su dung may tinh casio giai bai tap trac nghiem
TRANSCRIPT
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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
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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)
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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
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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.
Thchinphptnhvsphc Bm:MODE 2 MnhnhxuthinCMPLX
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
Nhpkhiugc BmSHIFT (-). Mnhnhhinth
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)
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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
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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
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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.
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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.
Thchinphptnhvsphc Bm:MODE 2 MnhnhxuthinCMPLX
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
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(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
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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
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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
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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
-
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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).
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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)
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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.
Thchinphptnhvsphc Bm:MODE 2 MnhnhxuthinCMPLX
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
-BmSHIFTMODE31:Citdngtocc:(a + bi).200 2 45
(2 0)
uZ
i:Nhp200 2 SHIFT (-) -45 : ( 2 SHIFT (-) 0 ) =Hinth:100-100i
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M ( ) L CZ R Z Z i .Suyra:R=100;ZC=100.Vyhpkn(en)chahaiphntR, C.
V d 3: Mthpkn(en)chchahaitrongbaphntR,L,Cmcnitip.Nutvohaiumchmt
inpxoaychiuu=200 6 cos(100t+6
)(V)thcngdnginquahpenl
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
-BmSHIFTMODE31:Citdngtocc:(a + bi).200 6 30
(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.
V d 4: 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 : BmMODE2mnhnhxuthinCMPLX.-Chnnvogcl(D),bm:SHIFTMODE 3 mnhnhhinthchD
-BmSHIFTMODE31:Citdngtocc:(a + bi).200 2 45
(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
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+ 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
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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.
Thchinphptnhvsphc Bm:MODE 2 MnhnhxuthinCMPLX
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
Nhpkhiugc BmSHIFT (-). Mnhnhhinth
- 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 )
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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
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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
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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
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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
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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
Cch gii truyn thng Hng dn bm my v kt qu
-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
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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
Cch gii truyn thng Hng dn bm my v kt qu
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.
Cch gii truyn thng Hng dn bm my v kt qu
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
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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.
Cch gii truyn thng Hng dn bm my v kt qu
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
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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
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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
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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
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c)V d 3:Tnhchiudicaconlcndaongnh,khibitchukT=2(s)vgiatctrngtrngg=
2(m/s2).Tadngbiuthc: 2l
Tg
Phng php truyn thng Phng php dng SOLVE
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
-Tiptcbm:SHIFTCALCSOLVE=
(chkhong6s)
Mnhnhhinth:
Xlilngl
Vy:l=1(m)
c)V d 4:Tnhgiatctrngtrngtinicconlcn,khibitchukT=2(s)vchiudicaconlc
ndaongnhl1m.Tadngbiuthc: 2l
Tg
Phng php truyn thng Phng php dng SOLVE
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
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c)V d 5:inptvohaiumtonmchR,L,Ckhngphnnhnh.inphiudnghaiumchl100V,haiucuncmthunLl120V,haibntCl60V.inphiudnghaiuRl:A.260VB.140VC.80VD.20V
Phng php truyn thng Phng php dng SOLVE
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
-Tiptcbm:SHIFTCALCSOLVE=
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.
Phng php truyn thng Phng php dng SOLVE
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
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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)
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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
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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
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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
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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.
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3.Sau tnhqungngvtictrongsnguynchukhocsnguynbnchuk,tngngviqungngtrongkhongthigianNTlS1 = 4nAhocmT/2lS1 = 2mA4.DngtchphnxcnhnhmytinhFx570Es,Fx570ESPlustmnhanhqungngitrongt