Ôn thi HSG THPT

8/2/2019 n thi HSG THPT

1/55

NGUYN VN TRUNG :0915192169

CHUYN N THI HC SINH GII LP 12Bi 1: (HSG B Sng Cu Long)

a.Tm thi gian ti thiu mt vn ng vin li mt vt qua mt khc quanh c di bng 1/3ng trn bn knh R. Cho h s ma st ngh gia bnh xe v mt ng l , mt ngc lmnghing mt gc so vi mt phng nm ngang.b.Tnh cng sut gii hn ca ng c lc y. Coi cc bnh xe u l bnh pht ng.

Giia.

msnma P R P N F (1)

Chiu ln Oy: 0 sin cosmsnmg F N cos sin sin

cos sin

msnmg N F N

mgN

(2)

Chiu ln Ox:

2

max cos sin cos sinmsnmV

F N N N R

(3)

T (2) v (3)

max1 1

gR tg gR tgV Vtg tg

Vy vn ng vin chy u vi tc ti a, ta c tmin l:

min max

12 1 2

3 3

R tgs R tgt

V gR tg g tg

b. Ta c: P = F.V

Pmax khi :max

max

msnF F N

V V

max

cos sin 1

gR tgmgP

tg

Bi 4: (Dao ng iu ha).T im A trong lng mt ci chn trn Mt trn mt sn phng nm ngang, ngi ta th mt vt m nh (hnh v).Vt m chuyn ng trong mt phng thng ng, n B th quay li. Bqua ma st gia chn M v m.a.Tm thi gian m chuyn ng t A n B. Bit A cch im gia Ica chn mt khong rt ngn so vi bn knh R. Chn ng yn.b.Tnh h s ma st ngh gia chn v sn.

Giia. Ta c: ma P N

* Chiu ln phng tip tuyn:sint

xma P mg

R

" 2 0x x Vi: 2

g

R

T cho thy m dao ng iu ho, thi gian i t A n B l1

2chu k dao ng.

2

T Rt

g

b. Chn ng yn nn: ' 0 M M msn

P N N F (1)

N

R

P

Fms

R

O x

y

m

I

M

A

NM

Fmsn

PMN'

N

O O

y

x


2/55

NGUYN VN TRUNG :0915192169* Chiu (1) ln phng Oy: ' cos 0M MP N N Vi N' = N (2)

gc lch , Vi m c:

2 2

2 2

0 0

cos cos

cos cos2 2

mV mV N mg N mg

R R

mV mV mgh mgh mgR

0

3cos 2cos N mg (3)

T (2) v (3) ta c: 0cos 3cos 2cosM N Mg mg (4)

* Chiu (1) ln Ox: ' sin 0 sinmsn msn N F N F N

max

min

( sin )sin

( )M M

NN

N N

0

0

sin 3cos 2cos sin

cos 3cos 2cosM

N mg

N Mg mg

(0 b; 0 )

minmaxsin ;( )

M

N N khi = 0

Vy: 2

sin2

2 cos

m

M m

Cu 4:(HSG Kin Giang):Ba qu cu c th trt khng ma st trn mt thanh cng,mnh nm

ngang.Bit khi lng 2 qu cu 1 v 2 l 1 2m m m ;l xo c cng K v khi lng khng ng

k.Qu cu 3 c khi lng3

2

mm .Lc u 2 qu cu 1,2 ng yn,l xo c di t nhin 0l .Truyn

cho 3m

vn tc 0v

n va chm n hi vo qu cu 1. Sau va

chm,khi tm G cu cc qu cu 1,2 chuyn ng nh th no?Tmvn tc cu G.Chng minh rng hai qu cu 1 v 2 dao ng iu hongc pha quanh v tr c nh i vi G.Tm chu k v bin dao ng cu cc vt.

P Na.Chuyn ng cu khi tm G:V qu cu 3 va chm n hi vi qu cu 1 v h kn nn ng lng(theo phng ngang) v ng

nng c bo ton.Gi 1 3,v v l vn tc qu cu 1 v 3 sau va chm,ta c:

0 1 32 2

m mv mv v (1)

2 22

0 31

2 2 2 2 2

v vmvm m (2) 2 23 0 3 03 2v v v v (3)

(3) c nghim 3 0v v (loi v v l) v0

3 3

v

v (4) a (4) vo (1) ta c:0

1

2

3

v

v H hai qu cu 1 v 2 l h c lp nn khi tm G chuyn ng thng u.T to khi tm,ta c :

1 1 2 2 1 1 2 2

1 2 1 2

GG G

dxm x m x m v m vx v

m m dt m m

(6)

Sau va chm: 01

2

3

vv v 2 0v nn (6) cho ta:

0 01

0

1 2

2 2

3 3

3G

v vm m

vv

m m m m

(7)

b.Dao ng cu qu cu 1 v 2+Chn trc to Ox nm ngang,gc O trng vi khi tm G cu hai qu cu

m

I

M

A

NM

Fmsn

PMMN'

N

O O

y

x

1 23 0v


3/55

NGUYN VN TRUNG :0915192169+Khi l xo cha bin dng,gi

1 20 ,0 l v tr cn bng cu hai qu cu.Lc 1 2,x x l to cu haiqu cu.To cu khi tm l :

1 1 2 2

1 2

0Gm x m x

xm m

Vi

1 2m m th

1 22

lx x

Phng trnh chuyn ng cu1

m m l:'

'' ' ''0

Kmx K x x x

m (8)

Do khi tm ng yn v lun c1 2

2

lx x nn ta coi G l ni buc cht cu hai con lcc khi

lng1 2,m m v chiu di l xo l

2

l

cng cu l xo t l nghch vi chiu di nn K = 2 K,nn (8) vit l: ''2

0K

x xm

Tn s gc cu dao ng l :1

2K

m

Chu k dao ng :1

1

22

2

mT

K

Tng t,m2 c chu k dao ng :2

22

mT

K

Hai dao ng ny ngc pha nhau

Vn tc cu qu cu 1 v 2 i vi khi tm:0 0 0

1 1

2

3 3 3G G

v v vv v v

0 02 2

03 3

G G

v vv v v

C nng bo ton nn bin dao ng c tnh:

2 2

1 1 011

2

2 2 3 2

Gm v vKA m

A K

2 2

2 2 02 2

2

2 2 3 2

Gm v vKA m

A K Cu 4 :(Tin Giang)Mt hnh tr c ng cht, c trng lng P, bn knh r t

trong mt mt lm bn knh cong R nh hnh v. im trn hnh tr ngi ta gn

hai l xo c cng nh nhau.Tm chu k dao ng nh ca hnh tru vi gi thit

hnh tr ln khng trt . Xt trng hp: khng c l xo, khi mt lm l mt

phng.

Gii:

Gi l gc quay quanh trc C ca tr, 1 l vn tc gc ca

chuyn ng quay quanh trc v V l vn tc tnh tin ca trc.

1

v'

r

Mt khc, ta c: v ' R r )()(. /1 rRrrRr

ng nng: 2

2 22

d 1

mv 1 3E I m R r '

2 2 4 vi 2

2

1mrI

Rk

Rk

A A

B1B

C

O


4/55


Th nng: 2

2

t

2kx 1E mg R r

2 2 x r (R r) 2 R r

Do :

2 22 2 2

t

1 mgE k.4 R r mg R r 4k R r

2 2 R r

C nng: E = Et + Ed = const . Ly o hm hai v

2 23 mgm ' 4k 0

4 2 R r

mg4k

2 R r 16k 2g

3 3m 3 R rm

4

Vy chu k dao ng T =

m

k

rR

g 16

)(3

2

22

Trng hp ring: - Khi k = 0 th

2g

3 R r

- Khi R th :16k

3m

Bi 4 (HSG Lao Cai): Con lc l xo t thng ng (nh hnh v 4), u di gn cht vo mt sn,u trn gn vt m1= 300g ang ng yn v tr cn bng, cng ca l xo l k = 200 N/m. T cao h = 3,75cm so vi m1, ngi ta th ri t do vt m2= 200 g, va chm mm vi m1. Sau va chm chai vt cng dao ng iu ho theo phng thng ng. Ly g = 10 m/s2, b qua mi ma st.a.Tnh vn tc ca m1ngay sau va chm.b. Hy vit phng trnh dao ng ca h hai vt m1 v m2.

GII

a. Vn tc ca m2ngay trc va chm : )/(866,0232 smglv

* Xt h hai vt m1 v m2 ngay trc v sau va chm, theo nh lut bo ton ng lng ta c :

22 1 2 0 0

1 2

. 3( ). ( / ) 20 3( / )

5

m vm v m m v v m s cm s

m m

V va chm mm nn ngay sau va chm c hai vt chuyn ng cng vn tc l:)/(3200 scmv

b. Chn trc to Ox c gc O trng vi VTCB ca hai vt, chiu dng thngng hng ln trn.Chn gc thi gian l lc hai vt bt u dao ng.

* bin dng ca l xo khi vt m1cn bng l :

)(5,111 cmk

gml

* bin dng ca l xo khi hai vt cn bng l : )(5,2)( 21

2 cmk

gmml

* Tn s gc : )/(2021

sradmm

k

m1

m2

h

k

Hnh v


5/55


* lc t = 0 ta c :

)/(320cos

)(1sin

scmcAv

cmAx

3

1 tg v 0sin v )(

6

50cos rad

Bin dao ng l : )(2

6

5sin

1cmA

* Vy phng trnh dao ng l : )(6

520sin2 cmtx

Bi 1 (HSG Lo Cai):Mt gi nh gn trn mt tm g khi lng M t trn bn nhn nm ngang ctreo mt qu cu khi lng m bng si dy di l (hnh v 1). Mt vin n nh khi lng m bayngang, xuyn vo qu cu v vng kt .a.Gi tr nh nht ca vn tc vin n bng bao nhiu si dy quay vng

nu tm g c gi cht.b.Vn tc s l bao nhiu nu tm g c th t do.

Giia. Vn tc ca qu cu v n sau khi va chm l

2

0V ( vi V0l vn tc l vn tc

ca n trc va chm)* dy quay mt vng, ti im cao nht vn tc ca qu cu l V phi tho

mn :l

VmmgT

2. ( T l lc cng ca dy) Do V = Vmin khi T = 0

lgV .min

* Theo nh lut bo ton c nng, vn tc nh nht V0ca n phi tho mn :2 2

0 min0

2 24 2 58 2

mV mVmgl V gl

b. Vn tc nh nht ca qu cu ti im cao nht ( i vi im treo) l : glu min * Xt trong HQC gn vi tri t : V1= uumin ( u l vn tc ca vt M )Ta c : )1)((2.'0 glumuMmV

Mt khc theo nh lut bo ton c nng :2' 2 2

02 ( . )2 ( ) .

4 (2)8 2 2

m u g lm V M umgl

* T (1) v (2) ta c : )

85(2'0

M

mglV

Cu 4 (ng Thp) Cho c h gm vt M, cc rng rc R1, R2 v dy treo c khi lngkhng ng k, ghp vi nhau nh hnh 1. Cc im A v B c gn cnh vo gi . VtM c khi lng m=250(g), c treo bng si dy buc vo trc rng rc R2. L xo c cng k=100 (N/m), khi lng khng ng k, mt u gn vo trc rng rc R2, cn u kiagn vo u si dy vt qua R1, R2u cn li ca dy buc vo im B. B qua ma st ccrng rc, coi dy khng dn. Ko vt M xung di v tr cn bng mt on 4(cm) ri bungra khng vn tc ban u. Chng minh rng vt M dao ng iu ho v vit phng trnhdao ng ca vt M .

Gii- Chn trc Ox thng ng hng xung, gc to O VTCB ca M.

M

m

0V

Hnh v 1

A

R1

R2

M


6/55


1)- Ti VTCB ca vt M ta c: 02 00 FTP

hay 03 0 FP

(1)

- T (1) suy ra: mg=3kl0 (2)- Ti v tr vt M c to x bt k ta c: amFTP

2 hay amFP

3 (3)- Chiu (3) ln trc to Ox ta c :

mg - 3k(l0+3x) = ma = mx (4)

- T (2) v (4) ta c : 09

'' xm

kx t

m

k92 ta c 0'' 2 xx (5)

- Phng trnh (5) c nghim :x = Acos( ) t trong A , , l nhng hng s2)- Chn gc thi gian l lc th vt. Ti thi im t =0 ta c:

4 = Acos

0 = -Asin.

suy ra A = 4 (cm) v = 0 m

k9 60(rad/s)

Vy phng trnh dao ng l x = 4cos 60 t (cm)

Bi 1 (HSG Lo Cai 06-07):Mt vt A chuyn ng vi vn tc v0n va chmhon ton n hi vi vt B ang ng yn ti C. Sau va chm vt B chuynng trn mng trn ng knh CD = 2R. Mt tm phng (E) t vung gc viCD ti tm O ca mng trn. Bit khi lng ca hai vt l bng nhau. B quami ma st. (Hnh v 1)

1.Xc nh vn tc ca vt B ti M m vt bt u ri khi mng.2.Bit Rgv 5,30 . Hi vt B c th ri vo tm (E) khng ? Nu c hy xc

nh v tr ca vt trn tm (E).Gii

1. V va chm n hi, khi lng hai vt bng nhau nn sau va chm vt B c/ vi

vn tc v0cn vt A ng yn.* nh lut bo ton c nng ( chn gc ...)

)sin1(22

22

0 mgRmvmv

)sin1(2202 gRvv (1)

* nh lut II N:R

mvNmg

2

sin

* Khi vt ri mng th N = 0Rg

Rgv

3

2sin

2

0

(2)

* Vn tc ca vt B khi bt u ri mng: Thay (2) vo (1) ta c :3

22

0 Rgvv

2. Khi Rgv 5,30 t (2) v tr vt ri mng c0

302

1sin . Vn tc ca vt lc :

2

2 Rgv

* Khi ri mng vt c/ ging nh vt b nm xin vi vn tc ban u l v.Chn trc to ...* phng trnh c/ ca vt :

cos)sin( Rtvx 22

1)cos(sin gttvRy

BAR

R

M

P

TTF

D

B

C

(E)

0v

O

Hnh v 1

D

B

C

(E)O

Hnh v 1

P

N


7/55


* vt ri vo vo tm (E) th : 0x v y =0. Vi 0xg

Rt

6 (*)

Vi y = 0 giiphng trnh c t1< 0 (**) So snh (*) v (**) thy vt B khng ri vo tm (E)Bi 4 (HSG Lo Cai 06-07): Cho h dao ng nh hnh v 4. L xo c khi lng khng ng k, cng k. Vt M = 400g c th trt khng ma st trn mt phng nm ngang. H ang trng thi cn

bng, dng vt m0= 100g bn vo M theo phng ngang vi vn tc v0= 1m/s, va chm l hon ton

n hi. Sau va chm vt M dao ng iu ho, chiu di cc i v cc tiu ca ca l xo ln lt l28cm v 20cm.

1.Tnh chu kdao ng ca vt v cng ca l xo.2.t mt vt m = 100g ln trn vt M, h gm hai vt m v M ang ng yn, vn dng vt m 0bn

vo vi vn tc v0. Va chm l hon ton n hi, sau va chm ta thy c hai vt cng dao ng iuho. Vit phng trnh dao ng ca h hai vt m v M. Chn gc to v tr cn bng v gc thigian l lc bt u va chm. Xc nh chiu v ln ca lc n hi cc i, cc tiu m l xo tcdng vo im c nh I trong qu trnh h hai vt dao ng.3. Chobit h s ma st gia vt M v vt m l = 0,4. Hi vn tc v0ca vt m0phi nh hn gi

tr bng bao nhiu vt m vn ng yn (khng b trt) trn vt M trong khi h dao ng. Cho g =

10m/s2

.Gii

1. Va chm n hi nn ng lng v ng nng c bo ton

Ta c : MVvmvm 000 (1)222

2

0

2

00 MVvmvm (2)

Vi v , V ln lt l vn tc ca cc vt m0v M ngay sau va chm

* Gii h (1), (2) c : )/(40)/(4,02

0

00 scmsmMm

vmV

* Sau v/c vt M dao ng iu ho, vn tc cc i ca vt l V = 40(cm/s)

Bin dao ng l : 2minmax ll

A

= 4(cm) Ta c: V = A. )/(10 sradA

V

=> chu k ca dao

ng l: T = )(5s

cng ca l xo : )/(40. 2 mNMk .

2.

a. Va chm n hi nn ng lng v ng nng c bo ton

Ta c : hVmMvmvm )(1000 (3)2

)(

22

22

10

2

00 hVmMvmvm (4)

Vi v1 , Vhln lt l vn tc ca cc vt m0v (M + m) ngay sau va chm

* Gii h (3), (4) c : )/(

3

1002

0

00 scm

mMm

vmVh

* Sau v/c vt (M + m) dao ng iu ho nn phng trnh dao ng c dng )sin( tAx .

Vn tc cc i ca h vt l : Vh =3

100(cm/s).

Tn s gc : )/(54 sradmM

k

Chn trc to c gc trng VTCB, chiu dng cng hng 0v

.

Mm0

0v

I k

Hnh v


8/55


Lc t = 0 ta c :

hVA

A

cos

0sin

)/(73,3cos.

0

0cos

0sin

scmV

A h

* Vy phng trnh dao ng ca vt l : ))(54sin(73,3 cmtx b. * Ti cc v tr bin lc n hi ca l xo tc dng vo im c nh l ln nht ta c

)(492,110.73,3.40. 2max

NAkF

Ti v tr bin bn tri lc n hi hng sang bn phiTi v tr bin bn phi lc n hi hng sang bn tri

* Ti VTCB lc n hi ca l xo c gi tr nh nht : Fmin = 0.3. vt m khng b trt trn M trong qu trnh dao ng th lc ma st ngh cc i phi c gi tr gi tr ca lc qun tnh cc i tc dng ln vt m (Xt trong h quy chiu gn vi vt M) :

(max)(max) qtmsn FF (*)

* Ta c :

Lc ma st ngh C : mgNFmsn .(max)

Lc qun tnh : )sin(. 2 tAmamFqt

lc qun tnh t cc i th AmFt qt2

(max) .1)sin(

* T biu thc (*) ta c :2

2

gAAmmg

* Mt khc: Mmm

vmVVA h

0

00max 2

)/(34,1

2

2

0

0

02

0

00 smm

Mmmgv

g

Mmm

vm

Vy v0 )/(34,1 sm th vt m khng b trt trn vt M trong qu trnh h dao ng.Cu 4 (HSG Hu Giang) .Mt con lc n c chiu di l thc hin dao ng iuho trn mt chicxe ang ln t do xung dc khng ma st. Dc nghing mt gc so vi phng nm ngang.a)Chng minh rng: V tr cn bng ca con lc l v tr c dy treo vung gc vi mt dc.

b)Tm biu thc tnh chu k dao ng ca con lc. p dng bng s l =1,73 m; =300; g = 9,8 m/s2.p n

+ Gia tc chuyn ng xung dc ca xe l a = gsin.Xt h quy chiu gn vi xe+ Tc dng ln con lc ti mt thi im no c 3 lc:Trng lng P,lc qun tnh Fv sc cng T ca dy treo.Ti v tr cn bng

Ta c: 0TFP

+ Chiu phng trnh trn xung phng OX song song vi mt dc ta c: Psin - F + TX = 0M F = ma = mgsinsuy ra TX = 0.

iu ny chng t v tr cn bng dy treo con lc vung gc vi Ox+ V tr cn bng nh trn th trng lc biu kin ca con lc l :

P' = Pcos. Tc l gia tc biu kin l g' = gcos.

+ Vy chuk dao ng ca con lc s l T = 2'

l

g= 2

cos

l

g 2,83 (s).

T

F

P

x


9/55

NGUYN VN TRUNG :0915192169Bi 1HSG Lo Cai 08-09Buc vo hai u mt si dy di 2lhai qu cu nh A vB ging nhau c cng khi lng m, chnh gia si dy gn mt qu cu nh khckhi lng M. t ba qu cu ng yn trn mt bn nm ngang nhn, dy c kocng.(Hnh v 1)

Truyn tc thi cho vt M mt vn tc 0V theo phng vung gc vi dy. Tnhlc cng ca dy khihai qu cu A v B sp p vo nhau.

Gii

H kn ng lng bo ton

0 1 2 MV mv mv M v

0 1 2

1 20

y y M

x x

MV mv mv Mv

mv mv

Ta lun c: 1 2 1 2; y y x xv v v v

Khi hai qu cu sp p vo nhau:

1 2 y y M yv v v v

02

yMVvm M

p dng nh lut bo ton nng lng:2 2 2 2

0

1 1 1 12 2

2 2 2 2 y x y

MV mv mv Mv ( xv ln vn tc ca hai qu cu A,B lc chng sp p vo

nhau)2

2 0

2x

mMVmv

m M

Gia tc ca qu cu M:

2Ta

M

Trong h quy chiu gn vi M hai qu cu m chuyn ng trn p dng nh lut 2 Niutn, chiuxung phng Oy:

2x

q

vT F m

l

202

(2 )

mMVTT m

M l m M

Lc cng ca dy khi :

2 20

2(2 )

mM VT

l m M

Bi 2 (HSG Lo Cai 08-09) Mt lxo l tng treo thng ng, u trn ca l xo c gi c nh,u di treo mt vt nh c khi lng m = 100g, l xo c cng k = 25N/m. T v tr cn bngnng vt ln theo phng thng ng mt on 2cm ri truyn cho vt vn tc 310 cm/s theo phngthng ng, chiu hng xung di. Chn gc thi gian l lc truyn vn tc cho vt, chn trc ta c gc trng v tr cn bng ca vt, chiu dng thng ng xung di. Cho g = 10m/s2; 102 .

1.Chng minh vt dao ng iu ha v vit phng trnh dao ng ca vt.2.Xc nh thi im lc vt qua v tr m l xo b gin 6cm ln th hai. Xc nh hng v ln

ca lc tc dng ln im treo ti thi im . Gii1. Chng minh vt dao ng iu ha* Vit phng trnh dao ng ca vt:Ti VTCB: 4l (cm) Tn s gc: 5 (rad/s). Ti thi im t = 0 ta c:

)/(310sin

)(2cos

scmAv

cmAx

V3

23tan;0cos;0sin

(rad) Bin dao ng : A = 4 (cm)

2yv

T T

T T

1yv

v T T

T

1yv

1xv

T

2yv

2xv

Ox

Hnh v 1

0V

B

A

M


10/55


Vy phng trnh dao ng ca vt l:

3

25cos4

tx (cm)

2. Khi vt qua v tr m l xo b gin 6cm ln th hai th vt c li x = 2cm v chuyn ng theochiu m ca trc ta .Ta c:

03

2

5sin

2

1

3

25cos

t

t Gii h phng trnh (ly gi tr nh nht) c kt qu: 2,0t (s)

* Xc nh hng v ln ca lc tc dng ln im treo ti thi im :- Hng: Phng thng ng, chiu t trn xung di.- ln: 5,110.6.25 21

lkF (N)

Cu 1:Hai vt1 v 2 u c khi lngbng m gn cht vo l xo c di l, cngk ng yn trn mt bn nm ngang tuyt i nhn.Vt th 3 cng c khi lng m chuyn ng

vi vn tc v n va chm hon ton n hi vi vt 1(xem hnh 1)

1. Chng t hai vt m1 v m2lun chuyn ng v cng mt pha.2. Tm vn tc ca hai vt 1 v 2 v khong cch gia chng vo thi im l xo bin dng ln nht.

GiiNgay sau lc va chm vt 1 c vn tc v (l xo cha bin dng, vn tc vt 2 bng khng). Gi v1,v2 l vn tc vt1,vt2 vo thi im sau va chm ca vt 3 vo 1 la v1, v2 . bin dng l k0 l x.

+ nh lut bo ton ng lng: mv = mv1 + mv2 . v = v1 + v2 (1)

+ nh lut bo ton c nng:2

1mv

2=

2

12

1mv + 22

2

1mv +

2

2

1kx

2

21

22

)( vvvm

kx (2). T (1) va

(2):m

kx

2

2

= v1v2 (3) vm

kx

2

2

> 0 v1v2 > 0 : tc l v1 v v2cng du ngha l sau khi va chm

hai vt 1 v 2 lun chuyn ng v cng mt pha.

2) v1 + v2 = v = const. Suy ra tch v1v2 cc i khi v1 = v2 =2

vngha l

m

kx

2

2

cc i

lc :4

2v

=m

kx

2

2

max xmax = vk

m

2 l xo bin dng ln nht khi v1 = v2 =

2

v lc ny khong

cch gia vt 1 v vt 2 l: l12 =k

mvlxl

2max

Bi 2( HSG Ngh An 07-08) Vt nng c khi lng m nm trn mt mtphng nhn nm ngang, c ni vi mt l xo c cng k, l xo c gn

vo bc tng ng ti im A nh hnh 2a. T mt thi im no , vtnng bt u chu tc dng ca mt lc khng i Fhng theo trc lxo nhhnh v.

a) Hy tm qung ng m vt nng i c v thi gian vt i ht qung ng y k t khi btu tc dng lc cho n khi vt dng li ln th nht.

b) Nu l xo khng khng gn vo im A m c ni vi mt vtkhi lngMnh hnh2b, h s ma st gia Mv mt ngang l . Hy xcnh ln ca lc F sau vt mdao ng iu ha.

GII

Fm

k

Hnh 2a

A

mk

Hnh 2b

M

3 21v


11/55

NGUYN VN TRUNG :0915192169a) Chn trc ta hng dc theo trc l xo, gc ta trng vo v tr

cn bng ca vt sau khi c lc Ftc dng nh hnh 1. Khi ,v tr banu ca vt c ta lx0. Ti v tr cn bng, l xo b bin dng mt lngx0 v:

.00k

FxkxF

Ti ta xbt k th bin dng ca l xo l (xx0), nn hp lc tc dng ln vt l:.)( 0 maFxxk Thay biu thc cax0vo, ta nhn c:

.0"2

xxmakxmaFk

Fxk

Trong mk . Nghim ca phng trnh ny l: ).sin( tAx

Nh vy vt dao ng iu ha vi chu kk

mT 2 . Thi gian k t khi tc dng lc Fln vt

n khi vtdng li ln th nht (ti ly cc i pha bn phi) r rng l bng 1/2 chu k dao ng,

vt thi gian l:.

2 k

mTt

Khi t=0 th:

0cos

,sin

Av

k

FAx

.2

,

k

FA

Vy vt dao ng vi bin F/k, thi gian t khivt chu tc dng ca lc Fn khi vt dng liln th nht l T/2 v n i c qung ng bng 2 ln bin dao ng. Do , qung ng vt ic trong thi gian ny l:

.22k

FAS

b) Theo cu a) th bin dao ng l .k

FA sau khi tc dng lc, vt mdao ng iu ha

th trong qu trnh chuyn ng ca m,Mphi nm yn.Lc n hi tc dng lnMt ln cc ikhi bin dng ca l xo t cc i khi vt m xaMnht (khi l xo gin nhiu nht v bng

AAx 20 ).

vtMkhng b trt th lc n hi cc i khng c vt qu ln ca ma st ngh cci:

..2.2. Mgk

F

kMgAk

T suy ra iu kin ca ln lc F: .2

mgF

Bi 3.HSG Ngh AN 07-08.Hai ngun sng kt hp S1 v S2 cch nhau 2m dao ng iu ha cngpha, pht ra hai sng c bc sng 1m. Mt im A nm khong cch lk t S1 v AS1S1S2 .

a)Tnh gi tr cc i ca l ti A c c cc i ca giao thoa.b)Tnh gi tr ca l ti A c c cc tiu ca giao thoa.

mk

Hnh 1

Ox0


12/55

NGUYN VN TRUNG :0915192169a) iu kin ti A c cc i giao thoa l hiu ng i t A n haingun sng phi bng s nguyn ln bc sng (xem hnh 2):

.22 kldl

Vi k=1, 2, 3...Khi lcng ln ng S1A ct cc cc i giao thoa c bc cng nh (kcng

b), vy ngvi gi tr ln nht ca l ti A c cc i ngha l ti A ng

S1A ct cc i bc 1 (k=1).Thay cc gi tr cho vo biu thc trn ta nhn c:

).(5,1142 mlll b) iu kin ti A c cc tiu giao thoa l:

.2

)12(22

kldl

Trong biu thc ny k=0, 1, 2, 3, ...

Ta suy ra :

)12(

2)12(

2

2

k

kd

l .

V l> 0 nn k = 0 hoc k = 1.T ta c gi tr ca l l :* Vi k =0 th l = 3,75 (m ).* Vi k= 1 th l 0,58 (m).Cu 1Cho c h nh hnh v 1. Hai thanh cng MA v NB khi lng khngng k, cng chiu di l = 50cm. u t do ca mi thanh u c gn mt qucu nh cng khi lng m =100g, u M v N ca mi thanh c th quay ddng. L xo rt nh c cng k = 100N/m c gn vo trung im C cathanh NB. Khi h cn bng l xo khng bin dng, hai qu cu tip xc nhau.Ko qu cu A sao cho thanh MA lch v bn tri mt gc nh ri th nh. Coiva chm gia cc qu cu l n hi xuyn tm. B qua mi ma st, lyg = 10m/s

2. Hy m t chuyn ng v xc nh chu k dao ng ca h .

+ Do A va chm vi B l n hi nn ng lng v ng nng h c bo ton.' '

1 1 2

2 ' 2 ' 2

1 1 2( ) ( )

2 2 2

mv mv mv

mv m v m v

+ Chn chiu dng cng chiu vi 1v suy ra:' '

1 1 2

2 ' 2 ' 2

1 1 2( ) ( )

2 2 2

mv mv mv

mv m v m v

' '

1 2 10,v v v

+Tng t cho va chm t qu cu B tr li qu cu A, ta c:'' ' ''

1 2 2, 0v v v

+ Sau va chm qu cu ny truyn hon ton vn tc cho qu cu kia. H thng dao ng tun hon, con lc tham gia mt na dao ng.

+ Chu k dao ng 1 21

( )2

T T T vi T1l chu k dao ng con lc n, T2l chu k dao ng ca con

gn vi thanh v l xo.

S1

S2

lA

dk

k

k

Hnh 2

A B

M

N

C

k

Hnh


13/55


+ Ta bit chu k dao ng ca con lc n1

2 1,4( )l

T sg

Ta tm T2bng phng php nng lng:+Chn mc th nng trng trng ti mt phng ngang qua m khi cn bng.+Xt vt m ti v tr c li x:

-ng nng ca qu cu E =

2

2

mv

-Th nng trng trng Et1=

2

2

mgx

l

-Th nng n hi: Et2 =2 2

1

2 8

kx kx

C nng ca h: E = E + Et1 + Et2 =2

2

mv-

2 2

2 8

mgx kx

l (1). Do khng c lc cn nn E = const.

+Ly o hm 2 v ca (1) theo thi gian t, ta c: mvv -' '

04

mgxx kxx

l Hay x+( )

4

k gx

m l

+Vy vt dao ng iu ha vi tn s gc4

k g

m l v chu k

2

20,4T s

+H dao ng tun hon vi chu k 1 21 ( )2

T T T = 0,7 + 0,2 = 0,9s

(HSG Hu Lc 05-06)a)Cho con lc lin hp nh hnh v 1 bit khi lng m1, m2v chiu di l1, l2. B qua khi lng

dy treo v lc cn mi trng. Tnh tn s dao ng.b)Nu mc thm vo h 3 l xo K1 = K2 = K3nh hnh v 2, h vn dao ng iu ho. Tnh tn s

dao ng ca h, cho nhn xt v tn s.

Cu a . Hc sinh c th lm theo nhiu cch cho kt qu: =2

22

2

11

2211 )(

lmlm

glmlm

Cu b .

HS lp lun c h gm c: (K1 nt K2) // K3 // Kh(vi Kh l K h cu a)

Hc sinh tnh c K(h mi) : K = hh KKKKK

K

2

3

2

2

Kt qu: =2

22

2

11

2

1

2211 )(

2

3

lmlm

l

glmlmK

M

K

hay =1

1

l 222

2

11

2211

2

1)(

2

3

lmlm

glmlmlK

o

Hnh

l2

l1m1

m2

K3K2K1

o

Hnh 2

m1

m2

l2

l1


14/55

NGUYN VN TRUNG :0915192169Bi 1(HSG Hai B Trng)Hai vt khi lng m0v m c ni vi nhau bng mt si dy mnh, bnkhng dn c chiu di L. Ti thi im ban u vt m0c nm t mt phng ngang vi vn tc banu v0thng ng hng ln. Hi cao cc i m m0c th t ti.Trng hp 1: Nu gLv 220 th dy cp khng b cng v cao cc i

Lg

vH

2

2

0

Trng hp 2:

+ Nu gLv 220 th ngay trc lc dy cng, vn tc ca m0 l gLvv 22

01 + Sau m0 v m c cng vn tc v

+ nh lut bo ton ng lng: m0v1 = (m + m0)v0

10

mm

vmv

+ cao h vt ln c k t lc dy cng:

g

ghv

mm

m

g

vh

2

2

2

2

0

2

0

0

2

+ Vy Hmax = L + h = L +

gghv

mm

m

2

22

0

2

0

0

I. C hc: HSG THANH HOA 06-071/. Mt ht thc hin dao ng iu ho vi tn s 0,25 (Hz) quanh im x = 0. Vo lc t = 0 n c di 0,37 (cm). Hy xc nh di v vn tc ca ht lc lc t = 3,0 (s) ?

2/.Mt con lc n c chiu di L thc hin dao ng iu ho trn mt chic xe ang ln t do xungdc khng ma st. Dc nghing mt gc so vi phng nm ngang.a) Hy chng minh rng: V tr cn bng ca con lc l v tr c dy treo vung gc vi mt dc.

b)Tm biu thc tnh chu k dao ng ca con lc. p dng bng s L=1,73 m; =300

; g = 9,8 m/s

2

.3/. Mt con lc n c ko ra khi v tr cn bng mt gc nh 0= 0,1 rad ri bung khng c vntc ban u. Coi rng trong qu trnh dao ng lc cn ca mi trng tc dng ln con lc khng iv bng 1/1000 trng lng ca con lc. Hi sau bao nhiu chu k dao ng th con lc dng hn li ?

4/. Mt ht khi lng 10 (g), dao ng iu ho theo qui lut hm sin vi bin 2.10-3 (m) v phaban u ca dao ng l -/3 (rad). Gia tc cc i ca n l 8.103 (m/s2). Hy:

a)Vit biu thc ca lc tc dng vo ht di dng hm ca thi gian.b)Tnh c nng ton phn ca dao ng ca ht.

Cu 1

+ Tn s dao ng = 2 = /2 (rad/s) ; Bin ca dao ng A = 0,37 (cm)

Vy x = 0,37sin( 2

t+ ) (cm).

+ Ti t = 0 th x = 0,37 => = /2. Vy phng trnh dao ng ca ht l

x = 0,37sin (2

t +2

) (cm) = 0,37cos2

t (cm).

+ Lc t = 3 (s) di l xt = = 0,37cos2

.3 = 0 v v = x't = - 0,37.2

. sin2

3 = 0,581 (cm/s).

Cu 2: a)

+ Gia tc chuyn ng xung dc ca xe l a = gsin.

m0

m

0v

T

F

P P'


15/55

NGUYN VN TRUNG :0915192169+ Tc dng ln con lc ti mt thi im no c3 lc:Trng lng P, lc qun tnh F (do xe ch g nh dn u)v sc cng T ca dy treo.V tr cn bng ca con lc l v tr c hp lc bng 0.

Tc l 0TFP

+ Chiu phng trnh trn xung phng OX songsong vi mt dc ta c: Psin - F + TX = 0

+ Ch rng ln lc qun tnh F = ma = mgsin suy ra TX= 0. iu ny chng t dy treo con lcvung gc vi OX khi trng thi cn bng. (pcm).b)

+ V tr cn bng nh trn th trng lc biu kin ca con lc l P' = Pcos. Tc l gia tc biu kin lg' = gcos.

+ Vy chu k dao ng ca con lc s l T = 2'g

L= 2

cosg

L 2,83 (s).

Cu 3(1,5 im):

+ Nng lng ban u ca con lc l E0 = mgl.(1-cos0) = 20mgl2

1 .

+ Gi 1 v 2l hai bin lin tip ca dao ng (mt ln con lc qua v tr cn bng). Ta c

gim th nng l ( 21mgl

2

1- 2

2mgl2

1).

+ gim ny bng cng ca lc cn mi trng A = Fc.S = Fc.l.(1 + 2).

+ Suy ra 21mg2

1 = Fc .

+ gim bin gc mi ln s l (1-2) = 2Fc/ mg = 2.10-3

mg/mg = 2.10-3

rad.

+ n khi con lc ngng dao ng th s ln i qua v tr cn bng s l N =0 /(1-2) = 50. Tngng vi 25 chu k.Cu 4(2,0 im):+ Gia tc a = x'' = -2x => gia tc cc i am =

2A => = (am/A)

1/2= 2.10

3(rad/s).

+ Vy ta c F = ma = - 0,01.(2.103)2. 2.10-3 sin(2.103.t -3

) = 80 sin(2.103t +

3

2) (N)

+ Vn tc cc i ca ht l vm = A = 4 (m/s)

+ C nng ton phn E0 =2

mv 2m = 0,08 (J).

Bi 2:a, (1) Khi cha t dy: 02 .mg k l ;

Ngay sau khi dy t: * Vt m: 0 1.k l mg ma 1 3 30a g (2/m s )

* Vt 2m: 0 2. 2 2k l mg ma 2 0a b,(3) Xt h quy chiu gn vi trng tm G ca h.G cch vt m mt khong bng 2/3

khong cch t vt m n vt 2m.* Xt vt m :

- Khi VTCB: 0qtmg F (1)

- Khi li x: l xo gin mt on bng 3x/2 . Suy ra: ''3

.2

qt

xmg F k m a mx (2)

T (1) v (2) : ''3

02

kx x

m '' 2 0x x vi

310

2

k

m (rad/s)

3.sin( . )

2

k x A t

m

m

2m


16/55


Ti 0t : 00

2.sin 0,2

3

lx A

(m) v 0 . .cosv A 0 0,2A (m); v

2

(rad)

0,2.sin(10. / 2)x t (m);- bin dng ca l xo: 3 / 2 0,3,sin(10. / 2)l x t ;

- L xo t trng thi khng bin dng ln u tin 0l 1,5720

t (s).

- Trng tm G chuyn ng vi gia tc g, khi trng tm G i c :2 2 / 2 /80h gt (m) vi vn tc . / 2Gv g t (m/s).

Ti thi im ta c: 2 os(10.t+ /2)=x c -2 (m/s) 2 / 2 3,57m Gv v x (m/s)

- Theo LBTNL: 2 2 20 2

1 1 1. 3 . .2 .

2 2 2m mk l mg h mv m v ;

Mt khc, ta c: 0. 2k l mg 2 1 0,572mv (m/s)

......................................................................................

DNG IN XOAY CHIU

I. TM TT L THUYT:1. Biu thc in p tc thi v dng in tc thi:u = U0cos(t + u) v i = I0cos(t + i)

* Vi = uil lch pha ca uso vi i, c2 2

2. Dng in xoay chiu :I= I0cos(2ft + i)* Mi giy i chiu 2f ln* Nu pha ban u I = 0 hoc I = th giy u tin ch i chiu 2f-1 ln.

3. Cng thc tnh thi gian n hunh quang sng trong mt chu k : * Khi t in p u = U0cos(t + u) vo hai u bng n, bit n ch sng ln khi u U1.

4t Vi 1

0

os UcU

, (0 < < /2)

4. Dng in xoay chiu trong on mch R,L,C* on mch ch c in tr thun R: uRcng pha vi i, ( = ui = 0)

UI

R v 0

0

UI

R

Lu :in tr R cho dng in khng i i qua v cU

IR

* on mch ch c cun thun cm L: uLnhanh pha hn i l /2, ( = ui = /2)

L

U

I Z v0

0

L

U

I Z vi ZL = L l cm khng

Lu :Cun thun cm L cho dng in khng i i qua hon ton (khng cn tr).* on mch ch c t in C: uCchm pha hn i l /2, ( = ui = -/2)

C

UI

Z v 00

C

UI

Z vi

1CZ

C l dung khng

Lu :T in C khng cho dng in khng i i qua (cn tr hon ton).* on mch RLC khng phn nhnh

2 2 2 2 2 2

0 0 0 0( ) ( ) ( ) L C R L C R L C Z R Z Z U U U U U U U U


17/55


tan ;sin ; os L C L C Z Z Z Z R

c R Z Z

vi2 2

+ Khi ZL > ZC hay1

LC > 0 th unhanh pha hn i

+ Khi ZL < ZC hay1

LC < 0 th uchm pha hn i

+ Khi ZL = ZC hay 1LC

= 0 th ucg pha vi i v I = Max UI =R

gi l hin tg cg

hg dng in5. Cng sut to nhit trn on mch RLC:

* Cng sut tc thi: P = UIcos + UIcos(2t + u+i)* Cng sut trung bnh:P = UIcos = I2R.

6. in p u = U1 + U0cos(t + )c coi gm mt in p khng i U1 v mt in p xoay chiuu = U0cos(t + ) ng thi t vo on mch.7. Tn s dng in do my pht in xoay chiu mt pha c P cp cc, rto quay vi vn tc nvng/giy

th my pht ra dng in c tn s l : f = pn ( Hz )* T thng gi qua khung dy ca my pht in = NBScos(t +) = 0cos(t + )

Vi 0= NBS l t thng cc i,N l s vng dy,B l cm ng t ca t, S l din tch cavng dy, = 2f

* Sut in ng trong khung dy: e = NSBcos(t + -2

) = E0cos(t + -

2

)

Vi E0 = NSB l sut inng cc i.8. Dng in xoay chiu ba phal h thng 3 dng in xoay chiu 1 pha c gy bi 3 sut in

ng xoay chiu cng tn s, cng bin nhng lch pha tng i mt l 23

.

*Cc pt ca sut in ng v dng in v cm ng t c dng : (Xt trng hp ti i xng )th

1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

e E c t

e E c t

e E c t

1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

i I c t

i I c t

i I c t

1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

B B c t

B B c t

B B c t

+Dng in xoay chiu 3 phac to ra t mt my pht in xoay chiu 3 pha

*My pht mc hnh sao: Ud = 3 Up v ti tiu th mc hnh sao: Id = Ip*My pht mc hnh tam gic: Ud = Up v ti tiu th mc hnh tam gic: Id = 3 IpLu : my pht vti tiu th thng chn cch mc tng ng vi nhau.

9. Cng sut hao ph trong qu trnh truyn ti in nng:2

2 2os

RU c

Trong : P l cng sut truyn i ni cung cp ; U l in p ni cung cpcosl h s cng sut ca dy ti in

*l

RS

l in tr tng cng ca dy ti in ( lu :dn in bng 2 dy)

* gim in p trn ng dy ti in: U = IR


18/55


* Hiu sut tiin: .100%H

10. on mch RLC c R thay i:

* Khi R=ZL-ZC th2 2

ax2 2

M

L C

U U

Z Z R

* Khi R=R1hoc R= R2 m P c cng gi tr

th ta c2

2

1 2 1 2; ( )

L C

U R R R R Z Z v khi 1 2 R R R th

2

ax

1 22M

U

R R

* Trng hp cun dy c in tr R0(hnh v)

+ Khi2 2

0 ax

02 2( )

L C M

L C

U U R R Z Z P

Z Z R R

+ Khi2 2

2 2

0 ax2 2

00 0

( )2( )2 ( ) 2

L C RM

L C

U U R R Z Z

R R R Z Z R

11. on mch RLC c L thay i:

* Khi2

1L

C th IMax URmax; PMax cn ULCMin **Lu :L v C mc lin tip nhau

* Khi2 2

CL

C

R ZZ

Z

th

2 2

ax

C

LM

U R ZU

R

v 2 2 2 2 2 2

ax ax ax; 0

LM R C LM C LM U U U U U U U U

* Vi L = L1hoc L = L2 th UL c cng gi tr th ULmax khi1 2

1 2

1 2

21 1 1 1( )

2 L L L

L LL

Z Z Z L L

* Khi2 24

2

C C

L

Z R Z Z

th ax

2 2

2 R

4

RLM

C C

UU

R Z Z

Lu :R v L mc lin tip nhau

12. on mch RLC c C thay i:

* Khi2

1C

L th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi2 2

L

C

L

R ZZ

Z

th

2 2

ax

L

CM

U R ZU

R

v 2 2 2 2 2 2ax ax ax; 0CM R L CM L CM U U U U U U U U

* Khi C = C1hoc C = C2 th UC c cng gi tr th UCmax khi1 2

1 21 1 1 1( )2 2C C C

C CC

Z Z Z

* Khi

2 24

2

L L

C

Z R Z Z

th ax 2 2

2 R

4RCML L

U

U R Z Z Lu :R v C mc lin tip nhau

13. Mch RLC c thay i:* Khi

1

LC th IMax URmax; PMax cn ULCMin Lu :L v C mc lin tip nhau

* Khi2

1 1

2

C L R

C

th ax 2 2

2 .

4LM

U LU

R LC R C

A B

CR L,R0


19/55


* Khi2

1

2

L R

L C th ax 2 2

2 .

4CM

U LU

R LC R C

* Vi = 1hoc = 2 m I hoc P hoc URc cng mt gi tr

th IMax hoc PMax hoc URMax khi

2

1 2

1

LC tn s2

1 2 f f f 14. Hai on mch AM gm R1L1C1ni tip v on mch MB gm R2L2C2ni tip mc ni tipvi nhau c UAB = UAM + UMB uAB ; uAMv uMB cng pha tan uAB = tan uAM= tan uMB16. Hai on mch R1L1C1 v R2L2C2 cng uhoc cng i c pha lch nhau

Vi 1 11

1

tanL CZ Z

R

v 2 2

2

2

tanL C

Z Z

R

(gi s 1 > 2)

C 12 = 1 2

1 2

tan tantan

1 tan tan

**Trnghpcbit = /2 (vung pha nhau) th tan1tan2 = -1.VD:* Mch in hnh 1 c uAB v uAMlch pha nhau

y 2 on mch AB v AM c cng i v uABchm phahn uAM

AMAB = tan tan

tan1 tan tan

AM AB

AM AB

Nu uABvung pha vi uAM th tan tan =-1 1L CL AM ABZ ZZ

R R

* Mch in hnh 2: Khi C = C1 v C = C2(gi s C1 > C2) th i1 v i2lch pha nhau

y hai on mch RLC1 v RLC2 c cng uABGi 1 v 2l lch pha ca uABso vi i1 v i2th c 1 > 21 - 2 =

Nu I1 = I2 th 1 = -2 = /2

Nu I1 I2 th tnh 1 21 2

tan tantan

1 tan tan

II. CC DANG TON:BI 1: (Nm hc 2007- 2008 tnh thi nguyn )Cho mch in nh hnh v. Cun dy c t cmL = 1,5/ (H), in tr thun R0; t c in dung C =2.10-4/9(F). Hiu in th tc thi gia hai im A vMlch pha mt gc 5 /6 so vi hiu in th gia haiim M v N, ng thi hiu in th gia hai im A vM c biu thc uAM = 100 6 sin(100t + /6)(V). Cng sut tiu th ca c mch l P = 100 3 (W).

a/Tnh R0; R.

b/Vit biu thc tc thi ca hiu in th gia hai im AB.

R L CMA B

Hnh

R L CMA B

Hnh

A B

M N

R

CL,R0


20/55

NGUYN VN TRUNG :0915192169Bi 2: (Nm hc2007 - 2008, Tnh Ngh An)

Cho mch in xoay chiu nh hnh v.Bit uAB = 180 2 sin(100t) (V), R1 = R2 = 100 , cun dy

thun cm c L = H3

, t in c in dung C bin i c.

1. Tm C hiu in th hiu dnggia hai im M, N t cc

tiu.

2. Khi C =100

F3

, mc vo M v N mt ampe k c in tr

khng ng k th s ch ampe k l bao nhiu?HNG DN GII:1.Gin vc t c v nh hnh bn..T gin suy ra UMNcc tiu khi M trngvi N ..Hay: UMN= 0 UR1 = UC I1R1 = I2ZC, UR2 = UL

= I2R2= I1ZL

LZ

R1 =2R

ZC ZC =LZ

RR 21 = 3

100 C = F

3100= 55( F )

2.Chp M v N thnh im E.Tng tr, lch pha gia hiu in th v cng dng in trongmi nhnh :

UEB

CI I A 2

LI

1 I

1RI AEU

22

1

2

1

111

CZRZ Z1 = 50 )(3 .Tg 1 = -

1R

C

I

I= -CZ

R1 = -3

1 1 = -

6

22

2

2

2

111

LZRZ Z2 = 50 )(3 . Tg 2 =

2R

L

I

I=LZ

R2 =3

1 2 =

6

.V Z1 = Z2 v cng hiu dng trong mch chnh nh nhau nn: UAE = UEB = U

.Mt khc AEU v EBU u lch v hai pha trc I mt gc6

nn:

UAE = UEB =

)6

cos(2

ABU

= 60 3 (V) :

B

R1 ML

AC

NR2

(Hnh 5)

UR1 UR2CU UL

N M

U AB


21/55

NGUYN VN TRUNG :0915192169.Chn chiu dng qua cc nhnh nh hnh v.

.Gin vc t biu dinLAR III 1 nh hnh bn.

.T ta c:

IA=

6

cos2 12

12

LRLR IIII = 0,6(A)

Bi 3: (Bc giang Nm hc 2006 - 2007)t hiu in th 100sin275u t (V) vo hai u mton mch gm cun dy ni tip vi mt t in. Dng vn k c in tr rt ln ln lt o hiu in

th gia hai u cun dy v ca t in ta c UCd = 100 (V) v UC= 35 (V). Bit L = 12

(H). Xc

nh in dung ca t in v vit biu thc cng dng in trong mch.

Bi 4: (NM HC 2007-2008. TNH DAKLAK )Cho mch in xoay chiu nh hnh v (h.1).

Hiu in th xoay chiu hai u mch c biu thc : uAB = U0.sin100t (V), b qua in tr cc dy

ni. Cc hiu in th hiu dng: UAN = 300 (V) , UMB = 60 3 (V). Hiu in th uAN lchpha so vi

uMB mt gc2

. Cun dy c h s t cm

1L

3

(H) vi in tr r, in dung ca t

in3

3.10C =

16

(F).

1) Tnh in tr r.2) Vit biu thc hiu in th uAN.

Bi 5: (Tnh Thanh Ha, nm hc 2010 - 2011)Cho mch in xoay chiu gm cun dy D c t cm L mc ni tip vi in tr thun R v t

in c in dung C (hnh v). Bit in p giahai u on mch AB c biu thc u =U0cos100t (V) khng i. Cc vn k nhit V1;V2c in tr rt ln ch ln lt l U1 = 120V; U2=80 3 V. in p tc thi gia hai u on mch MB lch pha so vi in p tc thi gia hai uon mch NB gc /6 v lch pha so vi in p tc thi gia hai u on mch AN gc /2. Ampeknhit c in tr khng ng k ch 3 A.

a. Xc nh cc gi tr ca R; L v C.b. Tnh U0v vit biu thc cng dng in tc thi qua mch.

Hng dn gii:

R

(h .1)

L , r C

A BM N

A BC

N

D RM

V1

V2

A

BC N R2

R1 M L

IAIL

A

IR1

300

600


22/55

NGUYN VN TRUNG :0915192169a. Xc nh gi tr R ; L ;CV gin vc t ng R = UR /I = U2cos60

0/ I = 40

ZC = UC /I = U2cos300

/I = 40 3 FC 510.59,4

ZL = UL /I = U1sin300 /I = 203

HL 11,0

b. Xc nh U0v vit biu thc iT GVT : U

= 1U

+ CU

. p dng nh l hm s cosin ta c :U

2= U1

2+ UC

2+ 2U1.UC. cos120

0

Thay s v tnh ton ta c: U = 120V => U0 = 120 2 (V)

Lp lun = -/6 i = 6 cos(100t + /6) (A)

Bi 6: (Tnh Thanh Ha, nm hc 2010 - 2011)Trong qu trnh truyn ti in nng i xa cn tng in p ca ngun ln bao nhiu ln gim

cng sut hao ph trn ng dy i 100 ln. Gi thit cng sut ni tiu th nhn c khng i, inp tc thi u cng pha vi dng in tc thi i. Bit ban u gim in th trn ng dy bng15% in p ca ti tiu th.HNG DN GII:t U, U1, U , I1, 1P l in p ngun, in p ti tiu th, gim in p trn ng dy,dng in hiu dng v cng sut hao ph trn ng dy lc u.U, U2, U' , I2, 2P l in p ngun, in p ti tiu th, gim in p trn ng dy, dngin hiu dng v cng sut hao ph trn ng dy lc sau.

Ta c:10

1'

10

1

100

1

1

2

2

1

2

1

2

U

U

I

I

I

I

P

P

Theo ra:1

U = 0,15.U10

15,0' 1

UU (1)

V u v i cng pha v cng sut ni tiu th nhn c khng i nn:2 1

1 1 2 2

1 2

U IU .I = U .I = = 10

U I U2 = 10U1 (2)

(1) v (2):

1 1

12 1 1

U = U + U = (0,15 + 1).U

0,15.U 0,15U' = U + U' = 10.U + = (10 + ).U

10 10

Do :

0,1510+

U' 10= = 8,7U 0,15+1


23/55


V

A BR,LC

E

hnh 2

Bi 7: (Tnh Thanh Ha, nm hc 2009 - 2010)Cho mch in xoay chiu nh hnh v 2. in p hai u

mch l uAB =

6

100cos260

t (V). iu chnh gi tr

in dung C ca t in vn k V ch gi tr cc i vbng 100V. Vit biu thc in p uAE.

HNG DN GII:V gin vc t biu din phng trnh

CLRAB UUUU

trc gc l I

Trn gin vc t ta c constZ

R

IZ

IR

U

Utan

LLL

R

p dng nh l hm sin vi OMNta c

sin

MN

sin

ON hay

sin

U

sin

U CAB

.sin

sin

UU ABC

UC max khi 1sin 090 : tam gic MON vung ti O

p dng nh l pitago cho OMNta c

80V60100UUU 222AB2

CmaxAE v UAEnhanh pha hn UAB 1 gc 900

Vy biu thc UAE l

80 2 cos 1003

AE

u t

(V)

Bi 8: (Tnh ng Nai, nm hc 2010 - 2011)p t mt in p xoay chiu n nh vo hai u on mch in nh hnh v. Bit 1/ ( )L H ;R v C c th thay i c.

a) Gi c nh gi tr C = C1 v thay i R , ta ccc kt qu sau :

+ S ch ca ampe k A lun bng 1A+ Khi R = R1 =100 th uAB v cng

dng in i trong mch chnh cng pha. Tnh C1v xc nh s ch ca cc ampe k lc nyb) Tm gi tr ca C phi tho khi iu chnh R ; in p tc thi uAB hai u mch in lun

lch pha vi cng dng in trong mch chnhBi 9: (TP HCM, nm hc 2010 - 2011)Mt on mch xoay chiu gm cun cm c in tr thunRv t cm L mc ni tip vi t in c in dung C thayi c nh hnh. in p hai u on mch cdng 2 cos 2ABu U ft , Uvafkhng i. Khi C = C1, in

p hiu dng hai u cun cm l Ud, hai u t in l1CU . Khi C = C2 = 2C1, in p hiu dng ha

u cun cm l Ud= Ud, hai u t in2C

U = U. Tm Udv1C

U theo U.

Bi 10: (Tnh Thanh Ha , nm hc 2008 - 2009)Mt on mch in gm 3 nhnh mc song song. Nhnh th nht l mt t in c dung khng Z Cnhnh th hai l mt cun dy thun cm c cm khng ZLv nhnh th ba l mt in tr R. Gi I

L,

A1

A2

A

O

M

N

UAE

UAB

URI

UL

UC


24/55

NGUYN VN TRUNG :0915192169IC, IL, IRl cng dng in hiu dng trn mch chnh v cc mch r tng ng, Z l tng tr caon mch. Hy chng minh cc h thc sau :

22 2

R L C I I I I v

2

2 2

1 1 1 1

C L Z R Z Z

HNG DN GII:+ Gi s u = U0cost. Ta c:

iR = I0Rcost ; iC = I0Ccos(t +2

) ; iL = I0Lcos(t -

2

)

+ Gin vc t (2 dao ng cng phng): iC+ iL=(I0C - I0L)cos(t +2

)

+ Vy i = iR+ iC+ iL = I0Rcost + (I0C - I0L)cos(t +2

). Hai dao ng ny vung gc nn I2 = IR

2+ (IC

- IL)2 (1) pcm.

+ Vi I = U/Z t (1) suy ra2

2 2

1 1 1 1

C L Z R Z Z

pcm.

Bi 11: (Tnh Tha Thin Hu, nm hc 2010 - 2011)Cho on mch RLC khng phn nhnh, cun dy L thun cm, in tr ca ampe k rt nh. t mtin p xoay chiu c gi tr hiu dng UAB= 150 V khng i vo hai u on mch, th thy h scng sut ca on mch AN

bng 0,6 v h s cng sut ca on mch AB bng 0,8.a,Tnh cc in p hiu dng UR, UL v UC, bit on

mch c tnh dung khng.b, Khi tn s dng in bng 100 Hz th thy in p hai u on mch AB lch pha /2 so vi in

p gia hai u on NB v s ch ca ampe k l 2,5A. Tnh cc gi tr ca R, L, C.Bi 12: (Tnh ng Thp, Trng THPT TP Cao Lnh ngh)

Cho mch in nh hnh v:Mt in tr thun R,mt tin C,hai cun cm l tng L1 = 2L, L2 = L v cc khaK1,K2 (RK= 0) c mc vo mt ngun in khng i (csut in ng ,in tr trong r = 0).Ban u K1 ng, K2ngt. Sau khi dng in trong mch n nh, ngi ta ng K2,ngt K1. Tnh hiu in th cc i t v IL2 max. ?HNG DN GII:

+K1ng, K2ngt, dng in n nh qua L1:R

I

0

K1ngt, K2ng: V 2 cun mc song song

u L1 = u L2 = uAB ==> - 2L (i1I0) = Li2 2L (I0i1) =Li2 (1) (0,5)

222

2

2

2 2222

1

2

0 CULiLiLI (2) (0,5)

IC = i1i2 UCmax IC = 0 i1 = i2= I (3) (0,25)(2) v (3) 220

2

2

2

1

2

0

2

0 3222 LILILiLiLICU (0,25)

(1) LILiLiLI 322 120 3

2 0II (0,25)

AA N B

R L C


25/55


C

L

RC

LIULICU

3

2

3

2

3

200

2

0

2

0

(0,25)

+Khi t in phng ht in th I1 v I2cc i

22

2

2

22

max2

2

max1

2

0 LILILI (4) (0,25)

(1) 2L (I0I1max) = LI2max I0I1max = 21

I2max (5) (0,25)(4) 2max2

2

max1

2

0 22 LILILI 2

max2

2

max1

2

0 22 III

2max2max10max10 ))((2 IIIII I0 + I1max = I2max (6) (0,25)

(5)(6) I2max = 03

4I =R3

4 (0,25)

Bi 13:Cho mch in c s nh hnh v bn.Cho bit: R1 = 3; R2 = 2; C = 100nF ; L l

cun dy thun cm vi L = 0,1H; RA 0;

21 VVRR . Ampe k v von k l ampe k v

von k nhit.t vo hai u A, B hiu in thuAB = 5 2 cost (V).

1. Dng cch v gin vect Frexnen tm biu thc ca cc hiu in th hiu dng

1RU , UCv cng dng in hiu dng qua R2theo hiu in th hiu dng U = UAB, R1, R2L, C v .

2. Tm iu kin ca ampe k c s ch ln nht c th. Tm s ch ca cc von k V1 v V2khi .

3. Tm iu kin ca cc von k V1 v V2c s ch nh nhau. Tm s ch ca ampe k v ccvon k khi .

HNG DN GII:1) MBAMAB UUU ; (1)

UMB = IR2; (2)

UAM = IR1. R1= IL

C

1L ; (3)

Chiu (1) ln 0x v 0y c:UAB .X = IR2cos = IR2.IL/I = R2IL;UAB.y = IR2sin + UAM

UAB.y = IL

C1L (R1+R2)/R1

Do U2 = 2 y.AB2

X.AB UU =

2

LI

22

21

21

2

1

21

C

1L

RR

RR

R

RR

t

21

21

RR

RRR (*), ch ti (3) c

A B

C

MAV1

V2

R1

R2

L

UAB

UR2

x

y

UAM

I

IL

IR1

UMB

UL

UC

0


26/55


IL =2

22

C

1LR

1

R

UR

; IR1 =2

221

C

1LR

C

1L

RR

UR

I =2

2

2

2

1

21

2

1

2

1

1

CLR

C

LR

RRURII RL (4)

UR1 = IR1R1 =2

22

C

1LR

C

1L

R

UR

(5)

UC = IL/C = 22

2

C

1LRC

1

R

UR

(6)

Vi R tnh bi (*)2) Xt biu thc ca I, ta thy biu thc di du cn (k hiu l y) l

22

22

1

22

22

1

)C/1L(R

RR1

)C/1L(R

)C/1L(Ry

Bi R1>R, y t cc i, tc l s ch ampe k kh d ln nht khi s/rad10LC

1 4 .

Khi theo (4), (5) v (6): Imax=U/R2=5/2=2,5(A)

S ch ca V2 l: UC=U/R2C= ))(!V(250010.10.2

547

3) Ta c UV1=UV2--> UR1 = UC --> L-1/C=1/(C)

--> s/rad10.41,1LC

2 4 .

222

222

1

21 L25,0R

L25,0R

RR

RUI

vi );A(1I)(10.2

C

L2L),(2,1

RR

RRR

3

21

21

).V(3)L5,0(R

L

R2

URUU22

2

C1R

Bi 14: (Tnh Tha Thin Hu, nm hc 2007- 2008)Mt on mch in xoay chiu AB gm mt in tr

thun, mt cun cm v mt t in ghp ni tip nh trnhnh v. Hiu in th hai u on mch c dng :

ABu = 175 2sin100t (V). Bit cc hiu in th hiu dng AM MNU = U = 25V , NBU = 175V . Tm h scng sut ca on mch AB.

HNG DN GII:

R C

A BM N


27/55


- Theo gi thit c :AB

175 2U = = 175

2(V).

- Gi r l in tr ni ca cun cm. Gi s r = 0, ta c :2 2 2 2

AB R L CU = U + (U - U ) = 25 + (25 - 175) = 25 37 175 r > 0.

- Ta c : 2 2 2 2MN L rU = U + U = 25 (1)

- Mt khc ta c :2 2 2 2 2 2 2

AB R r L C R R r r L C L CU = (U + U ) + (U - U ) = U + 2U U + U + U + U - 2U U 2 2 2R R r MN C L CU + 2U U + U + U - 2U U

2175

L r7U - U = 25 (2)

- Gii h phng trnh (1) v (2) :LU = 7(V) v rU = 24(V)

- H s cng sut ca on mch : R rAB

U + U 25 + 24cos = = = 0,28

U 175

Bi 15: (Tnh Thi Nguyn, nm hc 2009 - 2010)Cho mch in xoay chiu nh hnh v (h.1). Hiu in th xoaychiu hai u mch c biu thc: uAB = U0.sin100t (V), b qua

in tr cc dy ni. Cc hiu in th hiu dng: UAN = 300 (V),UMB = 60 3 (V). Hiu in th tc thi uAN lchpha so vi uMBmt gc

2

. Cun dy c h s t cm

1L

3

(H) vi in tr r, in dung ca t in

33.10

C =16

(F).

a/ Tnh in tr r. Vit biu thc hiu in th tc thi gia hai im A, N.b/ Thay i R n khi cng sut tiu th trn n cc i. Tnh gi tr ca R lc ny.

HNG DN GII:

a) Tnh r:L C

100 1 160Z .L ; Z .

C3 3

- Ta c: AN + MB = /2. Suy ra: ANMB

1tg

tg

, t : L

C L

Z r

R r Z Z

.

Vy : ZL.(ZCZL) = r.(R + r), hay: L C L r R rU (U U ) U (U U ) (1)

Mt khc: 2 2 2AN r R LU (U U ) U (2)

V:2 2 2

MB r L CU U (U U ) (3)

T (1), ta rt ra:2

2 2LR r C L2

r

U(U U ) (U U )

U (4)

Thay (4) vo (2):2 22 2 2 2 2L L

AN C L L C L r2 2

r r

U UU (U U ) U (U U ) U

U U (5)

Thay (3) vo (5), ta c:

2

2 2LAN MB

r

UU .U

U

Bin i ta c: Lr

U 300 5

U 60 3 3 , suy ra: r = ZL.

3 100 320

5 5 3 (6)

Biu thc uAN:- Ta c: AN 0AN uANu U sin(100 t ) .

R

(h .1)

L , r C

A BM N


28/55


+ Bin : U0AN = 300 2 (V)+ Pha ban u:

AN i AN u AN ANu (7)

M: L CZ Z

tgR r

(8)

T mc a/ ta c: R + r = L C LZ (Z Z )

r

=

100 160 100

3 3 3 10020

Suy ra: R = 80 (9)Thay vo (8), ta tnh c: tg = - 0,346 = -190 (10)

Ta li c: 0LAN AN

Z 100 1tg 30

R r 3 3100

(11)

Vy:AN

0 0 0 4919 30 49 (rad)u

180

(12)

- Biu thc:AN

49u 300 2 sin(100 t )(V)

180

(13)

Lu : HS c th gii bng gin vect.

b/ Cng sut tiu th trn R: PR2 2

2

2 2 2 2

L C L C

U R UI R

(R r) (Z Z ) r (Z Z )R 2r

R

Theo C si: PRmax khi2 2

L CR r (Z Z ) = 40.

Bi 16: (Tnh Bn Tre, nm hc 2008 - 2009)Mch in xoay chiu gm 3 phn t : in tr

thun R, cun thun cm c t cm L v t c in dungC mc ni tip nh hnh v (1).Bit uANnhanh pha so viuMB v MBAN tan2tan

Nu mc mch li nh hnh v (2) th cng hiu dng qua mch chnh lbao nhiu? Bit dung khngZC = 50v in p hiu dng hai u mch l 100V.

P NDo mch c ba phn tr R, L, C m uAN nhanh pha so

vi uMBth on mch AN gm c R, L v on mch MBgm c R v C x l cun thun cm L, Y l in tr

thun R v Z v t C. (0,5)

T MBAN tan2tan CLCL ZZ

R

Z

R

Z2

2 (0,5)

Hnh (2) c v li nh sau:

u

X Y Z A M N

B

(hnh 1)

u

Z A

B

(hnh 2)

X

Y

D

u

X Y Z

A M N

B

u

A

B

iR

L

R

D

C

IL

i


29/55

NGUYN VN TRUNG :0915192169 Gin vc t cho mch ny l: Ta c: cos2222 DBADDBAD UUUUU

I

IUUUUU LDBADDBAD 2

222

m

LLDBLCC

AD ZIUZZZI

U.;2;

nn 22222 ADDBDBAD UUUUU

U = UAD AZ

U

Z

UI

CC

AD 2

Bi 17: (Tnh Thanh Ha, nm hc 2007- 2008)Mt on mch in gm ba phn t R = 30, L = 0,2H, v C = 50F mc ni tip vi nhau v nitip vo 2 ngun in: Ngun in mt chiu U0= 12V v ngun in xoay chiu U = 120V, f = 50Hz.a) Tnh tng tr ca on mch v cng dng in i qua on mch.

b) Tnh lch pha gia hiu in th hai u on mch v dng in trong mch. Nhn xt v ktqu tm c.c) V gin vc t cc hiu in th gia hai u ca R, ca L, ca C v ca ton mch.d) Cun cm v t in y c vai tr g ? C th b i c khng ?HNG DN GII:

a) Ta c ZL = L = 62,8 ; ZC= 1/C = 63,7 . Suy ra Z = 2 2 2L C R Z Z = 30,01. Dng mtchiu khng qua t in nn I = U/Z 4A.

b) lch pha gia h..t v dng in ton mch l cos = RZ

1.

+ Suy ra 0. Trong mch c cng hng.c) Ta c: UR= IR 120V = U; UC = IZC = 255V; UL = IZL = 251V.Cc d liu trn cho gin vc t gm cc d liu tnh c t trn cng

thm hiu in th mt chiu U0. Hnh v bn.d) T C c tc dng ngn dng mt chiu i qua R. T C lm cho U v I lch pha. Cun L lm chomt s lch pha. Vi vai tr C, Lnh trn, khng th b i mt trong hai v hoc ng thi c hai.

Bi 18: (Tnh Bnh Thun, nm hc 2007- 2008)Hai u A, B ca mch in ni vi mt ngun in xoay

chiu c hiu in th hiu dng khng i U AB = 100 V v ctn s f thay i c. Hai vn k xoay chiu V 1 v V 2 c intr rt ln (coi nh ln v cng), ampe k A v dy ni c in trkhng ng k.

1. Mc vo hai cht A v D mt t in c in dung C vmc vo hai cht D, E mt cun cm c t cm L, in tr R v cho tn s f = f0 = 250 Hz

Ngi ta thy V 1 ch U 1 = 200 (V), vn k V 2 ch U 2 = 100 3 (V), ampe k ch 1 (A). Tnh ccgi tr C, L, R ca mch.

DBU

ADU

U

I

RI

LI

0 URU+U0 I

UL+U0

UC -UL+U0


30/55

NGUYN VN TRUNG :09151921692. Thay hai linh kin trn bng hai linh kin khc (thuc loi in tr, t in, cun cm)th s

ch ca cc dng c o vn nh trc v hn na khi thay i tn s f ca ngun in th s ch caampe k gim i.

a. Hi mc cc linh kin no vo cc cht ni trn v gii thch ti sao ? Tm cc gi tr R /

L/, C /(nu c) ca mchv lch pha gia u AD v u DE . b. Gi nguyn tn s f = f0 = 250 Hz v mc thm hai linh kin na ging ht hai linh kin

ca cu 2avo mch. Hi phi mc th no tha mn; s chca cc vn k vn nh trc, nhngs ch ca ampe k gim i mt na. Trong trng hp , nu thay i tn s f ca ngun in th sch ca ampe k thay i nh th no ?HNG DN GII:1. Ta c gin vc t nh hnh v :* Nhn xt :

- Dng i nhanh pha2

so vi u AD v chm pha

2

so vi u DF .

- Tam gic ADE c cc cnh 200 (V), 100 3 (V) v 100 (V)nn ADE l na tam gic u

+ U AE = I.Z = 1.Z Z = 100 ( ). 0.5 im

+ Sin A =AE

EF

U

U U EF = U AE .Sin A = 100.Sin60

0 = 100.2

3

I.R = 100.2

3 R = 50 3 ( ).

0.5 im

+ Ta c : CosD =DE

DF

U

U U DF = U DECosD = 100 3 .

2

3= 150 (V)

I.Z L = 150 Z L = 150 ( ).

L.2f0 = 150

L =500

150=

3,0(H).

+ U AD = I.Z C Z C = 200 (

M C =02

1

fZC =

500.200

1=

510 (F).

2. a :Tm cc gi tr R/, L/, C/(nu c) ca mch v lch pha gia u AD v u DE .* Khi tng hoc gim tn s f th dng in u gim,chng t dng in cc i tn s f0 , nga l c cnghng. Vy phi mc cun cm vo hai cht A, D vmc t in vo hai cht D, E c cng hng th tngtr rt v in tr R/.- Ta c gin vct nh hnh bn :

+ R/ =I

UAE =1

100= 100 ( ).

+ ZL / = ZC / =I

U2 =1

3100= 100 3 ( ).


31/55


+ L/ =02

/

f

ZL

=

500

3100=

5

3(H).

+ C/ =02

1

/ fZC

=

500.3100

1=

.35

10 4(F)

* lch pha gia u AD v u DE :

- Dng in nhanh pha 2 so vi u DE v chm pha 3

so vi u AD nn lch pha gia u AD

v u DE l6

5.

* Nu i v tr cun cm v t in th ta tr li s cu 1(khng c hin tng cng hng xyra).

2.b.Gi nguyn tn s f = f0 = 250 Hz v mc thm hai linh kin na ging ht hailinh kin ca cu 2avo mch. Hi phi mc th no tha mn; s ch ca cc vn k vnnh trc, nhng s ch ca ampe k gim imt na. Trong trng hp , nu thay i tns f ca ngun in th s ch ca ampe k thay i nh th no ?

* dng in gim i mt na ta mc cc linh kin theo s nh hnh v : 0.5imTheo s ny ta c :R = 2R /

L = 2L / Z L = 2L/2 0f

C =2

/C Z C =

0

/2

2

fC

V ZL / = ZC / nn trong mch xy ra cng hng Nu thay i tn s f th dng in s gim.Bi 19: (Tnh Gia Lai, nm hc 2008 - 2009)Mch in c s nh hnh v.Cun dy thun cm L. Ngi ta thay i L v C cng sut mch tun theo biu thc: 2 .L CP K Z Z .

a)Khi1

( )L H

th 2 4K , dng in trong mch cc i.

Tnh C v R.

b)Tnh lch pha gia uAE v uBD khi Imax. Tm lin h gia R, C, L I = K. Lc lch phagia uAE v uBDbng bao nhiu?HNG DN GII:

a)+ Ta c :1

.2 . 2 50 100L Z L f

+ Khi 2 4 4 L CK P Z Z (1)

+ V mch RLC ni tip c Imaxnn cng hng xy ra 100L CZ Z (2)

Do :4

1 1 10( )

100 100CC F

Z

+T (1) v (2), c : 4 400(W)LP Z

+ Mt khc : 2P R I , vi axmin R

m

U UI I

Z nn

2 2 2100

25400

U UP R

R P

A

R

D

L C

B E

f=50Hz~

U=100V


32/55


iLU

CU

AEU

BDU

O2

1

A BC

C

M

R

R

K

D

Bi 20: (Tnh Gia Lai, nm hc 2008 - 2009)Cho mch in xoay chiu (hnh v).

Bit in p n nh gia hai im A v B l

120 2 sin ( )ABu t V ;

1mR

C( m : tham s).

a) Khi kho K ng, tnh m h scng sut ca mch bng 0,5.

b) Khi kho K m, tnh m in p uABvung pha vi uMB v tnh gi tr in p hiu dng UMB.HNG DN GII:

b)+ Gin vc t v c :

+T gin vc t suy ra :1 2

Vi : 01 1

R

100tan 4 76

25

L LU Z

U R

+Suy ra : 01 2

38152

45 AE BDu u

+ Ta bit :2

2

L C

P R I

P K Z Z

nn khi I = K, ta suy ra :

2 2

L C L C

L R Z Z R Z Z R

C

+Lc ny c:1

1 2 2

2

tan

tan tan 1

tan

L

L C

C

Z

Z ZR

Z R

R

+Suy ra:1 2

2 AE BDu u

a)Tnh m os 0,5c +V khi K ng: mch in cu to : C nt (R // R) .

+Lc :2

2 2

2 2

12os2 4

( )2

C

C

R

Rc R Z

RZ


33/55


+Suy ra : 2 23 3 3 3

4 2 2 2C C

Z R Z R mR R m

b)+Nhnh (1) :

1 1 12 2 2 2

sin ; os ; 0C

C C

Z Rc

R Z R Z (1)

1 l gc lch pha ca DBU so vi 1I 1( )2

+Trong tam gic vect dng ta c : 2 2 21 2 1 2 12 os I I I I I c (2)

V 2 21 2 DB C U I R Z I R (3)

+Suy ra 212 2

C

RII

R Z

+Thay vo (2) c :22

2 2 2 22 22 2 2 2 2 2

2C C C

RIR R I I I

R Z R Z R Z

2 2 2 2

2 22 22 2 2 2

4 4( )C C

C C

R Z R Z I I I I

R Z R Z (4)

+p dng nh l hnh sin cho tam gic dng, ta c: 21

sin sin( )

I I

(5)

+p dng nh l hnh sin cho tam gic th, ta c:

11

sin ossin( )

2

DB AD ADU U U

c(6)

+T (5) v (6), suy ra: 2 1 1sin sin( ) cosDB

AD

I U

I U

2 2

2 2 2 2

C

CC C

Z I I R R

I IZ R Z R Z

+Suy ra: 1C Z R mR R m

+Khi m = 1 th ZC = R, ta c:

1

1 2 1os os( ) os os( )

2 2

MB

AB AD DB C

U I R

U U c U c IZ c I R c

(1)

MBU

1I I

2I DBU

DMU

ABU

ADU

1 O

( )


34/55


Bi 21: (Tnh An Giang, nm hc 2010 - 2011)

Mt mch in XC gm mt cun dy thun cm c L1mc ni tip vi cun dy L2 =2

1H; in tr

trong r = 50 . in p XC gia hai u on mch c dng u = t100cos2130 (V). Cng hiudng trong mch l 1A. Phi mc thm mt t c in dung C l bao nhiu in p gia hai ucun (L2, r) t gi tr cc i.HNG DN GII:

Ta c: Z = U/I = 130 . Mt khc: 22212 )( ZZZr LL 2

222

21 )(

rZLL

2,121 LL

Khi mc thm t C vo mch, lc ny:

22*2

222

)(.. day

CL

daydayday ZZZr

UZ

Z

UZIU

in p gia hai u cun dy 2 t cc tiu, tc l trong mch c cng hng

)(

121

* LLC

ZZ CL

Thay s tm c C=12

10 3F

............................................................................

NGHIN CU DAO NG BNG PHNG PHP NNG LNGPP nng lng dng:xc lp mt s tng ng gia nng lng ca h dao ng vi nng lng camt con lc n gin nht gm mt vt nng khi lng m treo trn mt l xo c cng k. Nu biu

thc c nng li x ca h a v dng E= 22

22' xkxm hdhd

th h s dao ng iu ho

x=Asin )( t vi tn s gc :hd

hd

m

k

minh ho cho PPNL, ta hy dng n d tm chu k ca con lc n: gm mt vt nng m treo trnsi dy mnh di l. c tham s x, ta chn dch chuyn ca vt nng theo cung trn tnh t v trcn bng. ng nng ca con lc l mx2/2, tc khi lng hiu dng ng bng khi lng ca vtnng. i vi nhng lch nh ca con lc (x


35/55


=2mglsin2

2

=mgl

2

2

=l

mg

2

2x (1) , trong

l

x l gc lch ca con lc. Do , cng hiu dng

by gi ll

mgv tn s gc ca dao ng l

hd

hd

m

k =

l

g

Sau y l mt s v d minh ho c th.Bi ton 1.Mt thanh di l=40cm c un thnh na vng trn nhcc nan hoa c khi lng khng ng k . Ngi ta gn vo na vngtrn ny mt trc quay nm ngang i qua tm vng trn. Hy tm tns gc ca nhng dao ng b ca na vng trn xung quanh v tr cn

bng nu trc quay vung gc vi mt phng . Ly g=9,8 m/s2.Gii:

c tham s x xc nh v tr ca h, ta chn dch chuyn ca ccim trn thanh ra khi v tr cn bng. Khi ng nng ca h bngmx2/2, vi khi lng hiu dng ng bng khi lng ca thanh. tnh th nng, ta coi rng ta dch chuyn mt mu ca thanh c chiu di x v khi lng mx/l tu ny n u kia ca thanh (nh minh ho trong hnh v). Khi tm ca u mu dch chuyn

mt on x, tc l bin thn th nng ca thanh bng Et= gxl

mx=l

mg2

2

2x(2) (ta quy cth nng

VTCB=0). iu ny c ngha l cng hiu dng ca h l khd=2mg/l. Do tn s gc ca dao ng

lhd

hd

m

k =

l

g2=7 (rad/s)

Bi ton 2.Mt ng ch U c thit din S=10cm2cha 400g nc. Tm tn s gc cadao ng theophng thng ng ca cht lng. B qua ma st v ly g =9,8 m/s2.

Gii: Nu nc dch chuyn mt on x t VTCB th, nh trong bi ton

trc, c th coi mt ct nc nh di x v c khi lng Sx dchchuyn tnhnh ny sang nhnh kia. bin thn th nng lc l:

Et= (Sx)gx=2gS2

2x

Vi =1g/cm3 l khi lng ring ca nc. Nh vy cng hiudng l 2gS, cn khi lng hiu dng ng bng khi lng m canc c trong ng. Do tn s gc ca dao ng l:

hd

hd

m

k =

m

gS2=7 (rad/s)

Bi tp 3.Mt thanh khng trng lng c un thnh 1/3 vng trn bn knh R=5cm. Nh cc nanhoa c khi lng khng ng k, ngi ta gn cung trn ny vo mt trc quay nn ngang i qua tmvng trn v vung gc vi mt phng ca n. Ngi ta gn vo 2 u cung trn 2 vt nng nh nhau.Hy tm tn s gc ca nhng dao ngb ca cung trn xung quanh v tr cn bng. Ly g=9,8 m/s2.

Gii: lm tham s xc nh lch ca cung trn ra khi v tr cn bng, ta chn gc lch nh . Khi ng nng ca h l:

Ek=22

22

2

2

2'2

222 mR

mRmv

Tc khi lng hiu dng l mhd=2mR2( y m l khi lng ca vt nng)

xx O

xx


36/55

NGUYN VN TRUNG :0915192169S l thun tin nu ta biu din th nng ca h qua s bin thin cao trng tm ca h. D thyrng trng tm ca h nm gia 2 vt nng v cch im treo mt khong l=Rcos600=R/2, do

Et=2mgl(1-cos)=2mgl2

2

=mgR2

2

. iu ny c ngha l cng hiu dng khd=mgR v tn s gc

ca dao ng b l:hd

hd

m

k =

R

g

2= 10 (rad/s)

(Coi nh l mt bi luyn tp nh, em hy lm li bi ton trn nhng by gi chn tham s xc nhv tr ca h khng phi l gc m l dch chuyn quen thuc x=R)

Bi ton 4.Mt thanh c khi lng M=20g v chiu di l=118cm c un thnh mt na vng trnv nh cc nan hoa c khi lng khng ng k ngi ta gn na vng trn ny vo mt trc quaynm ngang i qua tm vng trn v vung gc vi mt phng ca n. Ngi ta gn vo gia thanh mtvt nng m=100g. Hy tm tn s gc ca nhng dao ng b ca na vng trn xung quanh v tr cn

bng. Ly g=9,8 m/s2 v =3.14Gii:

C th coi bi ton ny l t hp ca v d v con lc n v bi ton 1. Do khi tnh th nng cnphi s dng ng thi phng php tnh cao khi gc lch b (cng thc (1))v phng php dch

chuyn mt mu ca thanh(cng thc (2)). Khi ta c

Et=(mgR(1-cos)+2

)2(

2

2

2

222 x

l

gMmx

l

Mgx

R

mggx

l

Mx

.

V ng nng ca h bng Ek=(m+M)2

2'x

nn tn s gc ca dao ng l:hd

hd

m

k =

l

g

Mm

Mm

2=5 (rad/s)

Bi ton 5. Mt thanh khng trng lng c chiu di l=3,5m c th quay t doquanh mt trc nm ngang i qua mt u ca thanh. Ngi ta gn vo u t do

ca thanh mt vt nng khi lng m v vo trung im ca thanh vt nng khilng 3m. Hy tm tn s gc ca nhng dao ng b thanh xung quanh v tr cnbng. Ly g=9,8 m/s2.

Gii: lm tham s xc nh lch ca thanh, ta chn dch chuyn ca thanh dctheo cung trn. Nh vy dch chuyn ca vt bn trn l x/2./ Khi ngnng ca h l:

Ek=2

75,12

)2/(3

2

2'2'2' xm

xmmx : Tc khi lng hiu dng ca h l mhd=1,75m.

Th nng ca h l Et=(mgl(1-cos)+ 3mg2

5,2)cos1(

2

2x

l

mgl .

Do khd=2,5mg/l. vy tn s dao ng b ca h l :hd

hd

m

k =

l

g

7

10=2 (rad/s)

Bi ton 6.Mt thanh khng trng lng di l=50cm c th quay t do quanh mttrc nm ngang i qua mt u ca thanh. Ngi ta gn vo u t do ca thanhmt vt nng khi lng m=0,5kg v vo trung im ca thanh mt l xo nmngang c cng k=32N/m. Khi thanh v tr thng ng, l xo khng bin dng.Hy tm tn s gc ca nhng dao ng b ca h xung quanh v tr cn bng. Lyg =10m/s

2.

Gii:

O

3m x/2

m x

O

k

xm


37/55

NGUYN VN TRUNG :0915192169 lm tham s xc nh lch ca h, ta chn dch chuyn ca vt nng theo cung trn. Vi dch chuyn , bin dng ca l xo l x/2. Khi th nng ca h l

Et=l

mg

2

2x+k

242

)2/(22 xk

l

mgx

. Hin nhin, khi lng hiu dng ca h ng bng khi lng

ca h. Khi tn s dao ng ca h l:hd

hd

m

k =

m

k

l

g

4 = 6 (rad/s)

Bi ton 7.Hai u mt thanh khng trnglng c chiu di l=10cm, ngi ta gn 2 qu cu nh, miqu c khi lng m=9g. Bit rng 2 qu cu tch in tri du v ln ca cc in tch bngq=3 C v ton b h thng c t trong mt in trng u c cng E=600V/m v c hngsong song vi thanh khi VTCB. Hy tm tn s gc ca nhng dao ng b ca h trn xung quangVTCB. B qua tc dng ca trng lc.

Gii:V tng cc lc tc dng ln h bng khng, nn h quy chiu gn vitm qun tnh ca h l mt h quy chiu qun tnh. Do ta c thgxem im gia ca thanh l ng yn. lm tham s c trng cho lch ca h khi v tr cn bng, ta chn gc quay ca h.

ng nng ca h l: Ek=2222

)2/( 2'22' mllm . iu ny c ngha l khi lng hiu dng ca h

bng mhd=ml2/2. bin thin th nng khi thanh quay mt gc b bng cng m lc in trng

thc hin ly vi du ngc li.

Et= 2qE cos12

l

=qEl2

2

. iu ny c ngha l cng hiu dng ca h bng khd=qEl. T suy

ra tn s gc ca dao ng ca thanh l:hd

hd

m

k =

ml

qE2=2 (rad/s)

Bi ton 8.Mt ng di c un thnh gc vung c t sao cho

mt nhnh thng ng. Trong nhnh thng ng ngi ta d mt sidy di l=90cm sao cho mt u ca n nm im un ca ng.Hi qua thi gian bao lu (tnh ra mili giy) sau khi bung tay ra thn trt c mt na vo nhnh nm ngang ca ng? B qua mast. ly g=10m/s2 v =3,14.

Gii:Nu ti thi im t phn dy cn li nhnh thng ng c chiu dix th nng lng ca h c dng:

E=22

'

22

' 222 x

l

mgmxxg

l

mxmx

Trong d mx/l l khi lng ca on dy nm trong nhnh thng ng, v x/2 l caotrng tm can. V biu thc ny ng nht vi biu thc ca nng lng ca dao ng iu ho vi tn s gc

l

g , v vn tc ban u bng khng, nn chuyn ng xy ra theo quy lut x=lcos t(v ti thi

imban u x0=l). Cng thc ny ch c hi lc trong chu k, khi m x cha =0, ngha l khi ton bsi dy cha nm trn ven trong ng nm ngang. tnh thi gian cn tm, ta ch cn t vo cng

thc trn x=l/2. Gii phng trnh ta nhn c: T =g

l

33

=314 ms

Bi tp

-q E

q

x


38/55

NGUYN VN TRUNG :0915192169Bi ton 9.Mt thanh di l=20 cm dc un thnh mt cung trn, c chiu di bng 1/6 vng trn.Nh cc nan hoa c khi lng khng ng k ngi ta gn cung trn ny vo mt trc quay nmngang i qua tm vng trnv vung gc vi mt phng ca n. Ngi ta gn vo 2 u cung trn 2vt nng nh nhau. Hy tm tn s gc ca nhng dao ng b ca cung trn xung quanh v tr cn

bng. Ly g=9,8 m/s2.

S:l

g = 7 (rad/s)

Bi ton 10.Mt bnh xe mnh c khi lng M=400g vi cc nan hoa khng trnh lng c th quayt do quanh trc nm ngang. Trn bnh xe ngi ta gn mt vt nng c khi lng m=100g. Hy tmtn s gc ca nhng dao ng b ca h xung quanh VTCB. Bit bn knh ca bnh xe R=50cm. Lyg=10m/s

2.

S:R

g

Mm

m

=2(rad/s)

Bi ton 11.Mt thanh khng trng lng di l=2,5m c un im gia thnh 2 nhnh lp vinhau mt gc 1200. Hai u ca thanh c gn 2 vt nng nh nhau. Bit rng thanh c ng chun vo mt chic inh ng trn tng. Hy tm tn s gc ca nhng dao ng b ca h xung quang

v tr cn bng. ly g=10m/s2.

S:l

g =2(rad/s)

Bi ton 12.Mt thanh khng trng lng di l=40cm c th quay t do xung quanh mt trc nmngang i qua mt u thanh. Ti im gia ca thanh ngi ta gn vt nng c khi lng m=0,5kg,cn u di ca thanh c gn vi mt u ca mt l xo nm ngang, cn u kia ca l xo cgi c nh. Bit rng cng ca l xo l k=30N/m v khi thanh v tr thng ng l xo khng bindng. Hy tm tn s gc ca nhng dao ng b ca h xung quang VTCB. Ly g=9,8m/s2.

S:m

k

l

g

4

2 =8 (rad/s)

Bi ton 13.Mt thanh di l=40cm nng m=0,5kg c un thnh mt na vng trn. Nh cc nanhoa c khi lng khng ng k ngi ta gn cung trn ny vo mt trcquay nm ngang i qua tm O ca vng trn v vung gc vi mt phng can. Mt u ca thanh c gn vo u di ca mt l xo thng ng, utrn ca l xo c gi c nh (nh hnh v). Bit cng ca l xok=16N/m v khi thanh VTCB l xo khng bin dng. Hy tm tn s gcca nhng dao ng b ca cung trn xung quanh VTCB. Ly g=9,8 m/s2.

S:m

k

l

g

2 =9 (rad/s)

Bi ton 14.Ngi ta d u trn ca mt sn xch mnh c chiu di l=45cm nm trn mt mtphng nghing lp vi mt phng ngang mt gc =300. Hi qua thi gian bao lu (tnh ra mili giy)sau khi bung u dy xch ra, th dy xch hon ton ri khi mt phng nghing? Bit rng thiim ban u, u di ca dy xch chm mp di ca mt phng nghing. Ly g=10m/s2 v=3,14.

S: t=

sin2 g

l=471 (ms)

p dng bt ng thc Bunhia Cpski:

O


39/55


Bi ton 1:Hai chuyn ng trn AO v BOcng hng v O vi 012

; 303

vv . Khi khong cch

gia hai vt cc tiu l dminth khong cch t vt mt n O l '1 30 3( )d cm . Hy tnh khong ccht vt hai n O.

BI GII

Gi d1, d2 l khong cch t vt mt v vt hain O lc u ta xt ( t = 0 ).p dng nh l hm sin ta c:

' '

1 2 1 1 2 2

sin sin sin sin sin sin

d d d v t d vd d

.

V 12

3

vv nn ta c:

1 1 2 1

0

3

sin30 sin 3sin

d v t d v t d

.

p dng tnh cht ca phn thc ta c:

1 1 2 1 2 1 1 1 2 13 ( 3 ) ( ) 3

sin 3sin 3sin sin 3sin sin

d v t d v t d v t d v t d d

2 1

0

3

sin30 3 sin sin

d dd

Mt khc, tac:0 0sin sin(180 ) sin( ) sin(30 ) 0 0 03sin 3sin(30 ) 3(sin30 cos cos30 sin )

3 3cos sin

2 2

2 1

0

3

sin30 3 1 1cos sin sin

2 2 2

d dd

0

2 1 2 1( 3 )sin30 3

3 1 3 cos sincos sin

2 2

d d d d

d

Vy 2 1 2 13 3

3cos sin

d d d d d

y

.

Khong cch gia hai vt dmin ymaxvi y = 2( 3 cos sin ) p dng bt ng thc Bunhia Cpski:

2 2 2 2 2( 3 cos sin ) (( 3) 1 ).(cos sin ) 2

ymax= 203 cos cot 3 30

1 sing

v 0120

Lc :' ' 0

' ' '1 22 1 10 0 0

sin120. 3 90( )

sin30 sin120 sin30

d dd d d m

Vy, khong ccht vt hai n O lc ny l:d2= 90(m)

Bi ton 2:Cho c h nh hnh v:Cho bit: H s ma st gia M v snl k2.H s ma st gia M v m l k1.Tc dng mt lc F ln M theo

phng hp vi phng ngang mt gc . Hy tm Fmin m thot khiM.tnh gc tng ng?

BI GII

A

OB

d d

d

F

Mm

O

y

1P

F

21ms

F 12ms

F

1N

2N

x


40/55


+ Xt vt m:1 1 21ms

P N F ma (1).

Chiu ln OX: Fms21= ma 211mn

Fa

m

Chiu ln OY: N1P1 = 0 N1 = P1 Fms21= k1.N1 = k1.mg

11 1

k mga k gm

. Khi vt bt u trt th th a1 = k1mg.

+ Xt vt M:2 1 2 12 2

( )ms ms

F P P N F F M m a .

Chiu ln trc OX: 12 2cos ( )ms msF F F M m a 12

2

cosms ms

F F Fa

M m

Chiu ln OY:1 2 2 2 1 2sin ( ) 0 sinF P P N N P P F

Ta c:12 1ms

F k mg

2 2 2 1 2( sin )msF k N k P P F

1 2 1 22

cos ( sin )F k mg k P P F a

M m

Khi vt trt1 2

a a 1 2 1 21

cos ( sin )F k mg k P P F k g

M m

1 2 1 2 1 2( ) (cos sin ) ( )k g M m F k k mg k P P

1 2 1 2 1 2 1 2

2

( ) (2 ) ( ) (2 )

cos sin

k k Mg k k mg k k Mg k k mgF

k y

Nhn xt: Fmin ymax. Theo bt ng thc Bunhia Cpski:2 2 2 2 2 2

2 2 2(cos sin ) (1 )(cos sin ) 1 y k k k 2

max 21y k .

Vy 1 2 1 2min

22

( ) (2 )

1

k k Mg k k mgF

k

Lc : 22

sin

cos 1

ktg k

p dng tam thc bc hai:Bi ton 1:Mt con kin bm vo u B ca mt thanh cng mnh AB c chiudi L ang dng ng cnh mt bc tng thng ng. Vo thi im m u Bca thanh bt u chuyn ng sang phi vi vn tc khng i v theo sn ngangth con kin bt u b dc theo thanh vi vn tc khng i u i vi thanh.Trong qu trnh b trn thanh , con kin t c cao cc i l bao nhiu ivi sn? Cho u A ca thanh lun t ln sn thng ng.

BI GIIKhi B di chuyn mt on s = v.t th con kin i c mt on l = u.t. cao m con kin t c:

sin sinh l ut vi2 2 2

sinL v t

L

2 2 2 4.

u uh L t v t y

L L

Vi y = 2 2 2 4. L t v t t X = t2 2 2 .y v X L X Nhn xt: max max.h y y l tam thc bc hai c a = - v

2< 0

A

B

h

B

u


41/55

NGUYN VN TRUNG :0915192169 ymaxti nh Parabol

2 4 4

max max 2 24 4( ) 4

L Ly y

a v v

4

max 24

Ly

v ti

2

22 2

b LX

a v

Vy cao m con kin t c l : max max.

2

u u L

h yL v p dng gi tr cc i ca hm s sin v hm s cosin:

Bi ton 1:Hai vt chuyn ng t A v B cng hng v im O vi cng vn tc . BitAO = 20km; BO = 30km; Gc 060 . Hy xc nh khong cchngn nht gia chng trong qu

chuyn ng?BI GII

Xt ti thi im t : Vt A A

Vt B B

Khong cch d = AB

Ta c: sin sin sin

d AO vt BO vt

10

sin sin sin sin sin

d BO AO

10

sin2cos .sin

2 2

d

vi 0120

0

0

10sin 60 5 3

2cos60 .sin sin2 2

d d

Nhn xt: dmin (sin ) 12

min 5 3( )d cm

Bi ton 2:

Cho mch in nh hnh v: Cho bit:0.9

( )L H

, UMN

khng i, C thay i, RA = 0, RVrt ln, tn sca dng in f = 50Hz ; r = 90( ). Hy chng t rngkhi iu chnh C hiu in th trn cc vn k lch pha

nhau mt gc2

th UCt gi tr cc i.

BI GIIMch in c v li :Ta c : 90( )LZ L

+ Gianr vc t:T gin vc t ta c:

+ 1 114

L L

r

U Ztg

U r

.

+ 1

1

.sin( )

sin sin( ) sin

MN C MN C

U U UU

AA O

B

B

C

L,

M

N

B

V1

A

V2

CL, B

NM

V1

A

V2LU

BMU


42/55


M1

2 2 4 4

11

sin( )2 sin( )

sin4

MNC MN

UU U

Nhn xt: UCcc i khi 1 1sin( ) 12

=1

Theo bi ra: Hiu in th trn cc vn k lch pha nhau2

1 2( , )2 2

BM MN U U

iu phi chng minh

5. Dng phng php o hm:Bi ton 1:Cho mch in nh hnh v:

410

200 2 cos100 ( )., 100( ); ( )2

ABu t V R C F

Cun dy thun cm v c t cm L thay i c.Tm L

UAMt gi tr cc i. Tm gi tr cc i .BI GII

Dung khng:1

200( )CZC

Tng tr : 2 2 2 2( ) ; L C AM L Z R Z Z Z R Z

Ta c : . . AM AM AM U

U I Z Z Z

2 2 2 2

2 2 2 2

2 21

AM

L C L C C C L

L L

U UU

R Z Z Z Z Z Z Z

R Z R Z

t y =2

2 2

21 C C L

L

Z Z Z

R Z

Nhn xt: UAMcc i miny y 2 2

'

2 2 2

2 (

( )

C L C L

L

Z Z Z Z Ry

R Z

. ' 2 20 0 L C L y Z Z Z R

2 24

241( )2

C C

L

Z Z RZ

hoc

2 24

02

C C

L

Z Z RZ

(loi).

Bng bin thin:ZL 0 241

+

y - 0 +y

ymin

Vy, khi ZL = 241( ) L = 0,767(H) th ymin UAMcc i.2 2

max

( 4 )482( ).

2

C C

AM

U R Z Z U

R

Bi ton 2: Cho mch in nh hnh v: 2 cosAB

u U t

M

CLRA B

M

C LRA B


43/55

NGUYN VN TRUNG :0915192169R khng i, cun dy thun cm c L khng i. T C c in dung thay i . Tm C UAMcc i?Tnh gi tr cc i ?

BI GII

2 2

..

( )

AM AM AM

L C

U ZU I Z

R Z Z

2

2 2

21

AM

L L C

C

U UU

y Z Z Z

R Z

UAMcc i khi y = ymin .

Tng t nhbi ton 1, ta tm c : Khi2 2

4

2

L L

C

R Z Z Z

th ymin v UAMcc i.

2 2

max

( 4 )

2

L L

AM

U R Z Z U

R

khi

2 2

2

( 4 L L

C R Z Z

Bi ton 3:Cho mch in nh hnh v:4

10200 2 cos100 ( ). 100( ); ( )

ABu t V R C F

Cun dy thun cm v c th thay i c t cm . Hy xcnh L hiu in th ULt cc i. Tnh gi tr cc i ?

BI GII

+ Cm khng: LZ L , dung khng1

100( )C

ZC

+ Tng tr: 2 2( )C L Z R Z Z

Ta c:2 2

. ..

( )

LL L

L C

U Z U Z U I Z

Z R Z Z

2 2

2

1 1( ). 2 . 1

L

C C

L L

U UU

y R Z Z

Z Z

+ Nhnxt: ULmax ymin, vi y l tam thc bc hai c a = R2+ZC

2> 0 nn

ymin ti nh Parabol

Ta nh2 2 2 2 2 2'

2 2

1 C C C C L

L C C C C

Z R Z R Z R Z b x Z L L

a Z R Z Z Z Z

Thay s :2 2

100 100 2( )

100.100L H

2 2

max 200 2( )C

L

U R ZU V

R

M rng: Nu L = cosnt , t C c in dung thay i tm C UCcc i ta lm tng t nhtrn v kt qu:

2 2

max

C

C

U R ZU

R

khi

2 2

LC

L

R ZZ

Z

p dng bt ng thc Csi:Bi ton 1:Cho mch in nh hnh v: Cho bit: 12V , r = 4 , R l mt bintr.Tm gi trca R cng sut mch ngoi t gi tr cc i.

BI GII

CLRA B

R


44/55


-Dng in trong mch:IR r

- Cng sut: P = I2.R = 22

.( )

RR r

2

2 22

RP

R rR r

=

2 2

22( )2

rrRR r

RR

.

t ( )r

y R R

2

2P y

Nhn xt: Pma x yminTheo bt ng thc Csi: Tch hai s khng i, tng nh nht khi hai s bng nhau => ymin

r

RR

R = r = 4 ( ) th2 2 2

max

129( )

2 4 4.4P W

r r r r

Bi ton 2:Cho mch in nh hnh v: 200 2 cos100 ( ).AB

u t V

1( )L H

,

410( ).

2C F

R thay i.

a. Tm R cng sut trn R cc i khi r = 0.

b. Tm R cng sut trn R cc i khi r = 50( ) BI GII

a. + Cm khng 100( )LZ L .

+ Dung khng:1

200( ).CZC

+ Tng tr: 2 2( )L C

Z R Z Z .

+ Cng sut : P = I2.R =2 2

2 2 2. .

( )L C

U UR R

Z R Z Z

2

2( )L C

U

P Z ZR

R

t

2( )L C

Z Z

y R R

2U

P y

+ Nhn xt: Theo btng thc csi ymin 100( )L C R Z Z , lc 2 2 2

max

200200(W)

2 2.100 200L C

U UP

Z Z

.

Vy Pma x = 200(W) khi R = 100 ( )

b. + Tng tr 2 2( ) ( )L C Z R r Z Z

+ Cng sut2 2

2

2 2 2. . .

( ) ( )L C

U UP I R R R

Z R r Z Z

2

2 2 2.

2 ( )L C

UP R

R Rr r Z Z

=

2

2 2( )

2 L C

U

r Z ZR r

R

t2 2

( )2 L C

r Z Z y R r

R

2U

Py

.

+Nhn xt: Pmax miny .

Theo bt ng thc Csi2 2

min

( )L Cr Z Zy RR

2 2( )L C R r Z Z

CL,

RA B


45/55


max2 2

2 2

2 2

( )( ) 2

( )

L CL C

C C

UP

r Z Zr Z Z r

r Z Z

2

max2 2 2 2

2 2

2 2 2 2

( ) . ( )( ) 2

( ) . ( )

L C L C

L C L C L C

UP

r Z Z r Z Z r Z Z r

r Z Z r Z Z

2

max2 2

2. ( ) 2L C

UP

r Z Z r

2

max2 2

200124( )

2.( 50 (100 200) 50)P W

Vy Pmax = 124(W) th2 2( ) 100( )

L C R r Z Z .

*M rng:Khi tnh P ca mch:+ Nu L C Z Z r th Pmax khi L C R Z Z r .

+NuL C

Z Z r th Pmax khi R = 0.

Bi ton 3: Vt m1chuyn ng vi vn tc 1v ti A v ng thi va chm vi vt m2ang nm yn

ti . Sau va chm, m1 cvn tc'

1v . Hy xc nh t s'

1

1

v

vca m1 gc lch gia 1v v

'

1v l ln

nhtmax . Cho m1 > m2, va chm l n hi v h c xem l h kn.

BI GII* ng lng ca h trc va chm:

1 1 1TP P m v

* ng lng ca h sau va chm : ' ' ' '1 2 1 1 2 2SP P P m v m v V h l kn nn ng lng c bo ton :

1S TP P P

Gi'

1 1 1( , ) ( , ).Sv v P P Ta c: '2 '2 22 1 1 1 22 cosP P P PP (1).

Mt khc, v va chm l n hi nn ng nng bo ton:2 '2 '2

1 1 1 1 2 2

2 2 2

m v m v m v

2 2 2 2 2 '2

1 1 1 1 2 2

1 1 22 2 2

m v m v m v

m m m

2 '2 '2

1 1 2

1 1 22 2 2

P P P

m m m

2 '2 '22 '2 '21 1 2 1

1 1 2

1 2 2

. . .2 2

P P P mP P P

m m m

2 '2'2 2 1 1

2

1

( )m P PP

m

(2).

T (1) v (2) ta suy ra:'

2 1 2 1

'

1 1 1 1

(1 ) (1 ) 2cosm P m P

m P m P

'

2 1 2 1

'

1 1 1 1

(1 ). (1 ). 2cosm v m v

m v m v

t'

1

1

0v

xv

2 2

1 1

1(1 ). (1 ). 2cos

m mx

m m x

max

thmin

(cos )

Theo bt ng thc Csi 2 2min

1 1 min

1(cos ) (1 ). (1 ).

m mx

m m x

Tch hai s khng i, tng nh nht khi hai s bng nhau

2 2

1 1

11 . 1 .

m mx

m m x

1 2

1 2

m mx

m m

sp

1p

2p


46/55


Vy khi'

1 1 2

1 1 2

v m m

v m m

th gc lch gia 1v v

'

1v cc i.Khi ,2 2

1 2max

1

cosm m

m

.

T TRNGBi 1.Mt dy dn cng c in tr khng ng k, c un thnh khung ABCD nm trong mtphng nm ngang,c AB v CD song song vi nhau, cch nhau mt khong l=0,5m, c t trongmt t trng u c cm ng t B=0,5Thng vung gc vi mt phng ca khung nh hnh 1. Mtthanh dn MN c in trR=0,5c th trt khng ma st dc theo hai cnhAB v CD.

a) Hy tnh cng sut c hc cn thit ko thanh MN trt u vi vn tcv=2m/sdc theo cc thanh AB v CD. So snh cng sut ny vi cng sutta nhit trn thanh MN v nhn xt. b) Thanh ang trt u th ngng tc dng lc. Sau thanh cn c th

trt thm c on ng bao nhiu nu khi lng ca thanh l m=5gam?Bi 2.Bit cm ng t gy bi mt dng in chy trong dy dn mnh,

thng ti mt im M (hnh 4): 7M 1 2

IB 10 . (sin sin )

R

Hy tnh cm ng t ti tm O ca dng in chy trong dydn mnh hnh trn bn knh R?

QUANG HNHCu 3.Cho quang h ng trc gm thu knh phn k O1v thu knh hi t O2. Mt im sng S nmtrn trc chnh ca h trc O1mt on 20cm. Mn E t vung gc trc chnh ca h sau O2 cch O2mt on 30cm. Khong cch gia hai thu knh l 50cm. Bit tiu c ca O2l 20cm v h cho nh r

nt trn mn. Thu knh phn k

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