Su Dung May Tinh Casio Giai Bai Tap Trac Nghiem

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DNG MY TNH : CASIO: Fx570ES & Fx-570ES Plus; VINA CAL Fx-570ES Plus GII NHANH mt s bi tp TRC NGHIM VT L 12!

PHN MT. NG DNG CA S PHC TRONG BI TON VT L -Dngsphctrongbitonvitphngtrnhdaongiuha -Dngsphctrongphptnghpcchmiuho. -Dngsphctrongccbitoninxoaychiu.

I. KHI NIM V S PHC:

1-S phc x lscdng x i a b

alphnthc: Re x a ; blphno: Im x b , i nvo: 2 1i

2- Biu din s phc x a bi trnmtphngphc:

r:mouncasphc, 2 2r a b . : acgumen casphc,Im

tanRe

b x

a x

3- Dng lng gic ca s phc:

(cos sin )x a bi r i * cos

* sin

a r

b r

Theo cng thcle: (cos sin ) .ix a bi r i r e A

4- Biu din mt hm iu ho di dng s phc:

Hmiuha cos( . )x A t biudinvectquaytit=0: 0| |

cos( . ) :( , )

tA OA A

x A t AOx OA

Tathy:a=Acos, b = Asin=> tit=0,biudinxbisphc: (cos sin ) . ix a bi A i Ae

Vy mt hm iu ha (xt ti t = 0) c th vit di cc dng s phc nh sau:

cos( . ) . (cos sin )t o jx A t x A e a bi A i A

Vi :

2 2

cos , sin ,tan

A a ba A b A b

a

IIVIT PHNG TRNH DAO NG IU HA:

1- C s l thuyt: (0)

(0)0

(0)(0)

coscoscos( . )

sin( . ) sin sin

t

x A ax Ax A t

vv A t v A A b

Vy

(0)

0

(0)cos( ) ,t

a x

x A t x a bi vb

2- Phng php gii: Bitlct=0c:

(0)(0)

(0)(0)cos( )

a x

Av

x x i x tvb

A

ybr O

Max

y

b A

O a x



3. Chn ch thc hin tnh s phc ca my: CASIO fx570ES, 570ES Plus,VINA CAL Fx-570ES Plus Chn ch Nt lnh ngha- Kt qu Chnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMath.

Thchinphptnhvsphc Bm:MODE 2 MnhnhxuthinCMPLX

Hinthdngtocc:r Bm:SHIFT MODE3 2 HinthsphcdngAHinthdngcc:a + ib. Bm:SHIFT MODE31 Hinthsphcdnga+bi

ChnnvogclRad(R) Bm:SHIFT MODE 4 MnhnhhinthchR Hoc(Chnnvogcl(D)) Bm:SHIFT MODE 3 (MnhnhhinthchD )

Nhpkhiugc BmSHIFT (-). Mnhnhhinth

-Thao tc trn my tnh (fx 570MS;570ES): Mode2,vdngnvR(radian), Bm nhp :(0 )

( 0 )

vx i

=

- Vi my fx 570ES:MunxuthinbinAvphabanu:Lmnhsau:

-Vi my fx 570MS :bmtipSHIFT+( ( )r A ),=(Re-Im):hinA,SHIFT=(Re-Im):hin.

Lu :NumyFx570EScilnhSHIFT MODE3 2dng:A thkhngcnbm SHIFT 2 3

4- Th d:

V d 1.Vtmdaongiuhavitns0,5Hz,tigcthigiannclix(0)=4cm,vntcv(0)=

12,56cm/s,ly 3,14 . Hyvitphngtrnhdaong.

Gii:Tnh=2f=2.0,5=(rad/s)

(0)

(0)

4

0 : 4 44

a x

t x ivb

. Nhp:4-4i= 23 co4 )4 4

s2 (4 2xSHIF t cmT

V d 2 . Vtmgnvoumtlxonh,daongiuhavichuk1s.ngitakchthchdaongbngcchkomkhivtrcnbngngcchiudngmton3cmribung.ChngctaVTCB,gcthigianlcbungvt,hyvitphngtrnhdaong.

Gii:=2/T=2/1=2(rad/s)

(0 )

(0 )

3

0 : 3;0

a x

t xvb

Nhp:-3,= 2 cos(2 )3 3 3xSHIF mT t c

V d 3. Vtnhm=250gctreovoudimtlxonh,thngngk=25N/m.TVTCBngitakchthchdaongbngcchtruynchommtvntc40cm/stheophngcatrclxo.ChngctaVTCB,gcthigianlcmquaVTCBngcchiudng,hyvitphngtrnhdaong.

Gii:

( 0 )

( 0 )

0

10 / ; 44

a xk

rad s x ivm b

.Nhp:4i,= 2 3 cos2

0 )4 (142

xSHIF mT t c

BmSHIFT 2 mnhnhxuthinnhhnhbn

Nubmtipphm 3 = kt qu dngcc(r ) Nubmtipphm 4 = kt qu dngphc(a+bi )

( angthchinphptnh)



5. Ch cc v tr c bit: (Hnh v bn phi) V tr ca vt lc u t=0

Phn thc: a

Phn o: bi

Kt qu: a+bi = A

Phng trnh: x=Acos(t+)

Bindng(I):x0=A;v0=0

a=A 0 A0 x=Acos(t)

Theochium(II):x0=0;v00

a=0 bi=-Ai A-/2 x=Acos(t-/2)

Vtrbtk: a=x00vbi i

A x=Acos(t+)

6. Tin li: Nhanh, HSchcntnh,vitngcciukinbanuvvithaotcbmmy.

III.GII NHANH TNG HP DAO NG: A.TNG HP CC DAO NG IU HA 1.Tng hp hai dao ng iu ho cng phng cng tn s :x1=A1cos(t+1)vx2=A2cos(t+2)th:x=x1+x2tacx=Acos(t+).Vi: A2 = A1

2+ A22+2A1A2cos (2 - 1);

tan = 2211

2211

coscos

sinsin

AA

AA

[12;nu 12]

2. Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cng tn s: x1=A1cos(t+1),x2=A2cos(t+2)vx3=A3cos(t+3)...thdaongtnghpcngldaongiuhocngphngcngtns:x=Acos(t+).ChiulntrcOxvtrcOytronghxOy.Tac:Ax=Acos=A1cos1+A2cos2+A3cos3+..vAy=Asin=A1sin1+A2sin2+A3sin3+..

Bin : :A =2 2x yA A v Pha ban u :

tan=y

x

A

A vi[Min,Max]

3. Khi bit dao ng thnh phnx1=A1cos(t+1)vdaongtnghpx=Acos(t+)thdaong

thnhphncnlilx2=x-x1.vi x2 = A2cos (t + 2).

Bin : A22=A2+ A1

2-2A1Acos( -1); Pha tan 2=1 1

1 1

sin sin

cos cos

A A

A A

vi12(nu12)

4.Nhc im ca phng php trn khi lm trc nghim: -XcnhAv cadaongtnghptheophngphptrnmtnhiuthigian.Vicbiudinginvctlphctpvinhngtnghpt3daongtrln,hayitmdaongthnhphn!-Xcnhgc hay 2thtskhkhnivihcsinhbivcngmtgitrtanluntntihaigitrca(v d: tan=1 th = /4 hoc -3/4). Vychngitrnochophhpvibiton!.-cbittrongphmvi:-1800



Bm:MODE 2 xuthinchCMPLX

B. GII PHP: Dng my tnh CASIO fx570ES, 570ES Plus hoc CASIO fx 570MS.

1. C s l thuyt: x = Acos(t + ) biudinbngvectquayA vibinAvphabanu,hoc

biudinbngsphc: (cos sin ) .ix i ib ea A A . (vimun:A= 2 2a b )

+TrongmytnhCASIO fx- 570ES; 570MS khiul:r (tahiul: A ). 2.Chn ch thc hin php tnh s phc ca my tnh: CASIO fx570ES, 570ES Plus Chn ch Nt lnh ngha- Kt qu Chnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMath.


Hinthdngtocc:r Bm:SHIFT MODE32 HinthsphcdngAHinthdngcc:a + ib. Bm:SHIFT MODE31 Hinthsphcdnga+bi

ChnnvogclRad(R) Bm:SHIFT MODE 4 MnhnhhinthchR Hoc Chnnvogcl(D) Bm:SHIFT MODE 3 MnhnhhinthchD


V d:Cchnhp:Cho:x= 8cos(t+ /3)scbiudinvisphc:8 600 hay 8

3 talmnhsau:

My CASIO fx 570ES; 570ES Plus Bm:MODE 2 xuthinCMPLX +Chnnvgcl(D)bm:SHIFTMODE 3 hinthD Nhpmy:8SHIFT(-)60hinth:860 +ChnnvgclRad(R)bm:SHIFTMODE 4 mnhnhhinthR

Nhpmy:8SHIFT(-)(:3shinthl:81

3

Kinh nghim:Nhpvinvnhanhhnnvrad (Vnhptheonvradphicdungocn(),hocphinhpdngphnsnnthaotcnhpluhn).

V d:Nhp90thnhanhhnnhp(/2)hay

2

Tuynhindnhnvthnthintannnhptheon v rad (R)

Bng chuyn i n v gc: (Rad)=(D).

180

nvgc() 15 30 45 60 75 90 105 120 135 150 165 180 360nvgc(Rad) 1

12

1

6

1

4

1

3

5

12

1

2

7

12

2

3

3

4

5

6

11

12

2

3.Lu : Kt qu c th hin th dng i s: a +bi (hoc dng cc: A ). -Chuyntdng:a + bi sangdng:A ,bmSHIFT 2 3 =

V d:Nhp:8SHIFT(-)(:3->Nuhinth: 4+ 4 3 i , mun chuynsangdngccA :

BmSHIFT 2 3 =kt qu:81

3

V d: Nhp:8SHIFT(-)(:3 ->Nuhinth:81

3

,munchuynsangdngphca+bi :

-BmSHIFT 2 4 =kt qu:4+4 3 i

BmSHIFT 2 mnhnhxuthinnhhnhbn

Nubmtipphm 3 = kt qu dngcc(r ) Nubmtipphm 4 = kt qu dngphc(a+bi )

( angthchinphptnh)



4. Tm dao ng tng hp xc nh A v bng cch thc hin php CNG: a.Vi my FX570ES; 570ES Plus : Bm:MODE2mnhnhxuthin:CMPLX.ChnnvgclRadbm:SHIFTMODE 4 mnhnhhinthR (hocchnnvgclbm:SHIFTMODE 3 mnhnhhinthD )

Thc hin php cng s phc: 1 1 2 2A A A Talmnhsau:

-Nhp:A1SHIFT(-)1+A2SHIFT(-)2=hinthktqu.:a+bi (hoc: A)(Nuhinthsphcdng:a+bi thbmSHIFT23=hinthktqu: A)b.Vi my FX570MS : BmMODE2mnhnhxuthinch:CMPLX.

Thc hin php cng s phc: 1 1 2 2A A A Talmnhsau: NhpA1SHIFT(-)1+A2SHIFT(-)2=BmtipSHIFT+=hinthktqu:A.SHIFT=hinthktqu:

c.Lu Ch hin th mn hnh kt qu: Saukhinhptandu=cthhinthktqudidng:phns, v t,hu t,...munktqudidngthp phntanSHIFT=(hocdngphmSD)chuyniktquHin th.d.Cc v d: V d 1: Mtvtthchinngthihaidaongiuhacngphng,cngtnscphngtrnh:x1=5cos( t+ /3)(cm);x2=5cos t(cm).Daongtnghpcavtcphngtrnh

A. x=5 3 cos( t- /4)(cm) B.x=5 3 cos( t+ /6)(cm)

C. x=5cos( t+ /4)(cm) D.x=5cos( t- /3)(cm)pnB

Phng php truyn thng Phng php dng s phc

Bin: 2 21 2 1 2 2 12. .cos( ) A A A A A

Phabanu:tan=2211

2211

coscos

sinsin

AA

AA

Ths:

A= 2 25 5 2.5.5.cos( / 3) 5 3 (cm)

tan=5.sin( / 3) 5.sin 0 5. 3 / 2 3

15cos( / 3) 5.cos 0 35. 12

=>

=/6.Vy:x=5 3 cos( t+ /6)(cm)

-Vi my FX570ES: Bm:MODE2

-n v gc l (D)Bm:SHIFTMODE 3

Nhp:5SHIFT(-)(60)+5SHIFT(-)0=

Hinth5 330=>:x=5 3 cos( t+ /6)(cm)

(NuHinthdngcc:15 5 3

2 2 i th

BmSHIFT23 =Hinth:5 330)

-n v o gc l Rad(R) bm:SHIFTMODE 4

Nhp:5SHIFT(-).(/3)+5SHIFT(-)0=

Hinth:5 31

6

V d 2: Mtvtthchinngthihaidaongiuhocngphng,cngtns

x1=cos(2t+)(cm),x2= 3 .cos(2t-/2)(cm).Phngtrnhcadaongtnghp

A.x=2.cos(2t-2/3)(cm) B.x=4.cos(2t+/3)(cm)C.x=2.cos(2t+/3)(cm) D.x=4.cos(2t+4/3)(cm) Gii: Vi FX570ES;570ES Plus: BmMODE2 , Chnnvgc(R):BmSHIFTMODE 4

-Nhpmy:1SHIFT(-)+ 3 SHIFT(-)(-/2=Hinth:2-2

3

.pnA



V d 3: MtvtdaongiuhaxungquanhvtrcnbngdctheotrcxOxcli

)()2

2cos(3

4))(

62cos(

3

4cmtcmtx

.Binvphabanucadaongl:

A. .3

;4 radcm

B. .6

;2 radcm

C. .6

;34 radcm

D. .3

;3

8radcm

pnA

Gii 1: Vi FX570ES , 570ES Plus: BmMODE2Chnnvgc(R):SHIFTMODE 4

Nhpmy:4

3

SHIFT(-).(/6)+

4

3

SHIFT(-).(/2=Hinth:4

1

3

V d 4: Badaongiuhacngphng,cngtnscphngtrnhlnlt:x1=4cos(t-/2)(cm),x2=6cos(t+/2)(cm)vx3=2cos(t)(cm).Daongtnghpca3daongnycbinvphabanul

A.2 2 cm;/4rad B.2 3 cm;-/4radC.12cm;+/2rad D.8cm;-/2rad

Gii: Vi FX570ES; 570ES Plus: BmMODE2.Chnnvgc(R).SHIFTMODE 4 Tmdaongtnghp,nhpmy:

4SHIFT(-)(-/2)+6SHIFT(-)(/2)+2SHIFT(-)0=Hinth:2 2 /4.ChnAV d 5: Daongtnghpcahaidaongiuhacngphng,cngtns

x1=a 2 cos(t+/4)(cm)vx2=a.cos(t+)(cm)cphngtrnhdaongtnghpl

A.x=a 2 cos(t+2/3)(cm) B.x=a.cos(t+/2)(cm)C.x=3a/2.cos(t+/4)(cm) D.x=2a/3.cos(t+/6)(cm)ChnB Gii: Vi FX570ES;570ES Plus: BmMODE2:CMPLX. Chnnvgc(D)Bm:SHIFTMODE 3

( Lu : Khng nhp a)Nhpmy: 2 SHIFT(-)45+1SHIFT(-)180=Hinth:190.

V d 6: TmdaongtnghpcabnDHcngphngsau:

1 210cos(20 )( ), 6 3 cos(20 )( )6 2

x t cm x t cm

3 44 3 cos(20 )( ), 10cos(20 )( )6

x t cm x t cm

Gii: 61 110cos(20 ) 106

i

x t x e

, 22 26 3 cos(20 ) 6 32

i

x t x e

3 14 3 cos(20 ) 4 3x t x , 6

4 41 0 co s( 2 0 ) 1 06

i

x t x e

Bm: 10 6 3 4 3 106 2 6

,SHIFT, 2, 3 =hinth: 6 6

4

cos(206 6 )

4)(x t cm

V d 7: Hai chtimM1,M2chuynngtrnhaingthngsongsong,theophngOxsongsongvi

haingthngtrn,chnglnltcccphngtrnh 1 3(cos2 . )2

x t cm

v 2 3 3 cos 2 . ( )x t cm . Tm

khongcchgiaM1vM2theophngOxtrn.

Gii: 1 3cos(2 )2

x t

, 2 3 3 cos(2 )x t Tac: 1 2 2 1| | | | 3 3 32

M M x x x x

Bm my: 2 36

3 63 3 ;2

SHIFT

Vy: 1 2 | 6cos(2 ) | ( )6

M M t cm

e. Trc nghim vn dng : Cu 1: Mtvtthchinngthihaidaongiuhacngphng,cngtnstheoccphngtrnh:x1

=acos(t+/2)(cm)vx2=a 3 cos(t)(cm).Phngtrnhcadaongtnghp

A. x=2acos(t+/6)(cm) B. x=2acos(t-/6)(cm) C. x=2acos(t-/3)(cm) D. x=2acos(t+/3)(cm)(Lukhngnhpa)pnA



5. Tm dao ng thnh phn ( xc nh A2 v 2 ) bng cch thc hin php TR: V d tm dao ng thnh phn x2: x2 =x - x1vi:x2 = A2cos(t + 2) Xc nh A2 v 2? a.Vi my FX570ES; 570ES Plus: BmMODE2mnhnhxuthin:CMPLX

Thc hin php tr s phc: 2 2 1 1A A A ;hoc 1 1 2 2A A A NhpASHIFT(-)-(chdutr),NhpA1SHIFT(-)1=kt qu.(NuhinthsphcthbmSHIFT23=ktqutrnmnhnh: A2 2

b.Vi my FX570MS BmMODE2mnhnhxuthin:CMPLX

Thc hin php tr s phc: 2 2 1 1A A A ;hoc 1 1 2 2A A A NhpASHIFT(-)-(chdutr),NhpA1SHIFT(-)1=BmtipSHIFT + =hinthktqu:A2. bm SHIFT =hinthktqu:2c.Cc v d :

V d 8: Mtchtimdaongiuhocphngtrnhdaongtnghpx=5 2 cos(t+5/12)(cm)viccdaongthnhphncngphnglx1=A1 cos(t+ 1)vx2=5cos(t+/6)(cm),Binvphabanucadaong1l:

A.5cm;1=2/3 B.10cm;1=/2 C.5 2 (cm)1=/4 D.5cm;1=/3 Gii: Vi FX570ES; 570ES Plus:BmMODE 2CMPLX. Chnnvgclrad:SHIFTMODE 4 .

-Nhpmy:5 2 SHIFT(-)(5/12)5SHIFT(-)(/6=Hinth:52

3

.chnA

V d 9: Mtvtngthithamgia3daongcngphng,cngtnscphngtrnhdaong:x1=2 3 cos(2t+/3)(cm),x2=4cos(2t+/6)(cm)vx2=A3 cos(t+ 3)(cm).Phngtrnhdaongtnghpcdngx=6cos(2t-/6)(cm).Tnhbindaongvphabanucadaongthnhphnth3: A. 8cmv-/2. B. 6cmv/3. C. 8cmv/6. D. 8cmv/2.ChnA Gii: Vi FX570ES;570ES Plus: BmchnMODE 2mnhnhxuthin:CMPLX Chnnvogclrad (R)SHIFTMODE 4 .Tmdaongthnhphnth3:x3 = x - x1 x2

Nhpmy:6SHIFT(-)(-/6)-2 3SHIFT(-)(/3)-4SHIFT(-)(/6=Hinth:8-1

2

.

d.Trc nghim vn dng:

Cu 1:Daongtnghpcahaidaongiuhacngphng,cphngtrnh5

3cos( )6

x t

(cm).Bitdao

ngthnhtcphngtrnhli 1 5cos( )6

x t

(cm).Daongthhaicphngtrnhlil

A. 2 8cos( )6

x t

(cm).B. 2 2cos( )6

x t

(cm).C. 25

2cos( )6

x t

(cm).D. 25

8cos( )6

x t

(cm).

Cu 2: Mtvtngthithamgia2daongcngphng,cngtnscphngtrnhdaong:x1=8cos(2t+/2)

(cm)vx2=A2 cos(t+ 2)(cm).Phngtrnhdaongtnghpcdngx=8 2 cos(2t+/4)(cm).Tnhbindaongvphabanucadaongthnhphnth2: A. 8cmv0. B. 6cmv/3. C. 8cmv/6.D. 8cmv/2.Cu 3: Mtvtngthithamgia3daongcngphng,cngtnscphngtrnhdaong:x1=8cos(2t+/2)

(cm),x2=2cos(2t-/2)(cm)vx3=A3 cos(2t+ 3)(cm).Phngtrnhdaongtnghpcdngx=6 2 cos(2t+/4)(cm).Tnhbindaongvphabanucadaongthnhphnth3: A. 6cmv0. B. 6cmv/3. C. 8cmv/6.D. 8cmv/2.Cu 4: Mtvtngthithamgia3daongcngphng,cngtnscphngtrnhdaong:x1=a.cos(2t+/2)

,x2=2a.cos(2t-/2)vx3=A3 cos(2t+ 3).Phngtrnhdaongtnghpcdngx=a 2 cos(2t-/4)(cm).Tnhbindaongvphabanucadaongthnhphnth3:

A. av0. B. 2av/3. C. a 2 v/6. D. 2a 2 v/2.



IV. BI TON CNG IN P TRONG MCH IN XOAY CHIU: 1.Cng in p:Xtonmchnitip:u=u1+u2.Viu1=U01 1os( )c t vu2=U02 2os( )c t a.Cch 1: Phng php gin vc t:Tactnghpccdaongiuho:-inptngtrongonmchnitip:u=u1+u2= 01 021 2os( ) os( ) U c t U c t

-inptngcdng:u=U0 s( )co t

Vi:U02 = U201+ U02

2 + 2.U02.U01. Cos( 1 2) ; 01 1 02 2

01 1 02 2

sin .sintan

cos cos

U U

U U

V d 1 :Chomchgm:onAMcha:R,CmcnitipvionMBchacuncmL,r.TmuAB=?

Bit:uAM=100 2 s os(100 )3

c t

(V) 0 1100 2( ),3

AMU V

uMB=100 2 os(100 )6

c t

(V)->U0MB=100 2 (V), 26

Bi gii: Dng cng thc tng hp dao ng: uAB =uAM +uMB

+U0AB=2 2(100 2) (100 2) 2.100. 2.100 2.cos( ) 200( )

3 6V

=>U0AB=200(V)

+100 2 sin( ) 100 2 sin( )

3 6tan100 2 cos( ) 100 2 cos( ) 12

3 6

.VyuAB=200 os(101

0 )2

c t

(V)

b.Cch 2: Dng my tnh CASIO fx 570ES ; 570ES Plus, VINA CAL Fx-570ES Plus: RT NHANH! Chn ch ca my tnh: Chn ch Nt lnh ngha- Kt qu

Citbanu(Resetall): Bm:SHIFT 9 3 = = Resetall(cthkhngcnthit)Chnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMath.


Dngtocc:r Bm:SHIFT MODE32 Hinthsphcdng:AHinthdngcc:a + ib. Bm:SHIFT MODE31 Hinthsphcdng:a+bi

ChnnvogclRad(R) Bm:SHIFT MODE 4 MnhnhhinthchR HocChnnvogcl(D) Bm:SHIFT MODE 3 MnhnhhinthchD

Nhpkhiugc BmSHIFT (-) Mnhnhhinth

2.V d cch nhp my:Cho:uAM=100 2 s os(100 )3

c t

(V),biudin100 2 -600 hoc 100 2 - 1

3

Chn ch :BmMODE 2 xuthinCMPLX,Chnnvgclbm:SHIFTMODE 3 hinthD

Nhpmy:100 2 SHIFT(-)-60hinth:100 2 -60 -ChnnvogclRad(R)bm:SHIFTMODE 4 mnhnhhinthchR

Nhpmy:100 2 SHIFT(-)(-:3hinth:100 2 -1

3

-Cnchnchmcnhtheodngtoccr (ta hiu l A ) -Chuyntdng:a + bi sangdngA ,tabmSHIFT23 =

3. Xc nh U0 v bng cch bm my tnh: FX570ES; 570ES Plus , VINACAL 570EsPlus uAM +uMB = uAB => 01 1 02 2 0U U U xc nh U0 v . +Vi my FX570ES; 570ES Plus ,VINACAL 570EsPlus: BmMODE 2xuthin:CMPLX.-NhpU01 SHIFT (-) 1 + U02 SHIFT (-) 2 =kt qu.

HnhuAM

B A R L,r

uMB

M C



(Nuhinthsphcdng:a+bi thbmSHIFT23 =hinthktqu : A +Vi my FX570MS : BmMODE 2mnhnhxuthin:CMPLX. NhpU01 SHIFT (-) 1 + U02 SHIFT (-) 2 =SaubmSHIFT + =hinthktqul:ASHIFT =hinthktqul: +Lu Ch hin th kt qu trn mn hnh: Saukhinhp,ndu=hinthktqudidngv t,munktqudidngthp phntanSHIFT=(hocdngphmSD)chuyniktquHin th.

4.V d 1 trn : TmuAB=?vi:uAM=100 2 os(100 )3

c t

(V) 0 1100 2( ),3

AMU V

uMB=100 2 os(100 )6

c t

(V)->U0MB=100 2 (V), 26

Gii 1: Vi my FX570ES ;570ES Plus,VINACAL 570Es Plus : BmMODE2xuthin:CMPLX ChnnvogclD ():SHIFTMODE 3

TmuAB?Nhpmy:100 2 SHIFT(-)(-60)+100 2 SHIFT(-)30=Hinthktqu :

200-15.VyuAB=2000os( 15 )c t (V)Hay:uAB=200 os(100 )

12c t

(V)

Gii 2: ChnnvogclR (Radian):SHIFTMODE 4

TmuAB?Nhpmy:100 2 SHIFT(-).(-/3)+100 2 SHIFT(-)(/6=Hinthktqu:

200-/12 .VyuAB=200 os(101

0 )2

c t

(V)

5. Nu cho u1 = U01cos(t + 1) v u = u1 + u2 = U0cos(t + ) . Tm dao ng thnh phn u2 :(Vdhnhminhhabn) u2 = u - u1.vi:u2 = U02cos(t + 2). Xc nh U02 v 2 *Vi my FX570ES;570ES Plus,VINACAL 570EsPlus : BmMODE2 Nhpmy:U0 SHIFT (-) -(tr)U01 SHIFT (-) 1 =kt qu.

(NuhinthsphcthbmSHIFT23= kt qu trn mn hnh l: U02 2*Vi my FX570MS : BmMODE 2 Nhpmy:U0 SHIFT (-) -(tr)U01 SHIFT (-) 1 =bmSHIFT (+) =,tacU02 ;bmSHIFT (=);tac2

V d 2:Nutvohaiumtmchinchamtintrthunvmtcuncmthunmcnitipmtinp

xoay chiu c biu thc u = 100 2 cos( t +4

) (V), th khi in p hai u in tr thun c biu thc

uR=100cos( t)(V).Biuthcinpgiahaiucuncmthunsl

A. uL=100cos( t+2

)(V). B. uL=100 2 cos( t+

4

)(V).

C. uL=100cos( t+4

)(V). D. uL=100 2 cos( t+

2

)(V).

Gii 1: Vi my FX570ES ;570ES Plus,VINACAL 570EsPlus : BmMODE 2xuthin:CMPLX ChnnvogclD():SHIFTMODE 3 mnhnhxuthinD

TmuL?Nhpmy:100 2 SHIFT(-).(45)-100SHIFT(-).0=

Hinthktqu :10090.VyuL=1002

os( )c t

(V)ChnA

Gii 2: ChnnvogclR (Radian):SHIFTMODE 4

TmuL?Nhpmy:100 2 SHIFT(-).(/4)-100SHIFT(-).0=

Hinthktqu: 100/2 .VyuL=1002

os( )c t

(V)ChnA

Hnhu1

B A X Y

u2

M



V d 3:Nutvohaiumtmchinchamtintrthunvmttinmcnitipmtinpxoaychiu

cbiuthcu=100 2 cos( t-4

)(V),khiinphaiuintrthuncbiuthcuR=100cos( t)(V).Biu

thcinpgiahaiutinsl

A. uC=100cos( t-2

)(V). B. uC=100 2 cos( t+

4

)(V).

C. uC=100cos( t+4

)(V). D. uC=100 2 cos( t+

2

)(V).

Gii 1: Vi my FX570ES ;570ES Plus,VINACAL 570EsPlus: BmMODE 2xuthinCMPLX Chnnvogcl(D):SHIFTMODE 3 Tmuc?Nhpmy:100 2 SHIFT(-).(-45)-100SHIFT(-).0=

Hinthktqu:100-90.VyuC=1002

os( )c t

(V)ChnA

Gii 2: Chnnvogcl Radian( R):SHIFTMODE 4 TmuC?Nhpmy:100 2 SHIFT(-).(-/4)-100SHIFT(-).0=

Hinthktqu: 100-/2 .VyuC=1002

os( )c t

(VChnA

V d 4:onmchABcintrthun,cundythuncmvtinmcnitip.MlmtimtrntrndonAB

viinpuAM=10cos100t(V)vuMB=10 3cos(100t-2

)(V).TmbiuthcinpuAB.?

A. u 20 2cos(100 t) (V)AB B. ABu 10 2cos 100 t (V)3

C. u 20.cos 100 t V)AB 3(

D. ABu 20.cos 100 t V)3

(

ChnD

Gii : ChnnvogclRadian(R):SHIFTMODE 4

TmuAB?Nhpmy:10SHIFT(-).0+10 3 SHIFT(-).(-/2=

Hinthktqu: 20-/3 .VyuC=20 os(100 )3

c t

(V) ChnD

6. Trc nghim vn dng : Cu 1:tinpxoaychiuvohaiuonmchR,Lthuncm,CmcnitipthinponmchchaLCl

1 60 cos 100 . ( )2

u t V

(A)vinphaiuRonmchl 2 60cos 100 . ( )u t V .inphaiuonmch

l:

A. 3/.100cos260 tu (V). B. 6/.100cos260 tu (V)C. 60 2 cos 100 . / 4u t (V). D. 6/.100cos260 tu (V).ChnC

Cu 2:Chomchinxoaychiunhhnhv.tvohaiuA,Bmtinpxoaychiu,inptcthigiacc

imAvM,MvBcdng: AMu 15 2 cos 200 t / 3 (V)

V MBu 15 2 cos 200 t (V) .BiuthcinpgiaAvBcdng:

A. ABu 15 6 cos(200 t / 6)(V) B. ABu 15 6 cos 200 t / 6 (V)

C. ABu 15 2 cos 200 t / 6 (V) D. ABu 15 6 cos 200 t (V) Cu 3(H2009): tinpxoaychiuvohaiuonmchcR,L,Cmcnitip.BitR=10,cuncmthun

cL=1/(10)(H),tincC= (F)vinpgiahaiucuncmthunluL=20 cos(100t+/2)(V).Biu

thcinpgiahaiuonmchl

A.u=40cos(100t+/4)(V). B. u=40 cos(100t/4)(V).

B

A M



C LA BR M

C.u=40 cos(100t+/4)(V). D. u=40cos(100t/4)(V).ChnD

Cu 4: HaiuonmchCRLnitipcmtinpxoaychiu:uAB=100 2 cos(100t)(V),inpgiahaiu

MBl:uMB=100cos(100t+4

)(V).

BiuthccainpgiahaiuonAMl:

A. uAM=100cos(100t+2

)V.B.uAM=100 2 cos(100t-

2

)V.

C. uAM=100cos(100t-4

)VD.uAM=100 2 cos(100t-

4

)V.ChnC

Cu 5: MtmchinxoaychiuRLC(hnhv)cR=100 ;

L=3

(H).inphaiuonmchAMchaRcdng:

u1=100cos100 t(V).VitbiuthctcthiinphaiuABcamchin.

A. 200 2 cos(100 )3

u t

(V) B. 200 2 cos(100 )4

u t

(V)

C. 200cos(100 )3

u t

(V) D. 200 2 cos(100 )4

u t

(V). ChnC

Cu 6:mchinhnhvbn,khitmtinpxoaychiuvoABth 120 2 os(100 )AMu c t V v

120 2 os(100 )3

MBu c t V

.BiuthcinphaiuABl:

A. 120 2 os(100 )4

ABu c t V

. B. 240 os(100 )6

ABu c t V

.

C. 120 6 os(100 )6

ABu c t V

. D. 240 os(100 )4

ABu c t V

.

V. TM BIU THC i HOC u TRONG MCH IN XOAY CHIU: 1.Phng php gii truyn thng: Cho R,L,Cnitip.Nuchou=U0cos(t+u),viti?Hocnuchoi=I0cos(t+i),vitu?

Bc 1:TnhtngtrZ:Tnh LZ L .;1 1

2CZ

C fC v 2 2( )L CZ R Z Z

Bc 2:nhlutm:UvIlinhvinhaubiU

IZ

;Io=Z

U o ;

Bc 3:Tnhlchphagiauhaiumchvi: tan L CZ Z

R

;Suyra

Bc 4:Vitbiuthcihocu:

a)Nuchotrcu=U0cos(t+u)thicdng:i=I0cos(t+u-).b)Nuchotrci=I0cos(t+i)thucdng:u=U0cos(t+i+).V d 1: Mch in xoay chiu gm mt in tr thun R = 50, mt cun thun cm c h s t cm

1( )L H

vmttincindung

42.10( )

C F

mcnitip.Bitrngdnginquamchcdng

5cos100i t A .Vitbiuthcinptcthigiahaiumchin.

Gii 1:Bc 1:Cmkhng: 1100 . 100 LZ L ;Dungkhng:

4

1 150

2.10100 .

CZC

Hnhu1

B A R L

u2

M

M

C A B

R L,r



Tngtr: 2 22 250 100 50 50 2 L CZ R Z Z

Bc 2:nhlutm:ViUo=IoZ=5.50 2 =250 2 V;

Bc 3:Tnhlchphagiauhaiumchvi:100 50

tan 150

L C

Z Z

R

4

(rad).

Bc 4:Biuthcinptcthigiahaiumchin: 250 2 cos 1004

u t

(V).

2.Phng php dng my tnh FX-570ES, FX-570ES Plus,VINA CAL Fx-570ES Plus a.Tm hiu cc i lng xoay chiu dng phc: Xembnglinh

I LNG IN CNG THC DNG S PHC TRONG MY TNH FX-570ES

CmkhngZL ZL ZLi(ChtrcicducnglZL)DungkhngZC ZC -ZCi(ChtrcicdutrlZc)Tngtr:

LZ L. ;1

CZ.C

;

22

L CZ R Z Z

( ) L CZ R Z Z i =a+bi(via=R;b=(ZL-ZC))

-NuZL>ZC:onmchctnhcmkhng-NuZL



1.... 50 CZ

C.VZL-ZC=50

-Vi my FX570ES;570ES Plus,VINACAL 570EsPlus : BmMODE2xuthin:CMPLX.

-BmSHIFTMODE32:Citdngtocc:(A ) -Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthD

Tac: iZZRXIZiu CLi )((... 0 5 0 50 50 X ( i ) ( Php NHN hai s phc)

Nhpmy:5SHIFT(-)0X(50+50ENGi)=Hinth:353.5533945=250 2 45

Vybiuthctcthiinpcahaiumch:u = 250 2 cos( 100t +/4) (V).

V d 2:MtmchinxoaychiuRLCkhngphnnhnhcR=100 ;C= 41

10. F

;L=2

H.Cng

dnginquamchcdng:i=2 2 cos100 t(A).Vitbiuthcinptcthicahaiumch?

Gii: 2

100 200LZ L. ;

4

1 1

10100

CZ.C

.

=100 .VZL-ZC=100

-Vi my FX570ES, 570ES Plus,VINACAL 570EsPlus: BmMODE2mnhnhxuthin:CMPLX.

-BmSHIFTMODE32:Citdngtocc:(r ) -Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthD

Tac: iZZRXIZiu CLi )((... 0 2 2 0 100 100 X ( i ) ( Php NHN hai s phc)

Nhpmy:2 2 SHIFT(-)0X(100+100ENGi)=Hinth:40045

Vybiuthctcthiinpcahaiumch:u = 400cos( 100t +/4) (V).

V d 3:ChoonmchxoaychiucR=40 ,L=

1(H),C=

6.0

10 4(F),mcnitipinp2umch

u=100 2 cos100 t(V),Cngdnginquamchl:

A. i=2,5cos(100 t+ )( )4

A

B. i=2,5cos(100 t- )( )4

A

C. i=2cos(100 t- )( )4

A

C. i=2cos(100 t+ )( )4

A

Gii: 1

100 100 LZ L. ;

4

1 1

10100

0 6

CZ

.C.

,

=60 .VZL-ZC=40

-Vi my FX570ES ,570ES Plus,VINACAL 570EsPlus: BmMODE2xuthin:CMPLX.

-BmSHIFTMODE32:Citdngtocc:(r ) -Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthD

Tac:i0

( ( )u

L C

Uu

R Z Z iZ

100 2 0

40 40

.

( i )( Php CHIA hai s phc)

Nhp100 2 SHIFT(-)0:(40+40ENGi)=Hinth:2,5-45Vy:Biuthctcthicngdnginquamchl:i = 2,5 cos(100t -/4) (A).ChnB V d 4:MtonmchingmintrR=50mcnitipvicunthuncmL=0,5/(H).tvohaiu

onmchmtinpxoaychiuu=100 2 cos(100t-/4)(V).Biuthccacngdnginquaonmchl:

A.i=2cos(100t-/2)(A).B.i=2 2 cos(100t-/4)(A).

C.i=2 2 cos100t(A). D.i=2cos100t(A).



Gii: 0 5

100 50 L,

Z L.

; .VZL-ZC=50 -0=50

-Vi my FX570ES ,570ES Plus , VINACAL 570EsPlus: BmMODE2xuthin:CMPLX. BmSHIFTMODE32:dngtocc:(r ). Chnnvogc(D),bm:SHIFTMODE 3 mnhnhhinthD

Tac:i0

( )u

L

Uu

R Z iZ

100 2 45

50 50

.

( i )( Php CHIA hai s phc)

Nhp100 2 SHIFT (-) - 45 : ( 50 + 50 ENG i ) =Hinth:2- 90Vy:Biuthctcthicngdnginquamchl:i = 2 cos( 100t - /2) (A).ChnA V d 5(H 2009): Khithiuinthkhngi30VvohaiuonmchgmintrthunmcnitipvicuncmthunctcmL=1/4(H)thcngdngin1chiul1A.Nutvohaiuonmchnyin

pu =150 2 cos120t (V)thbiuthccngdngintrongmchl:

A. 5 2cos(120 )( )4

i t A

B. 5cos(120 )( )4

i t A

C. 5 2cos(120 )( )4

i t A

D. 5cos(120 )( )4

i t A

Gii: Khithiuinthkhngi(hiuinth1chiu)thonmchchcncR:R=U/I=30

1

120 304

LZ L. ;i=

u 150 2 0

(30 30i)Z

( Php CHIA hai s phc)

-Vi my FX570ES ,570ES Plus,VINACAL 570EsPlus : BmMODE2xuthin:CMPLX. -BmSHIFTMODE32:Citdngtocc:(A ) -Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthD

Nhpmy:150 2 :(30+30ENGi)=Hinth:5- 45

Vy:Biuthctcthicngdnginquamchl:i = 5cos( 120t - /4) (A).ChnD 3. Trc nghim vn dng:

Cu 1:ChomchinxoaychiucR=30 ,L=

1(H),C=

7.0

10 4(F);inphaiumchlu=120 2 cos100 t

(V),thcngdngintrongmchl

A. 4cos(100 )( )4

i t A

B. 4cos(100 )( )4

i t A

C. 2cos(100 )( )4

i t A

D. 2cos(100 )( )4

i t A

Cu 2: Mt mch gm cun dy thun cm c cm khng bng 10 mc ni tip vi t in c in dung

42 .10C F

.Dnginquamchcbiuthc 2 2 cos(100 )3

i t A

.Biuthcinpcahaiuonmch

l:

A. 80 2 cos(100 )6

u t

(V) B. 80 2 cos(100 )6

u t

(V)

C. 120 2 cos(100 )6

u t

(V) D.2

80 2 cos(100 )3

u t

(V)

Cu 3:Mtmchgmcundythuncmccmkhngbng10 mcnitipvitincindung

42 .10C F

.Dnginquamchcbiuthc 2 2 cos100 )3

i t A

.Biuthcinpcahaiuonmchl:

A. 80 2 s(100 )6

u co t

(V) B. 80 2 cos(100 )6

u t

(V)

C. 120 2 s(100 )6

u co t

(V) D.2

80 2 s(100 )3

u co t

(V)



VI. XC NH HP EN TRONG MCH IN XOAY CHIU: 1.Chn ci dt my tnh Fx-570ES, 570ES Plus , VINACAL 570ES Plus : Chn ch lm vic Nt lnh ngha - Kt qu

Citbanu(Resetall): Bm:SHIFT 9 3 = = Clear?3:All(xattc)Chnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMath.


Dngtocc:r (A ) Bm:SHIFT MODE32 HinthsphcdngATnhdngtocc:a + ib. Bm:SHIFT MODE31Hinthsphcdnga+bi

HocchnnvgclRad(R) Bm:SHIFT MODE 4 MnhnhhinthchR Hoc(Chnnvgcl(D)) Bm:SHIFT MODE 3 MnhnhhinthchD

Nhpkhiugc Bm:SHIFT (-) Mnhnhhinthkhiu Chuynta + bi sangA , Bm:SHIFT 2 3 = MnhnhhinthdngA

ChuyntA sanga + bi Bm:SHIFT 2 4 = Mnhnhhinthdnga + bi

Sdngbnhclp Bm:M+hocSHIFT M+ MHxuthinM v ...M+ hoc...M-

Gibnhclp Bm:RCL M+ Mnhnhxuthin......M

Xabnhclp Bm:SHIFT 9 2 =AC ClearMemory?[=]:Yes(mtchM)

2. Xc nh cc thng s ( Z, R, ZL, ZC) bng my tnh:

-TnhZ: u

Zi

0

0( )

u

i

U

I

( Php CHIA hai s phc )

Nhpmy:U0 SHIFT (-) u : ( I0 SHIFT (-) i ) =

-Vitngtrphc: ( ) L CZ R Z Z i ,nghalcdng(a + bi).via=R;b=(ZL-ZC)-ChuyntdngA sangdng:a + bi :bmSHIFT 2 4 = 3.Cc V d: V d 1: Mthpkn(en)chchahaitrongbaphntR,L,Cmcnitip.tvohaiumchmtin

pxoaychiuu=100 2 cos(100t+4

)(V)thcngdnginquahpenli=2cos(100t)(A).on

mchchanhngphntno?Gitrcaccilng? Gii: Vi my FX570ES,570ES Plus, VINACAL 570EsPlus: BmMODE2bm:SHIFTMODE 3

-BmSHIFTMODE31:Citdngtocc:(a + bi).100 2 45

(2 0)

uZ

iNhp:100 2 SHIFT (-) 45 : ( 2 SHIFT (-) 0 ) = Hinth:50+50i

M ( ) L CZ R Z Z i .Suyra:R=50;ZL=50.Vyhpkn(en)chahaiphntR,L.

V d 2: Mthpkn(en)chchahaitrongbaphntR,L,Cmcnitip.Nutvohaiumchmt

inpxoaychiuu=200 2 cos(100t-4

)(V)thcngdnginquahpenl

i=2cos(100t)(A).onmchchanhngphntno?Gitrcaccilng? Gii: -Vi my FX570ES ,570ES Plus,VINACAL 570EsPlus : BmMODE2xuthin:CMPLX. -Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthD


(2 0)

uZ

i:Nhp200 2 SHIFT (-) -45 : ( 2 SHIFT (-) 0 ) =Hinth:100-100i



M ( ) L CZ R Z Z i .Suyra:R=100;ZC=100.Vyhpkn(en)chahaiphntR, C.


inpxoaychiuu=200 6 cos(100t+6


i=2 2 cos(100t-6

)(A).onmchchanhngphntno?Gitrcaccilng?

Gii: - Vi my FX570ES , 570ES Plus,VINACAL 570EsPlus : BmMODE2xuthin:CMPLX.-Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthchD


(2 2 30)

uZ

i:Nhp200 6 SHIFT (-) 30 : ( 2 2 SHIFT (-) (-30) =

Hinth:86,6+150i=50 3 +150i.Suyra:R=50 3 ;ZL=150.VyhpknchahaiphntR, L.


inpxoaychiuu=200 2 cos(100t+4


i=2cos(100t)(A).onmchchanhngphntno?Gitrcaccilng? Gii: - Vi my FX570ES,570ES Plus,VINACAL 570EsPlus : BmMODE2mnhnhxuthinCMPLX.-Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthchD


(2 0)

uZ

i:Nhp200 2 SHIFT (-) 45 : ( 2 SHIFT (-) 0 =

Hinth:141.42...45.bmSHIFT24 =Hinth:100+100iHay:R=100;ZL=100.VyhpknchahaiphntR, L.

V d 5: Chomchinnhhnhv:C=410

(F)

;L=

2(H)

BittvohaiumchinpxoaychiuuAB=200cos100t(V)thcngdngintrongmchli =4cos(100t)(A);Xlonmchgmhaitrongbaphnt(R0,L0(thun),C0)mcnitip.CcphntcahpXl:

A.R0=50;C0=410

(F)

B.R0=50;C0=

410(F)

2.

C.R0=100;C0=

410(F)

D.R0=50;L0=

410(F)

Gii Cch 1:TrctintnhZL=200;ZC=100- Vi my FX570ES,570ES Plus,VINACAL 570EsPlus: BmMODE2mnhnhxuthinCMPLX.-Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthchD

-BmSHIFTMODE31:Citdngtocc:(a + bi).+ Bc 1: VituAN=i Z =4x(i(200-100)):Thaotcnhpmy: 4 x ( ENG ( 200 - 100 ) ) shift 2 3 = M+(Sdngbnhclp)

Ktqul: 40090=>nghal uAN =400cos(100t+/2)(V)+ Bc 2: TmuNB =uAB - uAN : Nhpmy: 200 - RCL M+(gibnhclpuAN l 40090)SHIFT 23= Ktqul: 447,21359-63,4349.Bm: 4(bmchia4:xembndi)

LA B

NM

CX



+ Bc 3: TmZNB:NB

NBu

Zi

.Nhpmy: 4ktqu: 447,21359 63, 4349

4

=50-100i

=>HpXc2phntnnsl: R0=50;ZC0=100.Suyra:R0=50;C0=410

(F)

.pnA

Gii Cch 2:Nhnxt:Theochothuvicngphannmchcnghng=>Z=R0=U0/I0=200/4=50=>XcchaR0TnhZL=200;ZC=100,doZC=100,mchphichaC0saocho:ZC+ZC0=ZL=200

=>ZC0=ZL-ZC=200-100=100=>C0=410

(F)

.pnA

4.Trc nghim:

Cu 1: Choonmchnhhnhv,bit Vtu )100cos(2100 ,C= F

410.HpknXch chamtphn t (R

hoccundythuncm),dngintrongmchsmpha/3soviinpgiahaiuonmchAB.HpXchag

?intrhoccmkhngcgitrbaonhiu?

A.ChaR;R=100/ 3 B.ChaL;ZL=100/ 3

C.ChaR;R=100 3 D.ChaL;ZL=100 3

Cu 2: ChoonmchgmhaiphntX,Ymcnitip.TrongX,YcthlR,LhocC.Chobithiuinth

giahaiuonmchlu=200 2 cos100 t(V)vi=2 2 cos(100 t- /6)(A).ChobitX,Ylnhngphntnovtnhgitrcaccphnt?A.R=50 vL=1/H. B.R=50 vC=100/ F.

C.R=50 3 vL=1/2H. D.R=50 3 vL=1/H. Cu 3: Choonmchxoaychiugmhaiphntmcnitip.inpgiahaiumchvcngdnginquamchcbiuthc:u=200cos(100t-/2)(V),i=5cos(100t-/3)(A).Chnpnng? A.onmchc2phntRL,tngtr40. B.onmchc2phntLC,tngtr40.

C.onmchc2phntRC,tngtr40. D.onmchc2phntRL,tngtr20 2 .

Cu 4:Mtonmchxoaychiugm2trong3phntR,LhocCmcnitip.Biuthchiuinth2umch

vcngdnginquamch l 80cos 100 ( )2

u t V

v 8cos(100 )( )4

i t A

.Ccphnttrongmch

vtngtrcamchlA.RvL,Z=10 . B.RvL,Z=15 . C.RvC,Z=10 . D.LvC,Z=20 .Cu 5: MchinnitipR,L,Ctrongcundythuncm(ZLZLnntrongsphctac:ZL+ZC=-50i. Suyra:u 200 2 ( : 4)

i 450 50i 2Z

ChnA

Cu 6: MtonmchxoaychiuchaitrongbaphntR,Choccundythuncm.inphaiumchvcngdnginquamchcbiuthc:u=100 cos100t(V);i =2cos(100t-0,25)(A).intrhoctrkhngtngngl: A.L,C;ZC=100;ZL=50 B.R,L;R=40;ZL=30 C.R,L;R=50;ZL=50 D.R,C;R=50;ZC=50.

X B C A



VII. XC NH H S CNG SUT TRONG MCH IN XOAY CHIU: a.H s cng sut ca on mch:

-onmchRLC:Z

Rcos haycos= R

U

U

-onmchRrLC:cos=R r

Z

.haycos= R

U Ur

U

-anmchchacundy:cosd = d

r

Z=

2 2L

r

r Z

-Tngtr: 2 2L CZ R ( Z Z )

-Tngtrphccaonmch: L CZ R ( Z Z )i Lu :i y l s o!

-Dngcngthcny:u

Zi

vi Z Z ;i y l cng dng in!

-Tngtrphccacundy: ddu

Zi

vi d d dZ Z

-Vn l tnh Cos nh my tnh vi: Z Z ; v tnh Cos d vi : d d dZ Z

Nh MY TNH CM TAY:CASIO fx570ES ; 570ES Plus, VINACAL 570ES Plus b.Chn ci dt my tnh: Chn ch Nt lnh ngha- Kt qu Chnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMath.


Hinthdngtocc:r Bm:SHIFT MODE3 2 Hinthsphcdng:AHinthdngcc:a + ib. Bm:SHIFT MODE31 Hinthsphcdng:a+bi

ChnnvogclRad(R) Bm:SHIFT MODE 4 MnhnhhinthchR Hoc Chnnvogcl(D) Bm:SHIFT MODE 3 MnhnhhinthchD


- Vi my fx 570ES:Kt qu hin th:

c.Cc v d: V d 1:onmchABgmhaionmchAMvMBmcnitip.onAMgmintrthunR=100 mcni

tipvicuncmthun1

L ( H )

.onMBltincindungC.BiuthcinptrnonmchAMvMB

lnltl: 100 2 cos(100 )( )4

AMu t V

v 200cos(100 )( )2

MBu t V

.HscngsutcaonmchAB

l:

A.2

2cos B.

3

2cos C.0,5 D.0,75.

I

Ur

Ud

UL

d

I

UR

U

U UL C

I

R

Z

Nuangthchinphptnhsphc:BmSHIFT 2 mnhnhxuthinnhhnhbn

Nubmtipphm 1 = my hin th:arg ( hay ) Nubmtipphm 2 = my hin th:Conjg(a-bi )

Nubmtipphm 3 = my hin th:dngcc(r) Nubmtipphm 4 = my hin th:dngcc(a+bi )



Gi 1:ZL=100 ;ZAM=100 2 ;100 2

2100 2AM

AM

UI ( A )

Z ;

100 2 2200

2MB

C

U .Z

I

2 2L CZ R ( Z Z ) =100 2 =>

100 2

2100 2

Rcos

Z .ChnA

Gii 2: Tac:ZAM=(100+100i).

TngtrphccaonmchAB: ( ) (1 )AB AM MB MB

AB AM AM

AM AM

u u u uZ Z Z

i u u

Dng my Fx570ES, 570ES Plus,VINACAL 570EsPlus : BmMODE 2xuthin:CMPLX..bm:SHIFT MODE 4 xuthin: (R)

Nhp my:

2002(1 ) (100 100 )

100 24

X i

Bm du = . Hin th:c2trnghp:

A

a bi

(Takhngquantmndnghinthny:V d my hin th: 141,4213562 4

( Dng A ))

Tamunlygitrthbmtip:SHIFT 2 1 =Hin th:-1

4 (y l gi tr ca )

Bm tip: cos = cos( Ans -> Kt qu hin th : 2

2 y l gi tr ca cos cn tnh

2

2cos pnA

V d 2:onmchgm2onmchAMvMBmcnitip.onAMgmintrthun 1R nitipvi

cuncmthunL,onmchMBgmintrthun2

50R nitiptin42 10C F

.Bitinptc

thi7

200 2 cos(100 )( )12

AMu t V

80cos(100 )MBu t V .TnhhscngsutcaonmchAB.

Gii 1: Tng tr phc : ZMB=(50-50i).

Tacthtnhitrc(hoctnhgpnhbitrn):80 4 2

50 50 5 4MB

MB

ui

Z i

=> 0,8 2 cos(100 )( )

4i t A

.

Dng myFx 570ES, 570ES Plus,VINACAL 570EsPlus :

TngtrphccaonmchAB: ( )AB AM MB

AB

u u uZ

i i

Ci t my: BmMODE 2xuthin:CMPLX..bm:SHIFTMODE 4 ChnnvlRad (R)

Nhp my:

7200 2 80

12( )0,8 2

4

. Bm du = . Hin thc2trnghp:A

a bi

(Takhngquantmndnghinthny:V d my hin th: 241,556132 0,7605321591 ( A ) )Tamunlygitrthbmtip:SHIFT 2 1 =0,7605321591 . (y l gi tr ca ) Bm tip: cos = cos( Ans -> Kt qu hin th : 0,7244692923 y l gi tr ca cos cn tnh cos =0,72. V d 3:onmchABnitipgmchccphntnhintrthun,cuncmvtin.onmchABgmhaionmchAMvMBmcnitip.onAMgmintrthunR=50 mcnitipvitincindungkhng



50 .BiuthcinptrnonmchAMvMBlnltl: 80cos(100 )( )AMu t V v

100cos(100 )( )2

MBu t V

.HscngsutcaonmchABl:

A.0,99 B.0,84. C.0,86. D.0,95.

Gi : Dng my Fx -570ES, 570ES Plus,VINACAL 570EsPlus.

TngtrphccaonmchAB: ( ) (1 )AB AM MB MB

AB AM AM

AM AM

u u u uZ Z Z

i u u

Citmy: BmMODE2xuthin:CMPLX..bm:SHIFTMODE 4 ChnnvlRad (R)

Nhp my:100

2(1 ) (50 50 )80

X i

(ktquc2trnghp:225 25

+ i2 2

hoc 25 82

0,11065722122

.

Tamunc,thbmtip:SHIFT 2 1 Hinth: arg( Bm tip =Hinth: 0,1106572212 .(y l gi tr ca )

Bm tip: cos = Hinthgitrcacos:0,9938837347 = 0,99 pnA. V d 4 (H-2011):onmchABgmhaionmchAMvMBmcnitip.onAMgmintrthunR1=40

mcnitipvitincindungC=

4

10 3F,onmchMBgmintrthunR2mcvicunthuncm.t

voA,BinpxoaychiucgitrhiudngvtnskhngithinptcthihaiuonmchAMvMB

lnltl: )V)(12

7t100cos(250uAM

v )(100cos150 VtuMB .HscngsutcaonmchABl

A.0,84. B.0,71. C.0,86. D.0,95.

Gai cch 1 : (Truyn thng)

+TacZC=40;tanAM=4

11

AM

C

R

Z

+Thnhv:MB=3

tanMB= 33 2

2

RZR

ZL

L

*XtonmchAM: 2625,0240

50

AM

AM

Z

UI

*XtonmchMB: 360;602120 2222

2 LLMB

MB ZRRZRI

UZ

HscngsutcamchABl:Cos=22

21

21

)()( CL ZZRR

RR

0,84 pnA.

Gi cch 2 : Dng myFx570ES,570ES Plus,VINACAL 570EsPlus.

TngtrphccaonmchAB: ( ) (1 )AB AM MB MBAB AM AMAM AM

u u u uZ Z Z

i u u

Citmy: BmMODE2xuthin:CMPLX..bm:SHIFTMODE 4 ChnnvlRad (R)

Nhp my : 150(1 ) (40 40 )7

50 212

X i

Hinthc2trnghp:A

a bi

(Takhngquantmndnghinth

ny.Numyhindnga+bithcthbm:SHIFT 2 3 =Kt qu: 118,6851133 0,5687670898 ( A ) )

Tamunhin ththbm:SHIFT 2 1 Hin th: arg( Bm = Hin th: 0,5687670898 (y l gi tr ca ) Muntnhcos:Bm tip: cos = cos(Ans Hinth:0,842565653 = 0,84 lgitrcacospnA.

I

UAM

UMB

7/12

/4

/3



V d 5: MchingmmtcundycintrRmcnitipvimttC.Mchctdiinpulunnnh.BitgitrhiudngUC=3Ucd,lchphacainphaiucundysoviCdng

inquamchl/3.Tnhhscngsutcamch.

Gii:CoiUcdbng1(nv)=>UC= 3 vUcdnhanhphahndngingc/3: 1 3cdu

Vucchmphathuadngingc-/2: 32

Cu

.Tac: cd Cu u u

Dng myFx570ES,570ES Plus,VINACAL 570EsPlus: BmMODE2xuthin:CMPLX.bm:SHIFTMODE 4 .ChnnvlRad (R)

Nhpmy 2 3(1 ) ( 3 ) 1

3 2 3

SHIFT Tamunhin ththbm:SHIFT 2 1 Hin

th: arg( Bm = Hin th: 3

(y l gi tr ca ) .Muntnhcos:Bm tip: cos = cos(Ans

Hin th :0,5 , Hay cos = 0,5 .Vy /, cos 0,53

cd u iU U

V d 6 : Mtonmchxoaychiugm3phntmcnitip:intrthunR,cundyctcmLvintrthunr,tincindungC.tvohaiuonmchmtinpxoaychiu,khiinptcthihaiucun

dyvhaiutinlnltcbiuthc 80 6 cos / 6du t V , 40 2 os 2 / 3Cu c t V ,inphiudng

haiuintrlUR= 60 3 V.Hscngsutcaonmchtrnl

A. 0,862. B. 0,908. C. 0,753. D. 0,664.Gii 1:Nhnvogintac:

40 3 ; 120 os 0,908r LU V U V c .p n B

Gii 2:Dng my Fx570ES , 570ES Plus,VINACAL 570EsPlus:

Ta c 2

60 3 2 cos( )( ) 60 6 cos( )( )3 2 6

Ru t V t V

V 0 cos /6 ( )i I t A (Phacail 6

)

Tac: 02

60 6 80 6 40 26 6 3

R d C uu u u u U

.Vi6

u i u

BmMODE2xuthin:CMPLX..bm:SHIFTMODE 4 Chnnvl (R)

Cch 1:Nhpmy:2

60 6 80 6 40 26 6 3

Bm = Hin th: .....(khngquantm)

Bm:SHIFT 2 1 Hin th: arg( Bm = Hin th: - 0,09090929816 (y l gi tr ca u)

Bm - (6

)Bm = Hin th0,4326894774 (y l gi tr ca ) .

Muntnhcos:Bm tip: cos = cos(Ans Hinth:0,907841299 = 0,908 .Chn B

Cch 2:VkhngchoI0nntacthchobng1nv,nn: 0 16

ii I

=>u

Zi

Nhpmy:2

60 6 80 6 40 26 6 3

Bm : (1 )

6

Bm = Hin th: (khngquantm)

bm:SHIFT 2 1 Hin th: arg( Bm = Hin th : 0,4326894774(y l gi tr ca ) . Muntnhcos:Bm tip: cos = cos(Ans Hinth:0,907841299 = 0,908 lgitrcacos.Chn B



PHN HAI: DNG (MODE 7) GII BI TP GIAO THOA SNG Ci t my : Fx570ES , 570ES Plus,VINACAL 570EsPlus:

Bm:SHIFT 9 3 = = Resetall(cthkhngcnthit)

Bm:SHIFT MODE 1 Math(cthkhngcnthit)

HocBm:SHIFT MODE 2 LineIO(cthkhngcnthit)

Bm:MODE 7: TABLE

I. DNG (MODE 7) GII BI TP GIAO THOA SNG C

V d tachmsf(x)=2

12 x

Bc 1:(MODE 7) TABLE

Bc 2:Nhphmsvomytnh

Bc 3:bm=nhp1

Bc 4:bm=nhp5

Bc 5:bm=nhp1

Bc 6:bm=

Ta c bng bin thin: f(X)

a.V d 1: Sidydil=1mctreollnglnmtcnrung.Cnrungtheophngngangvitnsthayit100Hzn120Hz.Tctruynsngtrndyl8m/s.Trongqutrnhthayitnsrungthslnquanstcsngdngtrndyl:

A. 5 B. 4 C. 6 D. 15Cch gii truyn thng Hng dn bm my v kt qu

-l =(2k+1)4

=(2k+1)

f

v

4

f=(2k+1)l

v

4=(2k+1)2

Do100Hzf120Hz.Chok=0,1,2..

k=24f=98Hzk=25 f =102Hz k=26 f =106Hz k=27 f =110Hz k=28 f =114Hz k=29 f =118Hz k=30f=122HzchnA

MODE7:TABLE.

8

( ) (2 1)4 1

f x f kx

=(2X+1)x2=(2X+1)x2

.Nhpmy:(2xALPHA)X+1)x2

=START20=END30=STEP1=kt qu C 5 gi tr

D f(x)=

D

f(x)=x2+1 2

D

Start?

1 D

End?

5

D

Step?

1

D

x f(x)

1 2 3

1 2 3

1.5 4.5 9.5

1

x=k f(x) = f

24 98 25 26 27 28 29 30

102 106 110 114 118 122



b.V d 2:MtsidynhirtdicuAdaongvitnsfvtheophngvunggcvisidy.Bindaongl4cm,vntctruynsngtrnyl4(m/s).XtmtimMtrndyvcchAmton28cm,ngita

thyMlunlundaonglchphaviAmtgc (2 1)2

k

vik=0,1,2.Tnhbcsng?Bittnsf

cgitrtrongkhongt22Hzn26Hz. A.12cm B.8cm C.14cm D.16cm

Cch gii truyn thng Hng dn bm my v kt qu

2

)12(

k =

2d

d=(2k+1)4

=(2k+1)

f

v

4

Do22Hzf26Hzf=(2k+1)d

v

4

Chok=0,1,2.3.k=3

f=25Hz=v/f=16cmchnD

MODE7:TABLE

( ) (2 1)4

vf x f k

d =(2X+1)

4

4.0, 28

Nhpmy:(2xALPHA)X+1)x(1:0,28)

=START0=END10=STEP1=kt qu Chnf=25Hz

=v/f=25

40=16cm

c.V d 3: Cu 50 - thi tuyn sinh i hc khi A nm 2011 - M 817 Cu 50: MtsnghnhsintruyntheophngOxtngunOvi tn s20Hz,ctctruynsngnmtrongkhongt0,7m/sn1m/s.GiAvBlhaiimnmtrnOx,cngmtphasoviOvcchnhau10cm.HaiphntmitrngtiAvBlundaongngcphavinhau.Tctruynsngl A.100cm/s B.80cm/s C.85cm/s D.90cm/s


-d=(2k+1)2

=(2k+1)

f

v

2

Do0,7m/sv1m/s.12

2

k

dfv

Chok=0,1,2..v=80cm/schnB.vik=2

MODE7:TABLE

2 10 20( )

2 1

x xf x v

k

;Mauso=2xALPHA)+1

Nhpmy:...tngtnhtrn....(400:(2xALPHA)X+1)=START0=END10=STEP1=kt qu: 80

Ch :-ChnStart:Thngthnglbtut0hoctytheobi

-ChnEnd:Tythucvobicho(nunhpslnquthkhngbnh: InsufficientMEM)-ChnStep:1(vknguyn)d.Trc nghim vn dng : Cu 1.(H) TiimStrnmtncyntnhcngundaongiuhotheophngthngngvitnsf.KhitrnmtnchnhthnhhsngtrnngtmS.TihaiimM,Nnmcchnhau5cmtrnngthngiquaSlundaongngcphavinhau.Bittctruynsngtrnmtncl80cm/svtnscangundaongthayitrongkhongt48Hzn64Hz.Tnsdaongcangunl A.64Hz. B.48Hz. C.54Hz. D.56Hz.Cu 2.(H) TiimStrnmtncyntnhcngundaongiuhotheophngthngngvitns50Hz.KhitrnmtnchnhthnhhsngtrnngtmS.TihaiimM,Nnmcchnhau9cmtrnngthngiquaSlundaongcngphavinhau.Bit rng, tc truynsng thayi trongkhong t70cm/sn80cm/s.Tctruynsngtrnmtncl A.75cm/s. B.80cm/s. C.70cm/s. D.72cm/s.

x=k f(x) = f

0 3.517 1 2 3 4

10.71 17.85 25 32.42

x=k f(x) = v

0 400 1 2 3

133 80 57.142



II. DNG (MODE 7) GII BI TP GIAO THOA SNG NH SNG Ci t my : Fx570ES , 570ES Plus,VINACAL 570EsPlus:

Bm:SHIFT 9 3 = = Resetall(cthkhngcnthit)

Bm:SHIFT MODE 1 Math(cthkhngcnthit)

HocBm:SHIFT MODE 2 LineIO(cthkhngcnthit)

Bm:MODE 7: TABLE

Hoc Ch cn bm: MODE 7:TABLE a.V d 1: Cu 22 - thi tuyn sinh i hc khi A nm 2010 - M 136 Cu 22:TrongthnghimY-ngvgiaothoanhsng,haikhecchiubngnhsngtrngcbcsngt380nmn760nm.Khongcchgiahaikhel0,8mm,khongcchtmtphngchahaikhenmnquanstl2m.Trnmn,tivtrcchvntrungtm3mmcvnsngcaccbcxvibcsngA.0,48mv0,56m B.0,40mv0,60mC.0,45mv0,60m D.0,40mv0,64m


x=a

Dk ..

Do:0,380m0,760m.=Dk

xa

.

.

Chok=1,2..k=1=1.2m.

k=2 =0.6m. k=3 =0.4m.k=4=0.3m.chnB

MODE7TABLE Nhpmy:2

38.0)(

xmauso

xxf

Mauso=ALPHA)BinXlk

Nhpmy:.(0,8 x 3 ) : ( ALPHA ) X x 2 ) = START 1 = END 10 = STEP 1 = kt qu:

b.V d 2: Cu 30 - thi tuyn sinh i hc khi A nm 2009 - M 629 Cu 30:TrongthnghimY-ngvgiaothoanhsng,haikhecchiubngnhsngtrngcbcsngt0,38mn0,76m.Tivtrvnsngbc4canhsngnsccbcsng0,76mcncbaonhiuvnsngnacaccnhsngnsckhc? A.3. B.8. C.7. D.4.


k=k11Do0,40m0.76m.

=k

k 11

Chok=1,2..k=4=0.76m. (loi) k=5 =0.608m. k=6 =0.506m. k=7 =0.434m. k= 8 =0.38m.chnD

MODE7TABLE Nhpmy:mauso

xxf

76.04)(

Mauso=ALPHA)XBinXlkNhpmy:...tngtnhtrn....(4 x 0,76 ) : ALPHA ) X = START 0 = END 20 = STEP 1 = kt qu:

x=k f(x) =

1 2 3 4

1.2 0.6 0.4 0.3

x=k f(x) =

1 3.04 2 3 4 5 6 7 8 9

1.52 1.0133 0.76 0.608 0.506 0.434 0.38 0.3377



c.V d 3: Cu 43 - thi tuyn sinh cao ng khi A nm 2011 - M 142 Cu 43:trongthnghimY-ngvgiaothoanhsng,khongcchgiahaikhel2mm,khongcchtmtphngcha2khenmnquanstl2m.Ngunphtnhsnggmccbcxnsccbcsngtrongkhong0,40mn0.76m.Trnmn,tiimcchvntrungtm3,3mmcbaonhiubcxchovnti?

A.6bcx. B.4bcx. C.3bcx. D.5bcx.


Ccbcxchovntiti

x=(.

0,5) ;0,4 0,76( 0,5)

D a xk m m

a k D

.0,4 0,76 3,9 7,75

( 0,5)

a xm m k

k D

Vyk=4;5;6;7:c4bcx.

Hayx=( 0,5). .k D

a

;Do0,40m0.76m.

=.

( 0,5).

a x

k D

Chok=0,1,2..

k=4 =0.733m. k=5 =0.60m. k=6 =0.507m. k=7 =0.44m. ChnB:4bcx.

MODE7TABLE Nhpmy:2

3.32)(

xmauso

xxf

Mauso=ALPHA)X+ 0,5BinXlkNhpmy:...tngtnhtrn....(2 x 3,3 ) : ( ( ALPHA ) X + 0,5 ) x 2 )

= START 0 = END 10 = STEP 1 = kt qu

Ch :CchchnStart?End?VStep?-ChnStart?:Thngthnglbtut0hay1hoctytheobi

-ChnEnd:Tythucvobicho(nunhpslnquthkhngbnh: InsufficientMEM)-ChnStep:1(vknguyn)

d.Trc nghim vn dng : Cu 1: TrongthnghimYoung,cckhesngcchiubngnhsngnsc,khongcchgia2khela

=0,5mm,khongcchthaikhenmnquanstlD=1,5m.Tmnhngnhsngnscchovnsngti

im Mcch vn trung tm mt khong xM= 6mm. Bit nh sng trng c bc sng nm trong khong t

0,4mn0,75m

A.2bcx. B.3bcx. C.4bcx. D.5bcx.Cu 2:TrongthnghimYoung,cckhesngcchiubngnhsngtrng,khongcchgia2khe la=0,3mm,

khongcchthaikhenmnquanstlD=2m.Tnhxemtingvtrcavnsngbc4canhsngmuc

nhngvchsngcanhsng nscnotrngti.(bitnhsngtrngcbcsngnmtrongkhongt0,4m

n0,76m)

A.2bcx. B.3bcx. C.4bcx. D.5bcx.

x=k f(x) =

0 6.63 1 2 3 4 5 6 7 8

2.2 1.32 0.942 0.733 0.60 0.507 0.44 0.388



III. DNG MY TNH TM BI CHUNG NH NHT (BCNN) V C CHUNG LN NHT (UCLN) CA 2 S. (p dng trong bi tp tm khong vn trng khi giao thoa nh sng vi 2 hoc 3 bc x n sc). 1.Phng Php chung: Vi2s(a,b)tanhpmya:b=ktqul:c:d(c:dlphnstigincaa,b)

tm BCNN ta ly a*d ; tm UCLN ta ly: a:c Vd:TimBCNNvUCLNcahais50v20Nhpmy:50 : 20 = 5/2=>BCNN(50,20)=50*2=100; UCLN(50,20) =50: 5=10. 2.c bit my VINACAL fx-570ES Plus cn c thm chc nng SHIFT 6 nh sau: 1: Q,r (Chia tm phn nguyn v d) 2: LCM ( Tm bi chung nh nht: BCNN: The Least Common Multiple hay Lowest Common Multiple ) 3: GCD (Tm c chung ln nht: UCLN) 4: FACT( phn tch ra tha s nguyn t) V d:Tm bi chung nh nht (BCNN) ca 2 s 4 v 5: SHIFT 6 2 4 , 5 = 20

3.V d minh ha: TrongthnghimgiaothoanhsngvikheYoung,nhsngcdnglmthnghimgmchaithnhphnnsccbcsng1=0,75m()v2=0,6m(vng).Khongcch

giahaikhela=1mm,khongcchthaikhenmnlD=2m. a. M t hnh nh quan st c trn mn:

+Nudngringnhsngnscthtrnmnthucmthvn.+Nudngringnhsngnscvngthtrnmnthuchvnvng.+Khidngchaibcxtrnthtrnmnthucngthichvnvhvnvng.Vntrungtmcahaihvnnytrngnhau,toramutnghpcavvng,gilvntrng.Ngoivntrungtmlvntrng,cncccvtrcnglvntrng(vdnhtiM).Vytrnmnxuthin3loivnKHCnhau:mu,muvngvmutnghpcavvng.

b.Xc nh khong cch t vn sng trung tm n vn sng gn nht cng mu vi n:

-pdngcngthctnhkhongvngiaothoa,tatnhc:i1=1,2mmvi2=1,5mm.-Trnhnhv,khongvntrngcdibngonOM.

OMlbiscai1vOMcnglbiscai2.

VyOMchnhlbischungnhnht(BCNN)cai1vi2: itrung = BCNN(i1 , i2 )-Mun tm itrng, tacn tnh i1vi2,Sautnh bischung nh nht(BCNN)ca chng.

-tnhbischungnhnht(BCNN)cahais,tacthdungcc phng php sau:i.Phng Php chung : Cho hai s a v b. tmBCNN(a,b)vUCLN(a,b)vtalmnhsau:Talya/b=c/d(c/dlaphnstigincaa/b) tm BCNN ta ly a*d tm UCLN ta ly: a/c V d:TimBCNNvUCLNca50v20.Tac:50/20=5/2.BCNN(50;20)=50*2=100;UCLN(50;20)=50/5=10. Dng my tnh vi s liu ca bi tp v d trn:Tm(UCLN) v (BCNN)ca1,2v1,5Nhpmytnh:1,2:1,5=4:5.Sauly1,2X5=6Vy:BCNN(1,2;1,5)=1,2*5= 6 ii.Hoc dng chc nng (LCM) ca my tnh VINA CAL 570ES PLUS: SHIFT 6 2 a , b = kt qu



1,2 2 1,5 3 0,6 2 0,5 5 0,3 3 0,1 0,1

1,2 2 1,5 3 0,8 2 0,6 2 0,5 5 0,4 2 0,3 3 0,1 0,2 2 0,1 0,1

Lu my ch cho nhp s nguyn! Nhp my SHIFT 6 2 12 , 15 = 60 ta chia 10 kt qu: 6 iii. Hoc C th tnh BCNN ca i1 v i2 bng cch: phn tch cc snythnhtchcaccthasnguynt!

Tac: 1,2=22.3.0,1

1,5= 3.5.0,1

Bischungnhnhtca1,2v1,5l:22.3.0,1.5=6.

Vykhongvntrngtrongbitonnyl:itrng=6mm.KhongcchtvntrungtmnvngnnhtcngmuvinlOM=6mm.

c. Nu trong thnh phn ca nh sng th nghim t r n c thm nh sng tm c 3 = 0,4m th

khong cch t vn trung tm n vn sng gn nht, cng mu vi n l bao nhiu? Tatnhc:i1=1,2mm,i2=1,5mmvi3=0,8mm.

Lubischungnhnht(BCNN)cahais1,2 v 1,5 l 6Bygitatnhbischungnhnht(BCNN)cahais6v0,8,tacthlmnhsau:

+Dng My Tnh VINA CAL 570ES PLUS SHIFT 6 2 : (LCM) Lu my ch cho nhp s nguyn! Nhp my SHIFT 6 2 60 , 8 = 120 ta chia 10, kt qu: 12 +Hoc c th phn tch cc s ny thnh tch ca cc tha s nguyn s

nhbngsau:Tac:1,2=22.3.0,11,5=3.5.0,1

0,8=23.0,1

Bischungnhnhtca1,2;0,8v1,5l:3.0,1.5.23=12(ltchscanhngscsmlnnht)Vy,nucthmbcxtm,vntrungtmslstnghpca3mu:, vng, tm.Khongcchtvntrungtmnvncngmuvinvgnnnhtl12mm.Trongtrnghpny,trnmnquanstxuthin7loivngm:

3loinsc:,vng,tm3loivntnghpca2mu:(+vng),(+tm)v(vng+tm)1loivntnghpca3mu:+vng+tm.

4.Vn dng cng ly s liu tng t v d trn: TrongThNghimY-ngvgiaothoanhsng,chiuvo2khe1chmsngascgm3thnhphnnsccbcsng1=0.4m,2=0.6m,3=0.75m).Khongcchgiahaikhela=2mm,khongcchthaikhenmnlD=2m.Tmkhong cch tvntrung tmnvnsnggn nht, cngmu vin? Gii: Vtrccvncngmuvivntrungtm:x=k1i1=k2i2=k3i3hayk11=k22=k33Ths:0,4k1=0,6k2=0,75k3hay8k1=12k2=15k3Tm (BCNN) ca 8, 12 v 15: Nhp my 8: 12 = 2:3 => (BCNN) ca 8, 12 l: 8 x 3 =24 Nhp my 24: 15 = 8:5 => (BCNN) ca 24, 15 l: 24 x 5=120 Vy BCNN) ca 8, 12 v 15: 120 Suyra:k1=15n;k2=10n;k3=8n.Khonggia2vnsngcngmuvivntrungtmgnnhaunhtn=0vn=1(k1=15;k2=10vk3=8) Haykhong cch tvntrung tmnvnsnggn nht, cngmu vinl:

x=k1i1=k2i2=k3i3hay: , .

. . .M MDD D D

x k k k x n n n mma a a a

31 2 11 2 3

0 4 215 15 6

2

Vykhong cch tvntrung tmnvnsnggn nhtkhin=1=> xM = 6mm



PHN BA. TM NHANH I LNG CHA BIT TRONG BIU THC: (VI MY TNH : CASIO: Fx570ES & Fx-570ES Plus; VINA CAL Fx-570ES Plus)

1.S dng SOLVE ( Ch dng trong COMP: MODE 1 ) SHIFTMODE1 Mnhnh:Math

Chn ch lm vic Nt lnh ngha- Kt qu DngCOMP Bm:MODE 1 COMP l tnh ton chungChnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMathNhpbin X Bm:ALPHA ) MnhnhxuthinX.

Nhpdu= Bm:ALPHA CALC Mnhnhxuthin=Chcnng SOLVE: Bm:SHIFT CALC= hinthktqu X= .....

a)V d 1:Tnhkhilngmcaconlclxodaong,khibitchukT=0,1(s)vcngk=100N/m.Ta

dngbiuthc 2m

Tk

Phng php truyn thng Phng php dng SOLVE

Tac: 2m

Tk

=> 2 24

mT

k

Suyra:

2

2

.

4

k Tm

Ths:nhpmytnh:

m=

2

2

100.(0,1 )

4

=0,25

Vy:khilngmcaconlc0,25kg

-Bm:MODE1

-Bm:0.1 SHIFTX10X ALPHACALC = 2 SHIFT

X10X

ALPHA) X 100

Mnhnhxuthin: 0.1 2100

X

-Bmtip:SHIFTCALCSOLVE=(ch6s)

Mnhnhhinth:

Xlilngm

Vy:m= 0,25 kg

T v d ny chng ta c th suy lun cch dng cc cng thc khc!!! b)V d 2:Tnhcngcaconlclxodaong,khibitchukT=0,1(s)vkhilng=0,25kg.

.-Dngbiuthc 2m

Tk

lmnhtrn,cuicngmnhnhxuthin:0.25

0.1 2X

-Tiptcbm:SHIFTCALCSOLVE=

(chkhong6s),Mnhnhhinthnhhnhbn:

Xlilngkcntm.Vy:k =100N/m

0 .1 21 0 0

X

X= 0.25 L--R = 0

0 .2 50 .1 2

X

X= 100 L--R = 0



c)V d 3:Tnhchiudicaconlcndaongnh,khibitchukT=2(s)vgiatctrngtrngg=

2(m/s2).Tadngbiuthc: 2l

Tg


Tac: 2l

Tg

=> 2 24

lT

g

Suyra:

2

2

.

4

T gl

Ths:

2 2

2

21( )

4 . m

Vychiudicaconlcnl=1(m)

Bm:MODE1

Tac: 2l

Tg

ths: 22 2X

-Bm:2 ALPHACALC = 2 SHIFTX10X

ALPHA) X SHIFTX10X x2


(chkhong6s)

Mnhnhhinth:

Xlilngl

Vy:l=1(m)

c)V d 4:Tnhgiatctrngtrngtinicconlcn,khibitchukT=2(s)vchiudicaconlc

ndaongnhl1m.Tadngbiuthc: 2l

Tg


Tac: 2l

Tg

=> 2 24

lT

g

Suyra: 2

24

lg

T

Ths:

22

2

4 . .1

2 g

=9,869m/s2

Vygiatctrngtrngtinicconlcn

daongg==9,869m/s2

Bm:MODE1

Tac: 2l

Tg

ths:1

2 2X

-Bm:2 ALPHACALC = 2 SHIFTX10X

1 ALPHA) X .Tiptcbm:

SHIFTCALCSOLVE=(chthigian....)

Mnhnhhinth:

Xlilngg

Vy:g= 9,869m/s2

22 2

X

X= 1 L--R = 0

12 2

X

X= 9.869604401 L--R = 0



c)V d 5:inptvohaiumtonmchR,L,Ckhngphnnhnh.inphiudnghaiumchl100V,haiucuncmthunLl120V,haibntCl60V.inphiudnghaiuRl:A.260VB.140VC.80VD.20V


Gii:inphaiuR:Tac:2 2 2( )R L CU U U U .Binitac(=>)2 2 2( )R L CU U U U .Tiptcbini:

2 2( )R L CU U U U ths:

Nhpmy:2 2100 (120 60) 80V

Vy:inphiudnghaiuRl:80V

pnC.

-Bm:MODE1

Dngcngthc : 2 2 2( )R L CU U U U

-Bm:100 x2ALPHACALC =ALPHA) X x2

+ ( 120 -60 ) x2

Mnhnhxuthin: 1002 =X2 +(120-60)2


Mnhnhhinth:

XlURcntm

Vy:UR = 80V

c)V d 6:MtmchdaonggmmttincindungCvmtcuncmctcmL.Mchdao

ngctnsring100kHzvtincC=5nF.tcmLcamchdaongl:A.5.10-5H. B.5.10-4H.C.5.10-3H. D.2.10-4H.


Gii: Cngthctnsring:1

2f

LC

Binitac:2 2

1

4

L

f C

Thsbmmy:

2 5 2 9

1

4 .(10 ) .5.10

L =5.066.10-4(H)

pnB.

-Bm:MODE1 (COMP)

Bm:SHIFTMODE1 Mnhnhhinth:Math

Dngcngthc :1

2f

LC

-Bm: X10X 5 ALPHACALC =

1 2

SHIFTX10X ALPHA) X X 5 X10X - 9

Mnhnhxuthin: 59

110

2 5 10

X

Xx x

-Tiptcbm:SHIFTCALCSOLVE=(ch6giy)

Mnhnhhinth:

XlLcntm

Vy:L= 5.10-4H.

1002 = X2 + (120-60)2

X= 80 L--R = 0

5

9

110

2 5 10

X

Xx x

X= 5.0660 x 10

-4 L--R = 0



PHN BN: DNG CC HNG S CI T SN TRONG MY TINH: I. Cc hng s VT L v I N V VT L : 1.CC LNH: CchngsccisntrongmytinhFx570MS;Fx570ES;570ESPlus;VINACAL570ESPlusbngcclnh: [CONST] Number [0 40] (xemccmlnhtrnnpcamytnhcmtay). +Lu :Khitnhtondngmytnhcmtay,tytheoyucubicthnhptrctipcchngstbicho,hocnumunktquchnhxchnthnnnhpcchng sthngquaccmlnhCONST [0 40] ccitsntrongmytinh!(Xem thm bng HNG S VT L di y)

2..CC HNG S VTT L :

Vimytnhcmtay,ngoicctinchnhtnhtonthunli,thchinccphptnhnhanh,nginvchnhxcthphiktitinchtra cu mt s hng s vt lvimtsnvtrongvtl.CchngsvtlccisntrongbnhcamytnhvinvtronghnvSI.Cchngsthngdngl:

Hng s vt l M s Cch nhp my :

My 570MS bm: CONST 0 40 =

My 570ES bm: SHIFT7 0 40 =

Gi tr hin th

Khilngprton(mp) 01 Const[01]= 1,67262158.10-27(kg)

Khilngntron(mn) 02 Const[02]= 1,67492716.10-27(kg)

Khilnglectron(me) 03 Const[03]= 9,10938188.10-31(kg)

BnknhBo(a0) 05 Const[05]= 5,291772083.10-11(m)

HngsPlng(h) 06 Const[06]= 6,62606876.10-34(Js)

Khilng1u(u) 17 Const[17]= 1,66053873.10-27(kg)

intchlectron(e) 23 Const[23]= 1,602176462.10-19(C)

SAvgar(NA) 24 Const[24]= 6,02214199.1023(mol-1)

Th tchmolkhiukintiuchun(Vm)

26 Const[26]= 0,022413996(m3)

Tc nh sng trong chnkhng(C0)hayc

28 Const[28]= 299792458(m/s)

Gia tc trng trng ti mtt(g)

35 Const[35]= 9,80665(m/s2)

HngsRydbergRH(R) 16 Const[16]= 1,097373157.107(m-1)

Hngshpdn(G) 39 Const[39]= 6,673.10-11(Nm2/kg2)

-V d1: My 570ES:

Cc hng s Thao tc bm my Fx 570ES Kt qu hin th mn hnh Ghi ch Hng s Plng (h) SHIFT7CONST06 = 6.62606876 .10-34 J.s Tc nh sng trong chn khng (C0) hay c

SHIFT7CONST28 = 299792458 m/s

in tch lectron (e) SHIFT7CONST23 = 1.602176462 10-19 C Khi lng lectron (me) SHIFT7CONST03 = 9.10938188 .10-31 Kg

HngsRydbergRH(R) SHIFT7CONST16 = 1,097373157.10 7 (m-1)



II. I N V ( khng cn thit lm):Viccmlnhtacthtrabnginnpcamytnh.-My 570ESbmShift 8Conv[ms]=

-V d 2: T36km/hsang?m/s,bm: 36 Shift 8[Conv] 19= Mn hnh hin th : 10m/s

My 570MSbmShift Const Conv[ms]=

III. V D V CCH NHP CC HNG S: V d 1:Giihnquangincakmlo=0,35m.Tnhcngthotcalectronkhikm?

HD:Tcngthc:0

hc hcA0 A

34 86,625.10 .3.10

60,35.10

=5,67857.10-19J=3,549eV

BM MY:phns SHIFT7 06 hXSHIFT728 Co0,35 X10x -6 =5.6755584x10-19J

isangeV:Chiatipchoe:BmchiaSHIFT7 23 =Hin th:3,5424 eV Nhn xt: Hai kt qu trn khc nhau l do thao tc cch nhp cc hng s !!! V d 2:invtuc2sangMeV:1uc2 = 931,5MeV.My 570ES nhp nh sau: Nhp my: SHIFT7 17 x SHIFT7 28 x2 : SHIFT7 23 : X10X6= hin th 931,494... Vy: 1uc2 = 931,5MeV

IV. V D V CCH DNG LNH SOLVE:

V d 1:Bcsngcaccvchquangphcanguynthirctnhtheocngthc:2 2

1 1 1HR

m n

Vi 7 11,097.10HR m =hngsRittberg.Vchutincbc sng ln nht(ngvim=1->n=2)

cabcxtrongdyLymanl:Tadngbiuthc2 2

1 1 1HR

m n

Viilngchabitl: (bin X)

BM MY:2 2

1 1 1[ ] [7] [16] [ ][ ][ ]

1 2XSHIFT SHIFT CALC

Hinth: X= 1,215.10 -7 m =0,1215m

V d 2:Mtmu Na2411 tit=0ckhilng48g.Sauthigiant=30gi,mu Na2411 cnli12g.Bit Na

2411 lcht

phngx -tothnhhtnhnconl Mg2412 .Chukbnrca Na2411 l

A:15h B:15ngy C:15pht D:15giy

Tadngbiuthc:0

0 .2 :

2

t

Tt

T

mm m Hay m

Viilngchabitl: T ( T lbin X)

Nhpmy:

30

12 48.2 X

Bm:SHIFT CALC= (ch khong thi gian 6s) Hinth: X= 15 .Chn A

T nhng v d ny chng ta c th suy lun cch dng cc cng thc khc!!!

Nguyn tc thnh cng: Suy ngh tch cc; Cm nhn am m; Hnh ng kin tr ! Chc cc em hc sinh THNH CNG trong hc tp! Su tm v chnh l: GV: on Vn Lng Email: [email protected] ; [email protected] T: 0915718188 0906848238



PHN NM: DNG TCH PHN TNH QUNG NG TRONG DAO NG IU HA ( Nh my tnh Fx570ES ,Fx570ES Plus, VINACAL Fx570ES Plus)

I.Xt bi ton tng qut :Mt vt dao ng u ho theo quy lut: s( )x Aco t (1) Xc nh qung ng vt i c t thi im 1t n 2t :t=t2-t1

-giiquytbitonnytachiakhongthigianrtnhthnhnhngphndintchthhinqungngrtnh,trongkhongthigiandtcthcoivntccavtlkhngi:

, sin( t+ )v x A (2)

-Trongkhongthigiandtny,qungngdsmvticl: sin( t+ )ds v dt A dt

-Do,qungngScavttthiim 1t nthiim 2t l:

2 2

1 1

sin( t+ )t t

t t

S ds A dt (3)

-Tuynhin,victnh(3)nhmytnhFx570ESrtchm,tythucvohmsvphabanu(nhiupht).-Dovytacthchiakhongthigiannhsau:

t2- t1 = nT + t; Hoc: t2- t1 = mT/2 + t -Ta bit: +Qung ng vt i c trong 1 chu k l 4A. +Qung ng vt i c trong 1/2 chu k l 2A.

-Nu t 0hoc t 0th vic tnh qung ng l kh khn...-> Ta dng my tnh h tr! II.V d: Mtvtdaongiuhodctheotrc0xviphngtrnhx=6.cos(20t-/3)cm(tobnggiy).Qungngvtictthiimt=0nthiimt=0,7/6(s)lA.9cmB.15cm C.6cmD.27cm

Gii 1:ChukT=2

20 10T s

;Thigiani:t = t2- t1 = t2- 0

0,7 7

6 60s

70

7 160 1 6 6

1 0

n v T

.

T/6ngvigcquay/3tMnAdthyonX0A=3cm(Hnh1)Qungngvtic1chukl4Avtx0nAngvigcquay/3lx0A.Qung ng vt i c:4A+X0A=4.6+3=24+3=27cm.Chn D Gii 2: Dng tch phn xc nh nh my tnh Fx570ES ,Fx570ES Plus, VINACAL Fx570ES Plus:

Vntc: 120 in(20t- )(cm/s)3

v s

.

Qungngvtictrongkhongthigianchol:

2

1

7 / 6 0

0

1 2 0s in (2 0 x - )3

t

t

S d s d x

Nhpmy:Bm:SHIFT MODE 4 Bm

,bm:SHIFT hyp(Dngtrtuyti(Abs)).Vibiuthc

trongdutchphnlvntc,cntrnlthigiancui,cndilthigianu,.bintlx,tacbiuthc

nhsau:

7 / 6 0

0

1 2 0sin (2 0 x - )3

d x Bm = (ch khong trn 3 pht ) hin th: 27. Chn D

O

AA x0x

6

Hnh 1

M



Qu Lu! Sau y l cch khc phc v thi gian! III.Cc trng hp c th xy ra: t2- t1 = nT + t; hoc: t2- t1 = mT/2 + t 1.Trng hp 1: Nuchot2-t1=nT(nghalt=0)thqungngl:S = n.4A2.Trng hp 2:Nuchot2-t1=mT/2(nghalt=0)thqungngl:S = m.2A3.Trng hp 3: Nut0hoc:: t0 Dngtchphnxcnhtnhqungngvtictrongthigianthoct:

=>Tngqungng:S=S1+S2=4nA+S2vi

2 2

1 1

2 sin( t+ )nT

t t

t nT t

S ds A dt

Hoc:S=S1+S2=2mA+S2vi

2 2

1 1 /2

2

/2

' sin( t+ )mT

t t

t mT t

S ds A dt

Tnh S2 hoc S2 dng tch phn xc nh nh my tnh Fx 570ES; Fx570ES Plus sau y:

IV. Chn ch thc hin php tnh tch phn ca MT CASIO fx570ES, 570ES Plus Chn ch Nt lnh ngha- Kt qu Chnh dngnhp/xutton Bm:SHIFT MODE 1 MnhnhxuthinMath.

ChnnvogclRad(R) Bm:SHIFT MODE 4 MnhnhhinthchR

ThchinphptnhtichphnBm:Phm

Mnhnhhinth

dx

Dnghmtrtuyti(Abs) Bm:SHIFT hyp Mnhnhhinth dx

Chbintthaybngx Bm:ALPHA ) MnhnhhinthX

Nhphm sin( + )v A x Bm: sin( + )v A x Hinth sin( + )A x dx

NhpcccntchphnBm:

2

1

t

t nT . Hinth2

1

sin( + )t

t nTA x dx

Bmdubng(=) Bm:= ch hi lu Hin th kt qu:.....

V.CC BI TP : BI TP 1: Chophngtrnhdaongiuho 4 s(4 / 3)( )x co t cm .Tmtngqungngvtictrongkhong0,25sktlcu.

Gii 1:TacChuk2 2 1

0,54 2

T s s

.Dothigianicl0,25sbng1nachuknnqung

ngtngngl2A.=>Qung ng S = 2A = 2.4 = 8cm ( mt na chu k: m = 1 ) Gii 2:Tphngtrnhli,tacphngtrnhvntc: 16 sin(4 / 3)( / )v t cm s ,

Qungngvtictrongkhongthigianchol:

2

1

t

t

S ds 0 ,25

0

16 sin(4 )3

x dx

NhpmyFx570ES: Bm:SHIFT MODE 4 :Bm

,bm:SHIFT hyp Dnghmtrtuyti

(Abs).Vibiuthctrongdutchphnlphngtrnhvntc,cntrnlthigiancui,cndilthigian



u,.bintlx,tac:

0,25

0

16 sin(4 )3

x dx

Bm=ch kh lu...mn hnh hin th: 8=>Qung ng S = 8cmBI TP 2: Mtvtchuynngtheoquylut: 2 s(2 / 2)( ) x co t cm .Tnhqungngcansauthigiant=2,875sktlcbtuchuynng.GII:Vntc 4 sin(2 / 2)( / )v t cm s

*Chukdaong2

1T s

;*Sbnchuk:

2,8755, 75 5

0, 5m

(chlyphnnguyn)

*Qungngtrong5bnchuk: '1 2 2.5.2 20S mA cm

*Qungngvtictrongt: 2 21

2

' ( )mTS t t

Vi 15

0 2, 52 2

mTt s

Tac:

2

1

2,875

2

/2 2,5

' 4 sin(2 - )2

t

t mT

S ds t dt

NhpmytnhFx570ES: Bm:SHIFT MODE 4 Bm:

2,875

2,5

4 sin(2 - )2

x dx

=

Ch vi pht ..mn hnh hin th: 2,585786438=2,6 => Qung ng S = 2mA + S2 = 20 + 2,6 = 22,6cm BI TP 3:Mtvtdaonguhocphngtrnh: 2 s(4 / 3)( ) x co t cm Tnhqungngvtictlct1=1/12snlct2=2s.

GII:*Vntc 8 sin(4 / 3)( / )v t cm s *Chukdaong:2 1

2T s

*Sbnchukvtthchinc:

12

2312 71 34

m

(lyphnnguyn)=>m=7

*Qungngvtictrongmnachuk:1 /21 1

' ( ) 2 . 2. .2 87 2mT

S t t mA cm

*Qungngvtictrongt: 2 1 /2 2' ( )mTS t t Vi 11 7 22

/ 2)12 4 12

t mT s =11/6s

Tac:

2

1

2

2

/2 11/6

' 8 sin(4 t- )3

t

t mT

S ds dt

NhpmytinhFx570ES: Bm:SHIFT MODE 4 Bm:

2

11/ 6

8 s in (4 - )3

x d x

=

Ch vi giy..mn hnh hin th : 3 => Qung ng S= S1+ S2 = 2mA + S2 = 28+3 =31cm

VI. PHNG PHP CHUNG : Qua cc bi tp trn, chng ta c th a ra phng php chung gii cc bi ton tm qung ng vt i c trong khong thi gian t2-t1 :1.Cncvophngtrnhdaong,xcnhccilngA, vT.Vitphngtrnhvntccavt.

2.Chiakhongthigian:t2- t1 = nT + t hoc:t2- t1 = mT/2 + t.



3.Sau tnhqungngvtictrongsnguynchukhocsnguynbnchuk,tngngviqungngtrongkhongthigianNTlS1 = 4nAhocmT/2lS1 = 2mA4.DngtchphnxcnhnhmytinhFx570Es,Fx570ESPlustmnhanhqungngitrongt

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