cc tiu ca l xo ln gi treo hay gi . . 17 1.Trng hp l xo nm ngang . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.Trng hp l xo treo thng ng . . . . . . . . . . . . . . . . . . . . . . . 17 3.Ch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Ch 7. H hai l xo ghp ni tip: tm cng kh, t suy ra chu k T . . . . 18 Ch 8. H hai l xo ghp song song: tm cng kh, t suy ra chu k T . . . 18 Ch 9. H hai l xo ghp xung i: tm cng kh, t suy ra chu k T . . . 18 Ch 10. Con lc lin kt vi rng rc( khng khi lng): chng minh rng h dao ng iu ha, t suy ra chu k T . . . . . . . . . . . . . . . . . . . . 19 1.Hn bi ni vi l xo bng dy nh vt qua rng rc . . . . . . . . . . . . . . 19 2.Hn bi ni vi rng rc di ng, hn bi ni vo dy vt qua rng rc . . . . 19 3.L xo ni vo trc rng rc di ng, hn bi ni vo hai l xo nh dy vt qua rng rc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1 www.VNMATH.com

2. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 11.Lc hi phc gy ra dao ng iu ha khng phi l lc n hi nh: lc y Acximet, lc ma st, p lc thy tnh, p lc ca cht kh...: chng minh h dao ng iu ha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.F l lc y Acximet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.F l lc ma st . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3.p lc thy tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.F l lc ca cht kh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Phn2 . PHNG PHP GII TON V DAO NG IU HA CA CON LC N 22 Ch 1. Vit phng trnh dao ng iu ha ca con lc n . . . . . . . . . . . 22 Ch 2. Xc nh bin thin nh chu k T khi bit bin thin nh gia tc trng trng g, bin thin chiu di l . . . . . . . . . . . . . . . . . . . 22 Ch 3. Xc nh bin thin nh chu k T khi bit nhit bin thin nh t; khi a ln cao h; xung su h so vi mt bin . . . . . . . . . . . 23 1. Khi bit nhit bin thin nh t . . . . . . . . . . . . . . . . . . . . . . 23 2. Khi a con lc n ln cao h so vi mt bin . . . . . . . . . . . . . . . 23 3. Khi a con lc n xung su h so vi mt bin . . . . . . . . . . . . . 23 Ch 4. Con lc n chu nhiu yu t nh hng bin thin ca chu k: tm iu kin chu k khng i . . . . . . . . . . . . . . . . . . . . . . . . . . 24 1.iu kin chu k khng i . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.V d:Con lc n chu nh hng bi yu t nhit v yu t cao . . . 24 Ch 5. Con lc trong ng h g giy c xem nh l con lc n: tm nhanh hay chm ca ng h trong mt ngy m . . . . . . . . . . . . . . . . . . . 24 Ch 6. Con lc n chu tc dng thm bi mt ngoi lc F khng i: Xc nh chu k dao ng mi T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 1.F l lc ht ca nam chm . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.F l lc tng tc Coulomb . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.F l lc in trng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.F l lc y Acsimet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 5.F l lc nm ngang . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Ch 7. Con lc n treo vo mt vt ( nh t, thang my...) ang chuyn ng vi gia tc a: xc nh chu k mi T . . . . . . . . . . . . . . . . . . . . . . 26 1.Con lc n treo vo trn ca thang my ( chuyn ng thng ng ) vi gia tc a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.Con lc n treo vo trn ca xe t ang chuyn ng ngang vi gia tc a . 27 Th.s Trn AnhTrung 2 Luyn thi i hc www.VNMATH.com 3. Phng php gii ton Vt L 12 Trng THPT - Phong in 3.Con lc n treo vo trn ca xe t ang chuyn ng trn mt phng nghing mt gc : . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Ch 8. Xc nh ng nng E th nng Et, c nng ca con lc n khi v tr c gc lch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Ch 9. Xc nh vn tc di v v lc cng dy T ti v tr hp vi phng thng ng mt gc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.Vn tc di v ti C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.Lc cng dy T ti C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.H qa: vn tc v lc cng dy cc i v cc tiu . . . . . . . . . . . . . . 30 Ch 10. Xc nh bin gc mi khi gia tc trng trng thay i t g sang g 30 Ch 11. Xc nh chu k v bin ca con lc n vng inh (hay vt cn) khi i qua v tr cn bng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1.Tm chu k T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.Tm bin mi sau khi vng inh . . . . . . . . . . . . . . . . . . . . . . 31 Ch 12. Xc nh thi gian hai con lc n tr li v tr trng phng (cng qua v tr cn bng, chuyn ng cng chiu) . . . . . . . . . . . . . . . . . . 31 Ch 13. Con lc n dao ng th b dy t:kho st chuyn ng ca hn bi sau khi dy t? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 1.Trng hp dy t khi i qua v tr cn bng O . . . . . . . . . . . . . . . . 31 2.Trng hp dy t khi i qua v tr c li gic . . . . . . . . . . . . . . . . 32 Ch 14. Con lc n c hn bi va chm n hi vi mt vt ang ng yn: xc nh vn tc ca vin bi sau va chm? . . . . . . . . . . . . . . . . . . . . . . 32 Phn3 . PHNG PHP GII TON V DAO NG TT DN V CNG HNG C HC 33 Ch 1. Con lc l xo dao ng tt dn: bin gim dn theo cp s nhn li v hng, tm cng bi q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Ch 2. Con lc l n ng tt dn: bin gc gim dn theo cp s nhn li v hng, tm cng bi q. Nng lng cung cp duy tr dao ng . . . . . . . 33 Ch 3. H dao ng cng bc b kch thch bi mt ngoi lc tun hon: tm iu kin c hin tng cng hng . . . . . . . . . . . . . . . . . . . . . 34 Phn 4 . PHNG PHP GII TON V S TRUYN SNG C HC, GIAO THOA SNG, SNG DNG, SNG M 35 Ch 1. Tm lch pha gia hai im cch nhau d trn mt phng truyn sng? Tm bc sng khi bit lch pha v gii hn ca bc sng,( tn s, vn tc truyn sng). Vit phng trnh sng ti mt im . . . . . . . . . . . . . . . 35 1.Tm lch pha gia hai im cch nhau d trn mt phng truyn sng . . 35 Th.s Trn AnhTrung 3 Luyn thi i hc www.VNMATH.com 4. Phng php gii ton Vt L 12 Trng THPT - Phong in 2.Tm bc sng khi bit lch pha v gii hn ca bc sng,( tn s, vn tc truyn sng) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.Vit phng trnh sng ti mt im trn phng truyn sng . . . . . . . . 35 4.Vn tc dao ng ca sng . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Ch 2. V th biu din qu trnh truyn sng theo thi gian v theo khng gian 36 1.V th biu din qa trnh truyn sng theo thi gian . . . . . . . . . . . . 36 2.V th biu din qa trnh truyn sng theo khng gian ( dng ca mi trng...) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Ch 3. Xc nh tnh cht sng ti mt im M trn min giao thoa . . . . . . . 36 Ch 4. Vit phng trnh sng ti im M trn min giao thoa . . . . . . . . . . 37 Ch 5. Xc nh s ng dao ng cc i v cc tiu trn min giao thoa . . . 37 Ch 6. Xc nh im dao ng vi bin cc i ( im bng) v s im dao ng vi bin cc tiu ( im nt) trn on S1S2 . . . . . . . . . . . . . . 38 Ch 7.Tm qy tch nhng im dao ng cng pha (hay ngc pha) vi hai ngun S1, S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Ch 8.Vit biu thc sng dng trn dy n hi . . . . . . . . . . . . . . . . . 38 Ch 9.iu kin c hin tng sng dng, t suy ra s bng v s nt sng 39 1.Hai u mi trng ( dy hay ct khng kh) l c nh . . . . . . . . . . . . 39 2.Mt u mi trng ( dy hay ct khng kh) l c nh, u kia t do . . . . 39 3.Hai u mi trng ( dy hay ct khng kh) l t do . . . . . . . . . . . . . 40 Ch 10.Xc nh cng m (I) khi bit mc cng m ti im. Xc nh cng sut ca ngun m? to ca m . . . . . . . . . . . . . . . . . . . . . 40 1.Xc nh cng m (I) khi bit mc cng m ti im . . . . . . . . 40 2.Xc nh cng sut ca ngun m ti mt im: . . . . . . . . . . . . . . . . 40 3. to ca m: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Phn5 . PHNG PHP GII TON V MCH IN XOAY CHIU KHNG PHN NHNH (RLC) 42 Ch 1. To ra dng in xoay chiu bng cch cho khung dy quay u trong t trng, xc nh sut in ng cm ng e(t)? Suy ra biu thc cng dng in i(t) v hiu in th u(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Ch 2. on mch RLC: cho bit i(t) = I0 sin(t), vit biu thc hiu in th u(t). Tm cng sut Pmch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Ch 3. on mch RLC: cho bit u(t) = U0 sin(t), vit biu thc cng dng in i(t). Suy ra biu thc uR(t)?uL(t)?uC(t)? . . . . . . . . . . . . . . 42 Th.s Trn AnhTrung 4 Luyn thi i hc www.VNMATH.com 5. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 4. Xc nh lch pha gia hai ht tc thi u1 v u2 ca hai on mch khc nhau trn cng mt dng in xoay chiu khng phn nhnh? Cch vn dng? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Ch 5. .on mch RLC, cho bit U, R: tm h thc L, C, : cng dng in qua on mch cc i, hiu in th v cng dng in cng pha, cng sut tiu th trn on mch t cc i. . . . . . . . . . . . . . . . . . . 43 1.Cng dng in qua on mch t cc i . . . . . . . . . . . . . . . . 43 2.Hiu in th cng pha vi cng dng in . . . . . . . . . . . . . . . . 44 3.Cng sut tiu th trn on mch cc i . . . . . . . . . . . . . . . . . . . 44 4.Kt lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Ch 6. .on mch RLC, ghp thm mt t C :tm C : cng dng in qua on mch cc i, hiu in th v cng dng in cng pha, cng sut tiu th trn on mch t cc i. . . . . . . . . . . . . . . . . . . . . . 44 Ch 7. .on mch RLC: Cho bit UR, UL, UC: tm U v lch pha u/i. . . . 45 Ch 8.Cun dy (RL) mc ni tip vi t C: cho bit hiu in th U1 ( cun dy) v UC . Tm Umch v . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Ch 9. Cho mchRLC: Bit U, , tm L, hayC, hayR cng sut tiu th trn on mch cc i. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 1.Tm L hay C cng sut tiu th trn on mch cc i . . . . . . . . . . 46 2.Tm R cng sut tiu th trn on mch cc i . . . . . . . . . . . . . 46 Ch 10. .on mch RLC: Cho bit U, R, f: tm L ( hay C) UL (hay UC ) t gi tr cc i? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 1.Tm L hiu th hiu dng hai u cun cm cc i . . . . . . . . . . . 47 2.Tm C hiu th hiu dng hai u t in cc i . . . . . . . . . . . . 48 Ch 11. .on mch RLC: Cho bit U, R, L, C: tm f ( hay ) UR, UL hay UC t gi tr cc i? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 1.Tm f ( hay ) hiu th hiu dng hai u in tr cc i . . . . . . . 49 2.Tm f ( hay ) hiu th hiu dng hai u cun cm cc i . . . . . . 49 3.Tm f ( hay ) hiu th hiu dng hai u t in cc i . . . . . . . . 49 Ch 12. Cho bit th i(t) v u(t), hoc bit gin vect hiu in th: xc nh cc c im ca mch in? . . . . . . . . . . . . . . . . . . . . . . . . 50 1.Cho bit th i(t) v u(t): tm lch pha u/i . . . . . . . . . . . . . . . 50 2.Cho bit gin vect hiu in th: v s on mch? Tm Umch . . . . 51 Ch 13. Tc dng nhit ca dng in xoay chiu: tnh nhit lng ta ra trn on mch? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Th.s Trn AnhTrung 5 Luyn thi i hc www.VNMATH.com 6. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 14. Tc dng ha hc ca dng in xoay chiu: tnh in lng chuyn qua bnh in phn theo mt chiu? Tnh th tch kh Hir v Oxy xut hin cc in cc? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 1.Tnh in lng chuyn qua bnh in phn theo mt chiu ( trong 1 chu k T, trong t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.Tnh th tch kh Hir v Oxy xut hin cc in cc trong thi gian t(s) . 52 Ch 15. Tc dng t ca dng in xoay chiu v tc dng ca t trng ln dng in xoay chiu? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 1.Nam chm in dng dng in xoay chiu ( tn s f) t gn dy thp cng ngang. Xc nh tn s rung f ca dy thp . . . . . . . . . . . . . . 52 2.Dy dn thng cng ngang mang dng in xoay chiu t trong t trng c cm ng t B khng i ( vung gc vi dy): xc nh tn s rung ca dy f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Phn6 . PHNG PHP GII TON V MY PHT IN XOAY CHIU, BIN TH, TRUYN TI IN NNG 53 Ch 1. Xc nh tn s f ca dng in xoay chiu to bi my pht in xoay chiu 1 pha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.Trng hp roto ca mp c p cp cc, tn s vng l n . . . . . . . . . . . 53 2.Trng hp bit sut in ng xoay chiu ( E hay Eo) . . . . . . . . . . . . 53 Ch 2. Nh my thy in: thc nc cao h, lm quay tuabin nc v roto ca mp. Tm cng sut P ca my pht in? . . . . . . . . . . . . . . . . . . . . 53 Ch 3. Mch in xoay chiu ba pha mc theo s hnh : tm cng dng trung ha khi ti i xng? Tnh hiu in th Ud ( theo Up)? Tnh Pt (cc ti) 53 Ch 4. My bin th: cho U1, I1: tm U2, I2 . . . . . . . . . . . . . . . . . . . . 54 1.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp h 54 2.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp c ti 54 3.Trng hp cc in tr ca cun s cp v th cp khc 0: . . . . . . . . . 55 Ch 5.Truyn ti in nng trn dy dn: xc nh cc i lng trong qu trnh truyn ti . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Ch 6.Xc nh hiu sut truyn ti in nng trn dy? . . . . . . . . . . . . . . 55 Phn7 . PHNG PHP GII TON V DAO NG IN T DO TRONG MCH LC 57 Ch 1. Dao ng in t do trong mch LC: vit biu thc q(t)? Suy ra cng dng in i(t)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Ch 2. Dao ng in t do trong mch LC, bit uC = U0 sin t, tm q(t)? Suy ra i(t)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Th.s Trn AnhTrung 6 Luyn thi i hc www.VNMATH.com 7. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 3. Cch p dng nh lut bo ton nng lng trong mch dao ng LC . . 58 1.Bit Q0 ( hay U0) tm bin I0 . . . . . . . . . . . . . . . . . . . . . . . . 58 2.Bit Q0 ( hay U0)v q ( hay u), tm i lc . . . . . . . . . . . . . . . . . . 58 Ch 4. Dao ng in t do trong mch LC, bit Q0 v I0:tm chu k dao ng ring ca mch LC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Ch 5. Mch LC li vo ca my thu v tuyn in bt sng in t c tn s f (hay bc sng ).Tm L( hay C) . . . . . . . . . . . . . . . . . . . . . . . 59 1.Bit f( sng) tm L v C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 2.Bit ( sng) tm L v C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Ch 6. Mch LC li vo ca my thu v tuyn c t in c in dung bin thin Cmax Cmin tng ng gc xoay bin thin 00 1800 : xc nh gc xoay thu c bc x c bc sng ? . . . . . . . . . . . . . . . . . . . . . 59 Ch 7. Mch LC li vo ca my thu v tuyn c t xoay bin thin Cmax Cmin: tm di bc sng hay di tn s m my thu c? . . . . . . . . . . . 60 Phn8 . PHNG PHP GII TON V PHN X NH SNG CA GNG PHNG V GNG CU 61 Ch 1. Cch v tia phn x trn gng phng ng vi mt tia ti cho ? . . . . 61 Ch 2. Cch nhn bit tnh cht "tht - o" ca vt hay nh( da vo cc chm sng) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Ch 3. Gng phng quay mt gc (quanh trc vung gc mt phng ti): tm gc quay ca tia phn x? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 1.Cho tia ti c nh, xc nh chiu quay ca tia phn x . . . . . . . . . . . . 61 2.Cho bit SI = R, xc nh qung ng i ca nh S . . . . . . . . . . . . 61 3.Gng quay u vi vn tc gc : tm vn tc di ca nh . . . . . . . . . . 62 Ch 4. Xc nh nh to bi mt h gng c mt phn x hng vo nhau . . . 62 Ch 5. Cch vn dng cng thc ca gng cu . . . . . . . . . . . . . . . . . . 63 1.Cho bit d v AB: tm d v cao nh A B . . . . . . . . . . . . . . . . . 63 2.Cho bit d v A B : tm d v cao vt AB . . . . . . . . . . . . . . . . . 63 3.Cho bit v tr vt d v nh d xc nh tiu c f . . . . . . . . . . . . . . . 63 4.Ch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Ch 6. Tm chiu v di ca mn nh khi bit chiu v di ca vt. H qa? 64 1.Tm chiu v di ca mn nh khi bit chiu v di ca vt . . . . . . 64 2.H qa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Ch 7. Cho bit tiu c f v mt iu kin no v nh, vt: xc nh v tr vt dv v tr nh d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Th.s Trn AnhTrung 7 Luyn thi i hc www.VNMATH.com 8. Phng php gii ton Vt L 12 Trng THPT - Phong in 1.Cho bit phng i k v f . . . . . . . . . . . . . . . . . . . . . . . . . . 64 2.Cho bit khong cch l = AA . . . . . . . . . . . . . . . . . . . . . . . . . 64 Ch 8. Xc nh th trng ca gng ( gng cu li hay gng phng) . . . . . 65 Ch 9. Gng cu lm dng trong n chiu: tm h thc lin h gia vt sng trn trn mn ( chn chm tia phn x) v kch thc ca mt gng . . . . . . 65 Ch 10. Xc nh nh ca vt to bi h "gng cu - gng phng" . . . . . . . 65 1.Trng hp gng phng vung gc vi trc chnh . . . . . . . . . . . . . . 66 2.Trng hp gng phng nghing mt gc 450 so vi trc chnh . . . . . . . 66 Ch 11. Xc nh nh ca vt to bi h "gng cu - gng cu" . . . . . . . . 66 Ch 12. Xc nh nh ca vt AB xa v cng to bi gng cu lm . . . . . 67 Phn9 . PHNG PHP GII TON V KHC X NH SNG, LNG CHT PHNG ( LCP), BNG MT SONG SONG (BMSS), LNG KNH (LK) 69 Ch 1. Kho st ng truyn ca tia sng n sc khi i t mi trng chit quang km sang mi trng chit quang hn? . . . . . . . . . . . . . . . . . . 69 Ch 2. Kho st ng truyn ca tia sng n sc khi i t mi trng chit quang hn sang mi trng chit quang km? . . . . . . . . . . . . . . . . . . 69 Ch 3. Cch v tia khc x ( ng vi tia ti cho) qua mt phng phn cch gia hai mi trng bng phng php hnh hc? . . . . . . . . . . . . . . . . 70 1.Cch v tia khc x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 2.Cch v tia ti gii hn ton phn . . . . . . . . . . . . . . . . . . . . . . . 70 Ch 4. Xc nh nh ca mt vt qua LCP ? . . . . . . . . . . . . . . . . . . . . 70 Ch 5. Xc nh nh ca mt vt qua BMSS ? . . . . . . . . . . . . . . . . . . . 71 1. di nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2. di ngang ca tia sng . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Ch 6. Xc nh nh ca mt vt qua h LCP- gng phng ? . . . . . . . . . . 71 1.Vt A - LCP - Gng phng . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2.Vt A nm gia LCP- Gng phng . . . . . . . . . . . . . . . . . . . . . . 72 Ch 7. Xc nh nh ca mt vt qua h LCP- gng cu ? . . . . . . . . . . . . 72 Ch 8. Xc nh nh ca mt vt qua h nhiu BMSS ghp st nhau? . . . . . . 72 Ch 9. Xc nh nh ca mt vt qua h nhiu BMSS - gng phng ghp song song? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 1.Vt S - BMSS - Gng phng . . . . . . . . . . . . . . . . . . . . . . . . . 73 2.Vt S nm gia BMSS - Gng phng . . . . . . . . . . . . . . . . . . . . . 73 Ch 10. Xc nh nh ca mt vt qua h nhiu BMSS - gng cu? . . . . . . . 73 Th.s Trn AnhTrung 8 Luyn thi i hc www.VNMATH.com 9. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 11. Cho lng knh (A,n) v gc ti i1 ca chm sng: xc nh gc lch D? . 74 Ch 12. Cho lng knh (A,n) xc nh i1 D = min? . . . . . . . . . . . . . . 74 1.Cho A,n: xc nh i1 D = min,Dmin? . . . . . . . . . . . . . . . . . . . . 74 2.Cho Av Dmin: xc nh n? . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.Ch : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Ch 13. Xc nh iu kin c tia l ra khi LK? . . . . . . . . . . . . . . . 75 1.iu kin v gc chic quang . . . . . . . . . . . . . . . . . . . . . . . . . . 75 1.iu kin v gc ti . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Phn10 . PHNG PHP GII TON V THU KNH V H QUANG HC NG TRC VI THU KNH 76 Ch 1. Xc nh loi thu knh ? . . . . . . . . . . . . . . . . . . . . . . . . . . 76 1.Cn c vo s lin h v tnh cht, v tr, ln gia vt - nh . . . . . . . . 76 2.Cn c vo ng truyn ca tia sng qua thu knh . . . . . . . . . . . . . . 76 3.Cn c vo cng thc ca thu knh . . . . . . . . . . . . . . . . . . . . . . 76 Ch 2. Xc nh t ca thu knh khi bit tiu c, hay chic sut ca mi trng lm thu knh v bn knh ca cc mt cong. . . . . . . . . . . . . . . . 76 1.Khi bit tiu c f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 2.Khi bit chic sut ca mi trng lm thu knh v bn knh ca cc mt cong 76 Ch 3. Cho bit tiu c f v mt iu kin no v nh, vt: xc nh v tr vt d v v tr nh d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 1.Cho bit phng i k v f . . . . . . . . . . . . . . . . . . . . . . . . . . 77 2.Cho bit khong cch l = AA . . . . . . . . . . . . . . . . . . . . . . . . . 77 Ch 4. Xc nh nh ca mt vt AB xa v cc . . . . . . . . . . . . . . . . . 77 Ch 5. Xc nh nh ca mt vt AB xa v cc . . . . . . . . . . . . . . . . . 77 1.Cho bit khong cch "vt - nh" L, xc nh hai v tr t thu knh . . . . . 78 2.Cho bit khong cch "vt - nh" L, v khong cch gia hai v tr, tm f . . 78 Ch 6. Vt hay thu knh di chuyn, tm chiu di chuyn ca nh . . . . . . . . . 78 1.Thu knh (O) c nh: di vt gn ( hay xa) thu knh, tm chiu chuyn di ca nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 2.Vt AB c nh, cho nh A B trn mn, di thu knh hi t, tm chiu chuyn di ca mn . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 Ch 8. Lin h gia kch thc vt sng trn trn mn( chn chm l) v kch thc ca mt thu knh. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Ch 9. H nhiu thu knh mng ghp ng trc vi nhau, tm tiu c ca h. . . 79 Th.s Trn AnhTrung 9 Luyn thi i hc www.VNMATH.com 10. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 10. Xc nh nh ca mt vt qua h " thu knh- LCP". . . . . . . . . . . . 79 1.Trng hp: AB - TK - LCP . . . . . . . . . . . . . . . . . . . . . . . . . . 79 2.Trng hp: AB - LCP - TK . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Ch 11. Xc nh nh ca mt vt qua h " thu knh- BMSS". . . . . . . . . . . 80 1.Trng hp: AB - TK - BMSS . . . . . . . . . . . . . . . . . . . . . . . . . 80 2.Trng hp: AB - LCP - TK . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Ch 12. Xc nh nh ca mt vt qua h hai thu knh ghp ng trc. . . . . . 81 Ch 13. Hai thu knh ng trc tch ri nhau: xc nh gii hn ca a = O1O2( hoc d1 = O1A) nh A2B2 nghim ng mt iu kin no ( nh nh tht, nh o, cng chu hay ngc chiu vi vt AB). . . . . . . . . . . . . . . 82 1.Trng hp A2B2 l tht ( hay o ) . . . . . . . . . . . . . . . . . . . . . . . 82 2.Trng hp A2B2 cng chiu hay ngc chiu vi vt . . . . . . . . . . . . 82 Ch 14. Hai thu knh ng trc tch ri nhau: xc nh khong cch a = O1O2 nh cui cng khng ph thuc vo v tr vt AB. . . . . . . . . . . . . . . 82 Ch 15. Xc nh nh ca vt cho bi h "thu knh - gng phng". . . . . . . . 83 1.Trng hp gng phng vung gc vi trc chnh . . . . . . . . . . . . . . 83 2.Trng hp gng phng nghing mt gc 450 so vi trc chnh . . . . . . . 83 3.Trng hp gng phng ghp xc thu knh ( hay thu knh m bc) . . . . 84 4.Trng hp vt AB t trong khong gia thu knh v gng phng . . . . 84 Ch 16. Xc nh nh ca vt cho bi h "thu knh - gng cu". . . . . . . . . 84 1.Trng hp vt AB t trc h " thu knh- gng cu" . . . . . . . . . . . 85 2.Trng hp h "thu knh- gng cu" ghp st nhau . . . . . . . . . . . . . 85 3.Trng hp vt AB t gia thu knh v gng cu: . . . . . . . . . . . . . 85 Phn11 . PHNG PHP GII TON V MT V CC DNG C QUANG HC B TR CHO MT 89 Ch 1. My nh: cho bit gii hn khong t phim, tm gii hn t vt? . . . . 89 Ch 2. My nh chp nh ca mt vt chuyn ng vung gc vi trc chnh. Tnh khong thi gian ti a m ca sp ca ng knh nh khng b nho. . 89 Ch 3. Mt cn th: xc nh t ca knh cha mt? Tm im cc cn mi c khi eo knh cha? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Ch 4. Mt vin th: xc nh t ca knh cha mt? Tm im cc cn mi c khi eo knh cha? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Ch 5. Knh lp: xc nh phm vi ngm chng v bi gic. Xc nh kch thc nh nht ca vt ABmin m mt phn bit c qua knh lp . . . . . . 90 1.Xc nh phm vi ngm chng ca knh lp . . . . . . . . . . . . . . . . . . 90 Th.s Trn AnhTrung 10 Luyn thi i hc www.VNMATH.com 11. Phng php gii ton Vt L 12 Trng THPT - Phong in 2.Xc nh bi gic ca knh lp . . . . . . . . . . . . . . . . . . . . . . . 91 3.Xc nh kch thc nh nht ca vt ABmin m mt phn bit c qua knh lp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Ch 6. Knh hin vi: xc nh phm vi ngm chng v bi gic. Xc nh kch thc nh nht ca vt ABmin m mt phn bit c qua knh hin vi . . . . 92 1.Xc nh phm vi ngm chng ca knh hin vi . . . . . . . . . . . . . . . . 92 2.Xc nh bi gic ca knh hin vi . . . . . . . . . . . . . . . . . . . . . 93 3.Xc nh kch thc nh nht ca vt ABmin m mt phn bit c qua knh hin vi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Ch 7. Knh thin vn: xc nh phm vi ngm chng v bi gic? . . . . . . 94 1.Xc nh phm vi ngm chng ca knh thin vn . . . . . . . . . . . . . . . 94 2.Xc nh bi gic ca knh thin vn . . . . . . . . . . . . . . . . . . . . 94 Phn12 . PHNG PHP GII TON V HIN TNG TN SC NH SNG 95 Ch 1. S tn sc chm sng trng qua mt phn cch gia hai mi trng: kho st chm khc x? Tnh gc lch bi hai tia khc x n sc? . . . . . . . . . 95 Ch 2. Chm sng trng qua LK: kho st chm tia l? . . . . . . . . . . . . . . 95 Ch 3. Xc nh gc hp bi hai tia l ( , tm)ca chm cu vng ra khi LK. Tnh b rng quang ph trn mn? . . . . . . . . . . . . . . . . . . . . . . . . 95 Ch 4. Chm tia ti song song c b rng a cha hai bt x truyn qua BMSS: kho st chm tia l? Tnh b rng cc i amax hai chm tia l tch ri nhau? 95 Phn13 . PHNG PHP GII TON V GIAO THOA SNG NH SNG 97 Ch 1. Xc nh bc sng khi bit khong vn i, a,, D . . . . . . . . . . . . 97 Ch 2. Xc nh tnh cht sng (ti) v tm bc giao thoa ng vi mi im trn mn? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Ch 3. Tm s vn sng v vn ti quang st c trn min giao thoa . . . . . . 97 Ch 4. Trng hp ngun pht hai nh sng n sc. Tm v tr trn mn c s trng nhau ca hai vn sng thuc hai h n sc? . . . . . . . . . . . . . . 98 Ch 5. Trng hp giao thoa nh sng trng: tm rng quang ph, xc nh nh sng cho vn ti ( sng) ti mt im (xM ) ? . . . . . . . . . . . . . . . . 98 1.Xc nh rng quang ph . . . . . . . . . . . . . . . . . . . . . . . . . . 98 2.Xc nh nh sng cho vn ti ( sng) ti mt im (xM ) . . . . . . . . . . . 98 Ch 6. Th nghim giao thoa vi nh sng thc hin trong mi trng c chic sut n > 1. Tm khong vn mi i ? H vn thay i th no? . . . . . . . . . 98 Ch 7. Th nghim Young: t bn mt song song (e,n) trc khe S1 ( hoc S2). Tm chiu v dch chuyn ca h vn trung tm. . . . . . . . . . . . . . . . 98 Th.s Trn AnhTrung 11 Luyn thi i hc www.VNMATH.com 12. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 8. Th nghim Young: Khi ngun sng di chuyn mt on y = SS . Tm chiu, chuyn di ca h vn( vn trung tm)? . . . . . . . . . . . . . . . . 99 Ch 9.Ngun sng S chuyn ng vi vn tc v theo phng song song vi S1S2: tm tn s sut hin vn sng ti vn trung tm O? . . . . . . . . . . . . . . . 99 Ch 10.Tm khong cch a = S1S2 v b rng min giao thoa trn mt s dng c giao thoa? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 1.Khe Young . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 2.Lng lng knh Frexnen . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 3.Hai na thu knh Billet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 4.Gng Frexnen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Phn14 . PHNG PHP GII TON V TIA RNGHEN 101 Ch 1. Tia Rnghen: Cho bit vn tc v ca electron p vo i catot: tm UAK 101 Ch 2. Tia Rnghen: Cho bit vn tc v ca electron p vo i catot hot UAK : tm tn s cc i Fmax hay bc sng min? . . . . . . . . . . . . . . . . . . 101 Ch 3. Tnh lu lng dng nc lm ngui i catot ca ng Rnghen: . . . . . 101 Phn15 . PHNG PHP GII TON V HIN TNG QUANG IN 103 Ch 1. Cho bit gii hn quang in (0). Tm cng thot A ( theo n v eV )? . 103 Ch 2. Cho bit hiu in th hm Uh. Tm ng nng ban u cc i (Emax) hay vn tc ban u cc i( v0max), hay tm cng thot A? . . . . . . . . . . . 103 1.Cho Uh: tm Emax hay v0max . . . . . . . . . . . . . . . . . . . . . . . . . . 103 2.Cho Uh v (kch thch): tm cng thot A: . . . . . . . . . . . . . . . . . . 103 Ch 3. Cho bit v0max ca electron quang in v ( kch thch): tm gii hn quang in 0? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Ch 4. Cho bit cng thot A (hay gii hn quang in 0) v ( kch thch): Tm v0max ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Ch 5. Cho bit UAK v v0max. Tnh vn tc ca electron khi ti Ant ? . . . . . 104 Ch 6. Cho bit v0max v A.Tm iu kin ca hiu in th UAK khng c dng quang in (I = 0) hoc khng c mt electron no ti Ant? . . . . . . 104 Ch 7. Cho bit cng dng quang in bo ho (Ibh) v cng sut ca ngun sng. Tnh hiu sut lng t? . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Ch 8. Chiu mt chm sng kch thch c bc sng vo mt qa cu c lp v in. Xc nh in th cc i ca qa cu. Ni qu cu vi mt in tr R sau ni t. Xc nh cng dng qua R. . . . . . . . . . . . . . . . . 105 1.Chiu mt chm sng kch thch c bc sng vo mt qa cu c lp v in. Xc nh in th cc i ca qa cu: . . . . . . . . . . . . . . 105 Th.s Trn AnhTrung 12 Luyn thi i hc www.VNMATH.com 13. Phng php gii ton Vt L 12 Trng THPT - Phong in 2.Ni qu cu vi mt in tr R sau ni t. Xc nh cng dng qua R:105 Ch 9. Cho kch thch, in trng cn Ec v bc sng gii hn 0: tm on ng i ti a m electron i c. . . . . . . . . . . . . . . . . . . . . . . . 105 Ch 10. Cho kch thch, bc sng gii hn 0 v UAK : Tm bn knh ln nht ca vng trn trn mt Ant m cc electron t Katt p vo? . . . . . . . . . 105 Ch 11. Cho kch thch, bc sng gii hn 0 , electron quang in bay ra theo phng vung gc vi in trng (E). Kho st chuyn ng ca electron ?106 Ch 12. Cho kch thch, bc sng gii hn 0 , electron quang in bay ra theo phng vung gc vi cm ng t ca tr trng u (B). Kho st chuyn ng ca electron ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Phn16 . PHNG PHP GII TON V MU NGUYN T HIR THEO BO 108 Ch 1. Xc nh vn tc v tn s f ca electron trng thi dng th n ca nguyn t Hir? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Ch 2. Xc nh bc sng ca photon do nguyn t Hir pht ra khi nguyn t trng thi dng c mc nng lng Em sang En ( < Em )? . . . . . . . . . . 108 Ch 3. Tm bc sng ca cc vch quang ph khi bit cc bc sng ca cc vch ln cn? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Ch 4. Xc nh bc sng cc i (max) v cc tiu (min) ca cc dy Lyman, Banme, Pasen? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Ch 5. Xc nh qy o dng mi ca electron khi nguyn t nhn nng lng kch thch = hf? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Ch 6. Tm nng lng bc electron ra khi nguyn t khi n ang qy o K ( ng vi nng lng E1)? . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Phn17 . PHNG PHP GII TON V PHNG X V PHN NG HT NHN 110 Ch 1. Cht phng x A Z X c s khi A: tm s nguyn t ( ht) c trong m(g) ht nhn ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Ch 2. Tm s nguyn t N( hay khi lng m) cn li, mt i ca cht phng x sau thi gian t? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Ch 3. Tnh khi lng ca cht phng x khi bit phng x H? . . . . . . . 110 Ch 4. Xc nh tui ca mu vt c c ngun gc l thc vt? . . . . . . . . . 110 Ch 5. Xc nh tui ca mu vt c c ngun gc l khong cht? . . . . . . . 111 Ch 6. Xc nh nng lng lin kt ht nhn( nng lng ta ra khi phn r mt ht nhn)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Ch 7. Xc nh nng lng ta ra khi phn r m(g) ht nhn A ZX? . . . . . . . 111 Ch 8. Xc nh nng lng ta ( hay thu vo ) ca phn ng ht nhn? . . . . . 111 Th.s Trn AnhTrung 13 Luyn thi i hc www.VNMATH.com 14. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch 9. Xc nh nng lng ta khi tng hp m(g) ht nhn nh(t cc ht nhn nh hn)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Ch 10. Cch vn dng nh lut bo ton ng lng, nng lng? . . . . . . . 112 1.Cch vn dng nh lut bo ton ng lng: . . . . . . . . . . . . . . . . . 112 2.Cch vn dng nh lut bo ton nng lng: . . . . . . . . . . . . . . . . . 113 Ch 11. Xc nh khi lng ring ca mt ht nhn nguyn t. Mt in tch ca ht nhn nguyn t ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Th.s Trn AnhTrung 14 Luyn thi i hc www.VNMATH.com 15. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 1 PHNG PHP GII TON V DAO NG IU HA CA CON LC L XO CH 1.Lin h gia lc tc dng, gin v cng ca l xo: Phng php: 1.Cho bit lc ko F, cng k: tm gin l0, tm l: +iu kin cn bng: F + F0 = 0 hayF = kl0 hay l0 = F k +Nu F = P = mg th l0 = mg k +Tm l: l = l0 + l0, lmax = l0 + l0 + A; lmin = l0 + l0 A Ch : Lc n hi ti mi im trn l xo l nh nhau, do l xo gin u. 2.Ct l xo thnh n phn bng nhau ( hoc hai phn khng bng nhau): tm cng ca mi phn? p dng cng thc Young: k = E S l a. Ct l xo thnh n phn bng nhau (cng k): k k0 = l0 l = n k = nk0. b. Ct l xo thnh hai phn khng bng nhau: k1 k0 = l0 l1 v k2 k0 = l0 l2 CH 2.Vit phng trnh dao ng iu ha ca con lc l xo: Phng php: Phng trnh li v vn tc ca dao ng iu ha: x = Asin(t + ) (cm) v = Acos(t + ) (cm/s) Tm : + Khi bit k, m: p dng: = k m + Khi bit T hay f: = 2 T = 2f Tm A: + Khi bit chiu di qy o: d = BB = 2A A = d 2 + Khi bit x1, v1: A = x2 1 + v2 1 2 Th.s Trn AnhTrung 15 Luyn thi i hc www.VNMATH.com 16. Phng php gii ton Vt L 12 Trng THPT - Phong in + Khi bit chiu di lmax, lmin ca l xo: A = lmax lmin 2 . + Khi bit nng lng ca dao ng iu ha: E = 1 2 kA2 A = 2E k Tm : Da vo iu kin ban u: khi t0 = 0 x = x0 = A sin sin = x0 A Tm A v cng mt lc:Da vo iu kin ban u: t0 = 0 x = x0 v = v0 x0 = Asin v0 = Acos A Ch :Nu bit s dao ng n trong thi gian t, chu k: T = t n CH 3.Chng minh mt h c hc dao ng iu ha: Phng php: Cch 1: Phng php ng lc hc 1.Xc nh lc tc dng vo h v tr cn bng: F0k = 0. 2.Xt vt v tr bt k ( li x), tm h thc lin h gia F v x, a v dng i s: F = kx ( k l hng s t l, F l lc hi phc. 3.p dng nh lut II Newton: F = ma kx = mx, a v dng phng trinh: x + 2 x = 0. Nghim ca phng trnh vi phn c dng: x = Asin(t + ). T , chng t rng vt dao ng iu ha theo thi gian. Cch 2: Phng php nh lut bo ton nng lng 1.Vit biu thc ng nng E ( theo v) v th nng Et ( theo x), t suy ra biu thc c nng: E = E + Et = 1 2 mv2 + 1 2 kx2 = const () 2.o hm hai v () theo thi gian: (const) = 0; (v2 ) = 2v.v = 2v.x; (x2 ) = 2x.x = 2x.v. 3.T () ta suy ra c phng trnh:x + 2 x = 0. Nghim ca phng trnh vi phn c dng: x = Asin(t + ). T , chng t rng vt dao ng iu ha theo thi gian. CH 4.Vn dng nh lut bo ton c nng tm vn tc: Phng php: nh lut bo ton c nng: E = E + Et = 1 2 mv2 + 1 2 kx2 = 1 2 kA2 = Emax = Etmax () T () ta c: v = k m (A2 x2) hay v0max = A k m Th.s Trn AnhTrung 16 Luyn thi i hc www.VNMATH.com 17. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 5.Tm biu thc ng nng v th nng theo thi gian: Phng php: Th nng: Et = 1 2 kx2 = 1 2 kA2 sin2 (t + ) ng nng: E = 1 2 mv2 = 1 2 kA2 cos2 (t + ) Ch :Ta c: t = 2 T t CH 6.Tm lc tc dng cc i v cc tiu ca l xo ln gi treo hay gi : Phng php: Lc tc dng ca l xo ln gi treo hay gi chnh l lc n hi. 1.Trng hp l xo nm ngang: iu kin cn bng: P + N = 0, do lc ca l xo tc dng vo gi chnh l lc n hi.Lc n hi: F = kl = k|x|. v tr cn bng: l xo khng b bin dng: l = 0 Fmin = 0. v tr bin: l xo b bin dng cc i: x = A Fmax = kA. 2.Trng hp l xo treo thng ng: iu kin cn bng: P + F0 = 0, gin tnh ca l xo: l0 = mg k . Lc n hi v tr bt k: F = k(l0 + x) (*). Lc n gi cc i( khi qa nng bin di): x = +A Fmax = k(l0 + A) Lc n hi cc tiu: Trng hp A < l0: th F = min khi x = A: Fmin = k(l0 A) Trng hp A > l0: th F = min khi x = l0 (l xo khng bin dng): Fmin = 0 3.Ch : *Lc n hi ph thuc thi gian: thay x = A sin(t + ) vo (*) ta c: F = mg + kA sin(t + ) th: Th.s Trn AnhTrung 17 Luyn thi i hc www.VNMATH.com 18. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 7.H hai l xo ghp ni tip: tm cng kh, t suy ra chu k T: Phng php: v tr cn bng: + i vi h nm ngang: P + N = 0 + i vi h thng ng: P + F0 = 0 v tr bt k( OM = x): L xo L1 gin on x1: F = k1x1 x1 = F k1 L xo L2 gin on x2: F = k2x2 x2 = F k2 H l xo gin on x: F = khx x = F kh Ta c :x = x1 + x2, vy: 1 kh = 1 k1 + 1 k2 , chu k: T = 2 m kh CH 8.H hai l xo ghp song song: tm cng kh, t suy ra chu k T: Phng php: v tr cn bng: + i vi h nm ngang: P + N = 0 + i vi h thng ng: P + F01 + F02 = 0 v tr bt k( OM = x): L xo L1 gin on x: F1 = k1x L xo L2 gin on x: F2 = k2x H l xo gin on x: Fh = khx Ta c :F = F1 + F2, vy: kh = k1 + k2 , chu k: T = 2 m kh CH 9.H hai l xo ghp xung i: tm cng kh, t suy ra chu k T: Phng php: v tr cn bng: + i vi h nm ngang: P + N = 0 + i vi h thng ng: P + F01 + F02 = 0 v tr bt k( OM = x): L xo L1 gin on x: F1 = k1x L xo L2 nn on x: F2 = k2x H l xo bin dng x: Fh = khx Ta c :F = F1 + F2, vy: kh = k1 + k2 , chu k: T = 2 m kh CH 10.Con lc lin kt vi rng rc( khng khi lng): chng minh rng h Th.s Trn AnhTrung 18 Luyn thi i hc www.VNMATH.com 19. Phng php gii ton Vt L 12 Trng THPT - Phong in dao ng iu ha, t suy ra chu k T: Phng php: Dng 1.Hn bi ni vi l xo bng dy nh vt qua rng rc: p dng nh lut bo ton c nng:E = E + Et = 1 2 mv2 + 1 2 kx2 = const o hm hai v theo thi gian: 1 2 m2vv + 1 2 k2xx = 0. t: = k m , ta suy ra c phng trnh:x + 2 x = 0. Nghim ca phng trnh vi phn c dng: x = Asin(t+ ). T , chng t rng vt dao ng iu ha theo thi gian.Chu k: T = 2 Dng 2.Hn bi ni vi rng rc di ng, hn bi ni vo dy vt qua rng rc: Khi vt nng dch chuyn mt on x th l xo bin dng mt on x 2 . iu kin cn bng: l0 = F0 k = 2T0 k = 2mg k . Cch 1: v tr bt k( li x): ngoi cc lc cn bng, xut hin thm cc lc n hi |Fx| = kxL = k x 2 |Tx| = |Fx| 2 = k 4 x Xt vt nng:mg + T = ma mg (|T0| + |Tx|) = mx x + k 4m x = 0. t: 2 = k 4m , phng trnh tr thnh:x + 2 x = 0, nghim ca phng trnh c dng:x = Asin(t + ), vy h dao ng iu ho. Chu k: T = 2 hay T = 2 4m k Cch 2:C nng:E = E + Et = 1 2 mv2 + 1 2 kx2 L = 1 2 mv2 + 1 2 k( x 2 )2 = const o hm hai v theo thi gian: 1 2 m2vv + 1 2 k 4 2xx = 0 x + k 4m x = 0. t: 2 = k 4m , phng trnh tr thnh:x + 2 x = 0, nghim ca phng trnh c dng:x = Asin(t + ), vy h dao ng iu ho. Chu k: T = 2 hay T = 2 4m k Dng 3.L xo ni vo trc rng rc di ng, hn bi ni vo hai l xo nh dy vt qua rng rc: v tr cn bng: P = 2T0; F02 = 2T vi (F01 = T0) Th.s Trn AnhTrung 19 Luyn thi i hc www.VNMATH.com 20. Phng php gii ton Vt L 12 Trng THPT - Phong in v tr bt k( li x) ngoi cc lc cn bng ni trn, h cn chu tc dng thm cc lc: L1 gin thm x1, xut hin thm F1, m di x1. L2 gin thm x2, xut hin thm F2, m di 2x2. Vy: x = x1 + 2x2 (1) Xt rng rc: (F02 + F2) 2(T0 + F1) = mRaR = 0 nn: F2 = 2F1 k2x2 = 2k1x1, hay: x2 = 2k1 k2 x1 (2) Thay (2) vo (1) ta c: x1 = k2 k2 + 4k1 x Lc hi phc gy ra dao ng ca vt m l: Fx = F1 = k1x1 (3) Thay (2) vo (3) ta c: Fx = k2k1 k2 + 4k1 x, p dng: Fx = max = mx. Cui cng ta c phng trnh: x + k2k1 m(k2 + 4k1) x = 0. t: 2 = k2k1 m(k2 + 4k1) , phng trnh tr thnh:x + 2 x = 0, nghim ca phng trnh c dng:x = Asin(t + ), vy h dao ng iu ho. Chu k: T = 2 hay T = 2 k2k1 m(k2 + 4k1) CH 11.Lc hi phc gy ra dao ng iu ha khng phi l lc n hi nh: lc y Acximet, lc ma st, p lc thy tnh, p lc ca cht kh...: chng minh h dao ng iu ha: Dng 1.F l lc y Acximet: V tr cn bng: P = F0A V tr bt k ( li x): xut hin thm lc y Acximet: FA = V Dg. Vi V = Sx, p dng nh lut II Newton: F = ma = mx. Ta c phng trnh:x+2 x = 0, nghim ca phng trnh c dng:x = Asin(t+), vy h dao ng iu ho. Chu k: T = 2 , vi = SDg m Dng 2.F l lc ma st: V tr cn bng: P = (N01 + N02) v Fms01 = Fms02 V tr bt k ( li x):Ta c: P = (N1 + N2) nhng Fms1 = Fms2 Th.s Trn AnhTrung 20 Luyn thi i hc www.VNMATH.com 21. Phng php gii ton Vt L 12 Trng THPT - Phong in Hp lc: |F| = F1 F2 = (N1 N2) (*) M ta c: MN1/G = MN2/G N1(l x) = N2(l + x) N1 (l + x) = N2 (l x) = N1 + N2 2l = N1 N2 2x Suy ra: N1 N2 = (N1 + N2) x l = P x l = mg x l T (*) suy ra: |F| = mg x l , p dng nh lut II Newton: F = ma = mx. Ta c phng trnh:x+2 x = 0, nghim ca phng trnh c dng:x = Asin(t+), vy h dao ng iu ho. Chu k: T = 2 , vi = g l Dng 3.p lc thy tnh: v tr bt k, hai mc cht lng lch nhau mt on h = 2x. p lc thu tnh: p = Dgh suy ra lc thu tnh: |F| = pS = Dg2xS, gi tr i s:F = pS = Dg2xS, p dng nh lut II Newton: F = ma = mx. Ta c phng trnh:x + 2 x = 0, nghim ca phng trnh c dng:x = Asin(t+), vy h dao ng iu ho. Chu k: T = 2 , vi = 2SDg m Dng 4.F l lc ca cht kh: V tr cn bng: p01 = p02 suy ra F01 = F02; V0 = Sd V tr bt k ( li x):Ta c: V1 = (d + x)S; V2 = (d x)S p dng nh lut Bil-Marit: p1V1 = p2V2 = p0V0 Suy ra: p1 p2 = 2p0d d2 x2 x Hp lc: |F| = F2 F1 = (p1 p2)S = 2p0dS d2 x2 x 2p0dS d2 x i s: F = 2p0dS d2 x, p dng nh lut II Newton: F = ma = mx. Ta c phng trnh:x+2 x = 0, nghim ca phng trnh c dng:x = Asin(t+), vy h dao ng iu ho. Chu k: T = 2 , vi = md2 2p0V0 Th.s Trn AnhTrung 21 Luyn thi i hc www.VNMATH.com 22. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 2 PHNG PHP GII TON V DAO NG IU HA CA CON LC N GHI NH 1. bin thin i lng X:X = Xsau Xtrc a. Nu X > 0 th X tng. b. Nu X < 0 th X gim. 2.Cng thc gn ng: a. 1 ta c: (1 + )n 1 + n H qu: 1 + 1 1 + 2 (1 1 2 2)(1 + 1 2 1) = 1 1 2 (2 1) b. 100 ; 1(rad) Ta c: cos 1 2 2 ;sin tg (rad) CH 1.Vit phng trnh dao ng iu ha ca con lc n: Phng php: Phng trnh dao ng c dng: s = s0sin(t + ) hay = 0sin(t + ) (1) s0 = l0 hay 0 = s0 l : c xc nh bi: = g l Tm s0 v cng mt lc:Da vo iu kin ban u: t0 = 0 s = s1 v = v1 s1 = s0sin v1 = s0cos s0 Ch :Nu bit s dao ng n trong thi gian t, chu k: T = t n CH 2.Xc nh bin thin nh chu k T khi bit bin thin nh gia tc trng trng g, bin thin chiu di l: Phng php: Lc u: T = 2 l g ; Lc sau: T = 2 l g Lp t s: T T = l l . g g M T = T T g = g g l = l l T = T + T g = g + g l = l + l Th.s Trn AnhTrung 22 Luyn thi i hc www.VNMATH.com 23. Phng php gii ton Vt L 12 Trng THPT - Phong in Vy: T + T T = l + l l 1 2 g g + g 1 2 1 + T T = 1 + 1 2 l l 1 1 2 g g Hay: T T = 1 2 l l g g Ch : a. Nu g = const th g = 0 T T = 1 2 l l b. Nu l = const th l = 0 T T = 1 2 g g CH 3.Xc nh bin thin nh chu k T khi bit nhit bin thin nh t; khi a ln cao h; xung su h so vi mt bin: Phng php: 1.Khi bit nhit bin thin nh t: nhit t0 1C: T1 = 2 l1 g ; nhit t0 2C: T2 = 2 l2 g Lp t s: T2 T1 = l2 l1 = l0(1 + t2) l0(1 + t1) = 1 + t2 1 + t1 = 1 + t2 1 2 1 + t1 1 2 p dng cng thc tnh gn ng:(1 + )n 1 + n T2 T1 = 1 + 1 2 t2 1 1 2 t1 Hay: T T1 = 1 2 (t2 t1) = 1 2 t 2.Khi a con lc n ln cao h so vi mt bin: mt t : T = 2 l g ; cao h: Th = 2 l gh ; Lp t s: Th T = g gh (1). Ta c, theo h qa ca nh lut vn vt hp dn: g = G M R2 gh = G M (R + h)2 Thay vo (1) ta c: Th T = R + h R Hay: T T = h R 3.Khi a con lc n xung su h so vi mt bin: mt t : T = 2 l g ; su h: Th = 2 l gh ; Lp t s: Th T = g gh (2). Ta c, theo h qa ca nh lut vn vt hp dn: Th.s Trn AnhTrung 23 Luyn thi i hc www.VNMATH.com 24. Phng php gii ton Vt L 12 Trng THPT - Phong in g = G M R2 gh = G Mh (R h)2 Thay vo (2) ta c: Th T = (R h)2 R2 M Mh Ta li c: M = V.D = 4 3 R3 .D Mh = Vh.D = 4 3 (R h)3 .D Thay vo ta c: Th T = R R h 1 2 Hay: T T = 1 2 h R CH 4.Con lc n chu nhiu yu t nh hng bin thin ca chu k: tm iu kin chu k khng i: Phng php: 1.iu kin chu k khng i: iu kin l:"Cc yu t nh hng ln chu k l phi b tr ln nhau" Do : T1 + T2 + T3 + = 0 Hay: T1 T + T2 T + T3 T + = 0 (*) 2.V d: Con lc n chu nh hng bi yu t nhit v yu t cao: Yu t nhit : T1 T = 1 2 t; Yu t cao: T2 T = h R Thay vo (*): 1 2 t + h R = 0 CH 5.Con lc trong ng h g giy c xem nh l con lc n: tm nhanh hay chm ca ng h trong mt ngy m: Phng php: Thi gian trong mt ngy m: t = 24h = 24.3600s = 86400(s) ng vi chu k T1: s dao ng trong mt ngy m: n = t T1 = 86400 T1 . ng vi chu k T2: s dao ng trong mt ngy m: n = t T2 = 86400 T2 . chnh lch s dao ng trong mt ngy m: n = |n n| = 86400 1 T1 1 T2 Hay: n = 86400 |T| T2.T1 Th.s Trn AnhTrung 24 Luyn thi i hc www.VNMATH.com 25. Phng php gii ton Vt L 12 Trng THPT - Phong in Vy: nhanh ( hay chm) ca ng h trong mt ngy m l: = n.T2 = 86400 |T| T1 Ch :Nu T > 0 th chu k tng, ng h chy chm; Nu T < 0 th chu k gim, ng h chy nhanh. CH 6.Con lc n chu tc dng thm bi mt ngoi lc F khng i: Xc nh chu k dao ng mi T : Phng php: Phng php chung: Ngoi trng lc tht P = mg, con lc n cn chu tc dng thm mt ngoi lc F, nn trng lc biu kin l: P = P + F g = g + F m (1) S dng hnh hc suy ra c ln ca g , chu k mi T = 2 l g . Ch : chng ta thng lp t s: T T = g g 1.F l lc ht ca nam chm: Chiu (1) ln xx : g = g + Fx m ; Nam chm t pha di: Fx > 0 F hng xung g = g + F m . Nam chm t pha trn: Fx < 0 F hng ln g = g F m . Chu k mi T = 2 l g . Ch : chng ta thng lp t s: T T = g g . 2.F l lc tng tc Coulomb: Lc tng tc Coulomb: F = k |q1q2| r2 ; Tm g v chu k T nh trn. Hai in tch cng du: Flc y. ; Hai in tch tri du: Flc ht. 3.F l lc in trng F = qE: Trng lc biu kin l: P = P + qE g = g + qE m (2) Chiu (2) ln xx : g = g + qEx m ; Th.s Trn AnhTrung 25 Luyn thi i hc www.VNMATH.com 26. Phng php gii ton Vt L 12 Trng THPT - Phong in Chu k mi: T = 2 l g + qEx m = 2 l g 1 + qEx mg . Ch : chng ta thng lp t s: T T = 1 1 + qEx mg = 1 + qEx mg 1 2 = 1 1 2 qEx mg hay T T = 1 2 qEx mg 4.F l lc y Acsimet FA = V Dkkg: Trng lc biu kin l: P = P + FA g = g V Dkkg m = 1 V Dkk m g (3) Chiu (3) ln xx :g = 1 V Dkk m g; Vi: m = V.D, trong D l khi lng ring ca qa cu: g = 1 Dkk D g; Chu k mi: T = 2 l 1 Dkk D g . Ch : chng ta thng lp t s: T T = 1 1 Dkk D hay T T = 1 2 Dkk D 5.F l lc nm ngang: Trng lc biu kin: P = P + F hay mg = mg + F hng xin, dy treo mt gc so vi phng thng ng. Gia tc biu kin: g = g + F m . iu kin cn bng: P + T + F = 0 P = T. Vy = PO P ng vi v tr cn bng ca con lc n. Ta c: tg = F mg Tm T v g : p dng nh l Pitago: g = g2 + (F m )2 hoc: g = g cos . Chu k mi: T = 2 l g . Thng lp t s: T T = g g = cos CH 7.Con lc n treo vo mt vt ( nh t, thang my...) ang chuyn ng vi gia tc a: xc nh chu k mi T : Th.s Trn AnhTrung 26 Luyn thi i hc www.VNMATH.com 27. Phng php gii ton Vt L 12 Trng THPT - Phong in Phng php: Trong h quy chiu gn lin vi im treo( thang my, t..) con lc n cn chu tc dng thm mt lc qun tnh F = ma. Vy trng lc biu kin P = P ma hay gia tc biu kin: g = g a (1) S dng hnh hc suy ra c ln ca g , chu k mi T = 2 l g . Ch : chng ta thng lp t s: T T = g g 1.Con lc n treo vo trn ca thang my ( chuyn ng thng ng ) vi gia tc a: Chiu (1) ln xx : g = g ax (2) a.Trng hp a hng xung: ax > 0 ax = |a| (2) : g = g a chu k mi: T = 2 l g a Thng lp t s: T T = g g a l trng hp thang my chuyn ng ln chm dn u (v, a cng chiu) hay thang my chuyn ng xung nhanh dn u (v, a ngc chiu). b.Trng hp a hng ln: ax < 0 ax = |a| (2) : g = g + a chu k mi: T = 2 l g + a Thng lp t s: T T = g g + a l trng hp thang my chuyn ng ln nhanh dn u (v, a ngc chiu) hay thang my chuyn ng xung chm dn u (v, a cng chiu). 2.Con lc n treo vo trn ca xe t ang chuyn ng ngang vi gia tc a: Gc: = PO P ng vi v tr cn bng ca con lc n. Ta c: tg = F mg = a g Th.s Trn AnhTrung 27 Luyn thi i hc www.VNMATH.com 28. Phng php gii ton Vt L 12 Trng THPT - Phong in Tm T v g : p dng nh l Pitago: g = g2 + a2 hoc: g = g cos . Chu k mi: T = 2 l g . Thng lp t s: T T = g g = cos 3.Con lc n treo vo trn ca xe t ang chuyn ng trn mt phng nghing mt gc : Ta c iu kin cn bng: P + Fqt + T = 0 (*) Chiu (*)/Ox: T sin = ma cos (1) Chiu (*)/Oy: T cos = mg ma sin (2) Lp t s: 1 2 : tg = a cos g a sin T (1) suy ra lc cng dy: T = ma cos sin T(*) ta c: P = T mg = T hay g = a cos sin Chu k mi: T = 2 l g hay T = 2 l sin a cos Th.s Trn AnhTrung 28 Luyn thi i hc www.VNMATH.com 29. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 8.Xc nh ng nng E th nng Et, c nng ca con lc n khi v tr c gc lch : Phng php: Chn mc th nng l mt phng i qua v tr cn bng. Th nng Et: Ta c: Et = mgh1 , vi h1 = OI = l(1 cos ) Vy: Et = mgl(1 cos ) (1) C nng E: p dng nh lut bo ton c nng: E = EC = EB = mgh2 = mgl(1 cos ) Hay E = mgl(1 cos ) (2) ng nng E: Ta c: E = E + Et E = E Et Thay (1) , (2) vo ta c: E = mgl(cos cos ) (3) t bit: Nu con lc dao ng b: p dng cng thc tnh gn ng: cos 1 2 2 ; cos 1 2 2 (1) Et = 1 2 mgl2 (2) E = 1 2 mgl2 (3) E = 1 2 mgl(2 2 ) CH 9.Xc nh vn tc di v v lc cng dy T ti v tr hp vi phng thng ng mt gc : Phng php: 1.Vn tc di v ti C: Ta c cng thc tnh ng nng: E = 1 2 mv2 , thay vo biu thc (3) ch 8 ta c: v = 2gl(cos cos ) (1) 2.Lc cng dy T ti C: p dng nh lut II Newton: P + T = maht (2) Chn trc ta hng tm, chiu phng trnh (2) ln xx : Ta c: mg cos + T = m v2 l Th.s Trn AnhTrung 29 Luyn thi i hc www.VNMATH.com 30. Phng php gii ton Vt L 12 Trng THPT - Phong in Thay (1) vo ta c: T = m[3 cos 2 cos ]g (3) t bit: Nu dao ng ca con lc n l dao ng b Thay biu thc tnh gn ng vo ta c: (1) v = gl(2 2) (4) (2) T = m 1 + 2 3 2 2 g (5) 3.H qa: vn tc v lc cng dy cc i v cc tiu: (1), (4) v = max = 0(v tr cn bng), vmax = 2gl(1 cos ) vmax = gl v = min = (v tr bin) vmin = 0, (3), (5) T = max = 0(v tr cn bng), Tmax = m(3 2 cos )g Tmax = m[1 + 2 ]g T = min = (v tr bin) Tmin = mg cos Tmin = m[1 1 2 2 ]g CH 10.Xc nh bin gc mi khi gia tc trng trng thay i t g sang g : Phng php: p dng cng thc s (2) ch (8) Khi con lc ni c gia tc trng trng g: C nng ca con lc: E = 1 2 mgl2 . Khi con lc ni c gia tc trng trng g : C nng ca con lc: E = 1 2 mg l 2 . p dng nh lut bo ton c nng: E = E 1 2 mgl2 = 1 2 mg l 2 Hay: = g g CH 11.Xc nh chu k v bin ca con lc n vng inh (hay vt cn) khi i qua v tr cn bng: Phng php: 1.Tm chu k T: Chu k ca con lc n vng inh T = 1 2 chu k ca con lc n c chiu di l + 1 2 chu k ca con lc n c chiu di l Th.s Trn AnhTrung 30 Luyn thi i hc www.VNMATH.com 31. Phng php gii ton Vt L 12 Trng THPT - Phong in Ta c: T = 1 2 T1 + 1 2 T2 Trong : T1 = 2 l g T2 = 2 l g vi:l = l QI 2.Tm bin mi sau khi vng inh: Vn dng ch (10) ta c: 1 2 mgl2 = 1 2 mgl 2 Hay: = l l CH 12.Xc nh thi gian hai con lc n tr li v tr trng phng (cng qua v tr cn bng, chuyn ng cng chiu): Phng php: Gi s con lc th nht c chu k T1, con lc n th hai c chu k T2 ( T2 > T1). Nu con lc th nht thc hin c n dao ng th con lc th hai thc hin c n 1 dao ng. Gi t l thi gian tr li trng phng, ta c: t = nT1 = (n 1)T2 n = T2 T2 T1 Vy thi gian tr li trng phng: t = T1.T2 T2 T1 CH 13.Con lc n dao ng th b dy t:kho st chuyn ng ca hn bi sau khi dy t? Phng php: 1.Trng hp dy t khi i qua v tr cn bng O: Lc chuyn ng ca vt xem nh l chuyn ng vt nm ngang. Chn h trc ta Oxy nh hnh v. Theo nh lut II Newton: F = P = ma Hay: a = g (*) Chiu (*) ln Ox: ax = 0, trn Ox, vt chuyn ng thng u vi phng trnh: x = v0t t = x v0 (1) Chiu (*) ln Oy: ax = g, trn Oy, vt chuyn ng thng nhanh dn u vi phng trnh: Th.s Trn AnhTrung 31 Luyn thi i hc www.VNMATH.com 32. Phng php gii ton Vt L 12 Trng THPT - Phong in y = 1 2 ayt2 = 1 2 gt2 (2) Thay (1) vo (2), phng trnh qu o: y = 1 2 . g v2 0 x2 Kt lun: qu o ca qa nng sau khi dy t ti VTCB l mt Parabol.( y = ax2 ) 2.Trng hp dy t khi i qua v tr c li gic : Lc chuyn ng ca vt xem nh l chuyn ng vt nm xin hng xung, c vc hp vi phng ngang mt gc : vc = 2gl(cos cos 0). Chn h trc ta Oxy nh hnh v. Theo nh lut II Newton: F = P = ma Hay: a = g (*) Chiu (*) ln Ox: ax = 0, trn Ox, vt chuyn ng thng u vi phng trnh: x = vc cos t t = x v0 cos (1) Chiu (*) ln Oy: ax = g, trn Oy, vt chuyn ng thng bin i u, vi phng trnh: y = vc sin t 1 2 gt2 (2) Thay (1) vo (2), phng trnh qu o: y = g 2vc cos2 x2 + tg.x Kt lun: qu o ca qa nng sau khi dy t ti v tr C l mt Parabol.( y = ax2 +bx) CH 14.Con lc n c hn bi va chm n hi vi mt vt ang ng yn: xc nh vn tc ca vin bi sau va chm? Phng php: * Vn tc ca con lc n trc va chm( VTCB): v0 = 2gl(1 cos 0) *Gi v, v l vn tc ca vin bi v qa nng sau va chm: p dng nh lut bo ton ng nng: mv0 = mv + m1v (1) p dng nh lut bo ton ng lng: 1 2 mv2 0 = 1 2 mv2 + 1 2 m1v 2 (2) T (1) v (2) ta suy ra c v v v. Th.s Trn AnhTrung 32 Luyn thi i hc www.VNMATH.com 33. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 3 PHNG PHP GII TON V DAO NG TT DN V CNG HNG C HC CH 1.Con lc l xo dao ng tt dn: bin gim dn theo cp s nhn li v hng, tm cng bi q: Phng php: C nng ban u(cung cp cho dao ng): E0 = Et(max) = 1 2 kA2 1 (1) Cng ca lc masat (ti lc dng li): |Ams| = Fmss = mgs (2), vi s l on ng i ti lc dng li. p dng nh lut bo ton v chuyn ha nng lng: Ams = E0 s Cng bi q: v bin gim dn theo cp s nhn li v hn nn: q = A2 A1 = A3 A2 = = An A(n1) A2 = qA1, A3 = q2 A1 , An = qn1 A1(viq < 1) ng i tng cng ti lc dng li: s = 2A1 + 2A2 + + 2An = 2A1(1 + q + q2 + + qn1 ) = 2A1S Vi: S = (1 + q + q2 + + qn1 ) = 1 1 q Vy: s = 2A1 1 q CH 2.Con lc l n ng tt dn: bin gc gim dn theo cp s nhn li v hng, tm cng bi q. Nng lng cung cp duy tr dao ng: Phng php: Cng bi q: v bin gc gim dn theo cp s nhn li v hn nn: q = 2 1 = 3 2 = = n (n1) 2 = q1, 3 = q2 1 , n = qn1 1(viq < 1) Vy: q =n1 n 1 Nng lng cung cp ( nh ln dy ct) trong thi gian t duy tr dao ng: C nng chu k 1: E1 = EtB1max = mgh1, hay E1 = 1 2 mgl2 1 C nng chu k 2: E2 = EtB2max = mgh1, hay E2 = 1 2 mgl2 2 gim c nng sau 1 chu k: E = 1 2 mgl(2 1 2 2) Th.s Trn AnhTrung 33 Luyn thi i hc www.VNMATH.com 34. Phng php gii ton Vt L 12 Trng THPT - Phong in Hay : E = 1 2 mgl(2 1(1 q2 ), y chnh l nng lng cn cung cp duy tr dao ng trong mt chu k. Trong thi gian t, s dao ng: n = t T . Nng lng cn cung cp duy tr sau n dao ng: E = n.E. Cng sut ca ng h: P = E t CH 3.H dao ng cng bc b kch thch bi mt ngoi lc tun hon: tm iu kin c hin tng cng hng: Phng php: iu kin c hin tng cng hng: f = f0, vi f0 l tn s ring ca h. i vi con lc l xo: f0 = 1 T0 = 1 2 k m i vi con lc n: f0 = 1 T0 = 1 2 g l Th.s Trn AnhTrung 34 Luyn thi i hc www.VNMATH.com 35. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 4 PHNG PHP GII TON V S TRUYN SNG C HC , GIAO THOA SNG, SNG DNG, SNG M CH 1.Tm lch pha gia hai im cch nhau d trn mt phng truyn sng? Tm bc sng khi bit lch pha v gii hn ca bc sng,( tn s, vn tc truyn sng). Vit phng trnh sng ti mt im : Phng php: 1.Tm lch pha gia hai im cch nhau d trn mt phng truyn sng: lch pha gia hai im hai thi im khc nhau: = 2 T t = t lch pha gia hai im cch nhau d trn mt phng truyn sng = 2 d Vi Hai dao ng cng pha = 2k; k Z Hai dao ng ngc pha = (2k + 1); k Z 2.Tm bc sng khi bit lch pha v gii hn ca bc sng,( tn s, vn tc truyn sng): Gi s xt hai dao ng cng pha = 2k , so snh vi cng thc v lch pha: T suy ra c bc sng theo k: = d k Nu cho gii hn ca : ta c: 1 d k 2, c bao gi tr nguyn ca k thay vo ta suy ra c bc sng hay tn s, vn tc. Nu bi ton cho gii hn ca tn s hay vn tc, p dng cng thc: = V.T = V f . T suy ra cc gi tr nguyn ca k, suy ra c i lng cn tm. Ch : Nu bit lc cng dy F, v khi lng trn mi mt chiu di , ta c: V = F 3.Vit phng trnh sng ti mt im trn phng truyn sng: Gi s sng truyn t O n M:OM = d, gi s sng ti O c dng: uO = a sin t (cm). Sng ti M tr pha 2 d so vi O. Phng trnh sng ti M: uM = a sin(t 2 d) (cm) vi t d V 4.Vn tc dao ng ca sng: Vn tc dao ng: v = duM dt = a cos(t + 2 d) (cm/s) Th.s Trn AnhTrung 35 Luyn thi i hc www.VNMATH.com 36. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 2.V th biu din qu trnh truyn sng theo thi gian v theo khng gian: Phng php: 1.V th biu din qa trnh truyn sng theo thi gian: Xem yu t khng gian l khng i. Cch 1:( V trc tip) gc O: uO = a sin t = a sin 2 T t Xt im M(xM = OM = const): uM = a sin(t 2 xM ) iu kin t xM V Lp bng bin thin: t 0 T 4 T 2 3T 4 T uM a sin 2 xM X 0 X X V th biu din, ch ly phn biu din trong gii hn t xM V Cch 2:( V gin tip) -V th : u0 t 0 T 4 T 2 3T 4 T u0 0 A 0 A 0 Tnh tin th u0(t) theo chiu dng mt on = xM V ta c th biu din ng sin thi gian. Ch : Thng lp t s: k = T 2.V th biu din qa trnh truyn sng theo khng gian ( dng ca mi trng...): Xem yu t thi gian l khng i. Vi M thuc dy: OM = xM , t0 l thi im ang xt t0 = const Biu thc sng:uM = a sin(t 2 x) (cm) , vi chu k: ng sin khng gian l ng biu din u theo x. Gi s ti t0, sng truyn c mt on xM = V.t0, iu kin x xM .Ch : Thng lp t s: k = xM . Lp bng bin thin: x 0 4 2 3 4 u a sin t0 X X X X CH 3.Xc nh tnh cht sng ti mt im M trn min giao thoa: Th.s Trn AnhTrung 36 Luyn thi i hc www.VNMATH.com 37. Phng php gii ton Vt L 12 Trng THPT - Phong in Phng php: M : MS1 = d1; MS2 = d2 Tm hiu ng i: = d2 d1 v tm bc sng: = V.T = V f Lp t s: k = Nu p = k( nguyn) = k Mdao ng cc i Nu p = k + 1 2 ( bn nguyn) = (k + 1 2 ) Mdao ng cc tiu CH 4.Vit phng trnh sng ti im M trn min giao thoa: Phng php: Gi s:u1 = u2 = a sin t (cm) Sng tryn t S1 n M:sng ti M tr pha 2 d1 so vi S1:u1 = a sin(t 2 d1) (cm) Sng tryn t S2 n M:sng ti M tr pha 2 d2 so vi S2:u2 = a sin(t 2 d2) (cm) Sng ti M: uM = u1+u2 , thay vo, p dng cng thc: sin p+sin q = 2 sin p + q 2 cos p q 2 Cui cng ta c: uM = 2a cos (d2 d1) sin t d2 + d1 (*) Phng trnh (*) l mt phng trnh dao ng iu ha c dng: uM = A sin(t + ) Vi: Bin dao dng: A = 2a cos (d2 d1) Pha ban u: = d2 + d1 CH 5.Xc nh s ng dao ng cc i v cc tiu trn min giao thoa: Phng php: M : MS1 = d1; MS2 = d2, S1S2 = l Xt MS1S2 : ta c: |d2 d1| l l d2 d1 l (*) M dao ng vi bin cc i: = d2 d1 = k k Z Thay vo (*),ta c: l k l , c bao nhiu gi tr nguyn ca k th c by nhiu ng dao ng vi bin cc i ( k c ng trung trc on S1S2 ng vi k = 0) M dao ng vi bin cc tiu: = d2 d1 = k + 1 2 k Z Thay vo (*),ta c: l 1 2 k l 1 2 , c bao nhiu gi tr nguyn ca k th c by nhiu ng dao ng vi bin cc tiu. Th.s Trn AnhTrung 37 Luyn thi i hc www.VNMATH.com 38. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 6.Xc nh im dao ng vi bin cc i ( im bng) v s im dao ng vi bin cc tiu ( im nt) trn on S1S2: Phng php: M S1S2 : MS1 = d1; MS2 = d2, S1S2 = l Ta c: d1 + d2 = l (*) M dao ng vi bin cc i: = d2 d1 = k k Z (1) Cng (1) v (*) ta c: d2 = l 2 + k 2 , iu kin: 0 d2 l Vy ta c: l k l , c bao nhiu gi tr nguyn ca k th c by nhiu im bng ( k c im gia) M dao ng vi bin cc tiu: = d2 d1 = k + 1 2 k Z (2) Cng (2) v (*) ta c: d2 = l 2 + k + 1 2 2 , iu kin: 0 d2 l Vy ta c: l 1 2 k l 1 2 , c bao nhiu gi tr nguyn ca k th c by nhiu im nt. Ch : tm v tr cc im dao ng cc i ( hay cc tiu) ta thng lp bng: k cc gi tr m -1 0 1 cc gi tr dng d2 d2i 2 d20 d2i + 2 CH 7.Tm qy tch nhng im dao ng cng pha (hay ngc pha) vi hai ngun S1, S2: Phng php: Pha ban u sng ti M: M = (d2 + d1) Pha ban u sng ti S1 (hay S2): = 0 lch pha gia hai im: = M = (d2 + d1) (*) hai im dao ng cng pha = 2k, so snh (*): d2 + d1 = 2k. Vy tp hp nhng im dao ng cng pha vi hai ngun S1, S2 l h ng Ellip, nhn hai im S1, S2 lm hai tiu im. hai im dao ng ngc pha = (2k + 1), so snh (*): d2 + d1 = (2k + 1). Vy tp hp nhng im dao ng ngc pha vi hai ngun S1, S2 l h ng Ellip, nhn hai im S1, S2 lm hai tiu im ( xen k vi h Ellip ni trn). CH 8.Vit biu thc sng dng trn dy n hi: Phng php: Th.s Trn AnhTrung 38 Luyn thi i hc www.VNMATH.com 39. Phng php gii ton Vt L 12 Trng THPT - Phong in Gi: MC = d, AC = l th AM = l d. Cc bc thc hin: 1.Vit biu thc sng ti: Sng ti A: uA = a sin t Sng ti M: Ti M sng tr pha 2 (l d) so vi A uM = a sin t 2 (l d) (1) Ti C sng tr pha 2 l so vi A uC = a sin(t 2 l) (2) 2.Vit biu thc sng phn x: Sng ti C: Nu C c nh uC = uC = a sin(t 2 l) (3) Nu C t do uC = uC = a sin(t 2 l) (4) Sng ti M: Ti M sng tr pha 2 d so vi C: Nu C c nh uM = a sin(t 2 l 2 d) (5) Nu C t do uM = a sin(t 2 l 2 d) (6) 3.Sng ti M: u = uM + uM , dng cng thc lng gic suy ra c biu thc sng dng. CH 9.iu kin c hin tng sng dng, t suy ra s bng v s nt sng: Phng php: 1.Hai u mi trng ( dy hay ct khng kh) l c nh: + iu kin v chiu di: l s nguyn ln mi sng: l = k 2 + iu kin v tn s: = V f f = k V 2l + S mi: k = 2l , s bng l k v s nt l k + 1. 2.Mt u mi trng ( dy hay ct khng kh) l c nh, u kia t do: + iu kin v chiu di: l s bn nguyn ln mi sng: Th.s Trn AnhTrung 39 Luyn thi i hc www.VNMATH.com 40. Phng php gii ton Vt L 12 Trng THPT - Phong in l = k + 1 2 2 + iu kin v tn s: = V f f = k + 1 2 v 2l + S mi: k = 2l 1 2 , s bng l k + 1 v s nt l k + 1. 3.Hai u mi trng ( dy hay ct khng kh) l t do: + iu kin v chiu di: l s nguyn ln mi sng: l = k 2 + iu kin v tn s: = V f f = k v 2l + S mi: k = 2l , s bng l k v s nt l k 1. Ch : Cho bit lc cng dy F, mt chiu di : V = F Thay vo iu kin v tn s: F = 4l2 f2 k2 CH 10.Xc nh cng m (I) khi bit mc cng m ti im. Xc nh cng sut ca ngun m? to ca m: Phng php: 1.Xc nh cng m (I) khi bit mc cng m ti im: *Nu mc cng m tnh theo n v B: L = lg I I0 T : I = I0.10L * Nu mc cng m tnh theo n v dB:L = 10lg I I0 T : I = I0.10 L 10 Ch : Nu tn s m f = 1000Hz th I0 = 1012 Wm2 2.Xc nh cng sut ca ngun m ti mt im: Cng sut ca ngun m ti A l nng lng truyn qua mt cu tm N bn knh NA trong 1 giy. Ta c: IA = W S W = IA.S hay Pngun = IA.SA Nu ngun m l ng hng: SA = 4NA2 Nu ngun m l loa hnh nn c na gc nh l : Gi R l khong cch t loa n im m ta xt. Din tch ca chm cu bn knh R v Th.s Trn AnhTrung 40 Luyn thi i hc www.VNMATH.com 41. Phng php gii ton Vt L 12 Trng THPT - Phong in chiu cao h l S = 2Rh Ta c: h = R R cos , vy S = 2R2 (1 cos ) Vy, cng sut ca ngun m: P = I.2R2 (1 cos ) 3. to ca m: Ty tn s, mi m c mt ngng nghe ng vi Imin to ca m: I = I Imin to ti thiu m tai phn bit c gi l 1 phn Ta c: I = 1phn 10lg I2 I1 = 1dB Th.s Trn AnhTrung 41 Luyn thi i hc www.VNMATH.com 42. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 5 PHNG PHP GII TON V MCH IN XOAY CHIU KHNG PHN NHNH (RLC) CH 1.To ra dng in xoay chiu bng cch cho khung dy quay u trong t trng, xc nh sut in ng cm ng e(t)? Suy ra biu thc cng dng in i(t) v hiu in th u(t): Phng php: 1.Tm biu thc t thng (t): (t) = NBS cos(t) hay (t) = 0 cos(t) vi 0 = NBS. 2. Tm biu thc ca s cm ng e(t): e(t) = d(t) dt = NBS sin(t) hay e(t) = E0 sin(t) vi: E0 = NBS 3.Tm biu thc cng dng in qua R: i = e(t) R 4.Tm biu thc ht tc thi u(t): u(t) = e(t) suy ra U0 = E0 hay U = E. CH 2.on mch RLC: cho bit i(t) = I0 sin(t), vit biu thc hiu in th u(t). Tm cng sut Pmch? Phng php: Nu i = I0 sin(t) th u = U0 sin(t + ) (*) Vi: U0 = I0.Z, tng tr: Z = R2 + (ZL ZC )2 vi ZL = L ZC = 1 C tg = ZL ZC R , vi l lch pha ca u so vi i. Cng sut tiu th ca on mch: Cch 1: Dng cng thc: P = UI cos , vi U = U0 2 , I = I0 2 , cos = R Z Cch 2: Trong cc phn t in, ch c in tr R mi tiu th in nng di dng ta nhit: P = RI2 Ch : 1 = 0, 318 CH 3.on mch RLC: cho bit u(t) = U0 sin(t), vit biu thc cng dng in i(t). Suy ra biu thc uR(t)?uL(t)?uC(t)? Phng php: Nu u = U0 sin(t) th i = I0 sin(t ) (*) Th.s Trn AnhTrung 42 Luyn thi i hc www.VNMATH.com 43. Phng php gii ton Vt L 12 Trng THPT - Phong in I0 = U0 . Z, tng tr: Z = R2 + (ZL ZC )2 vi tg = ZL ZC R H qa: Hiu in th hai u in tr R cng pha vi cd: uR = U0R sin(t ). vi: U0R = I0.R. Hiu in th hai u cun cm L nhanh pha 2 so vi cd: uL = U0L sin(t + 2 ). vi: U0L = I0.ZL. Hiu in th hai u t in C chm pha 2 so vi cd: uC = U0C sin(t 2 ). vi: U0C = I0.ZC . Ch : Nu phn t in no b on mch hoc khng c trong on mch th ta xem in tr tng ng bng 0. Nu bit: i = I0 sin(t+i) v u = U0 sin(t+u) th lch pha: u/i = u i CH 4.Xc nh lch pha gia hai ht tc thi u1 v u2 ca hai on mch khc nhau trn cng mt dng in xoay chiu khng phn nhnh? Cch vn dng? Phng php: Cch 1:(Dng i s) lch pha ca u1 so vi i: tg1 = ZL1 ZC1 R1 1 lch pha ca u2 so vi i: tg2 = ZL2 ZC2 R2 2 Ta c: u1/u2 = u1 u2 = (u1 i) (u2 i) = u1/i u2/i = 1 2 lch pha ca u1 so vi u2: = 1 2 Cch 2:(Dng gin vect) Ta c: u = u1 + u2 U = U1 + U2 trc pha I. U1 U1 = I.Z1 tg1 = ZL1 ZC1 R1 1 ; U2 = I.Z2 tg2 = ZL2 ZC2 R2 1 lch pha ca u1 so vi u2: = 1 2 CH 5.on mch RLC, cho bit U, R: tm h thc L, C, : cng dng in qua on mch cc i, hiu in th v cng dng in cng pha, cng sut tiu th trn on mch t cc i. Phng php: 1.Cng dng in qua on mch t cc i: Th.s Trn AnhTrung 43 Luyn thi i hc www.VNMATH.com 44. Phng php gii ton Vt L 12 Trng THPT - Phong in p dng nh lut Ohm cho on mch: I = U Z = U R2 + (ZL ZC )2 () Ta c: I = max M = R2 + (ZL ZC )2 = min ZL ZC = 0 L = 1 C Hay LC2 = 1 () Imax = U R 2.Hiu in th cng pha vi cng dng in: u v i cng pha: = 0 hay tg = ZL ZC R = 0 ZL ZC = 0 L = 1 C Hay LC2 = 1 3.Cng sut tiu th trn on mch cc i: Ta c: P = UI cos , P = max cos = 1 Ta c: cos = R R2 + (ZL ZC )2 = 1 Hay R2 + (ZL ZC )2 = R2 Hay LC2 = 1 4.Kt lun: Hin tng cng hng in: LC2 = 1 I = max u, i cng pha ( = 0) cos = 1 H qa: 1.Imax = U R 2.Do ZL = ZC UL = UC vi L = C = 2 nn UL = UC uL = uC CH 6.on mch RLC, ghp thm mt t C :tm C : cng dng in qua on mch cc i, hiu in th v cng dng in cng pha, cng sut tiu th trn on mch t cc i. Phng php: Gi Cb l in dung tng ng ca b t, tng t ch 5, ta c: LCb2 = 1 Cb = 1 L2 Nu C ni tip vi C : 1 Cb = 1 C + 1 C Nu C song song vi C : Cb = C + C Th.s Trn AnhTrung 44 Luyn thi i hc www.VNMATH.com 45. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 7.on mch RLC: Cho bit UR, UL, UC: tm U v lch pha u/i. Phng php: Cch 1:( Dng i s) p dng cng thc: I = U Z = U R2 + (ZL ZC )2 U = I R2 + (ZL ZC )2 U = U2 R + (UL UC )2 Cch 2:( Dng gin vect) Ta c: u = uR + uL + uC U = UR + UL + UC trc pha I Da vo gin vect: ta c U = U2 R + (UL UC )2 lch pha: tg = ZL ZC R = IZL IZC IR Hay tg = UL UC UR CH 8.Cun dy (RL) mc ni tip vi t C: cho bit hiu in th U1 ( cun dy) v UC . Tm Umch v . Phng php: Ta c: u = u1 + uC U = U1 + UC () trc pha I Vi U1 +U1 = I.Z1 = I. R2 + Z2 L +(I, U1) = 1 vi tg1 = ZL R cos 1 = R R2 + Z2 L UC +UC = I.ZC vi ZC = 1 C +(I, UC ) = 2 Xt OAC: nh l hm cosin: U2 = U2 1 + U2 C 2U1UC cos( 2 1) Hay U = U2 1 + U2 C + 2U1UC sin 1 Vi: sin 1 = cos 1.tg1 = ZL R2 + Z2 L Chiu (*) ln OI: U cos = U1 cos 1 cos = U U1 cos 1 CH 9.Cho mchRLC: Bit U, , tm L, hayC, hayR cng sut tiu th trn on mch cc i. Phng php: Th.s Trn AnhTrung 45 Luyn thi i hc www.VNMATH.com 46. Phng php gii ton Vt L 12 Trng THPT - Phong in Trong cc phn t in, ch c in tr R mi tiu th in nng di dng ta nhit: P = RI2 Ta c: I = U Z = U R2 + (ZL ZC )2 Vy: P = RU2 R2 + (ZL ZC )2 (*) 1.Tm L hay C cng sut tiu th trn on mch cc i: D P = max t (*) M = R2 + (ZL ZC )2 = min ZL ZC = 0 hay LC2 = 1 C = 1 2L L = 1 2C () Pmax = U2 R a. th L theo P: L 0 1 2C P P0 Pmax 0 Vi P0 = RU2 R2 + Z2 C b. th C theo P: C 0 1 2L P 0 Pmax P1 Vi P1 = RU2 R2 + Z2 L 2.Tm R cng sut tiu th trn on mch cc i: Chia t v mu ca (*) cho R: P = U2 R + (ZL ZC )2 R = const M P = max khi v ch khi M = min. p dng bt ng thc Csin: M = R + (ZL ZC )2 R 2 R. (ZL ZC )2 R = 2|ZL ZC| Du = xy ra khi: R = (ZL ZC )2 R hay R = |ZL ZC | Vy: Pmax = U2 2|UL UC | Bng bin thin R theo P: R 0 |ZL ZC | P 0 Pmax 0 CH 10.on mch RLC: Cho bit U, R, f: tm L ( hay C) UL (hay UC ) t gi tr cc i? Th.s Trn AnhTrung 46 Luyn thi i hc www.VNMATH.com 47. Phng php gii ton Vt L 12 Trng THPT - Phong in Phng php: 1.Tm L hiu th hiu dng hai u cun cm cc i: Hiu in th hai u cun cm: UL = I.ZL = U.ZL R2 + (ZL ZC )2 (*) Cch 1:( Dng o hm) o hm hai v ca (*) theo ZL: UL ZL = (R2 + Z2 C ZLZC )U [R2 + (ZL ZC )2] 3 2 Ta c: UL ZL = 0 ZL = R2 + Z2 C ZC , ta c bng bin thin: ZL 0 R2 + Z2 C ZC UL ZL + 0 UL ULmax Vi ULmax = U R2 + Z2 C R Cch 2:( Dng i s) Chia t v mu ca (*) cho ZL, ta c: UL = U R2 Z2 L + (1 ZC ZL )2 = const y Vi y = R2 Z2 L + (1 ZC ZL )2 = (R2 + Z2 C) 1 Z2 L 2.ZC 1 ZL + 1 = (R2 + Z2 C)x2 2.ZC x + 1 Trong : x = 1 ZL ; Ta c: a = (R2 + Z2 C) > 0 Nn y = min khi x = b 2a = ZC R2 + Z2 C , ymin = 4a = R2 R2 + Z2 C Vy: ZL = R2 + Z2 C ZC v ULmax = U R2 + Z2 C R Cch 3:( Dng gin vect) Ta c: u = uRC + uL U = URC + UL () trc pha I , t AOB = Xt OAB: nh l hm sin: UL sin AOB = U sin OAB UL sin = U sin( 2 1) = U cos 1 Hay: UL = U cos 1 sin vy: UL = max khi sin = 1 = 900 AOB O T : 1 + |u/i| = 2 , v 1 < 0, u/i > 0 nn: tg1 = cotgu/i = 1 tgu/i Th.s Trn AnhTrung 47 Luyn thi i hc www.VNMATH.com 48. Phng php gii ton Vt L 12 Trng THPT - Phong in ZC R = R ZL ZC hay ZL = R2 + Z2 L ZC , vi ULmax = U cos 1 hay ULmax = U R2 + Z2 C R 2.Tm C hiu th hiu dng hai u t in cc i: Hiu in th hai u t in: UC = I.ZC = U.ZC R2 + (ZL ZC )2 (**) Cch 1:( Dng o hm) o hm hai v ca (*) theo ZC : UC ZC = (R2 + Z2 L ZLZC )U [R2 + (ZL ZC )2] 3 2 Ta c: UC ZC = 0 ZC = R2 + Z2 L ZL , ta c bng bin thin: ZC 0 R2 + Z2 L ZL UC ZC + 0 UC UCmax Vi UCmax = U R2 + Z2 L R Cch 2:( Dng i s) Chia t v mu ca (*) cho ZC , ta c: UC = U R2 Z2 C + ( ZL ZC 1)2 = const y Vi y = R2 Z2 C + ( ZL ZC 1)2 = (R2 + Z2 L) 1 Z2 C 2.ZL 1 ZC + 1 = (R2 + Z2 L)x2 2.ZLx + 1 Trong : x = 1 ZC ; Ta c: a = (R2 + Z2 L) > 0 Nn y = min khi x = b 2a = ZL R2 + Z2 L , ymin = 4a = R2 R2 + Z2 L Vy: ZC = R2 + Z2 L ZL v UCmax = U R2 + Z2 L R Cch 3:( Dng gin vect) Ta c: u = uRL + uC U = URL + UC () trc pha I , t AOB = Xt OAB: nh l hm sin: UC sin AOB = U sin OAB UC sin = U sin( 2 1) = U cos 1 Hay: UC = U cos 1 sin vy: UC = max Th.s Trn AnhTrung 48 Luyn thi i hc www.VNMATH.com 49. Phng php gii ton Vt L 12 Trng THPT - Phong in khi sin = 1 = 900 AOB O T : 1 + |u/i| = 2 , v 1 > 0, u/i < 0 nn: tg1 = cotgu/i = 1 tgu/i ZL R = R ZL ZC hay ZC = R2 + Z2 L ZL , vi UCmax = U cos 1 hay UCmax = U R2 + Z2 L R CH 11.on mch RLC: Cho bit U, R, L, C: tm f ( hay ) UR, UL hay UC t gi tr cc i? Phng php: 1.Tm f ( hay ) hiu th hiu dng hai u in tr cc i: Hiu in th hai u in tr R: UR = I.R = UR R2 + (ZL ZC )2 = const M UR = max M = min ZL ZC = 0 hay 0 = 1 LC (1)( Vi 0 = 2f ) Vy URmax = U 2.Tm f ( hay ) hiu th hiu dng hai u cun cm cc i: Hiu in th hai u in tr L: UL = I.ZL = UZL R2 + (ZL ZC )2 = UL R2 + L 1 C 2 = U R2 2L2 + 1 1 2CL 2 Hay UL = const y , UL cc i khi y = min. Ta c: y = R2 2L2 + (1 1 2CL )2 = 1 C2L2 1 4 + R2 L2 2 1 CL 1 2 + 1 Hay: y = 1 C2L2 x2 + R2 L2 2 1 CL x + 1 vi x = 1 2 Ta c: a = 1 C2L2 > 0 Nn y = min khi x = b 2a = 2 CL R2 L2 . L2 C2 2 = 2LC R2 C2 2 Vy 1 = 2 2LC R2C2 (2) 3.Tm f ( hay ) hiu th hiu dng hai u t in cc i: Th.s Trn AnhTrung 49 Luyn thi i hc www.VNMATH.com 50. Phng php gii ton Vt L 12 Trng THPT - Phong in Hiu in th hai u in tr C: UC = I.ZC = UZC R2 + (ZL ZC )2 = U 1 C R2 + L 1 C 2 = U R2C22 + (LC 1)2 Hay UL = const y , UL cc i khi y = min. Ta c: y = R2 C2 2 + (LC 1)2 = C2 L2 4 + (R2 C2 2CL)2 + 1 Hay: y = C2 L2 x2 + (R2 L2 2CL)x + 1 vi x = 2 Ta c: a = C2 L2 > 0 Nn y = min khi x = b 2a = 2CL R2 C2 2C2L2 Vy 2 = 2CL R2 C2 2C2L2 Hay: 2 = 1 LC . 2CL R2 C2 2 (3) Ch : Ta c: 2 0 = 1.2 Hiu in th cc i hai u cun cm v t in u c dng UCmax = ULmax = 2L R U 4LC R2C2 CH 12.Cho bit th i(t) v u(t), hoc bit gin vect hiu in th: xc nh cc t im ca mch in? Phng php: 1.Cho bit th i(t) v u(t): tm lch pha u/i: Gi l lch pha v thi gian gia u v i ( o bng khong thi gian gia hai cc i lin tip ca u v i) Lch thi gian T lch pha 2 Lch thi gian lch pha u/i Vy: u/i = 2 T Th.s Trn AnhTrung 50 Luyn thi i hc www.VNMATH.com 51. Phng php gii ton Vt L 12 Trng THPT - Phong in 2.Cho bit gin vect hiu in th: v s on mch? Tm Umch Quy tc: UR nm ngang phn t R UL thng ng hng ln phn t L UC thng ng hng xung phn t C Umch +gcO; +ngn: cui UR; u/i = (I, U) CH 13.Tc dng nhit ca dng in xoay chiu: tnh nhit lng ta ra trn on mch? Phng php: Bit I: p dng cng thc Q = RI2 t Bit U: T cng thc I = U Z Q = R U2 Z2 t Nu cun dy (RL) hoc in tr dm trong cht lng: tm t0 Ta c: Qta = RI2 t; Qthu = Cmt0 t0 = RI2 t Cm CH 14.Tc dng ha hc ca dng in xoay chiu: tnh in lng chuyn qua bnh in phn theo mt chiu? Tnh th tch kh Hir v Oxy xut hin cc in cc? Phng php: 1.Tnh in lng chuyn qua bnh in phn theo mt chiu ( trong 1 chu k T, trong t): Xt dng in xoay chiu i = I0 sin t(A) qua bnh in phn cha dung dch axit hay baz long. Trong thi gian dt ( b): in lng qua bnh in phn: dq = idt = I0 sin tdt Trong 1 chu k T: dng in ch qua bnh in phn trong T 2 theo mt chiu: q1 = T 2 0 idt = T 2 0 I0 sin tdt = 1 I0 cos t T 2 0 hay q1 = 2I0 Vi = 2 T do ta c: q1 = I0T Trong thi gian t, s dao ng n = t T , in lng qua bnh in phn theo mt chiu l: q = nq1 = t T .q1 , vy: q = 2I0 t T = I0t Th.s Trn AnhTrung 51 Luyn thi i hc www.VNMATH.com 52. Phng php gii ton Vt L 12 Trng THPT - Phong in 2.Tnh th tch kh Hir v Oxy xut hin cc in cc trong thi gian t(s): C 96500C gii phng A n = 1g tng ng 11, 2(l)H ktc. Vy qC :th tch kh H: vH = q 96500 .11, 2(l) Th tch ca kh O: vO = vH 2 Vy mi in cc xut hin hn hp kh vi th tch v = vO + vH CH 15.Tc dng t ca dng in xoay chiu v tc dng ca t trng ln dng in xoay chiu? Phng php: 1.Nam chm in dng dng in xoay chiu ( tn s f) t gn dy thp cng ngang. Xc nh tn s rung f ca dy thp: Trong mt chu k, dng in i chiu hai ln. Do nam chm ht hay nh dy thp hai ln trong mt chu k. Nn tn s dao ng ca dy thp bng hai ln tn s ca dng in: f = 2f 2.Dy dn thng cng ngang mang dng in xoay chiu t trong t trng c cm ng t B khng i ( vung gc vi dy): xc nh tn s rung ca dy f : T trng khng i B tc dng ln dy dn mang dng in mt lc t F = Bil( c chiu tun theo quy tc bn tay tri ). V F t l vi i , nn khi i i chiu hai ln trong mt chu k th F i chiu hai ln trong mt chu k, do dy rung hai ln trong mt chu k. f = f Th.s Trn AnhTrung 52 Luyn thi i hc www.VNMATH.com 53. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 6 PHNG PHP GII TON V MY PHT IN XOAY CHIU, BIN TH, TRUYN TI IN NNG CH 1.Xc nh tn s f ca dng in xoay chiu to bi my pht in xoay chiu 1 pha Phng php: 1.Trng hp roto ca mp c p cp cc, tn s vng l n: Nu n tnh bng ( vng/s) th: f = np Nu n tnh bng ( vng/pht) th: f = n 60 p Ch : S cp cc: p = s cc ( bc+ nam) 2 2.Trng hp bit sut in ng xoay chiu ( E hay Eo): p dng: Eo = NBS vi = 2f , nn: f = Eo 2NBS = E 2 2NBS Ch : Nu c k cun dy ( vi N1 vng) th N = kN1 Thng thng: my c k cc ( bc + nam) th phn ng c k cun dy mc ni tip. CH 2. Nh my thy in: thc nc cao h, lm quay tuabin nc v roto ca mp. Tm cng sut P ca my pht in? Phng php: Gi: HT l hiu sut ca tuabin nc; HM l hiu sut ca my pht in; m l khi lng nc ca thc nc trong thi gian t. Cng sut ca thc nc: Po = Ao t = mgh t = gh; vi = m t l lu lng nc ( tnh theo khi lng) Cng sut ca tuabin nc: PT = HT Po Cng sut ca my pht in: PM = HM PT = HM HT Po CH 3. Mch in xoay chiu ba pha mc theo s hnh : tm cng dng trung ha khi ti i xng? Tnh hiu in th Ud ( theo Up)? Tnh Pt (cc ti) Phng php: Th.s Trn AnhTrung 53 Luyn thi i hc www.VNMATH.com 54. Phng php gii ton Vt L 12 Trng THPT - Phong in Tm ith: i1 = I0 sin t i2 = I0 sin(t + 2 3 ) i3 = I0 sin(t 2 3 ) ith = i1 + i2 + i3 = 0 Suy ra:I1 = I23 Ith = 0 Tm Ud: Ta c: Ud = UA1A2 = UA2A3 = UA3A1 : hiu in th gia hai dy pha Up = UA1O = UA2O = UA3O : hiu in th gia dy pha v dy trung ha Ta c:ud = uA1A2 = uA1O + uOA2 = uA1O uA2O UA1A2 = UA1O UA1O T hnh ta c: Ud = Up 3 Tm Pti: Do hiu in th ca cc ti bng nhau (Up) nn: Iti = Up Zti Cng sut tiu th ca mi ti: Pt = UpIt cos t = RtI2 t CH 4. My bin th: cho U1, I1: tm U2, I2 Phng php: 1.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp h: Lc : I2 = 0 p dng: U2 U1 = N2 N1 U2 2.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp c ti: a. Trng hp hiu sut MBT H = 1: Ta c: P1 = P2 U1I1 = U2I2 Hay: U2 U1 = I1 I2 hay I2 = I1 N1 N2 b. Trng hp hiu sut MBT l H : Ta c: U2 U1 = N2 N1 hay I2 = HI1 N1 N2 Th.s Trn AnhTrung 54 Luyn thi i hc www.VNMATH.com 55. Phng php gii ton Vt L 12 Trng THPT - Phong in 3.Trng hp cc in tr ca cun s cp v th cp khc 0: Sut in ng qua cun s cp: e1 = N1 d dt (1); Sut in ng qua cun th cp: e2 = N2 d dt (2); Lp t: e1 e2 = N1 N2 k (3) Cun s cp ng vai tr nh mt my pht: u1 = e1 + r1i1 e1 = u1 r1i1 (4) Cun s cp ng vai tr nh mt my thu: u2 = e2 r2i2 e2 = u2 + r2i2 (5) Lp t: e1 e2 = u1 r1i1 u2 + r2i2 k u1 r1i1 = ku2 + kr2i2 (6) Ta c e1i1 = e2i2 hay e1 e2 = i1 i2 = 1 k i1 = i2 k v i2 = u2 R (7) Thay (7) vo (6), thc hin bin i ta c: u2 = kR k2(R + r2) + r1 u1 Hay: U2 = kR k2(R + r2) + r1 U1 CH 5. Truyn ti in nng trn dy dn: xc nh cc i lng trong qu trnh truyn ti Phng php: Sn xut: U2A U1A = I1A I2A = N2A N1A PA = U1AI1A = U2AI2A Tuyn ti: Cng d.in : I = I2A = I1B in tr : R = 2l S (l = AB) gim th : UAB = U2B U2A = IR Cng sut hao ph : P = PA PB = RI2 S dng: U2B U1B = I1B I2B = N2B N1B PB = U1BI1B = U2BI2B CH 6. Xc nh hiu sut truyn ti in nng trn dy? Phng php: Cng thc nh ngha hiu sut: H = PB PA ; Xc nh theo cng sut: H = PB PA = PA P PA = 1 P P ; Th.s Trn AnhTrung 55 Luyn thi i hc www.VNMATH.com 56. Phng php gii ton Vt L 12 Trng THPT - Phong in Xc nh theo ht: H = UB UA = UA U UA = 1 U U Th.s Trn AnhTrung 56 Luyn thi i hc www.VNMATH.com 57. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 7 PHNG PHP GII TON V DAO NG IN T DO TRONG MCH LC K hiu: qmax = Q0 ( bin in tch) umax = U0 ( bin hiu in th) imax = I0 ( bin dng in) GHI NH Dao ng c hc ( con lc l xo) Dao ng in ( mch LC) Li : x in tch : q Vn tc: v = dx dt = x Cng dng in : i = dq dt Cc i lng t trng Khi lng: m t cm : L cng: k Nghch o in dung : 1 C Lc tc dng : F Hiu in th : u Phng trnh ng lc hc x + k m x = 0 q + 1 LC q = 0 x + 2 x = 0 q + 2 q = 0 Nghim ca pt vi phn x = A sin(t + ) q = Q0 sin(t + ) Tn s gc ring = k m = 1 LC Chu k dao ng T = 2 m k T = 2 LC Th nng n hi : Nng lng in trng : Et = 1 2 kx2 W = 1 2 q2 C = 1 2 Cu2 = 1 2 qu ng nng : Nng lng t trng : Nng lng dao ng E = 1 2 mv2 Wt = 1 2 Li2 C nng : Nng lng in t : E = 1 2 mv2 + 1 2 kx2 W = 1 2 Li2 + 1 2 q2 C = 1 2 kA2 = 1 2 m2 A2 = 1 2 Q2 0 C = 1 2 LI2 0 Bng so snh dao ng iu ha ca con lc l xo v dao ng in t do Th.s Trn AnhTrung 57 Luyn thi i hc www.VNMATH.com 58. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 1.Dao ng in t do trong mch LC: vit biu thc q(t)? Suy ra cng dng in i(t)? Phng php: q(t) c dng tng qut: q = Q0 sin(t + ) vi: Q0 = CU0 = 1 LC hoc = 2 T = 2f c xc nh nh iu kin ban u ( t = 0) ca q. i(t) c xc nh: i = dq dt = q = Q0 cos(t + ) = I0 cos(t + ) Vi I0 = Q0 = Q0 LC CH 2.Dao ng in t do trong mch LC, bit uC = U0 sin t, tm q(t)? Suy ra i(t)? Phng php: Ta c: q = Cu = Q0 sin t viQ0 = CU0 i(t) c xc nh: i = dq dt = q = Q0 cos t = I0 cos t hay i = I0 sin t + 2 CH 3.Cch p dng nh lut bo ton nng lng trong mch dao ng LC. Phng php: p dng nh lut bo ton v chuyn ha nng lng: W = W + Wt = Wmax = Wtmax = const hay 1 2 Li2 + 1 2 Cu2 1 2 q2 C = 1 2 LI2 0 = 1 2 CU2 0 1 2 Q2 0 C () 1.Bit Q0 ( hay U0) tm bin I0 : T (*) ta c: 1 2 CU2 0 1 2 Q2 0 C = 1 2 LI2 0 Suy ra I0 = Q0 LC I0 = U0 L C 2.Bit Q0 ( hay U0) v q ( hay u), tm i lc : Th.s Trn AnhTrung 58 Luyn thi i hc www.VNMATH.com 59. Phng php gii ton Vt L 12 Trng THPT - Phong in T (*) ta c: 1 2 Li2 + 1 2 Cu2 1 2 q2 C = 1 2 CU2 0 1 2 Q2 0 C Suy ra i = Q2 0 q2 LC i = C L (U2 0 u2) CH 4.Dao ng in t do trong mch LC, bit Q0 v I0:tm chu k dao ng ring ca mch LC. Phng php: p dng cng thc Thomson: T = 2 LC (1) Ta c: I0 = Q0 LC LC = Q2 0 I2 0 , thay vo (1): T = 2 Q0 T0 CH 5.Mch LC li vo ca my thu v tuyn in bt sng in t c tn s f (hay bc sng ).Tm L( hay C)? Phng php: iu kin bt c sng in t l tn s ca sng phi bng tn s ring ca mch dao ng LC: f(sng) = f0(mch ) () 1.Bit f( sng) tm L v C: T (**) f = 1 2 LC L = 1 42f2C C = 1 42f2L 2.Bit ( sng) tm L v C: T (**) c = 1 2 LC L = 2 42c2C C = 2 42c2L CH 6.Mch LC li vo ca my thu v tuyn c t in c in dung bin thin Cmax Cmin tng ng gc xoay bin thin 00 1800 : xc nh gc xoay thu c bc x c bc sng ? Phng php: Lp lun nh ch 5: C = 2 42c2L Khi C0 = Cmax Cmin 0 = 1800 0 = 1800 Khi C = C Cmin Vy: = 1800 C Cmin Cmax Cmin Th.s Trn AnhTrung 59 Luyn thi i hc www.VNMATH.com 60. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 7.Mch LC li vo ca my thu v tuyn c t xoay bin thin Cmax Cmin: tm di bc sng hay di tn s m my thu c? Phng php: Lp lun nh ch 5, ta c: = 2c LCv min Cmin max Cmax min max f = 1 2 LCv Cmin fmax Cmax fmin fmin f fmax Th.s Trn AnhTrung 60 Luyn thi i hc www.VNMATH.com 61. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 8 PHNG PHP GII TON V PHN X NH SNG CA GNG PHNG V GNG CU CH 1.Cch v tia phn x trn gng phng ng vi mt tia ti cho ? Phng php: 1.Cch 1:( p dng nh lut phn x nh sng) + V php tuyn IN ti im ti I, vi gc ti i = SIN. + V tia phn x IR i xng vi SI: i = NIR = i 2.Cch 2:( Da vo mi lin h gia vt v nh) + Nu tia ti SI pht xut t im S th tia phn x c phng qua nh o S ( i xng vi S qua gng). + Nu tia ti SI c phng qua vt o S ( sau gng) th tia phn x trc tip qua nh tht ( trc gng). CH 2.Cch nhn bit tnh cht "tht - o" ca vt hay nh( da vo cc chm sng) Phng php: Nhn bit tnh cht "tht - o" ca vt: da vo tnh cht ca chm tia ti. + Chm tia ti phn k th vt tht.( vt trc gng). + Chm tia ti hi t th vt o.( vt sau gng). Nhn bit tnh cht "tht - o" ca nh: da vo tnh cht ca chm tia phn x. + Chm tia phn x hi t th nh tht.( nh trc gng). + Chm tia phn x phn k th nh o.( nh sau gng). Ch : i vi gng phng, vt tht cho nh o v ngc li. CH 3.Gng phng quay mt gc (quanh trc vung gc mt phng ti): tm gc quay ca tia phn x? Phng php: nh l:( v gng quay):Khi gng quay mt gc quanh mt trc mp ti th tia phn x quay mt gc = 2 cng chiu quay ca gng." 1.Cho tia ti c nh, xc nh chiu quay ca tia phn x: Dng hnh hc: i2 = i2 = i1 + Suy ra, gc quay: = RIR = 2(i2 i1) = 2 2.Cho bit SI = R, xc nh qung ng i ca nh S : ng i S S, ng vi gc quay = 2 ca tia phn x. Th.s Trn AnhTrung 61 Luyn thi i hc www.VNMATH.com 62. Phng php gii ton Vt L 12 Trng THPT - Phong in Vy: S S = Rrad = 2Rrad 3.Gng quay u vi vn tc gc : tm vn tc di ca nh? v = S S t = 2Rrad t = 2R CH 4.Xc nh nh to bi mt h gng c mt phn x hng vo nhau Phng php: Da vo hai nguyn tc: 1.Nguyn tc phn on: Chia qu trnh to nh thnh tng giai on, mi giai on ch xt to nh trn mt gng. 2.Nguyn tc to nh lin tip: nh ca gng ny l vt ca gng kia. C hai nhm lin tip Nhm nh 1: S G1 S1 G2 S2 G1 S3 Nhm nh 2: S G2 S1 G1 S2 G2 S3 S nh l tng tt c cc nh ca hai h H qa: i vi h hai gng song song th s nh l v hn nu mt t ngoi hai gng v hu hn nu mt t gia hai gng. Nu hai gng hp nhau mt gc Mi nhm nh, nu nh no nm sau gng th khng to nh na. Th.s Trn AnhTrung 62 Luyn thi i hc www.VNMATH.com 63. Phng php gii ton Vt L 12 Trng THPT - Phong in Ch : Ta chng minh c rng nu = 3600 n vi n l s nguyn dng th h c n 1 nh. CH 5.Cch vn dng cng thc ca gng cu Phng php: Xt s to nh: ABd=OA G A Bd =OA p dng cc cng thc: 1 d + 1 d = 1 f (1) vi f = R 2 Cng thc v phng i nh : k = A B AB = d d (2) Hay: k = f d f = d f f 1.Cho bit d v AB: tm d v cao nh A B T (1): d = df d f , nu d > 0: nh tht; d < 0 nh o. T (2): ta suy ra c gi tr ca k, nu k > 0 nh vt cng chiu; k < 0 nh vt ngc chiu. cao ca nh: A B = |k|AB 2.Cho bit d v A B : tm d v cao vt AB T (1): d = d f d f , nu d > 0: vt tht; d < 0 vt o. cao ca vt: AB = A B |k| 3.Cho bit v tr vt d v nh d xc nh tiu c f: T (1): f = d d d + d , nu f > 0: gng cu lm; f < 0 gng cu li. 4.Ch : *i vi gng cu li: Vt tht lun cho nh o, cng chiu, nh hn vt, gn gng hn vt. *i vi gng cu lm: Vt tht nm trong OF lun cho nh o, cng chiu, nh hn vt, xa gng hn vt.Vt tht nm ngoi OF lun cho nh tht, ngc chiu vi vt. Th.s Trn AnhTrung 63 Luyn thi i hc www.VNMATH.com 64. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 6.Tm chiu v di ca mn nh khi bit chiu v di ca vt. H qa? Phng php: 1.Tm chiu v di ca mn nh khi bit chiu v di ca vt: Cch 1: Ta c: 1 d + 1 d = 1 f = const (*) Do : khi d tng th d gim v ngc li. Cch 2: (*) d = df d f hay y = ax a x o hm theo x: y = a2 (a x)2 < 0, vy hm s y = f(x) l hm nghch bin. Kt lun: Khi dch chuyn vt li gn gng cu mt on d = d1 d2 th dch chuyn m ra xa gng cu mt on d = d2 d1, v ngc li. 2.H qa: Ln 1: k1 = d1 d1 = f d1 f = d1 f f T ta suy ra d1 ( hay d1) theo k1 v f Ln 2: k2 = d2 d2 = f d2 f = d2 f f T ta suy ra d2 ( hay d2) theo k2 v f Thay vo dch chuyn ca vt ( hay dch chuyn ca nh) suy ra c f. CH 7.Cho bit tiu c f v mt iu kin no v nh, vt: xc nh v tr vt dv v tr nh d Phng php: 1.Cho bit phng i k v f: T (2) ta c: d = kd, thay vo (1): 1 d + 1 kd = 1 f , ta suy ra c phng trnh theo d, t suy ra d . 2.Cho bit khong cch l = AA : Trong mi trng hp: l = AA = |d d| d = d l Thay vo (1) ta c phng trnh: 1 d + 1 d l = 1 f , ta suy ra c phng trnh theo d, Th.s Trn AnhTrung 64 Luyn thi i hc www.VNMATH.com 65. Phng php gii ton Vt L 12 Trng THPT - Phong in t suy ra d . Ch :nh trn mn l nh tht, nh nhn thy trong gng l nh o. CH 8.Xc nh th trng ca gng ( gng cu li hay gng phng) Phng php: Gi M l nh ca mt M qua gng, ta c s to nh: Md=OM G Md =OM Th trng ca gng l phn khng gian trc gng, gii hn bi mt phng gng v cc ng sinh v t M ta ln chu vi ca gng. 1.i vi gng cu li: 1 d + 1 d = 1 f d = df d f 2. i vi gng phng: M v M i xng nhau qua gng phng: d = d. Gi l gc na hnh nn ca th trng: ta c : tg = OM |d | = r |d | , r l bn knh ca gng. Ch : 1 = 1 3500 rad CH 9.Gng cu lm dng trong n chiu: tm h thc lin h gia vt sng trn trn mn ( chn chm tia phn x) v kch thc ca mt gng Phng php: Gi S l nh ca mt S( bng n) qua gng, ta c s to nh: Sd=OS G Sd =OS 1 d + 1 d = 1 f d = df d f = OS S dng hnh hc: xt cc tam gic ng dng suy ra mi quan h gia Dv D0 Gi D0, D ln lt l ng knh ca gng v ca vc sng trn. 1.S l nh o chm phn x l chm phn k. D D0 = |d | + L |d | 2.S l nh tht chm phn x l chm hi t. D D0 = L d d 3.Chm phn x l chm song song ( nh v cng) D = D0 CH 10.Xc nh nh ca vt to bi h "gng cu - gng phng" Phng php: Th.s Trn AnhTrung 65 Luyn thi i hc www.VNMATH.com 66. Phng php gii ton Vt L 12 Trng THPT - Phong in Xt 2 ln to nh: ABd1=O1A G1( g.cu ) d1=O1A1 A1B1d2=O2A1 G2( g. phng ) A2B2d2=O2A2 1.Trng hp gng phng vung gc vi trc chnh: Ln 1: 1 d1 + 1 d1 = 1 f 1 d1 = d1f1 d1 f1 phng i: k1 = A1B1 AB = d1 d1 = f1 d1 f1 Ta c: d2 = a d1 ( lun nh vy) Ln 2: Ta c A2B2 i xng vi A1B1 qua gng phng, do d2 = d2 = d1 + a phng i k2 = A2B2 A1B2 = d2 d2 = 1 (2) Vy: A2B2 = A1B1 2.Trng hp gng phng nghing mt gc 450 so vi trc chnh: Ln 1: 1 d1 + 1 d1 = 1 f 1 d1 = d1f1 d1 f1 phng i: k1 = A1B1 AB = d1 d1 = f1 d1 f1 Ta c: d2 = a d1 ( lun nh vy) Ln 2: Ta c A2B2 i xng vi A1B1 qua gng phng, do : O2A2 = O2A1; A1O2A2 = 2 450 = 900 Vy: A2B2 song song vi trc chnh v A2B2 = A1B1 CH 11.Xc nh nh ca vt to bi h "gng cu - gng cu" Phng php: Xt 2 ln to nh: ABd1=O1A1 G1 d1=O1A1 A1B1d2=O2A1 G2 A2B2d2=O2A2 Ln 1: 1 d1 + 1 d1 = 1 f 1 d1 = d1f1 d1 f1 phng i: k1 = A1B1 AB = d1 d1 = f1 d1 f1 = d1 f1 f1 (1) Ta c: d2 = a d1 (2)( lun nh vy) Th.s Trn AnhTrung 66 Luyn thi i hc www.VNMATH.com 67. Phng php gii ton Vt L 12 Trng THPT - Phong in Ln 2: 1 d2 + 1 d2 = 1 f 2 d2 = d2f2 d2 f2 phng i: k2 = A2B2 A1B1 = d2 d2 = f2 d2 f2 = d2 f2 f2 (3) Ch : phng i nh cui cng: kh = A2B2 AB = A2B2 A1B1 A1B1 AB = k2k1 = f2 (d2 f2) f1 (d1 f1) = (d2 f2) f2 (d1 f1) f1 CH 12.Xc nh nh ca vt AB xa v cng to bi gng cu lm? Phng php: Xt s to nh:AB()d= O A B d V d = nn 1 d = 0, t cng thc cart: 1 d + 1 d = 1 f d = f Vy nh A B nm trn mt phng tiu din ca gng cu lm. Gi l gc trng ca vt qua gng. Ta c: CA B : A B = CA tg hay A B = f.tg f.rad Th.s Trn AnhTrung 67 Luyn thi i hc www.VNMATH.com 68. Phng php gii ton Vt L 12 Trng THPT - Phong in PH LC: CCH XC NH TNH CHT NH CA VT QUA GNG CU 1.i vi gng cu lm: 2.i vi gng cu li: Th.s Trn AnhTrung 68 Luyn thi i hc www.VNMATH.com 69. Phng php gii ton Vt L 12 Trng THPT - Phong in PHN 9 PHNG PHP GII TON V KHC X NH SNG, LNG CHT PHNG ( LCP) BNG MT SONG SONG (BMSS), LNG KNH (LK) CH 1. Kho st ng truyn ca tia sng n sc khi i t mi trng chit quang km sang mi trng chit quang hn? Phng php: Lun c tia khc x gn php tuyn hn so vi tia ti 1.Mt phn cch l mt phng: p dng cng thc: n1 sin i = n2 sin r sin r = n1 sin i n2 Khi: i = 0th r = 0: Tia ti vung gc vi mt phn cch th tia l i thng. 2.Mt phn cch l mt cong: php tuyn ti im ti I l bn knh i qua im I. CH 2. Kho st ng truyn ca tia sng n sc khi i t mi trng chit quang hn sang mi trng chit quang km? Phng php: C th c tia khc x nhng cng c th c tia phn x tan phn 1.Mt phn cch l mt phng: p dng cng thc: n1 sin i = n2 sin r sin r = n1 sin i n2 Ta c: sin igh = chit quang b chit quang ln = n1 n2 Nu i < igh th c hin tng khc x nh sng Khi: i = 0th r = 0: Tia ti vung gc vi mt phn cch th tia l i thng. Nu i igh : Th c hin tng phn x ton phn : i = i 2.Mt phn cch l mt cong: php tuyn ti im ti I l bn knh i qua im I. Th.s Trn AnhTrung 69 Luyn thi i hc www.VNMATH.com 70. Phng php gii ton Vt L 12 Trng THPT - Phong in CH 3. Cch v tia khc x ( ng vi tia ti cho) qua mt phng phn cch gia hai mi trng bng phng php hnh hc? Phng php: 1.Cch v tia khc x a. V tia khc x thng :(n1 < n2) *Trong mi trng khc x (n2) v hai na ng trn: (I, n1); (I, n2) * Ni di SI ct vng trn (I, n1) ti J. H JHmp(P), ct vng trn (I, n2) K. Tia IK chnh l tia khc x, Tht vy: IJH : IH = IJ sin i = n1 sin i IKH : IH = IK sin r = n2 sin r Vy: n1 sin i = n2 sin r b. V tia khc x gii hn : Ta c: IH0K0 : sin igh = IH IK0 = n1 n2 2.Cch v tia ti gii hn ton phn *Trong mi trng ti (n1) v hai na ng trn: (I, n1); (I, n2) * T H0 v ng vung gc mp(P) , ct (I, n1) S0 *S0I chnh l tia ti gii hn ton phn( ng vi tia l IK0 l st mt phn cch) Ta c: S0IH0 : sin igh = IH0 IS0 = n2 n1 CH 4. Xc nh nh ca mt vt qua LCP ? Phng php: Lng cht phng (LCP) l mt phn cch gia hai mi trng c chit sut n1, n2 t: d = SH: khong cch t mt phn cch n vt; d = S H :khong cch t mt phn cch n nh. Ta c: SHI : tgi = HI SH sin i = HI d S HI : tgr = HI S H sin r = HI d Vy: sin i sin r = d d Ta c: n1 sin i = n2 sin r sin i sin r = n2 n1 Vy ta c cng thc: d d = n2 n1 (*) Th.s Trn AnhTrung 70 Luyn thi i hc www.VNMATH.com 71. Phng php gii ton Vt L 12 Trng THPT - Phong in Nu n1 > n2: nh sng i t mi trng chic quang hn sang mi trng chic quang km: (*) d < d , nh S nm di vt S. Nu n1 < n2: nh sng i t mi trng chic quang km sang mi trng chic quang hn: (*) d > d , nh S nm trn vt S. CH 5. Xc nh nh ca mt vt qua BMSS ? Phng php: Bn mng song song (BMSS) l h thng hai LCP. 1. di nh Gi S l nh ca S qua BMSS, di nh l : = SS Ta c: = SS = II = IH I H = e I H M: JH = I Htgi = IHtgr hay I H sin i = IH sin r IH I H = sin i sin r = n I H = IH n = e n Vy: = SS = e 1 1 n Ch : Khong di nh khng ph thuc vo v tr t vt. nh lun di theo chiu nh sang ti. 2. di ngang ca tia sng Khi tia sng qua BMSS th khng i phng, nhng di ngang. di ngang ca tia sng l khong cch gia tia ti v tia l: d = IM Xt: IJM : d = IM = IJ sin(i r) Ta c:IJN : cos r = IN IJ IJ = IN cos r = e cos r Vy: d = e sin(i r) cos r CH 6. Xc nh nh ca mt vt qua h LCP- gng phng ? Phng php: 1.Vt A - LCP - Gng phng Xt 3 ln to nh: Ln 1: HA1 HA = n n0 = n HA1 = nHA Ln 2: A2 i xng vi A1 qua gng phng: Ta c: KA2 = KA1 = KH + HA1 = e + nHA Ln 3: HA3 HA2 = n0 n = 1 n Vi: HA2 = HK + KA2 = 2e + nHA HA3 = 2e n + HA Th.s Trn AnhTrung 71 Luyn thi i hc www.VNMATH.com 72. Phng php gii ton Vt L 12 Trng THPT - Phong in 2.Vt A nm gia LCP- Gng phng Xt hai