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Mc lc
Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Phn1 . PHNG PHP GII TON V DAO NG IU HA CA CON LCL XO 15Ch 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 . . . . . . . . . . 15Ch 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 . . . . . . . . . . . . 16Ch 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 . . 171.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 . . . . 18Ch 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 hdao ng iu ha, t suy ra chu k T . . . . . . . . . . . . . . . . . . . . 191.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 quarng rc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1
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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: lcy Acximet, lc ma st, p lc thy tnh, p lc ca cht kh...: chng minhh 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 LCN 22Ch 1. Vit phng trnh dao ng iu ha ca con lc n . . . . . . . . . . . 22Ch 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 . . . . . . . . . . . 231. 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: tmiu 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 nhanhhay chm ca ng h trong mt ngy m . . . . . . . . . . . . . . . . . . . 24Ch 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 ngvi 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 giatc a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.Con lc n treo vo trn ca xe t ang chuyn ng ngang vi gia tc a . 27
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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 phngnghing mt gc : . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Ch 8. Xc nh ng nng E th nng Et, c nng ca con lc n khi v trc gc lch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Ch 9. Xc nh vn tc di v v lc cng dy T ti v tr hp vi phng thngng mt gc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.Vn tc di v ti C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.Lc cng dy T ti C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.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 30Ch 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 (cngqua 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 bisau 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: xcnh vn tc ca vin bi sau va chm? . . . . . . . . . . . . . . . . . . . . . . 32
Phn3 . PHNG PHP GII TON V DAO NG TT DN V CNG HNGC HC 33Ch 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 liv 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: tmiu kin c hin tng cng hng . . . . . . . . . . . . . . . . . . . . . 34
Phn 4 . PHNG PHP GII TON V S TRUYN SNG C HC, GIAOTHOA SNG, SNG DNG, SNG M 35Ch 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 tctruyn 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
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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, vntc 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 . . . . . . . . . . . . 362.V th biu din qa trnh truyn sng theo khng gian ( dng ca mitrng...) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Ch 3. Xc nh tnh cht sng ti mt im M trn min giao thoa . . . . . . . 36Ch 4. Vit phng trnh sng ti im M trn min giao thoa . . . . . . . . . . 37Ch 5. Xc nh s ng dao ng cc i v cc tiu trn min giao thoa . . . 37Ch 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 haingun S1, S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Ch 8.Vit biu thc sng dng trn dy n hi . . . . . . . . . . . . . . . . . 38Ch 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 nhcng 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 KHNGPHN NHNH (RLC) 42Ch 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 dngin 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 thu(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
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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 mchkhc nhau trn cng mt dng in xoay chiu khng phn nhnh? Cch vndng? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Ch 5. .on mch RLC, cho bit U, R: tm h thc L,C, : cng dngin 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 . . . . . . . . . . . . . . . . 432.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 inqua on mch cc i, hiu in th v cng dng in cng pha, cngsut 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 ( cundy) v UC. Tm Umch v . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Ch 9. Cho mchRLC: Bit U, , tm L, hayC, hayR cng sut tiu th trnon 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) tgi 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 hayUC 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: xcnh cc c im ca mch in? . . . . . . . . . . . . . . . . . . . . . . . . 50
1.Cho bit th i(t) v u(t): tm lch pha u/i . . . . . . . . . . . . . . . 502.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 trnon mch? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
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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 quabnh in phn theo mt chiu? Tnh th tch kh Hir v Oxy xut hin ccin cc? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
1.Tnh in lng chuyn qua bnh in phn theo mt chiu ( trong 1 chu kT, 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 dngin xoay chiu? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
1.Nam chm in dng dng in xoay chiu ( tn s f) t gn dy thp cngngang. Xc nh tn s rung f ca dy thp . . . . . . . . . . . . . . 52
2.Dy dn thng cng ngang mang dng in xoay chiu t trong t trngc cm ng t B khng i ( vung gc vi dy): xc nh tn s rungca dy f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Phn6 . PHNG PHP GII TON V MY PHT IN XOAY CHIU, BIN
TH, TRUYN TI IN NNG 53Ch 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 camp. Tm cng sut P ca my pht in? . . . . . . . . . . . . . . . . . . . . 53
Ch 3. Mch in xoay chiu ba pha mc theo s hnh : tm cng dngtrung 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 . . . . . . . . . . . . . . . . . . . . 541.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 trnhtruyn 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 TRONGMCH LC 57Ch 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)? Suyra i(t)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
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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 . . 581.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 ngring ca mch LC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Ch 5. Mch LC li vo ca my thu v tuyn in bt sng in t c tn sf (hay bc sng ).Tm L( hay C) . . . . . . . . . . . . . . . . . . . . . . . 591.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 binthin CmaxCmin 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 GNGPHNG V GNG CU 61Ch 1. Cch v tia phn x trn gng phng ng vi mt tia ti cho ? . . . . 61Ch 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): tmgc 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 . . . 62Ch 5. Cch vn dng cng thc ca gng cu . . . . . . . . . . . . . . . . . . 63
1.Cho bit d v AB: tm d v cao nh AB . . . . . . . . . . . . . . . . . 63
2.Cho bit d v AB: tm d v cao vt AB . . . . . . . . . . . . . . . . . 63
3.Cho bit v tr vt d v nh d xc nh tiu cf . . . . . . . . . . . . . . . 63
4.Ch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Ch 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 cf v mt iu kin no v nh, vt: xc nh v tr vtdv v tr nh d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
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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) . . . . . 65Ch 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" . . . . . . . 651.Trng hp gng phng vung gc vi trc chnh . . . . . . . . . . . . . . 662.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" . . . . . . . . 66Ch 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 CHTPHNG ( 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 chitquang km sang mi trng chit quang hn? . . . . . . . . . . . . . . . . . . 69Ch 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 cchgia 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 ? . . . . . . . . . . . . . . . . . . . 711. di nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
2. di ngang ca tia sng . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Ch 6. Xc nh nh ca mt vt qua h LCP- gng phng ? . . . . . . . . . . 711.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 songsong? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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
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Ch 11. Cho lng knh (A,n) v gc ti i1 ca chm sng: xc nh gc lch D? . 74Ch 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? . . . . . . . . . . . . . . . 751.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 HCNG TRC VI THU KNH 76Ch 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 . . . . . . . . . . . . . . 763.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 mitrng lm thu knh v bn knh ca cc mt cong. . . . . . . . . . . . . . . . 76
1.Khi bit tiu cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 cf v mt iu kin no v nh, vt: xc nh v tr vtd v v tr nh d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
1.Cho bit phng i k v f . . . . . . . . . . . . . . . . . . . . . . . . . . 772.Cho bit khong cch l = AA . . . . . . . . . . . . . . . . . . . . . . . . . 77
Ch 4. Xc nh nh ca mt vt AB xa v cc . . . . . . . . . . . . . . . . . 77Ch 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 . . . . . . . . . 781.Thu knh (O) c nh: di vt gn ( hay xa) thu knh, tm chiu chuyn di
ca nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782.Vt AB c nh, cho nh AB 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 kchthc ca mt thu knh. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
Ch 9. H nhiu thu knh mng ghp ng trc vi nhau, tm tiu c ca h. . . 79
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Ch 10. Xc nh nh ca mt vt qua h " thu knh- LCP". . . . . . . . . . . . 791.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". . . . . . . . . . . 801.Trng hp: AB - TK - BMSS . . . . . . . . . . . . . . . . . . . . . . . . . 80
2.Trng hp: AB - LCP - TK . . . . . . . . . . . . . . . . . . . . . . . . . . 81Ch 12. Xc nh nh ca mt vt qua h hai thu knh ghp ng trc. . . . . . 81Ch 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 nhtht, 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". . . . . . . . 831.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". . . . . . . . . 841.Trng hp vt AB t trc h " thu knh- gng cu" . . . . . . . . . . . 85
2.Trng hp h "thu knh- gng cu" ghp st nhau . . . . . . . . . . . . . 853.Trng hp vt AB t gia thu knh v gng cu: . . . . . . . . . . . . . 85
Phn11 . PHNG PHP GII TON V MT V CC DNG C QUANG HCB TR CHO MT 89Ch 1. My nh: cho bit gii hn khong t phim, tm gii hn t vt? . . . . 89Ch 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
ckhi eo knh cha? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Ch 4. Mt vin th: xc nh t ca knh cha mt? Tm im cc cn mic khi eo knh cha? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
Ch 5. Knh lp: xc nh phm vi ngm chng v bi gic. Xc nh kchthc 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
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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 knhlp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
Ch 6. Knh hin vi: xc nh phm vi ngm chng v bi gic. Xc nh kchthc 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 knhhin vi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
Ch 7. Knh thin vn: xc nh phm vi ngm chng v bi gic? . . . . . . 941.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: khost 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? . . . . . . . . . . . . . . 95Ch 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 97Ch 1. Xc nh bc sng khi bit khong vn i, a,, D . . . . . . . . . . . . 97Ch 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 . . . . . . 97Ch 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 nhnh sng cho vn ti ( sng) ti mt im (xM) ? . . . . . . . . . . . . . . . . 98
1.Xc nh rng quang ph . . . . . . . . . . . . . . . . . . . . . . . . . . 982.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 chicsut 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
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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. Tmchiu, chuyn di ca h vn( vn trung tm)? . . . . . . . . . . . . . . . . 99
Ch 9.Ngun sng Schuyn 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 dngc giao thoa? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
1.Khe Young . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 992.Lng lng knh Frexnen . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
3.Hai na thu knh Billet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.Gng Frexnen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
Phn14 . PHNG PHP GII TON V TIA RNGHEN 101Ch 1. Tia Rnghen: Cho bit vn tc v ca electron p vo i catot: tm UAK 101Ch 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? . . . . . . . . . . . . . . . . . . 101Ch 3. Tnh lu lng dng nc lm ngui i catot ca ng Rnghen: . . . . . 101
Phn15 . PHNG PHP GII TON V HIN TNG QUANG IN 103Ch 1. Cho bit gii hn quang in (0). Tm cng thot A ( theo n v eV)? . 103Ch 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: . . . . . . . . . . . . . . . . . . 103Ch 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): Tmv0max ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Ch 5. Cho bit UAK v v0max. Tnh vn tc ca electron khi ti Ant ? . . . . . 104Ch 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 ngunsng. Tnh hiu sut lng t? . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Ch 8. Chiu mt chm sng kch thch c bc sng vo mt qa cu c lpv in. Xc nh in th cc i ca qa cu. Ni qu cu vi mt in trR 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 vin. Xc nh in th cc i ca qa cu: . . . . . . . . . . . . . . 105
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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 nhnnh hn)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
Ch 10. Cch vn dng nh lut bo ton ng lng, nng lng? . . . . . . . 1121.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 tchca ht nhn nguyn t ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
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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 = ES
l
a. Ct l xo thnh n phn bng nhau (cng k):k
k0=
l0l
= n k = nk0.
b. Ct l xo thnh hai phn khng bng nhau:k1k0
=l0l1
vk2k0
=l0l2
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 =
x21 +
v212
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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
2kA2 A =
2E
k
Tm : Da vo iu kin ban u: khi t0 = 0 x = x0 = A sin sin = x0A
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 = tn
CH 3.Chng minh mt h c hc dao ng iu ha:Phng php:Cch 1: Phng php ng lc hc1.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 + 2x = 0. Nghim ca phng trnh vi phn c dng: x = Asin(t + ). T , chng trng vt dao ng iu ha theo thi gian.
Cch 2: Phng php nh lut bo ton nng lng1.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
2mv2 +
1
2kx2 = 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 + 2x = 0. Nghim ca phng trnh vi phnc 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
2mv2 +
1
2kx2 =
1
2kA2 = Emax = Etmax ()
T() ta c: v =
k
m(A2 x2) hay v0max = A
k
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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
2kx2 =
1
2kA2sin2(t + )
ng nng: E =1
2mv2 =
1
2kA2cos2(t + )
Ch :Ta c: t = 2T 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 (lxo 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:
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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 = Fk1L xo L2 gin on x2: F = k2x2 x2 = F
k2
H l xo gin on x: F = khx x = Fkh
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 = k1xL xo L2 gin on x: F2 = k2xH l xo gin on x: Fh = khx
Ta c :F = F1 + F2, vy: kh = k1 + k2 , chu k: T = 2
mkh
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 = k2xH 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
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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
2mv2 +
1
2kx2 = const
o hm hai v theo thi gian:
1
2m2vv
+
1
2k2xx
= 0.t: =
k
m, ta suy ra c phng trnh:x + 2x = 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 x2 .
iu kin cn bng: l0 = F0k= 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 x2
|Tx| = |Fx|2
=k
4x
Xt vt nng:mg + T = ma mg (|T0| + |Tx|) =mx x + k
4mx = 0.
t: 2 =k
4m, phng trnh tr thnh:x + 2x = 0,
nghim ca phng trnh c dng:x = Asin(t + ), vyh dao ng iu ho.
Chu k: T =2
hay T = 2
4m
k
Cch 2:C nng:E = E + Et =1
2mv2 +
1
2kx2L =
1
2mv2 +
1
2k(
x
2)2 = const
o hm hai v theo thi gian:1
2m2vv +
1
2
k
42xx = 0 x + k
4mx = 0.
t: 2 =k
4m, phng trnh tr thnh:x + 2x = 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 quarng rc:
v tr cn bng: P = 2 T0; F02 = 2 T vi ( F01 = T0)
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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 cclc:
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 =
2k1k2
x1 (2)
Thay (2) vo (1) ta c: x1 =k2
k2 + 4k1x
Lc hi phc gy ra dao ng ca vt m l:Fx = F1 = k1x1 (3)Thay (2) vo (3) ta c: Fx =
k2k1k2 + 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 + 2x = 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 daong iu ha:Dng 1. F l lc y Acximet:
V tr cn bng: P = F0AV 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+2x = 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 = Fms02V tr bt k ( li x):Ta c: P = ( N1 + N2) nhng Fms1 = Fms2
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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 N22x
Suy ra: N1 N2 = (N1 + N2)x
l = Px
l = mgx
lT (*) suy ra: |F| = mg x
l, p dng nh lut II Newton:
F = ma = mx.
Ta c phng trnh:x+2x = 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 onh = 2x.p lc thu tnh: p = Dgh suy ra lc thu tnh: |F| =
pS = Dg2xS, gi tr i s:F = pS = Dg2xS, pdng nh lut II Newton: F = ma = mx.Ta c phng trnh:x + 2x = 0, nghim ca phngtrnh 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)Sp dng nh lut Bil-Marit: p1V1 = p2V2 = p0V0
Suy ra: p1 p2 = 2p0dd2 x2x
Hp lc: |F| = F2 F1 = (p1 p2)S = 2p0dSd2 x2x
2p0dS
d2x
i s: F = 2p0dSd2
x, p dng nh lut II Newton:
F = ma = mx.
Ta c phng trnh:x+2x = 0, nghim ca phng trnh c dng:x = Asin(t+),
vy h dao ng iu ho. Chu k: T =2
, vi =
md2
2p0V0
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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 NH1. 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 + 11 + 2
(1 12
2)(1 +1
21) = 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 = s0l
: 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 = tn
CH 2.Xc nh bin thin nh chu k T khi bit bin thin nh gia tctrng trng g, bin thin chiu di l:
Phng php:
Lc u: T = 2
l
g; Lc sau: T = 2
l
gLp t s:
T
T=
l
l.
g
g
M
T = T Tg = g gl = l l
T = T + T
g = g + g
l = l + l
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Vy:T + T
T=
l + l
l
12
g
g + g
12
1 + TT
=
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 TT = 12 llb. 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 nht; 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 = 2l1
g ; nhit t0
2C: T2 = 2l2
g
Lp t s:T2T1
=
l2l1
=
l0(1 + t2)
l0(1 + t1)=
1 + t21 + t1
=
1 + t2
12
1 + t1
1
2
p dng cng thc tnh gn ng:(1 + )n 1 + nT2T1
=
1 +
1
2t2
1 1
2t1
Hay:
T
T1=
1
2(t2 t1) = 1
2t
2.Khi a con lc n ln cao h so vi mt bin:
mt t : T = 2l
g ; cao h: Th = 2l
gh ; Lp t s:Th
T =g
gh (1).
Ta c, theo h qa ca nh lut vn vt hp dn:
g = GM
R2
gh = GM
(R + h)2
Thay vo (1) ta c:ThT
=R + h
RHay:
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:
ThT
=
g
gh(2).
Ta c, theo h qa ca nh lut vn vt hp dn:
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Phng php gii ton Vt L 12 Trng THPT - Phong in
g = GM
R2
gh = GMh
(R h)2
Thay vo (2) ta c:ThT
=
(R h)2
R2M
Mh
Ta li c:
M = V.D =4
3R3.D
Mh = Vh.D =4
3(R h)3.D
Thay vo ta c:ThT
=
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: tmiu 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+
T2T
+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
2t +
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 1T1 1T2
Hay: n = 86400
|T|T2.T1
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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: Xcnh 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 +
Fxm
;
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 = 2lg. 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 = q E:Trng lc biu kin l: P = P + q E g = g + q
E
m(2)
Chiu (2) ln xx: g = g +qExm
;
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Chu k mi: T = 2
lg +
qExm
= 2
lg
1 +
qExmg
.Ch : chng ta thng lp t s:T
T=
1
1 +qEx
mg
=
1 +
qExmg
1
2
= 1 12
qExmg
hayT
T= 1
2
qExmg
4. F l lc y Acsimet FA = V Dkkg:Trng lc biu kin l:
P = P + FA g = g V Dkkgm
=
1 V Dkk
m
g (3)
Chiu (3) ln xx:g =
1 V Dkkm
g;
Vi: m = V.D, trong D l khi lng ring ca qacu: g =
1 Dkk
D
g;
Chu k mi: T = 2
l1 Dkk
D
g
.
Ch : chng ta thng lp t s:T
T=
11 Dkk
D
hay TT
=1
2
DkkD
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 = P OP 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 + ( Fm)2hoc: 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 ngvi gia tc a: xc nh chu k mi T:
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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 tcbiu kin:
g = g a (1)
S dng hnh hc suy ra c ln ca g, chu k mi T = 2lg. 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 tca:
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 = 2l
g
a
Thng lp t s: T
T=
gg a
l trng hp thang my chuyn ng ln chm dn u (v, acng 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 + aThng lp t s:
T
T=
g
g + a
l trng hp thang my chuyn ng ln nhanh dn u (v, a ngc chiu) hay thangmy 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: = P OP ng vi v tr cn bng ca con lc n.
Ta c: tg =F
mg=
a
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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 nghingmt gc :
Ta c iu kin cn bng: P + Fqt + T = 0 (*)
Chiu (*)/Ox: Tsin = ma cos (1)
Chiu (*)/Oy: Tcos = 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
ghay T = 2
l sin
a cos
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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 trc 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 EtThay (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 22
;cos 1 22
(1) Et = 1
2mgl2
(2) E = 12
mgl2
(3) E = 12
mgl(2 2)
CH 9.Xc nh vn tc di v v lc cng dy T ti v tr hp vi phng thngng mt gc :
Phng php:1.Vn tc di v ti C:
Ta c cng thc tnh ng nng: E =1
2mv2, 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 = mv
2
l
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Thay (1) vo ta c: T = m[3 cos 2cos ]g (3)t bit: Nu dao ng ca con lc n l dao ng bThay biu thc tnh gn ng vo ta c:
(1) v =
gl(2 2) (4)
(2) T = m1 + 2 3
22g (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 2cos )gTmax = m[1 + 2]g
T = min = (v tr bin) Tmin = mg cos Tmin = m[1 122]g
CH 10.Xc nh bin gc mi khi gia tc trng trng thay i tg sangg:
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
2mgl2.
Khi con lc ni c gia tc trng trng g: C nng ca con lc: E = 12mgl2.
p dng nh lut bo ton c nng: E = E 12
mgl2 =1
2mgl2
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
2chu k ca con lc n c chiu di l +
1
2chu k ca con lc n c chiu di l
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Ta c: T =1
2T1 +
1
2T2
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
2mgl2 =
1
2mgl2
Hay: =
l
l
CH 12.Xc nh thi gian hai con lc n tr li v tr trng phng (cngqua 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 = T2T2 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 bisau 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
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Phng php gii ton Vt L 12 Trng THPT - Phong in
y =1
2ayt2 =
1
2gt2 (2)
Thay (1) vo (2), phng trnh qu o:
y =1
2.
g
v20x2
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 vtxem 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 = maHay: 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 12
gt2 (2)
Thay (1) vo (2), phng trnh qu o:
y = g2vc 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: xcnh 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
2mv20 =
1
2mv2 +
1
2m1v
2 (2)
T (1) v (2) ta suy ra c v v v.
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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 liv hng, tm cng bi q:
Phng php: C nng ban u(cung cp cho dao ng): E0 = Et(max) = 1
2kA21 (1)
Cng ca lc masat (ti lc dng li): |Ams| = Fmss = mgs (2), vi s lon 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 =A2A1
=A3A2
= = AnA(n1)
A2 = qA1, A3 = q2A1 , An = qn1A1(viq < 1)
ng i tng cng ti lc dng li:
s = 2A1 + 2A2 + + 2An = 2A1(1 + q + q2 + + qn1) = 2A1SVi: 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 =21
=32
= = n(n1)
2 = q1, 3 = q21 , n = qn11(viq < 1)
Vy: q =n1
n1
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
2mgl21
C nng chu k 2: E2 = EtB2max = mgh1, hay E2 =1
2mgl22
gim c nng sau 1 chu k: E =1
2mgl(21 22)
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Hay : E =1
2mgl(21(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
tCH 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
2g
l
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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 tctruyn 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
Tt = t
lch pha gia hai im cch nhau d trn mt phng truyn sng
=2
d Vi
Hai dao ng cng pha = 2k; k ZHai 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 truynsng):
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 dk
2, c bao gi tr nguyn ca k thayvo 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 tO 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(t2
d) (cm)
vi t dV
4.Vn tc dao ng ca sng:
Vn tc dao ng: v =duM
dt= a cos(t +
2
d) (cm/s)
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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 khnggian:
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
Tt
Xt im M(xM = OM = const): uM = a sin(t 2
xM) iu kin t xMV
Lp bng bin thin:
t 0 T4
T2
3T4
T
uM a sin2
xM
X 0 X X
V th biu din, ch ly phn biu din trong gii hn t xMV
Cch 2:( V gin tip)-V th : u0
t 0 T4
T2
3T4
Tu0 0 A 0 A 0
Tnh tin th u0(t) theo chiu dng mt on =xMV
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 mton xM = V.t0, iu kin x
xM.Ch : Thng lp t s: k =
xM
.
Lp bng bin thin:
x 0 4
2
34
u
a sin t0
X X X X
CH 3.Xc nh tnh cht sng ti mt im M trn min giao thoa:
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Phng php: M : MS1 = d1; MS2 = d2Tm 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 iNu p = k + 12
( bn nguyn) = (k + 12
) 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 tS1 n M:sng ti M tr pha2
d1 so vi S1:u1 = a sin(t2
d1) (cm)
Sng tryn tS2 n M:sng ti M tr pha
2
d2 so vi S2:u2 = a sin(t2
d2) (cm)
Sng ti M: uM = u1+u2 , thay vo, p dng cng thc: sinp+sin q = 2 sinp + q
2cos
p q2
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 = lXt 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
12
k l
12
, 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
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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 imdao ng vi bin cc tiu ( im nt) trn on S1S2:
Phng php: M S1S2 : MS1 = d1; MS2 = d2, S1S2 = lTa 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 + 12
2
, iu kin: 0
d2
l
Vy ta c: l
12
k l
12
, 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 + 2CH 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 hpnhng im dao ng cng pha vi hai ngun S1, S2 l h ng Ellip, nhn hai im S1, S2lm hai tiu im.
hai im dao ng ngc pha = (2k + 1), so snh (*):d2 + d1 = (2k + 1). Vy tp hp nhng im dao ng ngcpha vi hai ngun S1, S2 l h ng Ellip, nhn hai im S1, S2lm hai tiu im ( xen k vi h Ellip ni trn).
CH 8.Vit biu thc sng dng trn dy n hi:Phng php:
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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 pha2
(l
d) so vi A uM = a sint
2
(l
d) (1)
Ti C sng tr pha2
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:
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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 =4l2f2
k2
CH 10.Xc nh cng m (I) khi bit mc cng m ti im. Xc nhcng 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 = lgI
I0
T : I = I0.10L
* Nu mc cng m tnh theo n v dB:L = 10lgI
I0
T : I = I0.10L
10
Ch : Nu tn s m f = 1000Hz th I0 = 1012W m2
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 NAtrong 1 giy.
Ta c: IA = WS
W = IA.Shay Pngun = IA.SANu ngun m l ng hng: SA = 4N A2
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
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Phng php gii ton Vt L 12 Trng THPT - Phong in
chiu cao h l S = 2RhTa 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 I2I1
= 1dB
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Phng php gii ton Vt L 12 Trng THPT - Phong in
PHN 5
PHNG PHP GII TON V MCH IN XOAY CHIUKHNG 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 ini(t) v hiu in th u(t):Phng php:1.Tm biu thc t thng (t):
(t) = NBScos(t) hay (t) = 0 cos(t) vi 0 = NBS.
2. Tm biu thc ca s cm ng e(t):
e(t) = d(t)dt
= NBSsin(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 thu(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 = UIcos , vi U =U0
2, I =
I02
, cos =R
Z
Cch 2: Trong cc phn t in, ch c in tr R mi tiu th in nng di dng tanhit: 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
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Phng php gii ton Vt L 12 Trng THPT - Phong in
I0 =U0.
Z, tng tr: Z =
R2 + (ZL ZC)2 vi tg = ZL ZCR
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 xemin tr tng ng bng 0.
Nu bit: i = I0 sin(t+i) v u = U0 sin(t+u) th lch pha: u/i = uiCH 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 ZC1R1
1 lch pha ca u2 so vi i: tg2 =
ZL2 ZC2R2
2Ta c: u1/u2 = u1 u2 = (u1 i) (u2 i)= u1/i u2/i = 1 2
lch pha ca u1 so vi u2: = 1 2Cch 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 dngin qua on mch cc i, hiu in th v cng dng in cng pha, cng suttiu th trn on mch t cc i.
Phng php:1.Cng dng in qua on mch t cc i:
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Phng php gii ton Vt L 12 Trng THPT - Phong in
p dng nh lut Ohm cho on mch: I =U
Z=
UR2 + (ZL ZC)2
()
Ta c:
I = max M = R2 + (ZL ZC)2 = min ZL ZC = 0 L = 1C
Hay LC2 = 1 (
)
Imax =
U
R2.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 = UIcos , P = max cos = 1
Ta c: cos =R
R2 + (ZL ZC)2 = 1Hay R2 + (ZL ZC)2 = R2Hay 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
R2.Do ZL = ZC UL = UC vi L = C =
2nn UL = UC uL = uC
CH 6.on mch RLC, ghp thm mt t C :tm C : cng dng inqua on mch cc i, hiu in th v cng dng in cng pha, cng sut tiu thtrn on mch t cc i.
Phng php:Gi Cb l in dung tng ng ca b t, tng t ch 5, tac:LCb2 = 1 Cb = 1
L2
Nu C ni tip vi C: 1Cb
=1
C+
1
C
Nu C song song vi C: Cb = C+ C
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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 = IR2 + (ZL ZC)2U =
U2R + (UL UC)2
Cch 2:( Dng gin vect)
Ta c: u = uR + uL + uC U = UR + UL + UC trc pha IDa vo gin vect: ta c U =
U2R + (UL UC)2
lch pha: tg =ZL ZC
R=
IZL IZCIR
Hay tg =UL UC
UR
CH 8.Cun dy (RL) mc ni tip vi t C: cho bit hiu in th U1 ( cundy) 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 + Z2L
+(I, U1) = 1 vi
tg1 =ZLR
cos 1 =
RR2 + Z2L
UC
+UC = I.ZC vi ZC =
1
C+(I, UC) =
2
Xt OAC: nh l hm cosin:
U2 = U21 + U2C 2U1UC cos(
2 1) Hay U =
U21 + U
2C + 2U1UC sin 1
Vi: sin 1 = cos 1.tg1 =ZL
R
2
+ Z
2
L
Chiu (*) lnOI: Ucos = U1 cos 1 cos = U
U1cos 1
CH 9.Cho mchRLC: Bit U, , tm L, hayC, hayR cng sut tiu th trnon mch cc i.
Phng php:
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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=
UR2 + (ZL ZC)2
Vy: P =RU2
R2 + (ZL ZC)2 (*)
1.Tm L hay C cng sut tiu th trn on mch cc i:
DP = 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 01
2C
P P0 Pmax 0
Vi P0 =RU2
R2 + Z2Cb. th C theo P:
C 01
2L
P 0 Pmax P1Vi P1 =
RU2
R2 + Z2L
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) tgi tr cc i?
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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:ULZL =
(R2 + Z2C
ZLZC)U
[R2 + (ZL ZC)2] 32
Ta c:ULZL
= 0 ZL = R2 + Z2C
ZC, ta c bng bin thin:
ZL 0R2 + Z2C
ZC
ULZL
+ 0 UL ULmax
Vi ULmax =U
R2 + Z2CR
Cch 2:( Dng i s)Chia t v mu ca (*) cho ZL, ta c: UL =
UR2
Z2L+ (1 ZC
ZL)2
=const
y
Vi y =R2
Z2L+ (1 ZC
ZL)2 = (R2 + Z2C)
1
Z2L 2.ZC 1
ZL+ 1 = (R2 + Z2C)x
2 2.ZCx + 1
Trong : x =1
ZL; Ta c: a = (R2 + Z2C) > 0
Nn y = min khi x = b
2a =
ZC
R2 + Z2C, ymin =
4a =
R2
R2 + Z2C
Vy: ZL =R2 + Z2C
ZCv ULmax =
U
R2 + Z2CR
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 1sin vy: UL = max
khi sin = 1 = 900 AOB O
T : 1 + |u/i| = 2
, v 1 < 0, u/i > 0 nn: tg1 = cotgu/i = 1tgu/i
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Phng php gii ton Vt L 12 Trng THPT - Phong in
ZCR
= RZL ZC hay ZL =
R2 + Z2LZC
, vi ULmax =U
cos 1
hay ULmax =U
R2 + Z2CR
2.Tm C hiu th hiu dng hai u t in cc i:
Hiu in th hai u t in: UC = I.ZC = U.ZCR2 + (ZL ZC)2 (**)
Cch 1:( Dng o hm)
o hm hai v ca (*) theo ZC:UCZC
=(R2 + Z2L ZLZC)U[R2 + (ZL ZC)2] 32
Ta c:UCZC
= 0 ZC = R2 + Z2L
ZL, ta c bng bin thin:
ZC 0R2 + Z2L
ZL UCZC
+ 0 UC UCmax
Vi UCmax = U
R2 + Z2LR
Cch 2:( Dng i s)Chia t v mu ca (*) cho ZC, ta c: UC =
UR2
Z2C+ (
ZLZC
1)2=
consty
Vi y =R2
Z2
C
+ (ZL
ZC 1)2 = (R2 + Z2L)
1
Z2
C 2.ZL
1
ZC+ 1 = (R2 + Z2L)x
2
2.ZLx + 1
Trong : x =1
ZC; Ta c: a = (R2 + Z2L) > 0
Nn y = min khi x = b2a
=ZL
R2 + Z2L, ymin =
4a=
R2
R2 + Z2L
Vy: ZC =R2 + Z2L
ZLv UCmax =
U
R2 + Z2LR
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
UCsin
=U
sin(2 1)=
U
cos 1
Hay: UC =U
cos 1sin vy: UC = max
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Phng php gii ton Vt L 12 Trng THPT - Phong in
khi sin = 1 = 900 AOB OT : 1 + |u/i| =
2, v 1 > 0, u/i < 0 nn: tg1 = cotgu/i = 1
tgu/i
ZLR
= RZL ZC hay ZC =
R2 + Z2LZL
,
vi UCmax =U
cos 1
hay UCmax =U
R2 + Z2LR
CH 11.on mch RLC: Cho bit U,R,L,C: tm f ( hay ) UR, UL hay UCt 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 =
U RR2 + (ZL ZC)2
=const
M
UR = max M = min ZL ZC = 0 hay 0 = 1LC
(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 = UR2
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
C2L21
4+
R2
L2 2 1
CL
1
2+ 1
Hay: y =1
C2L2x2 +
R2
L2 2 1
CLx + 1 vi x =1
2Ta c: a =
1
C2L2> 0
Nn y = min khi x = b2a
=
2
CL R
2
L2
.L2C2
2=
2LC R2C22
Vy 1 =
2
2LC R2C2 (2)
3.Tm f ( hay ) hiu th hiu dng hai u t in cc i:
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Hiu in th hai u in tr C:
UC = I.ZC =U ZC
R2 + (ZL ZC)2=
U1
CR2 +
L 1
C
2 = UR2C22 + (LC 1)2
Hay UL =
const
y , UL cc i khi y = min.Ta c: y = R2C22 + (LC 1)2 = C2L24 + (R2C2 2CL)2 + 1Hay: y = C2L2x2 + (R2L2 2CL)x + 1 vi x = 2
Ta c: a = C2L2 > 0 Nn y = min khi x = b2a
=
2CL R2C2
2C2L2
Vy 2 =
2CL R2C22C2L2
Hay: 2 =
1
LC.
2CL R2C2
2(3)
Ch : Ta c: 20 = 1.2Hiu in th cc i hai u cun cm v t in u c dng
UCmax = ULmax =2L
R
U4LC R2C2
CH 12.Cho bit th i(t) v u(t), hoc bit gin vect hiu in th: xcnh 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 bngkhong 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
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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 RUL 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 trnon mch?
Phng php:Bit I: p dng cng thc Q = RI2t
Bit U: T cng thc I = UZ
Q = RU2
Z2t
Nu cun dy (RL) hoc in tr dm trong cht lng: tm t0
Ta c: Qta = RI2t; Qthu = Cmt0 t0 = RI2t
Cm
CH 14.Tc dng ha hc ca dng in xoay chiu: tnh in lng chuynqua bnh in phn theo mt chiu? Tnh th tch kh Hir v Oxy xut hin cc incc?
Phng php:
1.Tnh in lng chuyn qua bnh in phn theo mt chiu ( trong 1 chu k T, trongt):
Xt dng in xoay chiu i = I0 sin t(A) qua bnh in phn cha dung dch axit haybaz 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 T2 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
Tdo 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
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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):
C96500C gii phngA
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 =vH2
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 lndng 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 chmht hay nh dy thp hai ln trong mt chu k. Nn tn s daong 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 cmng 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 mtlc tF = 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 kth F i chiu hai ln trong mt chu k, do dy rung hai lntrong mt chu k. f = f
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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 phaPhng 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
60p
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 camp. 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 =Aot
=mgh
t= gh; vi =
m
tl lu lng nc ( tnh
theo khi lng)
Cng sut ca tuabin nc: PT = HTPo
Cng sut ca my pht in: PM = HMPT = HMHTPo
CH 3. Mch in xoay chiu ba pha mc theo s hnh : tm cng dngtrung ha khi ti i xng? Tnh hiu in th Ud ( theo Up)? Tnh Pt (cc ti)
Phng php:
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Phng php gii ton Vt L 12 Trng THPT - Phong in
Tm ith:
i1 = I0 sin t
i2 = I0 sin(t +23 )
i3 = I0 sin(t 23 ) 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 UA1OT hnh ta c: Ud = Up
3
Tm Pti:
Do hiu in th ca cc ti bng nhau (Up) nn: Iti =UpZti
Cng sut tiu th ca mi ti: Pt = UpIt cos t = RtI2t
CH 4. My bin th: cho U1, I1: tm U2, I2Phng 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:U2U1
=N2N1
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: U2U1
=I1I2
hay I2 = I1N1N2
b. Trng hp hiu sut MBT l H :
Ta c:U2U1
=N2N1
hay I2 = HI1N1N2
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