PP Giai Bai Tap Vat Li

Mc lc

Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Phn1 . PHNG PHP GII TON V DAO NG IU H`A CA CON LCL` XO 15

Ch 1. Lin h gia lc tc dng, gin v cng ca l xo . . . . . . . . . . 15

1.Cho bit lc ko F , cng k: tm gin l0, tm l . . . . . . . . . . . . . 15

2.Ct l xo thnh n phn bng nhau ( hoc hai phn khng bng nhau): tm cng ca mi phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Ch 2. Vit phng trnh dao ng iu ha ca con lc l xo . . . . . . . . . . 15

Ch 3. Chng minh mt h c hc dao ng iu ha . . . . . . . . . . . . . . . 16

1.Phng php ng lc hc . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.Phng php nh lut bo ton nng lng . . . . . . . . . . . . . . . . . . 16

Ch 4. Vn dng nh lut bo ton c nng tm vn tc . . . . . . . . . . . . 16

Ch 5. Tm biu thc ng nng v th nng theo thi gian . . . . . . . . . . . . 17

Ch 6. Tm lc tc dng cc i v cc tiu ca l xo ln gi treo hay gi . . 17

1.Trng hp l xo nm ngang . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.Trng hp l xo treo thflng ng . . . . . . . . . . . . . . . . . . . . . . . 17

3.Ch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Ch 7. H hai l xo ghp ni tip: tm cng kh, t suy ra chu k T . . . . 18

Ch 8. H hai l xo ghp song song: tm cng kh, t suy ra chu k T . . . 18

Ch 9. H hai l xo ghp xung i: tm cng kh, t suy ra chu k T . . . 18

Ch 10. Con lc lin kt vi rng rc( khng khi lng): chng minh rng hdao ng iu ha, t suy ra chu k T . . . . . . . . . . . . . . . . . . . . 19

1.Hn bi ni vi l xo bng dy nh vt qua rng rc . . . . . . . . . . . . . . 19

2.Hn bi ni vi rng rc di ng, hn bi ni vo dy vt qua rng rc . . . . 19

3.L xo ni vo trc rng rc di ng, hn bi ni vo hai l xo nh dy vt 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: lc'y 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 H`A CA CON LCN 22

Ch 1. Vit phng trnh dao ng iu ha ca con lc n . . . . . . . . . . . 22

Ch 2. Xc nh bin thin nh chu k T khi bit bin thin nh gia tctrng trng g, bin thin chiu di l . . . . . . . . . . . . . . . . . . . 22

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 . . . . . . . . . . . 23

1. Khi bit nhit bin thin nh t . . . . . . . . . . . . . . . . . . . . . . 23

2. Khi a con lc n ln cao h so vi mt bin . . . . . . . . . . . . . . . 23

3. Khi a con lc n xung su h so vi mt bin . . . . . . . . . . . . . 23

Ch 4. Con lc n chu nhiu yu t nh hng bin thin ca chu k: 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 . . . . . . . . . . . . . . . . . . . 24

Ch 6. Con lc n chu tc dng thm bi mt ngoi lc ~F khng i: Xc nhchu k dao ng mi T 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 0 . . . . . . . . . . . . . . . . . . . . . . 26

1.Con lc n treo vo trn ca thang my ( chuyn ng thflng 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


3.Con lc n treo vo trn ca xe t ang chuyn ng trn mt phflngnghing 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 thflngng mt gc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.Vn tc di v ti C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.Lc cng dy T ti C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.H qa: vn tc v lc cng dy cc i v cc tiu . . . . . . . . . . . . . . 30

Ch 10. Xc nh bin gc 0 mi khi gia tc trng trng thay i t g sang g0 30

Ch 11. Xc nh chu k v bin ca con lc n vng inh (hay vt cn)khi i qua v tr cn bng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.Tm chu k T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.Tm bin mi sau khi vng inh . . . . . . . . . . . . . . . . . . . . . . 31

Ch 12. Xc nh thi gian hai con lc n tr li v tr trng phng (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 33

Ch 1. Con lc l xo dao ng tt dn: bin gim dn theo cp s nhn li vhng, 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 35

Ch 1. Tm lch pha gia hai im cch nhau d trn mt phng truyn sng?Tm bc sng khi bit lch pha v gii hn ca bc sng,( tn s, vn 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



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 . . . . . . . . . . . . 36

2.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 . . . . . . . 36

Ch 4. Vit phng trnh sng ti im M trn min giao thoa . . . . . . . . . . 37

Ch 5. Xc nh s ng dao ng cc i v cc tiu trn min giao thoa . . . 37

Ch 6. Xc nh im dao ng vi bin cc i ( im bng) v s im daong 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 . . . . . . . . . . . . . . . . . 38

Ch 9.iu kin c hin tng sng dng, t suy ra s bng v s nt sng 39

1.Hai u mi trng ( dy hay ct khng kh) l c nh . . . . . . . . . . . . 39

2.Mt u mi trng ( dy hay ct khng kh) l c nh, u kia t do . . . . 39

3.Hai u mi trng ( dy hay ct khng kh) l t do . . . . . . . . . . . . . 40

Ch 10.Xc nh cng m (I) khi bit mc cng m ti im. Xc 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) 42

Ch 1. To ra dng in xoay chiu bng cch cho khung dy quay u trong ttrng, 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



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 . . . . . . . . . . . . . . . . 43

2.Hiu in th cng pha vi cng dng in . . . . . . . . . . . . . . . . 44

3.Cng sut tiu th trn on mch cc i . . . . . . . . . . . . . . . . . . . 44

4.Kt lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Ch 6. .on mch RLC , ghp thm mt t C 0 :tm C 0 : 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 . . . . . . . . . . . . . . . 50

2.Cho bit gin vect hiu in th: v s on mch? Tm Umch . . . . 51

Ch 13. Tc dng nhit ca dng in xoay chiu: tnh nhit lng ta ra trnon mch? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51



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 0 ca dy thp . . . . . . . . . . . . . . 52

2.Dy dn thflng 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 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Phn6 . PHNG PHP GII TON V MY PHT IN XOAY CHIU, BINTH, TRUYN TI IN NNG 53

Ch 1. Xc nh tn s f ca dng in xoay chiu to bi my pht in xoaychiu 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 . . . . . . . . . . . . . . . . . . . . 54

1.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp h 54

2.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp c ti 54

3.Trng hp cc in tr ca cun s cp v th cp khc 0: . . . . . . . . . 55

Ch 5.Truyn ti in nng trn dy dn: xc nh cc i lng trong qu 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 57

Ch 1. Dao ng in t do trong mch LC: vit biu thc q(t)? Suy ra cng dng in i(t)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Ch 2. Dao ng in t do trong mch LC, bit uC = U0 sin !t, tm q(t)? Suyra i(t)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58



Ch 3. Cch p dng nh lut bo ton nng lng trong mch dao ng LC . . 58

1.Bit Q0 ( hay U0) tm bin I0 . . . . . . . . . . . . . . . . . . . . . . . . 58

2.Bit Q0 ( hay U0)v q ( hay u), tm i lc . . . . . . . . . . . . . . . . . . 58

Ch 4. Dao ng in t do trong mch LC, bit Q0 v I0:tm chu k dao 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) . . . . . . . . . . . . . . . . . . . . . . . 59

1.Bit f( sng) tm L v C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.Bit ( sng) tm L v C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Ch 6. Mch LC li vo ca my thu v tuyn c t in c in dung binthin Cmax Cmin tng ng gc xoay bin thin 00 1800: xc nh gc xoay thu c bc x c bc sng ? . . . . . . . . . . . . . . . . . . . . . 59

Ch 7. Mch LC li vo ca my thu v tuyn c t xoay bin thin Cmax Cmin: tm di bc sng hay di tn s m my thu c? . . . . . . . . . . . 60

Phn8 . PHNG PHP GII TON V PHN X NH SNG CA GNGPHNG V GNG CU 61

Ch 1. Cch v tia phn x trn gng phflng ng vi mt tia ti cho ? . . . . 61

Ch 2. Cch nhn bit tnh cht "tht - o" ca vt hay nh( da vo cc chmsng) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Ch 3. Gng phflng quay mt gc (quanh trc vung gc mt phflng 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 0 . . . . . . . . . . . . 61

3.Gng quay u vi vn tc gc !: tm vn tc di ca nh . . . . . . . . . . 62

Ch 4. Xc nh nh to bi mt h gng c mt phn x hng vo nhau . . . 62

Ch 5. Cch vn dng cng thc ca gng cu . . . . . . . . . . . . . . . . . . 63

1.Cho bit d v AB: tm d0 v cao nh A0B0 . . . . . . . . . . . . . . . . . 63

2.Cho bit d0 v A0B0: tm d v cao vt AB . . . . . . . . . . . . . . . . . 63

3.Cho bit v tr vt d v nh d0 xc nh tiu c f . . . . . . . . . . . . . . . 63

4.Ch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Ch 6. Tm chiu v di ca mn nh khi bit chiu v di ca vt. H qa? 64

1.Tm chiu v di ca mn nh khi bit chiu v di ca vt . . . . . . 64

2.H qa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Ch 7. Cho bit tiu c f v mt iu kin no v nh, vt: xc nh v tr vtdv v tr nh d0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64



1.Cho bit phng i k v f . . . . . . . . . . . . . . . . . . . . . . . . . . 64

2.Cho bit khong cch l = AA0 . . . . . . . . . . . . . . . . . . . . . . . . . 64

Ch 8. Xc nh th trng ca gng ( gng cu li hay gng phflng) . . . . . 65

Ch 9. Gng cu lm dng trong n chiu: tm h thc lin h gia vt sngtrn 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 phflng" . . . . . . . 65

1.Trng hp gng phflng vung gc vi trc chnh . . . . . . . . . . . . . . 66

2.Trng hp gng phflng nghing mt gc 450 so vi trc chnh . . . . . . . 66

Ch 11. Xc nh nh ca vt to bi h "gng cu - gng cu" . . . . . . . . 66

Ch 12. Xc nh nh ca vt AB xa v cng to bi gng cu lm . . . . . 67

Phn9 . PHNG PHP GII TON V KHC X NH SNG, LNG 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? . . . . . . . . . . . . . . . . . . 69

Ch 2. Kho st ng truyn ca tia sng n sc khi i t mi trng chitquang hn sang mi trng chit quang km? . . . . . . . . . . . . . . . . . . 69

Ch 3. Cch v tia khc x ( ng vi tia ti cho) qua mt phflng 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 ? . . . . . . . . . . . . . . . . . . . 71

1. di nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2. di ngang ca tia sng . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Ch 6. Xc nh nh ca mt vt qua h LCP- gng phflng ? . . . . . . . . . . 71

1.Vt A - LCP - Gng phflng . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2.Vt A nm gia LCP- Gng phflng . . . . . . . . . . . . . . . . . . . . . . 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 phflng ghp songsong? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

1.Vt S - BMSS - Gng phflng . . . . . . . . . . . . . . . . . . . . . . . . . 73

2.Vt S nm gia BMSS - Gng phflng . . . . . . . . . . . . . . . . . . . . . 73

Ch 10. Xc nh nh ca mt vt qua h nhiu BMSS - gng cu? . . . . . . . 73



Ch 11. Cho lng knh (A,n) v gc ti i1 ca chm sng: xc nh gc lch D? . 74

Ch 12. Cho lng knh (A,n) xc nh i1 D = min? . . . . . . . . . . . . . . 74

1.Cho A,n: xc nh i1 D = min,Dmin? . . . . . . . . . . . . . . . . . . . . 74

2.Cho Av Dmin: xc nh n? . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.Ch : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Ch 13. Xc nh iu kin c tia l ra khi LK? . . . . . . . . . . . . . . . 75

1.iu kin v gc chic quang . . . . . . . . . . . . . . . . . . . . . . . . . . 75

1.iu kin v gc ti . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Phn10 . PHNG PHP GII TON V THU KNH V H QUANG HCNG TRC VI THU KNH 76

Ch 1. Xc nh loi thu knh ? . . . . . . . . . . . . . . . . . . . . . . . . . . 76

1.Cn c vo s lin h v tnh cht, v tr, ln gia vt - nh . . . . . . . . 76

2.Cn c vo ng truyn ca tia sng qua thu knh . . . . . . . . . . . . . . 76

3.Cn c vo cng thc ca thu knh . . . . . . . . . . . . . . . . . . . . . . 76

Ch 2. Xc nh t ca thu knh khi bit tiu c, hay chic sut ca mitrng lm thu knh v bn knh ca cc mt cong. . . . . . . . . . . . . . . . 76

1.Khi bit tiu c f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2.Khi bit chic sut ca mi trng lm thu knh v bn knh ca cc mt cong 76

Ch 3. Cho bit tiu c f v mt iu kin no v nh, vt: xc nh v tr vtd v v tr nh d0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

1.Cho bit phng i k v f . . . . . . . . . . . . . . . . . . . . . . . . . . 77

2.Cho bit khong cch l = AA0 . . . . . . . . . . . . . . . . . . . . . . . . . 77

Ch 4. Xc nh nh ca mt vt AB xa v cc . . . . . . . . . . . . . . . . . 77

Ch 5. Xc nh nh ca mt vt AB xa v cc . . . . . . . . . . . . . . . . . 77

1.Cho bit khong cch "vt - nh" L, xc nh hai v tr t thu knh . . . . . 78

2.Cho bit khong cch "vt - nh" L, v khong cch gia hai v tr, tm f . . 78

Ch 6. Vt hay thu knh di chuyn, tm chiu di chuyn ca nh . . . . . . . . . 78

1.Thu knh (O) c nh: di vt gn ( hay xa) thu knh, tm chiu chuyn dica nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.Vt AB c nh, cho nh A0B0 trn mn, di thu knh hi t, tm chiuchuyn 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



Ch 10. Xc nh nh ca mt vt qua h " thu knh- LCP". . . . . . . . . . . . 79

1.Trng hp: AB - TK - LCP . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2.Trng hp: AB - LCP - TK . . . . . . . . . . . . . . . . . . . . . . . . . . 80

Ch 11. Xc nh nh ca mt vt qua h " thu knh- BMSS". . . . . . . . . . . 80

1.Trng hp: AB - TK - BMSS . . . . . . . . . . . . . . . . . . . . . . . . . 80

2.Trng hp: AB - LCP - TK . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Ch 12. Xc nh nh ca mt vt qua h hai thu knh ghp ng trc. . . . . . 81

Ch 13. Hai thu knh ng trc tch ri nhau: xc nh gii hn ca a = O1O2(hoc d1 = O1A) nh A2B2 nghim ng mt iu kin no ( nh 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 phflng". . . . . . . . 83

1.Trng hp gng phflng vung gc vi trc chnh . . . . . . . . . . . . . . 83

2.Trng hp gng phflng nghing mt gc 450 so vi trc chnh . . . . . . . 83

3.Trng hp gng phflng ghp xc thu knh ( hay thu knh m bc) . . . . 84

4.Trng hp vt AB t trong khong gia thu knh v gng phflng . . . . 84

Ch 16. Xc nh nh ca vt cho bi h "thu knh - gng cu". . . . . . . . . 84

1.Trng hp vt AB t trc h " thu knh- gng cu" . . . . . . . . . . . 85

2.Trng hp h "thu knh- gng cu" ghp st nhau . . . . . . . . . . . . . 85

3.Trng hp vt AB t gia thu knh v gng cu: . . . . . . . . . . . . . 85

Phn11 . PHNG PHP GII TON V MT V CC DNG C QUANG HCB TR CHO MT 89

Ch 1. My nh: cho bit gii hn khong t phim, tm gii hn t vt? . . . . 89

Ch 2. My nh chp nh ca mt vt chuyn ng vung gc vi trc chnh.Tnh khong thi gian ti a m ca sp ca ng knh nh khng b nho. . 89

Ch 3. Mt cn th: xc nh t ca knh cha mt? Tm im cc cn mi 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



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? . . . . . . 94

1.Xc nh phm vi ngm chng ca knh thin vn . . . . . . . . . . . . . . . 94

2.Xc nh bi gic ca knh thin vn . . . . . . . . . . . . . . . . . . . . 94

Phn12 . PHNG PHP GII TON V HIN TNG TN SC NH SNG 95

Ch 1. S tn sc chm sng trng qua mt phn cch gia hai mi trng: 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? . . . . . . . . . . . . . . 95

Ch 3. Xc nh gc hp bi hai tia l ( , tm)ca chm cu vng ra khi LK.Tnh b rng quang ph trn mn? . . . . . . . . . . . . . . . . . . . . . . . . 95

Ch 4. Chm tia ti song song c b rng a cha hai bt x truyn qua BMSS:kho st chm tia l? Tnh b rng cc i amax hai chm tia l tch ri nhau? 95

Phn13 . PHNG PHP GII TON V GIAO THOA SNG NH SNG 97

Ch 1. Xc nh bc sng khi bit khong vn i, a;, D . . . . . . . . . . . . 97

Ch 2. Xc nh tnh cht sng (ti) v tm bc giao thoa ng vi mi im trnmn? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Ch 3. Tm s vn sng v vn ti quang st c trn min giao thoa . . . . . . 97

Ch 4. Trng hp ngun pht hai nh sng n sc. Tm v tr trn mn cs 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 . . . . . . . . . . . . . . . . . . . . . . . . . . 98

2.Xc nh nh sng cho vn ti ( sng) ti mt im (xM) . . . . . . . . . . . 98

Ch 6. Th nghim giao thoa vi nh sng thc hin trong mi trng c chicsut n > 1. Tm khong vn mi i0? 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



Ch 8. Th nghim Young: Khi ngun sng di chuyn mt on y = SS 0. Tmchiu, chuyn di ca h vn( vn trung tm)? . . . . . . . . . . . . . . . . 99

Ch 9.Ngun sng S chuyn ng vi vn tc ~v theo phng song song vi S1S2:tm tn s sut hin vn sng ti vn trung tm O? . . . . . . . . . . . . . . . 99

Ch 10.Tm khong cch a = S1S2 v b rng min giao thoa trn mt s dngc giao thoa? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

1.Khe Young . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

2.Lng lng knh Frexnen . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

3.Hai na thu knh Billet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4.Gng Frexnen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Phn14 . PHNG PHP GII TON V TIA RNGHEN 101

Ch 1. Tia Rnghen: Cho bit vn tc v ca electron p vo i catot: tm UAK 101

Ch 2. Tia Rnghen: Cho bit vn tc v ca electron p vo i catot hot UAK :tm tn s cc i Fmax hay bc sng min? . . . . . . . . . . . . . . . . . . 101

Ch 3. Tnh lu lng dng nc lm ngui i catot ca ng Rnghen: . . . . . 101

Phn15 . PHNG PHP GII TON V HIN TNG QUANG IN 103

Ch 1. Cho bit gii hn quang in (0). Tm cng thot A ( theo n v eV )? . 103

Ch 2. Cho bit hiu in th hm Uh. Tm ng nng ban u cc i (Emax)hay vn tc ban u cc i( v0max), hay tm cng thot A? . . . . . . . . . . . 103

1.Cho Uh: tm Emax hay v0max . . . . . . . . . . . . . . . . . . . . . . . . . . 103

2.Cho Uh v (kch thch): tm cng thot A: . . . . . . . . . . . . . . . . . . 103

Ch 3. Cho bit v0max ca electron quang in v ( kch thch): tm gii hnquang 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 ? . . . . . 104

Ch 6. Cho bit v0max v A.Tm iu kin ca hiu in th UAK khng cdng 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



2.Ni qu cu vi mt in tr R sau ni t. Xc nh cng dng qua R:105

Ch 9. Cho kch thch, in trng cn Ec v bc sng gii hn 0: tm onng i ti a m electron i c. . . . . . . . . . . . . . . . . . . . . . . . 105

Ch 10. Cho kch thch, bc sng gii hn 0 v UAK : Tm bn knh ln nhtca vng trn trn mt Ant m cc electron t Katt p vo? . . . . . . . . . 105

Ch 11. Cho kch thch, bc sng gii hn 0 , electron quang in bay ratheo phng vung gc vi in trng ( ~E). Kho st chuyn ng ca electron ?106

Ch 12. Cho kch thch, bc sng gii hn 0 , electron quang in bay ratheo phng vung gc vi cm ng t ca tr trng u ( ~B). Kho st chuynng ca electron ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

Phn16 . PHNG PHP GII TON V MU NGUYN T HIR THEO BO 108

Ch 1. Xc nh vn tc v tn s f ca electron trng thi dng th n canguyn t Hir? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Ch 2. Xc nh bc sng ca photon do nguyn t Hir pht ra khi nguyn t trng thi dng c mc nng lng Em sang En ( < Em )? . . . . . . . . . . 108

Ch 3. Tm bc sng ca cc vch quang ph khi bit cc bc sng ca ccvch ln cn? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Ch 4. Xc nh bc sng cc i (max) v cc tiu (min) ca cc dy Lyman,Banme, Pasen? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Ch 5. Xc nh qy o dng mi ca electron khi nguyn t nhn nng lngkch thch " = hf? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Ch 6. Tm nng lng bc electron ra khi nguyn t khi n ang qy oK ( ng vi nng lng E1)? . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Phn17 . PHNG PHP GII TON V PHNG X V PHN NG HTNHN 110

Ch 1. Cht phng x AZX c s khi A: tm s nguyn t ( ht) c trong m(g)ht nhn ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

Ch 2. Tm s nguyn t N( hay khi lng m) cn li, mt i ca cht phngx sau thi gian t? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

Ch 3. Tnh khi lng ca cht phng x khi bit phng x H? . . . . . . . 110

Ch 4. Xc nh tui ca mu vt c c ngun gc l thc vt? . . . . . . . . . 110

Ch 5. Xc nh tui ca mu vt c c ngun gc l khong cht? . . . . . . . 111

Ch 6. Xc nh nng lng lin kt ht nhn( nng lng ta ra khi phn r mtht nhn)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

Ch 7. Xc nh nng lng ta ra khi phn r m(g) ht nhn AZX? . . . . . . . 111

Ch 8. Xc nh nng lng ta ( hay thu vo ) ca phn ng ht nhn? . . . . . 111



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? . . . . . . . 112

1.Cch vn dng nh lut bo ton ng lng: . . . . . . . . . . . . . . . . . 112

2.Cch vn dng nh lut bo ton nng lng: . . . . . . . . . . . . . . . . . 113

Ch 11. Xc nh khi lng ring ca mt ht nhn nguyn t. Mt in tchca ht nhn nguyn t ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113



PHN 1

PHNG PHP GII TON V DAO NG IU H`A 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 =Fk

+Nu F = P = mg th l0 =mgk

+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 cngca mi phn?

p dng cng thc Young: k = ESl

a. Ct l xo thnh n phn bng nhau (cng k):kk0

=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: ! =

rkm

+ Khi bit T hay f : ! =2T

= 2f

Tm A:

+ Khi bit chiu di qy o: d = BB0 = 2A ! A =d2

+ Khi bit x1, v1: A =

r

x21 +v21!2



+ Khi bit chiu di lmax; lmin ca l xo: A =lmax lmin

2.

+ Khi bit nng lng ca dao ng iu ha: E =12

kA2 ! A =

r2Ek

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 = x0v = v0

$

(x0 = Asinv0 = !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 hc

1.Xc nh lc tc dng vo h v tr cn bng:P ~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 lng

1.Vit biu thc ng nng E ( theo v) v th nng Et ( theo x), t suy ra biu thcc nng:

E = E + Et =12

mv2 +12

kx2 = const ()

2.o hm hai v () theo thi gian: (const)0 = 0; (v2)0 = 2v:v0 = 2v:x"; (x2)0 =2x:x0 = 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 =12

mv2 +12

kx2 =12

kA2 = Emax = Etmax ()

T () ta c: v =

rkm

(A2 x2) hay v0max = A

rkm



CH 5.Tm biu thc ng nng v th nng theo thi gian:

Phng php:

Th nng: Et =12

kx2 =12

kA2sin2(!t + )

ng nng: E =12

mv2 =12

kA2cos2(!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 = kjxj.

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 thflng ng:

iu kin cn bng: ~P + ~F0 = 0, gin tnh ca l xo: l0 =

mgk

.

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:



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 thflng ng: ~P + ~F0 = 0

v tr bt k( OM = x):

L xo L1 gin on x1: F = k1x1 ! x1 = Fk1

L xo L2 gin on x2: F = k2x2 ! x2 = Fk2

H l xo gin on x: F = khx ! x = Fkh

Ta c :x = x1 + x2, vy:1

kh=

1k1

+1k2

, chu k: T = 2r 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 thflng 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 = 2r m

kh

CH 9.H hai l xo ghp xung i: tm cng kh, t suy ra chu k T :

Phng php:

v tr cn bng:+ i vi h nm ngang: ~P + ~N = 0+ i vi h thflng ng: ~P + ~F01 + ~F02 = 0

v tr bt k( OM = x):L xo L1 gin on x: F1 = k1xL 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 = 2r m

kh

CH 10.Con lc lin kt vi rng rc( khng khi lng): chng minh rng h



dao ng iu ha, t suy ra chu k T :

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 =12

mv2 +12

kx2 = const

o hm hai v theo thi gian:12

m2vv0 +12

k2xx0 = 0.

t: ! =

rkm

, 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

=2T0k

=2mg

k.

Cch 1: v tr bt k( li x): ngoi cc lc cn bng, xut hin thm cc lc n hi

jFxj = kxL = kx2

, jTxj =jFxj

2=

k4

x

Xt vt nng:m~g + ~T = m~a , mg (jT0j + jTxj) =

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

r4mk

Cch 2:C nng:E = E + Et =12

mv2 +12

kx2L =12

mv2 +12

k(x2

)2 = const

o hm hai v theo thi gian:12

m2vv0 +12

k4

2xx0 = 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

r4mk

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)



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 =k2k1

k2 + 4k1x,

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 = 2r

k2k1m(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 D~g. 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 ! =

rSDg

m

Dng 2. ~F l lc ma st:

V tr cn bng: ~P = ( ~N01 + ~N02) v ~Fms01 = ~Fms02

V tr bt k ( li x):Ta c: ~P = ( ~N1 + ~N2) nhng ~Fms1 6= ~Fms2



Hp lc: jF j = F1 F2 = (N1 N2) (*)

M ta c: M ~N1=G = M ~N2=G

, N1(l x) = N2(l + x) ,N1

(l + x)=

N2(l x)

=

N1 + N22l

=N1 N2

2xSuy ra: N1 N2 = (N1 + N2)

xl

= Pxl

= mgxl

T (*) suy ra: jF j = mgxl

, 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 ! =r

gl

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: jF j =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 ! =

r2SDg

m

Dng 4. ~F l lc ca cht kh:

V tr cn bng: p01 = p02 suy ra F01 = F02; V0 = Sd

V tr bt k ( li x):Ta c: V1 = (d + x)S; V2 = (d x)S

p dng nh lut Bil-Marit: p1V1 = p2V2 = p0V0

Suy ra: p1 p2 =2p0d

d2 x2x

Hp lc: jF j = F2 F1 = (p1 p2)S =2p0dS

d2 x2x

2p0dSd2

x

i s: F = 2p0dS

d2x, 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 ! =

smd2

2p0V0



PHN 2

PHNG PHP GII TON V DAO NG IU H`A CA CON LC N

GHI NH

1. bin thin i lng X:X = Xsau Xtrc

a. Nu X > 0 th X tng.

b. Nu X < 0 th X gim.

2.Cng thc gn ng:

a.8" 1 ta c: (1 + ")n 1 + n"

H qu:r

1 + "11 + "2

(1 12

"2)(1 +12

"1) = 1 12

("2 "1)

b.8 100; 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: ! =r

gl

Tm s0 v cng mt lc:Da vo iu kin ban u:

t0 = 0 $

(s = s1v = v1

$

(s1 = s0sinv1 = !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 = 2r

lg

; Lc sau: T 0 = 2r

l0

g0Lp t s:

T 0

T=

rl0

l:

gg0

M

8>:

T = T 0 Tg = g0 gl = l0 l

,

8>:

T 0 = T + Tg0 = g + gl0 = l + l



Vy:T + T

T=

l + l

l

12

gg + g

12

, 1 +TT

=

1 +12

ll

1

12

gg

Hay:TT

=12

ll

gg

Ch :

a. Nu g = const th g = 0 )TT

=12

ll

b. Nu l = const th l = 0 )TT

= 12

gg

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 t01C: T1 = 2r

l1g

; nhit t02C: T2 = 2r

l2g

Lp t s:T2T1

=r

l2l1

=

sl0(1 + t2)l0(1 + t1)

=r

1 + t21 + t1

=

1 + t2

12

1 + t1

12

p dng cng thc tnh gn ng:(1 + ")n 1 + n"

T2T1

=

1 +12

t2

1

12

t1

Hay:

TT1

=12

(t2 t1) =12

t

2.Khi a con lc n ln cao h so vi mt bin:

mt t : T = 2r

lg

; cao h: Th = 2r

lgh

; Lp t s:ThT

=r

ggh

(1).

Ta c, theo h qa ca nh lut vn vt hp dn:8>:

g = GMR2

gh = GM

(R + h)2

Thay vo (1) ta c:ThT

=R + h

RHay:

TT

=hR

3.Khi a con lc n xung su h so vi mt bin:

mt t : T = 2r

lg

; su h: Th = 2r

lgh

; Lp t s:ThT

=r

ggh

(2).

Ta c, theo h qa ca nh lut vn vt hp dn:



8>:

g = GMR2

gh = GMh

(R h)2

Thay vo (2) ta c:ThT

=

r(R h)2

R2MMh

Ta li c: 8>:

M = V:D =43

R3:D

Mh = Vh:D =43

(R h)3:D

Thay vo ta c:ThT

=

RR h

12

Hay:TT

=12

hR

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=

12

t; Yu t cao:T2

T=

hR

Thay vo (*):12

t +hR

= 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=

86400T1

.

ng vi chu k T2: s dao ng trong mt ngy m: n0 =t

T2=

86400T2

.

chnh lch s dao ng trong mt ngy m: n = jn0 nj = 86400

1T1

1T2

Hay: n = 86400jT jT2:T1



Vy: nhanh ( hay chm) ca ng h trong mt ngy m l: = n:T2 = 86400jT j

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 0:

Phng php:

Phng php chung: Ngoi trng lc tht ~P = m~g, con lc n cn chu tc dng thm

mt ngoi lc ~F , nn trng lc biu kin l: ~P 0 = ~P + ~F , ~g0 = ~g +~Fm

(1)

S dng hnh hc suy ra c ln ca g0, chu k mi T 0 = 2r

lg0

. Ch : chng

ta thng lp t s:T 0

T=

rgg0

1. ~F l lc ht ca nam chm:

Chiu (1) ln xx0: g0 = g +Fxm

;

Nam chm t pha di: Fx > 0 , ~F hng xung

, g0 = g +Fm

:

Nam chm t pha trn: Fx < 0 , ~F hng ln

, g0 = g Fm

:

Chu k mi T 0 = 2r

lg0

. Ch : chng ta thng lp t

s:T 0

T=

rgg0

.

2. ~F l lc tng tc Coulomb:

Lc tng tc Coulomb: F = kjq1q2j

r2; Tm g0 v chu k T 0

nh trn.Hai in tch cng du: ~F lc 'y: ;Hai in tch tri du: ~F lc ht:

3. ~F l lc in trng ~F = q ~E:

Trng lc biu kin l: ~P 0 = ~P + q ~E , ~g0 = ~g +q ~Em

(2)

Chiu (2) ln xx0: g0 = g +qExm

;



Chu k mi: T 0 = 2

vuutl

g +qExm

= 2

vuuutl

g

1 +qExmg

.

Ch : chng ta thng lp t s:T 0

T=

vuut1

1 +qExmg

=

1 +qExmg

12= 1

12

qExmg

hayTT

= 12

qExmg

4. ~F l lc 'y Acsimet ~FA = V Dkk~g:

Trng lc biu kin l:

~P 0 = ~P + ~FA , ~g0 = ~g V Dkk~g

m=

1

V Dkkm

~g (3)

Chiu (3) ln xx0:g0 =

1 V Dkk

m

g;

Vi: m = V:D, trong D l khi lng ring ca qa

cu: g0 =

1 DkkD

g;

Chu k mi: T 0 = 2

vuuutl

1 DkkD

g

.

Ch : chng ta thng lp t s:T 0

T=

vuuut1

1 DkkD

hayTT

=12

DkkD

5. ~F l lc nm ngang:

Trng lc biu kin: ~P 0 = ~P + ~F hay m~g0 = m~g + ~F hng xin, dy treo mt gc so

vi phng thflng ng. Gia tc biu kin: ~g0 = ~g +~Fm

.

iu kin cn bng: ~P + ~T + ~F = 0 , ~P 0 = ~T .Vy = \P O0P 0 ng vi v tr cn bng ca con lc n.

Ta c: tg =F

mgTm T 0 v g0: p dng nh l Pitago: g0 =

qg2 + ( Fm )

2

hoc: g0 =g

cos .

Chu k mi: T 0 = 2r

lg0

. Thng lp t s:T 0

T=

rgg0

=p

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 0:



Phng php:

Trong h quy chiu gn lin vi im treo( thang my, t..) con lc n cn chu tcdng thm mt lc qun tnh ~F = m~a. Vy trng lc biu kin ~P 0 = ~P m~a hay gia tcbiu kin:

~g0 = ~g ~a (1)

S dng hnh hc suy ra c ln ca g0, chu k mi T 0 = 2r

lg0

. Ch : chng ta

thng lp t s:T 0

T=

rgg0

1.Con lc n treo vo trn ca thang my ( chuyn ng thflng ng ) vi gia tc~a:

Chiu (1) ln xx0: g0 = g ax (2)

a.Trng hp ~a hng xung: ax > 0 ! ax = jaj

(2) : g0 = g a chu k mi: T 0 = 2r

lg a

Thng lp t s:T 0

T=

rg

g 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 = jaj

(2) : g0 = g + a chu k mi: T 0 = 2r

lg + a

Thng lp t s:T 0

T=

rg

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 O0P 0 ng vi v tr cn bng ca con lc n.

Ta c: tg =F

mg=

ag



Tm T 0 v g0: p dng nh l Pitago: g0 =p

g2 + a2 hoc: g0 =g

cos .

Chu k mi: T 0 = 2r

lg0

. Thng lp t s:T 0

T=

rgg0

=p

cos

3.Con lc n treo vo trn ca xe t ang chuyn ng trn mt phflng nghingmt gc :

Ta c iu kin cn bng: ~P + ~Fqt + ~T = 0 (*)

Chiu (*)/Ox: T sin = ma cos (1)

Chiu (*)/Oy: T cos = mg ma sin (2)

Lp t s:12

: tg =a cos

g a sin

T (1) suy ra lc cng dy: T =ma cos

sin

T(*) ta c: P 0 = T $ mg0 = T hay g0 =a cos sin

Chu k mi: T 0 = 2r

lg0

hay T 0 = 2

rl sin a cos



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 phflng i qua v tr cn bng.

Th nng Et:Ta c: Et = mgh1 , vi h1 = OI = l(1 cos )Vy: Et = mgl(1 cos ) (1)

C nng E: p dng nh lut bo ton c nng:E = EC = EB = mgh2 = mgl(1 cos )

Hay E = mgl(1 cos ) (2)

ng nng E: Ta c: E = E + Et ! E = E Et

Thay (1) , (2) vo ta c: E = mgl(cos cos ) (3)

t bit: Nu con lc dao ng b: p dng cng thc tnh gn ng:

cos 1 2

2; cos 1

2

28>>>>>>>:

(1) ! Et =12

mgl2

(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 thflngng mt gc :

Phng php:

1.Vn tc di v ti C:

Ta c cng thc tnh ng nng: E =12

mv2, thay vo biu thc (3) ch 8 ta c:

v =p

2gl(cos cos ) (1)

2.Lc cng dy T ti C:

p dng nh lut II Newton: ~P + ~T = m~aht (2)

Chn trc ta hng tm, chiu phng trnh (2) ln xx0:

Ta c: mg cos + T = mv2

l



Thay (1) vo ta c: T = m[3 cos 2 cos ]g (3)

t bit: Nu dao ng ca con lc n l dao ng bThay biu thc tnh gn ng vo ta c:

8>>>>>>>>>>>>>>>>>>>>:

(1); (4) !

8>:

v = max $ = 0(v tr cn bng); !

(vmax =

p2gl(1 cos )

vmax = p

glv = min $ = (v tr bin) ! vmin = 0;

(3); (5) !

8>>>>>>>:

T = max $ = 0(v tr cn bng); !

(Tmax = m(3 2 cos )gTmax = m[1 + 2]g

T = min $ = (v tr bin) !

(Tmin = mg cos Tmin = m[1 12

2]g

CH 10.Xc nh bin gc 0 mi khi gia tc trng trng thay i t g sangg0:

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 =12

mgl2.

Khi con lc ni c gia tc trng trng g0: C nng ca con lc: E0 =12

mg0l02.

p dng nh lut bo ton c nng: E = E0 $12

mgl2 =12

mg0l02

Hay: 0 = r

gg0

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 =12

chu k ca con lc n c chiu di l +12

chu k ca con lc n c chiu di l0



Ta c: T =12

T1 +12

T2

Trong :

8>>>:

T1 = 2r

lg

T2 = 2r

l0

g

vi:l0 = l QI

2.Tm bin mi sau khi vng inh:

Vn dng ch (10) ta c:12

mgl2 =12

mgl002

Hay: 0 =

rll0

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 1dao ng. Gi t l thi gian tr li trng phng, ta c:

t = nT1 = (n 1)T2 ! n =T2

T2 T1

Vy thi gian tr li trng phng: t =T1:T2

T2 T1

CH 13.Con lc n dao ng th b dy t:kho st chuyn ng ca hn bisau khi dy t?

Phng php:

1.Trng hp dy t khi i qua v tr cn bng O: Lc chuyn ng ca vt xemnh l chuyn ng vt nm ngang. Chn h trc ta Oxy nh hnh v.

Theo nh lut II Newton: ~F = ~P = m~aHay: ~a = ~g (*)Chiu (*) ln Ox: ax = 0,trn Ox, vt chuyn ng thflng u vi phng trnh:x = v0t ! t =

xv0

(1)

Chiu (*) ln Oy: ax = g,

trn Oy, vt chuyn ng thflng nhanh dn u vi phng trnh:Th.s Trn AnhTrung 31 Luyn thi i hc


y =12

ayt2 =12

gt2 (2)

Thay (1) vo (2), phng trnh qu o:

y =12

:gv20

x2

Kt lun: qu o ca qa nng sau khi dy t ti VTCB l mt Parabol.( y = ax2)

2.Trng hp dy t khi i qua v tr c li gic : Lc chuyn ng ca vtxem nh l chuyn ng vt nm xin hng xung, c ~vc hp vi phng ngang mt gc :vc =

p2gl(cos cos 0). Chn h trc ta Oxy nh hnh v.

Theo nh lut II Newton: ~F = ~P = m~aHay: ~a = ~g (*)Chiu (*) ln Ox: ax = 0,trn Ox, vt chuyn ng thflng u vi phng trnh:x = vc cos t ! t =

xv0 cos

(1)

Chiu (*) ln Oy: ax = g,

trn Oy, vt chuyn ng thflng bin i u, vi phng trnh:

y = vc sin t 12

gt2 (2)

Thay (1) vo (2), phng trnh qu o:

y = g

2vc cos2 x2 + tg:x

Kt lun: qu o ca qa nng sau khi dy t ti v tr C l mt Parabol.( y = ax2 +bx)

CH 14.Con lc n c hn bi va chm n hi vi mt vt ang ng yn: xcnh vn tc ca vin bi sau va chm?

Phng php:

* Vn tc ca con lc n trc va chm( VTCB): v0 =p

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: m~v0 = m~v + m1~v0 (1)

p dng nh lut bo ton ng lng:12

mv20 =12

mv2 +12

m1v02 (2)

T (1) v (2) ta suy ra c v v v.



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) =12

kA21 (1)

Cng ca lc masat (ti lc dng li): jAmsj = 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

= =An

A(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) = 2A1S

Vi: S = (1 + q + q2 + + qn1) =1

1 q

Vy: s =2A1

1 q

CH 2.Con lc l n ng tt dn: bin gc gim dn theo cp s nhn liv 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 =n1r

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 =12

mgl21

C nng chu k 2: E2 = EtB2max = mgh1, hay E2 =12

mgl22

gim c nng sau 1 chu k: E =12

mgl(21 22)



Hay : E =12

mgl(21(1 q2), y chnh l nng lng cn cung cp duy tr daong trong mt chu k.

Trong thi gian t, s dao ng: n =tT

. Nng lng cn cung cp duy tr sau n daong: E = n:E.

Cng sut ca ng h: P =Et

CH 3.H dao ng cng bc b kch thch bi mt ngoi lc tun hon: tmiu 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 =1T0

=1

2

rkm

i vi con lc n: f0 =1T0

=1

2

rgl



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 truynsng? 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:

=2T

t = !t

lch pha gia hai im cch nhau d trn mt phng truyn sng

=2

d Vi

(Hai dao ng cng pha = 2k; k 2 ZHai dao ng ngc pha = (2k + 1); k 2 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: =dk

Nu cho gii hn ca : ta c: 1 dk

2, c bao gi tr nguyn ca k thay

vo ta suy ra c bc sng hay tn s, vn tc.

Nu bi ton cho gii hn ca tn s hay vn tc, p dng cng thc: = V:T =Vf

.

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 =r

F

3.Vit phng trnh sng ti mt im trn phng truyn sng:

Gi s sng truyn t O n M :OM = d, gi s sng ti O c dng: uO = a sin !t (cm).

Sng ti M tr pha2

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)Th.s Trn AnhTrung 35 Luyn thi i hc


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 sin2T

t

Xt im M(xM = OM = const): uM = a sin(!t 2

xM) iu kin t xMV

Lp bng bin thin:

t 0 T4T2

3T4 T

uM a sin 2 xMX 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 T4T2

3T4 T

u0 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 42

34

ua sin !t0

X X X X

CH 3.Xc nh tnh cht sng ti mt im M trn min giao thoa:



Phng php:

8 M : MS1 = d1; MS2 = d2

Tm hiu ng i: = d2 d1 v tm bc sng: = V:T =Vf

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 t S1 n M :sng ti M tr pha2

d1 so vi S1:u1 = a sin(!t2

d1) (cm)

Sng tryn t S2 n M :sng ti M tr pha2

d2 so vi S2:u2 = a sin(!t2

d2) (cm)

Sng ti M : uM = u1+u2 , thay vo, p dng cng thc: sin p+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:

8>:

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:

8 M : MS1 = d1; MS2 = d2; S1S2 = l

Xt MS1S2 : ta c: jd2 d1j l , l d2 d1 l (*)

M dao ng vi bin cc i: = d2 d1 = k k 2 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 +12

k 2 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


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:

8 M 2 S1S2 : MS1 = d1; MS2 = d2; S1S2 = l

Ta c: d1 + d2 = l (*)

M dao ng vi bin cc i: = d2 d1 = k k 2 Z (1)

Cng (1) v (*) ta c: d2 =l2

+ k2

, 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 +12

k 2 Z (2)

Cng (2) v (*) ta c: d2 =l2

+

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 dngd2 d2i 2 d20 d2i +

2

CH 7.Tm qy tch nhng im dao ng cng pha (hay ngc pha) vi haingun 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:



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 sin

!t 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:

8>:

Nu C c nh u0C = uC = a sin(!t 2

l) (3)

Nu C t do u0C = uC = a sin(!t 2

l) (4)

Sng ti M:

Ti M sng tr pha2

d so vi C:

8>:

Nu C c nh u0M = a sin(!t 2

l 2

d) (5)

Nu C t do u0M = a sin(!t 2

l 2

d) (6)

3.Sng ti M: u = uM + u0M , dng cng thc lng gic suy ra c biu thc sngdng.

CH 9.iu kin c hin tng sng dng, t suy ra s bng v s ntsng:

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 = k2

+ iu kin v tn s: =Vf

! f = kV2l

+ 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:



l =

k +12

2


! f =

k +12

v2l

+ S mi: k =2l

12

, 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 = k2


! f = kv2l

+ 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 =r

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 = lgII0

T : I = I0:10L

* Nu mc cng m tnh theo n v dB:L = 10lgII0

T : I = I0:10L10

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 flng hng: SA = 4NA2

Nu ngun m l loa hnh nn c na gc nh l :

Gi R l khong cch t loa n im m ta xt. Din tch ca chm cu bn knh R v



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 $ 10lgI2I1

= 1dB



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 trongt 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) = NBS cos(!t) hay (t) = 0 cos(!t) vi 0 = NBS.

2. Tm biu thc ca s cm ng e(t):

e(t) = d(t)

dt= !NBS sin(!t) hay e(t) = E0 sin(!t) vi: E0 = !NBS

3.Tm biu thc cng dng in qua R: i =e(t)R

4.Tm biu thc ht tc thi u(t): u(t) = e(t) suy ra U0 = E0 hay U = E.

CH 2.on mch RLC: cho bit i(t) = I0 sin(!t), vit biu thc hiu in 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 =p

R2 + (ZL ZC)2 vi

8>>>>>>>>>>>>>>>>:

I = max u; i cng pha ( = 0) cos = 1

H qa:

8>>>>>:

1:Imax =UR

2:Do ZL = ZC ! UL = UC vi L = C = 2

nn ~UL = ~UC $ uL = uC

CH 6.on mch RLC , ghp thm mt t C 0 :tm C 0 : 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:LCb!2 = 1 ! Cb =

1L!2

Nu C ni tip vi C 0:1

Cb=

1C

+1

C 0

Nu C song song vi C 0: Cb = C + C 0



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 =UZ

=U

pR2 + (ZL ZC)2

! U = Ip

R2 + (ZL ZC)2

U =p

U2R + (UL UC)2

Cch 2:( Dng gin vect)

Ta c: u = uR + uL + uC $ ~U = ~UR + ~UL + ~UC trc pha ~I

Da vo gin vect: ta c U =p

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

8>>>>>>>>>>>>>>>>>>>:

~U1

8>>>>>>>:

+U1 = I:Z1 = I:p

R2 + Z2L

+(~I; ~U1) = 1 vi

8>:

tg1 =ZLR

cos 1 =R

pR2 + Z2L

~UC

8:

C =1

!2LL =

1!2C

() ! Pmax =U2

R

a. th L theo P :

L 01

!2C1

P P0 Pmax 0

Vi P0 =RU2

R2 + Z2Cb. th C theo P :

C 01

!2L1

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 flng thc Csin:

M = R +(ZL ZC)2

R 2

r

R:(ZL ZC)2

R= 2jZL ZCj

Du " = " xy ra khi: R =(ZL ZC)2

R

hay R = jZL ZC j

Vy: Pmax =U2

2jUL UC jBng bin thin R theo P :

R 0 jZL ZC j 1P 0 Pmax 0

CH 10.on mch RLC: Cho bit U; R; f: tm L ( hay C) UL (hay UC) tgi tr cc i?



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:ZLp

R2 + (ZL ZC)2(*)

Cch 1:( Dng o hm)

o hm hai v ca (*) theo ZL:@UL@ZL

=(R2 + Z2C ZLZC)U

[R2 + (ZL ZC)2]32

Ta c:@UL@ZL

= 0 $ ZL =R2 + Z2C

ZC, ta c bng bin thin:

ZL 0R2 + Z2C

ZC1

@UL@ZL

+ 0

UL % ULmax &

Vi ULmax =U

pR2 + Z2C

R

Cch 2:( Dng i s)

Chia t v mu ca (*) cho ZL, ta c: UL =U

sR2

Z2L+ (1

ZCZL

)2=

constp

y

Vi y =R2

Z2L+ (1

ZCZL

)2 = (R2 + Z2C)1

Z2L 2:ZC

1ZL

+ 1 = (R2 + Z2C)x2 2:ZCx + 1

Trong : x =1

ZL; Ta c: a = (R2 + Z2C) > 0

Nn y = min khi x = b

2a=

ZCR2 + Z2C

, ymin = 4a

=R2

R2 + Z2C

Vy: ZL =R2 + Z2C

ZCv ULmax =

Up

R2 + Z2CR


Ta c: u = uRC + uL $ ~U = ~URC + ~UL () trc pha ~I ,

t [AOB =

Xt OAB: nh l hm sin:UL

sin AOB=

Usin OAB

$UL

sin =

Usin(2 1)

=U

cos 1

Hay: UL =U

cos 1sin vy: UL = max

khi sin = 1 ! = 900 ! AOB ? O

T : 1 + ju=ij =2

, v 1 < 0, u=i > 0 nn: tg1 = cotgu=i = 1

tgu=i



$ ZCR

= R

ZL ZChay ZL =

R2 + Z2LZC

, vi ULmax =U

cos 1

hay ULmax =U

pR2 + Z2C

R

2.Tm C hiu th hiu dng hai u t in cc i:

Hiu in th hai u t in: UC = I:ZC =U:ZCp

R2 + (ZL ZC)2(**)

Cch 1:( Dng o hm)

o hm hai v ca (*) theo ZC :@UC@ZC

=(R2 + Z2L ZLZC)U

[R2 + (ZL ZC)2]32

Ta c:@UC@ZC

= 0 $ ZC =R2 + Z2L

ZL, ta c bng bin thin:

ZC 0R2 + Z2L

ZL1

@UC@ZC

+ 0

UC % UCmax &

Vi UCmax =U

pR2 + Z2L

R

Cch 2:( Dng i s)

Chia t v mu ca (*) cho ZC , ta c: UC =U

sR2

Z2C+ (

ZLZC

1)2=

constp

y

Vi y =R2

Z2C+ (

ZLZC

1)2 = (R2 + Z2L)1

Z2C 2:ZL

1ZC

+ 1 = (R2 + Z2L)x2 2:ZLx + 1

Trong : x =1

ZC; Ta c: a = (R2 + Z2L) > 0


2a=

ZLR2 + Z2L

, ymin = 4a

=R2

R2 + Z2L

Vy: ZC =R2 + Z2L

ZLv UCmax =

Up

R2 + Z2LR


Ta c: u = uRL + uC $ ~U = ~URL + ~UC () trc pha ~I , t [AOB = Xt OAB:

nh l hm sin:UC

sin AOB=

Usin OAB

$UC

sin =

Usin(2 1)

=U

cos 1

Hay: UC =U

cos 1sin vy: UC = max



khi sin = 1 ! = 900 ! AOB ? O

T : 1 + ju=ij =2

, v 1 > 0, u=i < 0 nn: tg1 = cotgu=i = 1

tgu=i

$ZLR

= R

ZL ZChay ZC =

R2 + Z2LZL

,

vi UCmax =U

cos 1

hay UCmax =U

pR2 + Z2L

R

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 =UR

pR2 + (ZL ZC)2

=const

M

UR = max $ M = min $ ZL ZC = 0 hay !0 =1

pLC

(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 =UZLp

R2 + (ZL ZC)2=

U!Ls

R2 +

!L 1

!C

2=

Us

R2

!2L2+

1

1!2CL

2

Hay UL =constp

y, UL cc i khi y = min.

Ta c: y =R2

!2L2+ (1

1!2CL

)2 =1

C2L21

!4+

R2

L2 2

1CL

1

!2+ 1

Hay: y =1

C2L2x2 +

R2

L2 2

1CL

x + 1 vi x =

1!2

Ta c: a =1

C2L2> 0


2a=

2

CL

R2

L2

:L2C2

2=

2LC R2C2

2

Vy !1 =r

22LC R2C2

(2)

3.Tm f ( hay !) hiu th hiu dng hai u t in cc i:



Hiu in th hai u in tr C:

UC = I:ZC =UZCp

R2 + (ZL ZC)2=

U1

!Cs

R2 +

!L 1

!C

2=

Up

R2C2!2 + (LC! 1)2

Hay UL =constp

y, UL cc i khi y = min.

Ta c: y = R2C2!2 + (LC! 1)2 = C2L2!4 + (R2C2 2CL)!2 + 1

Hay: y = C2L2x2 + (R2L2 2CL)x + 1 vi x = !2

Ta c: a = C2L2 > 0 Nn y = min khi x = b

2a=

2CL R2C2

2C2L2

Vy !2 =

2CL R2C2

2C2L2

Hay: !2 =

1LC

:

r2CL R2C2

2(3)

Ch : Ta c: !20 = !1:!2

Hiu in th cc i hai u cun cm v t in u c dng

UCmax = ULmax =2LR

Up

4LC 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 = 2T



2.Cho bit gin vect hiu in th: v s on mch? Tm Umch

Quy tc:

8>:

~UR nm ngang $ phn t R~UL thflng ng hng ln $ phn t L~UC thflng ng hng xung $ phn t C

~Umch

8>:

+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 =RI2tCm

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 =

T2Z

0

idt =

T2Z

0

I0 sin !tdt = 1!

I0 cos !t

T2

0

hay q1 =2I0!

Vi ! =2T

do ta c: q1 =I0T

Trong thi gian t, s dao ng n =tT

, in lng qua bnh in phn theo mt chiu l:

q = nq1 =tT

:q1 , vy: q =2I0!

tT

=I0t



2.Tnh th tch kh Hir v Oxy xut hin cc in cc trong thi gian t(s):

C 96500C gii phngAn

= 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 0 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 0 = 2f

2.Dy dn thflng 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 0:

T trng khng i ~B tc dng ln dy dn mang dng in mtlc t F = Bil( c chiu tun theo quy tc bn tay tri ).V F t l vi i , nn khi i i chiu hai ln trong mt chu kth F i chiu hai ln trong mt chu k, do dy rung hai lntrong mt chu k. f 0 = f



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 xoaychiu 1 pha

Phng php:

1.Trng hp roto ca mp c p cp cc, tn s vng l n:

Nu n tnh bng ( vng/s) th: f = np

Nu n tnh bng ( vng/pht) th: f =n60

p

Ch : S cp cc: p =s cc ( bc+ nam)

2

2.Trng hp bit sut in ng xoay chiu ( E hay Eo):

p dng: Eo = NBS! vi ! = 2f , nn: f =Eo

2NBS=

Ep

22NBS

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 =

mt

l lu lng nc ( tnh

theo khi lng)

Cng sut ca tuabin nc: PT = HT Po

Cng sut ca my pht in: PM = HM PT = HMHT Po

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:



Tm ith:8>:

i1 = I0 sin !ti2 = 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 ~UA1O

T hnh ta c: Ud = Upp

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; I2

Phng php:

1.Trng hp cc in tr ca cun s cp v th cp bng 0, cun th cp h:

Lc : I2 = 0 p dng: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



3.Trng hp cc in tr ca cun s cp v th cp khc 0:

Sut in ng qua cun s cp: e1 = N1ddt

(1);

Sut in ng qua cun th cp: e2 = N2ddt

(2);

Lp t:e1e2

=N1N2

k (3)

Cun s cp ng vai tr nh mt my pht: u1 = e1 + r1i1 ! e1 = u1 r1i1 (4)

Cun s cp ng vai tr nh mt my thu: u2 = e2 r2i2 ! e2 = u2 + r2i2 (5)

Lp t:e1e2

=u1 r1i1u2 + r2i2

k $ u1 r1i1 = ku2 + kr2i2 (6)

Ta c e1i1 = e2i2 haye1e2

=i1i2

=1k

! i1 =i2k

v i2 =u2R

(7)

Thay (7) vo (6), thc hin bin i ta c: u2 =kR

k2(R + r2) + r1u1

Hay: U2 =kR

k2(R + r2) + r1U1

CH 5. Truyn ti in nng trn dy dn: xc nh cc i lng trong qutrnh truyn ti

Phng php:

Sn xut:U2AU1A

=I1AI2A

=N2AN1A

PA = U1AI1A = U2AI2A

Tuyn ti:Cng d.in : I = I2A = I1B

in tr : R = 2lS

(l = AB)

gim th : UAB = U2B U2A = IR

Cng sut hao ph : P = PA PB = RI2

S dng:U2BU1B

=I1BI2B

=N2BN1B

PB = U1BI1B = U2BI2B

CH 6. Xc nh hiu sut truyn ti in nng trn dy?

Phng php:

Cng thc nh ngha hiu sut: H =PBPA

;

Xc nh theo cng sut: H =PBPA

=PA P

PA= 1

PP

;



Xc nh theo ht: H =UBUA

=UA U

UA= 1

UU



PHN 7

PHNG PHP GII TON V DAO NG IN T DO TRONG MCH LC

K hiu:

qmax = Q0 ( bin in tch)

umax = U0 ( bin hiu in th)

imax = I0 ( bin dng in)

GHI NH Dao ng c hc ( con lc l xo) Dao ng in ( mch LC)Li : x in tch : q

Vn tc: v =dxdt

= x0 Cng dng in : i = dqdt

Cc i lng t trng Khi lng: m t cm : L

cng: k Nghch o in dung :1C

Lc tc dng : F Hiu in th : u

Phng trnh ng lc hc x" +km

x = 0 q" +1

LCq = 0

$ x" + !2x = 0 $ q" + !2q = 0Nghim ca pt vi phn x = A sin(!t + ) q = Q0 sin(!t + )

Tn s gc ring ! =

rkm

! =

r1

LC

Chu k dao ng T = 2r

mk

T = 2p

LC

Th nng n hi : Nng lng in trng :

Et =12

kx2 W =12

q2

C=

12

Cu2 =12

qu

ng nng : Nng lng t trng :

Nng lng dao ng E =12

mv2 Wt =12

Li2

C nng : Nng lng in t :

E =12

mv2 +12

kx2 W =12

Li2 +12

q2

C=

12

kA2 =12

m!2A2 =12

Q20C

=12

LI20

Bng so snh dao ng iu ha ca con lc l xo v dao ng in t do



CH 1.Dao ng in t do trong mch LC: vit biu thc q(t)? Suy ra cng dng in i(t)?

Phng php:

q(t) c dng tng qut: q = Q0 sin(!t + ) vi: Q0 = CU0

! =1

pLC

hoc ! =2T

= 2f

c xc nh nh iu kin ban u ( t = 0) ca q.

i(t) c xc nh: i = dqdt

= q0 = !Q0 cos(!t + ) = I0 cos(!t + )

Vi I0 = !Q0 =Q0pLC

CH 2.Dao ng in t do trong mch LC, bit uC = U0 sin !t, tm q(t)? Suyra i(t)?

Phng php:Ta c: q = Cu = Q0 sin !t viQ0 = CU0i(t) c xc nh: i =

dqdt

= q0 = !Q0 cos !t = I0 cos !t

hay i = I0 sin

!t +2

CH 3.Cch p dng nh lut bo ton nng lng trong mch dao ng LC .

Phng php:

p dng nh lut bo ton v chuyn ha nng lng:

W = W + Wt = Wmax = Wtmax = const

hay12

Li2 +

8>:

12

Cu2

12

q2

C

=12

LI20 =

8>:

12

CU2012

Q20C

()

1.Bit Q0 ( hay U0) tm bin I0 :

T (*) ta c:

8>:

12

CU2012

Q20C

=12

LI20 Suy ra

8>>>:

I0 =Q0pLC

I0 = U0

rLC

2.Bit Q0 ( hay U0) v q ( hay u), tm i lc :



T (*) ta c:

12

Li2 +

8>:

12

Cu2

12

q2

C

=

8>:

12

CU2012

Q20C

Suy ra

8>>>:

i =

rQ20 q

2

LC

i =

rCL

(U20 u2)

CH 4.Dao ng in t do trong mch LC, bit Q0 v I0:tm chu k dao ngring ca mch LC .

Phng php:

p dng cng thc Thomson: T = 2p

LC (1)

Ta c: I0 =Q0pLC

! LC =Q20I20

, thay vo (1): T = 2Q0T0

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)?

Phng php:

iu kin bt c sng in t l tn s ca sng phi bng tn s ring camch dao ng LC:

f(sng) = f0(mch ) ()

1.Bit f( sng) tm L v C:

T (**) ! f =1

2p

LC$

8>:

L =1

42f2C

C =1

42f2L

2.Bit ( sng) tm L v C:

T (**) !c

=1

2p

LC$

8>:

L =2

42c2CC =

2

42c2L

CH 6.Mch LC li vo ca my thu v tuyn c t in c in dung binthin Cmax Cmin tng ng gc xoay bin thin 00 1800: xc nh gc xoay thuc bc x c bc sng ?

Phng php:

Lp lun nh ch 5: C =2

42c2LKhi C0 = Cmax Cmin $ 0 = 1800 0 = 1800

Khi C = C Cmin $

Vy: = 1800C Cmin

Cmax Cmin



CH 7.Mch LC li vo ca my thu v tuyn c t xoay bin thin Cmax Cmin: tm di bc sng hay di tn s m my thu c?

Phng php:

Lp lun nh ch 5, ta c:8>>>>>>>:

= 2cp

LCv $

(min $ Cminmax $ Cmax

! min max

f =1

2p

LCv$

(Cmin $ fmaxCmax $ fmin

! fmin f fmax



PHN 8

PHNG PHP GII TON V PHN X NH SNG CA GNG PHNGV GNG CU

CH 1.Cch v tia phn x trn gng phflng ng vi mt tia ti cho ?

Phng php:

1.Cch 1:( p dng nh lut phn x nh sng)

+ V php tuyn IN ti im ti I , vi gc ti i = [SIN .+ V tia phn x IR i xng vi SI: i0 = [NIR = i

2.Cch 2:( Da vo mi lin h gia vt v nh)

+ Nu tia ti SI pht xut t im S th tia phn x cphng qua nh o S 0 ( i xng vi S qua gng).

+ Nu tia ti SI c phng qua vt o S ( sau gng) thtia phn x trc tip qua nh tht ( trc gng).

CH 2.Cch nhn bit tnh cht "tht - o" ca vt hay nh( da vo cc chmsng)

Phng php:

Nhn bit tnh cht "tht - o" ca vt: da vo tnh cht ca chm tia ti.+ Chm tia ti phn k th vt tht.( vt trc gng).+ Chm tia ti hi t th vt o.( vt sau gng).

Nhn bit tnh cht "tht - o" ca nh: da vo tnh chtca chm tia phn x.

+ Chm tia phn x hi t th nh tht.( nh trc gng).+ Chm tia phn x phn k th nh o.( nh sau gng).

Ch : i vi gng phflng, vt tht cho nh o v ngc li.

CH 3.Gng phflng quay mt gc (quanh trc vung gc mt phflng ti):tm gc quay ca tia phn x?

Phng php:

nh l:( v gng quay):Khi gng quay mt gc quanh mt trc ? mp ti th tiaphn x quay mt gc = 2 cng chiu quay ca gng."

1.Cho tia ti c nh, xc nh chiu quay ca tia phn x:

Dng hnh hc: i02 = i2 = i1 +

Suy ra, gc quay: = [RIR0 = 2(i02 i1) = 2

2.Cho bit SI = R, xc nh qung ng i ca nh S 0:

ng i S 0S", ng vi gc quay = 2 ca tia phn x.Th.s Trn AnhTrung 61 Luyn thi i hc


Vy: S 0S" = Rrad = 2Rrad

3.Gng quay u vi

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