161 Chu de LTDH Mon Vat Ly

8/6/2019 161 Chu de LTDH Mon Vat Ly

1/113

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

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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

Th.s Trn AnhTrung 2 Luyn thi i hc

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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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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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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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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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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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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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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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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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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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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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+ 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

mTh.s Trn AnhTrung 16 Luyn thi i hc

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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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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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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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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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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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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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23/113


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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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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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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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: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

gTh.s Trn AnhTrung 27 Luyn thi i hc

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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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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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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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31/113


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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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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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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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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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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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: 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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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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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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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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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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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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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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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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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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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: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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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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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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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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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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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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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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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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161 Chu de LTDH Mon Vat Ly