Hệ thống lý thuyết + công thức giải nhanh vật lý 12

5/25/2018 H th ng l thuy t + c ng th c gi i nhanh v t l 12

1/

Ex: Nguy n H ng Khnh_ HKPH THNG CNG THC - L THUYT GII NHANH VT L 12 Mobile: 09166.01248

TI LIU CHUN LUYN THI I HC 2012 Email: [email protected]

Gio Dc Hng Phc -Ni Khi u c M! HP 1

CHNG I: DAO NG CBI 1: I CNG DAO NG IU HA

I: PHNG PHP1.KHI NIM

Dao ng l chuyn ng c gii hn trong khng gian lp i lp li quanh v tr cn bng. Dao ng iu ha l dao ng trong li ca vt l mt hm cosin( hay sin) ca thi gian.2.PHNG TRNH DAO NG IU HA. x= Acos(t+)

Trong :

x: Li , li l khong cch t vt n v tr cn bngA: Bin ( li cc i): vn tc gc( rad/s)t + : Pha dao ng ( rad/s ): Pha ban u ( rad)., A l nhng hng s dng; ph thuc vo cch chn gc thi gian, gc ta .

3.PHNG TRNH GIA TC, VN TC.v = - Asin( t +) =Acos( t ++

2) = x v max=A.

a = - 2Acos( t +) = - 2x =

2Acos( t ++ ) a

max=

2A

=a maxv max

; A =v2maxa max

.

4.CHU K, TN S.A. Chu k:T =

2

=t

N( s)Trong :

t: l thigianN: l sdaongthchinctrong khongthigian t

Thi gian vt thc hin c mt dao ng hoc thi gian ngn nht trng thi dao ng lp li nh c.

B. Tn s: f =2

=Nt( Hz) Trong :

t: l thigianN: l sdao ngthchinctrong khongthigian t

Tn s l s dao ng vt thc hin c trong mt giy( s chu l vt thc hin trong mt giy).

5.CNG THC C LP THI GIAN:+ x = Acos( t +) cos( t+ ) =

xA

cos2 ( t +) = (xA

)2 (1)

+ v = -A. sin ( t +) sin ( t +) = -v

A.

sin2 ( t +) = (v

A.

)2 = (v

V

max

)2 (2)

+ a = - 2.Acos( t +) cos ( t +) = -a

2A cos2 ( t +) = (

a2A

)2 = (a

a max)2 (3)

T (1)v (2)cos2 ( t + ) + sin2( t +) = (

xA

)2 + (v

A.)2 = 1 A

2 = x

2 + (

v)2 ( Cng thc s 1)Ta c: a = - 2.x x = -

a2

x2 =a24

A2 =a24 + (v)2 ( Cng thc s 2)

T (2)v (3)ta c: sin2( t + ) + cos2 ( t +) = (

vV max

)2 + (a

a max)2 = 1. ( Cng thc s 3)

6.M HNH DAO NG

7. CNG THC LNG GIC QUAN TRNG

V > 0

(+)A- A

a < 0a > 0

V T CB

Xt x

Xt V

Xt a

x < 0

V max

a = 0


2/




1.- sin = sin( +)- cos = cos( +)

2.sin= cos(-2)

cos = sin (+2)

3.cos (a+ b) = cosa.cosb - sina .sinbcos(a - b) = cosa.cosb + sina .sinb

4.

cos a + cosb = 2 cos

a+ b

2 cos

a - b

2

5.sin ( + k2) = sin cos( + k2) = cos

6.Cos2x = 1 + cos2x2

Sin2 x =1 - cos2x

2

7. tan(a + b) = tana + tanb1 - tana.tanb

8.MT S TH C BN.

BI 2: BI TON VIT PHNG TRNH DAO NG IU HAI. PHNG PHPBc 1:Phng trnh dao ng c dng x = Acos(t +)Bc 2:Gii A, , .

- Tm A:A = x2+

v22

=a24

+v22

=v max

=a max2

=L2

=S4=

v2maxa max

Trong :o L l chiu di qu o ca dao ngo S l qung ng vt i c trong mt chu k

- Tm :

x

t

A

-A

th ca li theo thi gian th x - t

th ca vn tctheo thi gian th v - t

v

t

A

-A th ca gia tcthi gian

th a - t

a

x

A-A

A .2

- A .2

x

v

A.

- A.

A- A v

a

A. 2

- A. 2

- A. - A.

th ca gia t c theo li th a -x

th ca vn t c theo li th x -v

th ca gia t c theo vn t c th v -a

t

2A

2A

a


3/




=2T

= 2f =a maxA

=v maxA

=a maxv max

=v2

A2 - x2

- Tm:Cn c vo t = 0 ta c h sau:

x = Acos= x

o

v = - Asinv > 0 nuchuynngtheo chiudngv < 0 nuchuynngtheo chium.

cos =xo

A

sin > 0 nuv 0

Bc 3:Thay s vo phng trnh

BI 3: NG DNG VLG TRONG GII TON DAO NG IU HA1. BI TON TM THI GIANNGN NHT VT I T AB.Bc 1:Xc nh gc .

Bc 2:t = = 2 .T =

O

360O.T

Trong :- : L tn s gc- T : Chu k- : l gc tnh theo rad; 0 l gc tnh theo

A

AB

B

2.BI TON XC NH THI IM VT QUA V TR M CHO TRC.V d: Mt vt dao ng iu ha vi phng trnh x = 4cos( 6t + 3) cm.

A.Xc nh thi im vt qua v tr x = 2 cm theo chiu dng ln th 2 k t thi im ban u. Hng dn:

- Vt qua v tr x = 2cm ( +):6t +

3 = -

3 + k.2

6t = -23

+ k2

t = -19 +

k3 0 Vy k ( 1,2,3)

V t 0 t = -19 +

k3 0 Vy k =( 1,2,3)

- 4 42 (+)

= - /3

-Vt i qua ln th 2, ng vi k = 2.

t = -19 +

23 =

59s

B.Thi im vt qua v tr x = 2 3 cm theo chiu m ln 3 k t t = 2s.


4/




Hng dn:- Vt qua v tr x = 2 3 theo chiu m:

6t +3 =

6 + k2

6t = - 6 + k2

t = -136

+k3

V t 2

t = - 136

+k3 2 vy k = ( 7,8,9)

- 4 4

2 3

= /6

- Vt i qua ln th 3, ng vi k = 9

t = -136

+93 = 2,97s.

3.BI TON XC NH QUNG NG.Loi 1: Bi ton xc nh qung ng vt i c trong khong thi gian t.Bc 1:Tm t, t = t2- t

1.

Bc 2: t = a.T + t3

Bc 3:Tm qung ng. S = n.4.A + S

3.Bc 4:Tm S 3: tmc S3ta tnh nh sau:

- Ti t= t1: x1 = ? v >0v < 0- Ti t = t2; x2= ? v >0v < 0.

Cn c vo v tr v chiu chuyn ng ca vt ti t1v t2 tm ra

S3Bc 5:thay S 3vo S tm ra c qung ng.

AB

n.T S 1 = n.4.A

t 3

S 3

Loi 2: Bi ton xc nh S max- S

minvt i c trong khong thi gian t ( t t >T2

)


5/




A- A S max

A. Tm S max

S max = 2 A + A.cos 2-2 Vi[ ]=.t

A- A

S min

B. Tm Smin

S min = 4A - 2.A sin2-

2 Vi[ ]=.t

4.BI TON TNH TC TRUNG BNH.A. Tng qut:

v =St Trong

- S: l qung ngictrong khongthigian t- t: l thigian vt icqung ngS

- Tc trung bnh trong mt chu k v = 4AT

= 2vmax

B. Bi ton tnh tc trung bnh cc i ca vt trong khong thi gian t:

v max =S max

t

C. Bi ton tnh tc trung bnh nh nht vt trong khong thi gian t.

v min =S min

t

5.BI TON TNH VN TC TRUNG BNH.v

tb =

x

t Trong :

x: l binthindicavt

t: thigian vtthchincdix

6.BI TON XC NH S LN VT QUA V TR X CHO TRC TRONG KHONG THI GIANt

V d: Mt vt dao ng iu ha vi phng trnh x = 6cos( 4t + 3) cm.

A.Trong mt giy u tin vt qua v tr cn bng bao nhiu ln:Hng dn:Cch 1:

Mi dao ng vt qua v tr cn bng 2 ln ( 1 ln theo chiu m - 1ln theo chiu dng)

1 s u tin vt thc hin c s dao ng l: f = 2 = 2Hz

S ln vt qua v tr cn bng trong s u tin l: n = 2.f = 4 ln.Cch 2:Vt qua v tr cn bng

4t +3 =

2 + k

4t =6 + k

t =124

+k4

- A A

t = 0


6/




Trong mt giy u tin ( 0 t 1)

0 124

+k4 1

- 0,167 k 3,83 Vy k = (0;1;2;3)7. BI TON XC NH PHA BAN U CA DAO NG

- A A

v < 0

v > 0

= 0

- A A

VTB( +)= 0 rad

A/2( -)

- A A

= /3

A/2 ( -) =/3 rad

- A AA/2 (+)

= - /3

A/2 ( +) = - /3 rad

- A A- A/2 (+)

= - 2/3

- A/2 (+)= - 2/3 rad

- A AA 3 /2 (+)

= - /6

A. 3 /2 ( +)= - 6rad

BI 4: CON LC L XOI. PHNG PHP

1.CU TOGm mt l xo c cng K, khi lng l xo khng ng k.Vt nng khi lng mGi

2.TH NGHIM- Th nghim c thc hin trong iu kin chun, khng ma st vi mi trng.- Ko vt ra khi v tr cn bng mt khong A v th khng vn tc u, ta c:Vt thc hin dao ng iu ha vi phng trnh: x = Acos(t +)

Trong :- x: l li (cm hoc m)- A: l bin ( cm hoc m).- t +: pha dao ng ( rad)- l pha ban u (rad).- : Tn s gc ( rad/s)

3.CHU K - TN SA. Tn s gc -( rad/s)=

km

( rad/s). Trong :K:cngcal xo( N/m)m: Khilngcavt( kg)

B. Chu k - T (s): Thi gian con lc thc hin mt dao ng

T =2

= 2mk

( s);

C. Tn s - f( Hz): S dao ng con lc thc hin c trong 1s

f =2

=12

km

( Hz).

K m


7/




4. BI TON

K Gn m 1 T1

Gn m2 T

2

Gn m =(m 1+ m2)

Gn m =(m 1+ m 2) f = f1.f2f

12 + f

22

Bi ton 1

T2 = T

12 + T

22

Bi ton 2Vi con lc l xo treo thng ng ta c cng thc sau:

( P = F dhmg = kl mk

=lg

= 2)

T = 2 lg

s; f = 12 gl Hz

BI 5: CT- GHP L XOI. PHNG PHP

1.CT GHP L XOCho l xo koc di l

o, ct l xo lm n on, tm cng ca mi

on. Ta c cng thc tng qutsau:Kol

o= K

1l

1= K

2l

2= .= K

nl

n

Trng hp ct lm hai on:Kolo= K

1l

1= K

2l

2

K1K2

= l2

l 1

Nhn xt: L xo c di tng bao nhiu ln th cng gim i bynhiu ln v ngc li.

l

o, K

o

l 1, K1L

2, K

2L

3, K

3

2.GHP L XOa.Trng hp ghp ni tip:

K 1

K 2

m

K1 K2

Bi ton lin quan thng gp

Ta c: 1K

= 1K1

+ 1K2

K = K1. K2K1+ K

2

T = 2 m( K1+ K2)K1.K

2

( s)

f = 12 K

1.K

2

m(K 1+ K2)

( Hz)

m K1 T1

K2 T2

K1nt K2 K1nt K2 f = f1.f2

f12 + f

22

Bi ton 1

T2 = T12 + T

22


8/




b.Trng hp ghp song songK1

K2

K1 K

2

K1 K

2

Bi ton lin quan thng gp

Khi ghp song song ta c: K = K 1+ K2

T = 2 mK1+ K

2( s )

f = 12 K

1+ K

2

m(Hz)

m K1 T

1

K2 T

2

K1// K2

K1nt K2 f2 = f12 + f22

Bi ton 2

T = T

1.T

2

T12 + T

22

BI 6: CHIU DI L XO - LC N HI - LC PHC HII. PHNG PHP1. CON LC L XO TREO THNG NG

TH1: l >A

+

F dh= 0 V tr l xo khng bin dng

F ph= 0 V tr cn bng

l o

l

l

-A

A

-A

A

TH2: l A

l

o

A. Chiu di l xo:- Gi lol chiu di t nhin ca l xo- l l chiu di khi con lc v tr cn bng: l = lo+l- Al bin ca con lc khi dao ng.


9/




- Gc ta ti v tr cn bng, chiu dng hng xung di.

L max=l

o+l+A

L min=l0+l-A

B. Lc n hi: F dh= - Kx ( N)( Nu xt v ln ca lc n hi).F dh = K.( l + x)- F dhmax= K(l + A)

- F dhmin=K ( l - A) Nul > A0 Nul A

(F dhminti v tr l xo khng bin dng)

C. Lc phc hi ( lc ko v):F ph= ma = m (- 2.x) = - K.x

Nhn xt: Trng hp l xo treo thng ng lc n hi v lc phc hi khc nhau.

Ch :Trong trng hp A > l th l xo s b nn.- F nn = - K( |x| - l) vi |x| l.- F nenmax = K.( A - l)Tm thi gian l xo b nn, gin trong mt chu k.- Gi nnl gc nn trong mt chu k.

- nn= 2.Trong : cos.=lA

- t nn = nn

t gin=

dn

= 2-

nn

= T - t dn

2. XT CON LC L XO NM NGANG.

i vi con lc l xo nm ngang ta gii bnh thng nh con lc l xo treo thng ng nhng:- l = 0.

l = l

o

l max= l + Al min= l - A

F dhmax= K.AF dhmin= 0

- ln lc phc hi bng vi ln lc n hi. F

ph= F

dh = K.x.

BI 7: NNG LNG CON LC L XOI.PHNG PHP

Nng lng con lc l xo: W = Wd+ Wt

Trong :W: l c nng ca con lc l xo

Wd: ng nng ca con lc ( J ) Wd =

12m.v2

Wt: Th nng ca con lc ( J ) Wt=

12K.x2

K m

M hnh CLLX

*** Wd=12mv2 = 12m(-Asin(t+))2 =12m2A2 sin2(t +).

w dmax=12m2A

2 =

12m.vo

2

*** Wt=12Kx2 =

12K( Acos (t +) )2 =

12KA2cos

2(t +).

Wt

max=12kA2

W =Wd+ Wt=

12m2A

2 sin

2(t + ) +

12KA2cos

2(t +) =

12m2A

2( sin

2(t + ) + cos

2(t +) )

=12m2A

2 = const. C nng lun bo ton.


10




*** Tng kt:

W = Wd+ Wt=

12m.v2 +

12K.x2

= W dmax =12m2A

2 =

12m.vo

2

= W tmax=12kA2

W

W0=1/2KA

2

W0/2

t(s)0

W

Wt

th nng lng ca CLLXTa li c:

Wd=12m2A

2 sin

2(t +) =

12m2A

2(

1-cos(2t+2)2

)

=14m2A

2 +

14m2A

2cos(2t+2)

t Td l chu k ca ng nng

T =2

=22

=T2. Chu kngnng= chu kthnng=

T2

t fdl tn s ca ng nng:

fd=1Td

=2T

= 2f. Tnsngnng= tnscathnng= 2f

Thigian lin tipngnngv thnngbngnhau: t = T4.

Mt s ch trong gii nhanh bi ton nng lng:

Cng thc 1: V tr c W d= n.Wt x =

A

n + 1

Cng thc 2:T s gia tc cc i v gia tc ti v tr c W d= n.Wt

a maxa

= n + 1

Cng thc 3: Vn tc ti v tr c Wt= n.Wd v =

Vo

n + 1

BI 8: CON LC NI. PHNG PHP

1. CU TOGm si dy nh khng dn, u trn c treo c nh u di c gn vi vt nng c khi lng m2. TH NGHIMKo con lc lch khi v tr cn bng gc ori bung tay khng vn tc u trong mi trng khng c ma st ( mi lc cnkhng ng k) th con lc n dao ng iu ha vi bin gc o (

0 10

o).

3. PHNG TRNH DAO NG:

o

S o

ll


11




Ta c phng trnh dao ng ca con lc n c dng:s=Scos(t+)= ocos(t+) s = l.Trong :

- s: cung dao ng ( cm, m ..)-S:bin cung ( cm, m ..)-: li gc ( rad)- o:bin gc ( rad)-= g

l( rad/s) vig l gia tctrngtrng(m/s

2)

l l chiudi dy treo ( m)

4. PHNG TRNH VN TC - GIA TC.A. Phng trnh vn tc.

v = s = - Ssin(t +) ( m/s)v max = S

B. Phng trnh gia tca = v = x = - 2 .Scos( t +) (cm/s) = -

2.s ( m/s

2 )

a max= 2.S

5. CHU K - TN S.

A. Chu k. T =2 = 2 lg(s).

Bi ton:

Con lc n c chiu di l1th dao ng vi chu k T

1

Con lc n c chiu di l2th dao ng vi chu k T

2.

Hi con lc n c chiu di l = |l1l2| th dao ng vi chu k T l bao nhiu?T = |T12 T22|

B. Tn s: f =2 = gl (Hz).

Bi ton:

Con lc n c chiu di l1th dao ng vi tn s f1.

Con lc n c chiu di l2th dao ng vi tn s f

2.

Hi con lc n c chiu di l = |l1l2| th dao ng vi tn s l bao nhiu?f-2 = | |f1-2 f2-2 Hoc f = f1.f2| |f

12 f

22

6. CNG THC C LP THI GIAN

S2 = s

2 +

v22 = a24 + v

22

o2 =2 + v22l27. MT S BI TON QUAN TRNG

Bi ton 1: Bi ton con lc n vng inh v mt pha:

T = T1+ T

2

2 l1

l2

T 2/2T1/2


12




Bi ton 2: Con lc n trng phng

= n.T1= (n + 1).T2= T1.T2

| |T 1- T2

Trong :- T1l chu k ca con lc ln hn- T2l chu k ca con lc nh hn- l thi gian trng phng- n: l s chu k n lc trng phng m con lc ln thc hin- n + 1: l s chu k con lc nh thc hin trng phng

l 1

l2

VT

CBVTCB

BI 9: NNG LNG CON LC NI. PHNG PHP1. NNG LNG CON LC N

W = Wd+ Wt

Trong :W: l c nng ca con lc nWd: ng nng ca con lc ( J )Wt: Th nng ca con lc ( J )

- Wd=

12mv

2

w dmax=12m2S

2 =

12.m.Vo

2

- Wt= mgh = mgl( 1 - cos )Wt

max= mgl( 1 - cos

o)

M hnh CLTng t con lc l xo, Nng lng con lc nlun bo ton.

W = Wd+ Wt=

12m.v2 + mgl( 1 - cos)

= W dmax =12m2S

2 =

12m.Vo

2

= W tmax= mgl( 1 - cos o).

W

W0=1/2KA

2

W0/2

t(s)0

W

Wt

th nng lng con lc nTa li c:

Chu kngnng= chu kcathnng=T2

Tnsngnng= tnscathnng= 2f.

Khongthigian ngnngbngthnnglin tipl t =T4

.

2. VN TC - LC CNG DYA. Vn tc:

V = 2gl ( cos - cos o) v max= 2gl( 1 - cos

o) Tivtr cn bng

v min= 0 Tibin

B. Lc cng dy: T

T = mg ( 3cos - 2cos o) T max= mg ( 3 - 2cos

o) Vtr cn bng

T min= mg (cos o) Vtr bin

Mt s ch trong gii nhanh bi ton nng lng:

Nu con lc n dao ng iu ha o 10o th ta c h thng cng thc lm trn sau:( tnh theo rad).

Virt nh ta c: sin =cos = 1 - 2sin22

= cos = 1 -22


13




Thay vo cc biu thc c cha cos ta c:

Wt = mgl.22

=mgs2

2l

W tmax= mglo

2

2 =

mgS22l

v = gl( o2 -

2) V

max=

o gl

T = mg( 1 -3

22 +

o2) T

max= mg( 1 +

o2) > P T

min= mg( 1 -

o2

2) < P

BI 10: S THAY I CHU K CON LC NV BI TON NHANH CHM CA NG H QU LC

I. PHNG PHP

Ta c: T =2 = 2 g( s).

T cng thc trn ta thy c c hai nguyn nhn dn n bin i chu k con lc n l: thay i g hoc .1.THAY IL:1.1. Thay i ln: T = 2

g

1.2.Thay i nh: thay i do nhit :

- Chu k ca con lc nhit t l : T = 2 (1 + t)

g

Trong : - : l chiu di ca con lc n 0oC- : h s n di ca dy treo- t : l nhit ca mi trng

Bi ton 1:Bi ton tm thi gian nhanh hay chm ca ng qu lc trong khong thi gian t. = .2

| t2- t1|

Trong :- t2: nhit mi trng lc ng h chy sai- t1: nhit mi trng ng h chy ng

- : h s n di ca dy treo.- : l thi gian nghin cu( thng thng l 1 ngy: = 86400s)

2. THAY I DO G:2.1. Thay i ln ( di tc dng ca lc khc trng lc)A.Con lc trong thang my:

F qt

v

a

P

a

v

P

F qt

TMLn nhanh dn

TMXung chm dn

Khi thang my ln nhanh dn, xung chm dn:g hd= g + a

T = 2 g hd

= 2 g + a

TMXung nhanh dn

a

v

P

F qt

TMLn chm dn

P

F qt

v

a

Khi thang my xung nhanh dn, ln chm dn:g hd= g - a.

T = 2 g hd

= 2 g - a


14




B.Con lc trn xe di chuyn nhanh dn u hoc chm dn u trn mt phng ngang

F

P

F qt

a

v

Xe t chuyn ng chm dn vi gia tc a

F

P

F qt

a

v

Xe t chuyn ng nhanh dn vi gia tc a

g hd= g2+a2T = 2

g hd= 2

g2 + a2

tan = ag

C.Con lc t trong in trng u:(+) Vt mang in dng - in trng hng t trn xunghoc (vt mang in m -in trng t di hng ln):

E

P

F d

E

P

F d

g hd = g + a = g + | |q Em T = 2 g +

| |q Em

(+) Vt mang in dng -in trng hng t di lnhoc vt mang in m - in trng hng t trn xung

E

P

F

d

E

P

F d


15




g hd = g - a = g - | |q . Em T = 2 g -

| |q .Em

(+) in trng u theo phngnm ngang:

F

P

Fd

E

F

P

F

d

E

g hd= g2+a

2 = g

2 + (

q.Em

)2

ql in tch ca vt ( C ) El in trng ( V/m) ml khi lng ca vt ( kg)

T = 2 g hd

= 2 g2 + (

q.Em

)2

D.Con lc n chu tc dng ca lc y Aximet.Lc y Acximet: F A = .V.g g hd= g + a = g + F Am = g + .V.gm = g + .gD

2.2. Thay i nh:Do thay i chiu cao

T = 2 gh

Trong :gh= G.M

(R+h)2nu ti mt nc bin h = 0.

2.3. Bi ton tnh thi gian nhanh hay chm ca ng h con lc:Bi ton 2:

R

h

ng h qu lc c a ln cao h

Bi ton 3:

ng h qu lc c a xung suh

R

h

R - h


16




A. Khi a ng h ln cao h so vi mt t:

ng h s chy chm hn so vi mt t: = .hR

B. Khi a ng h xung su h:

ng h s chy chm so vi mt t: = .h

2R

C. bi ton nhanh chm ca ng h khi c s thay i ca c cao v nhit :

(+)Ln cao: = .hR

+.2

( t2- t1)

ng h vn chy ng khi t = 0

(+) Xung su: = . h2R

+ .2( t2- t

1)

Hng dn v cc bi ton sai s ca ng h:Gi T1l chu k ca ng h khi ng h chy ngT2l chu k ca ng h khi ng h chy sai.Mi chu k ng h chy sai l: T = T2- T

1

Gi N l s chu k m ng h sai ch trong mt ngy: N =

T2.

Thi gian ch sai trong mt ngy l: = N.( T2- T1) =

T 2

( T2- T1) = (1 -

T1T2

).

Ch :

- Nu = 0: ng h chy ng- Nu > 0: ng h chy chm- Nu < 0: ng h chy nhanh.

Bi ton 1: ( sai s do s thay i ca nhit )Ta c:

T1= 2

1

g = 2

( 1 + t1)g

T2= 2

2

g = 2

( 1 + t2)g

T1T2 =

1 +t11 +t2 1 +

2( t

1- t

2).( v


17




M = D. V =43. .( R - h)3.D g

2= G.

43. .( R - h)3.D

(R - h)2= G.

43.( R - h).D

M = D.V =43. . R3.D g

1= G.

43. . R3.D

R2= G.

43.R.D

T1T2

=g2g1

=G.

43.( R - h).D

G.43.R.D

=R - h

R = 1 -

h2R

( v h


18




bi:Mt vt thc hin ng thi n dao ng thnh phn vi:x1= A

1cos(t +

1)

x2= A2cos(t +

2)

xn= A

ncos(t +

n) tm dao ng tng hp

Bi lmPhng trnh dao ng tng hp c dng: x = Acos( t +)

Bc 1:A X= A

1cos

1+ A

2cos

2++ A

ncos

n

A Y= A1sin

1+ A

2sin

2++ A

nsin

n

Bc 2: A = A X2+A Y2; tan=A YA X

Bc 3:Hon chnh phng trnh x = Acos( t +)

4. TNG HP DAO NG BNG MY TNH B TIa my v Radian hoc ( gc thng nht vi nhau, cng rad hoc , hm cng sin hoc cos)

A. My tnh 750 MS

MODE 2A1SHIFT (-) ( NHPGC 1) +A2SHIFT (-) ( NHPGC 2 ) +AnSHIFT (-) ( NHPGC n ) ly bin A ta nhn : SHIFT + = ly ta nhn: SHIFT =B. My tnh 570 ES + 570ES - PLUSNhp s tng t my tnh 570 MS, nhng khi ly kt qu ta lm nh sau:SHIFT 2 3 =

5. TM DAO NG THNH PHNBi ton: Mt vt thc hin ng thi 2 dao ng iu ha x1, x

2. ta bit x

1= A

1cos(t +

1) v dao ng tng hp ca chng

l: x = Acos( t +). Tm dao ng x2.

Bi lmPhng trnh dao ng tng hp x 2c dng: x2= A

2cos(t +

2)

Cch 1:

A2= A2+A

12-2A.A

1cos(-

1) ; tan

2=

Asin-A1sin1

Acos-A1cos1

Cch2: Casiox = x1+ x

2 x2 = x - x1

MODE 2ASHIFT (-) ( NHPGC ) -A1 SHIFT (-) ( NHPGC 1 ) ly bin A ta nhn : SHIFT + =

ly ta nhn: SHIFT =BI 11: L THUYTCC LOI DAO NG

I. PHNG PHP1.CC LOI DAO NG

Dao ng tun hon:l dao ng m trng thi dao ng lp li nh c sau nhng khong thi gian nh nhauDao ng t do:l dao ng m chu k ca h ch ph thuc vo c tnh bn trong ca hDao ng tt dn:l dao ng c bin gim dn theo thi gian, nguyn nhn ca s tt dn l do ma st vi

mi trng. Ma st cng ln th tt dn cng nhanh.Dao ng duy tr:l dao ng c bin khng i theo thi gian trong s cung cp thm nng lng b

A X1 AX

2

A Y2

A Y1

A Y3

Y

XAX

3 A

X

A 2

A 3

A 1


19




li s tiu hao do ma st ma khng lm thay i chu k ring ca n th dao ng ko di mi mi v gi l dao ng duytr.

Dao ng cng bc: l dao ng chu s tc dng ca ngoi lc bin i iu ha F=Focost- Dao ng cng bc l iu ha c dng hm cos(t).- Tn s ca dao ng cng bc bng tn s gc ca ngoi lc- Bin ca dao ng cng bc ca ngoi lc t l thun vi bin F oca ngoi lc ph thuc vo tn s

gc ca ngoi lc v lc cn mi trng.- Hin tng cng hng: khi bin A ca dao ng cng bc t gi tr cc i. ngi ta ni rng c hin

tng cng hng.

Gi tr cc i ca bin A ca dao ng t c khi tn s gc ca ngoi lc bng tn s gc ring

0ca hdao ng tt dnHin tng cng hng cng r nt khi lc cn cng nh.

Phn bit dao ng duy tr v daong cng bc:Dao ng cng bc Dao ng duy tr

Dao ng cng bc l dao ng xy ra di tcdng ca ngoi lc tun hon c tn s gc bt k.sau giai on chuyn tip th dao ng cng bc ctn s gc ca ngoi lc.

Dao ng duy tr cng xy ra di tc dng ca ngoi lc,nhng y ngoi lc c iu khin c tn s gc bngtn s gc o ca dao ng t do ca h

Dao ng xy ra xy ra trong h di tc dng di

tc dng ca ngoi lc c lp i vi h

Dao ng duy tr l l dao ng ring l dao ng ring ca

h c b thm nng lng do mt lc iu khin bi chnhdao ng y thng qua mt h c cu no .

2.BI TP V DAO NG TTDN CA CON LC L XOBi ton:Mt vt c khi lng m, gn vo l xo c cng k. ko l xo ra khiv tr cn bng mt on A ri bung tay ra cho vt dao ng. Bit h s ma st cavt vi mt sn l

a.Tm qung ng vt i c n kh dng hn?n khi vt dng hn th ton b c nng ca con lc l xo b cng ca lc ma st lm trit tiu:

A ms= W mgS =12kA2 S =

kA22mg

b. gim bin sau na chu k, sau mt chu kGi A1l bin ban u ca con lc l xo, A

2l bin sau na chu k

Ta s c: W = mg( A1+A2) =

12( kA1

2 - kA

22) =

12k( A1+ A

2)(A

1- A

2)

A1- A2=

2.mgk

= A1

A1gi l gim bin trong na chu k.

gim bin sau mt chu k l: A = 2.2.mgk

= 4.mgk

.

c.S dao ng n lc dng hnN = AA

d.Thi gian n lc dng hn t = T.N = T.AA

e.Bi ton tm vn tc ca vt khi vt i c qung ng STa c: W = Wd+ W

t+ A

ms


20




Wd= W - A

ms- Wt

12mv2 =

12K A2 - F

ms. S -

12kx2 v =

K(A2 - x2) - 2F

ms.S

m

Vt s t c vn tc cc khi F hl= 0 ln u tin tix = mg

K

S = A - x

3. BI TP V DAO NG TT DN CA CON LC NCon lc n c chiu di l dao ng tt dn vi mt lc cn u l Fc, bin gc ban u l

o

1.

A. Hy xcnh qung ng m con lc thc hin n lc tt hn ca con lcn.

Ta c: W =12mgl 201 = F

c. S

S =mgl 201.F

c

2

B. Xc nh gim bin trong mt chu k.

Ta c: nng lng ban u ca con lc l: W1=12mgl 201

Nng lng cn li ca con lc khi bin 02. W2=

12mgl 202 S

o

1S

o

2

l o1

o2

Nng lng mt i W = W1- W2=

12mgl 201-

12mgl 202 =

12mgl(201-

202) = F

c.( S

01+ S

02)

12mgl( 01-

02)(

01+

02) = F

c.l. (

01+

02)

01-

02=

2.Fcmg

= 1( const)

gim bin trong mt chu k l: =4Fcmg

C. S dao ng n lc tt hn.N = 01

D. Thi gian n lc tt hn: t = N.TE. S ln i v tr cn bng n lc tt hn: n = 2.N

4.BI TP V CNG HNG.- iu kin cng hng: Tr= T cb Trong :T

rgil chu kring

T cbgil chu kcngbc

- Cng thc xc nh vn tc ca xe la conlc dao ng mnh nht. v = LTr

- Trong :L l chiudi thanh rayTrl chu kring cacon lc

BI 13: CC BI TON NNG CAO.BI TON VA CHM - H VT

1.BI TON VA CHMA.

Va chm mn:- Sau va chm 2 vt dnh vo nhau v cng chuyn ng

- ng lng c bo ton, ng nng khng bo ton.m1.v

1 + m

2.v

2 = ( m

1+ m

2).V

Trong :- m1: l khi lng ca vt 1- m2: l khi lng ca vt 2- m = (m1 + m2) l khi lng ca hai vt khi dnh vo nhau:- v1l vn tc ca vt 1 trc va chm- v2l vn tc vt 2 trc va chm- V l vn tc ca hai vt khi dnh sau va chm


21




B.Va chm n hi ( xt va chm n hi xuyn tm)- Sau va chm hai vt khng dnh vo nhau, chuyn ng c lp vi nhau- ng nng c bo ton

CT1: Bo ton ng lng m1. v1+ m

2. v

2= m

1. v

1 + m

2. v

2 (1)

CT2: Bo ton ng nng:1

2

m1. v12 +

1

2

m2. v22 =

1

2

m1. (v1)

2 +

1

2

m2. (v2)

2 (2)

Gii phng trnh 1 v 2 ta c:

v1 =(m1- m

2).v

1+ 2m

2. v

2

m1+ m2

v2 =( m2- m

1)v

2+ 2.m

1v

1

m1+ m2

2.BI TON XC NH IU KIN BIN DY TREO KHNG TRNG

m

M

A

K

M

A

Xc nh bin ln nht trong qu trnh M daong dy treo khng b trng

A ( M + m)gK

Xc nh bin ln nht trong qu trnh M daong dy treo khng b trng

A M.gK

2.BI TON KHNG DI VT

K

M

A

m

K

M

A

m

K

M

m

Xc nh bin dao ng ln nhtca m vt M khng b nhy ln

khi mt t.A ( M + m)g

K

Bin dao ng nh ln nhtca M vt m khng b nhy ra

khi vt MA ( M + m)g

K

Bin dao ng ln nht ca M m khng b trt ra khi M.

A ( M + m). .gK


22




CHNG II: SNG CBI 1: S TRUYN SNG

I. PHNG PHP.1.CC NH NGHA C BNa.nh ngha sng c:Sng c l dao ng lan truyn trong mt mi trng rn, lng, kh.b.Sng ngang:l sng c trong cc phn t ca mi trng dao ng theo phng vung gc vi phng truyn

sng. Sng ngang truyn trong cht rn v mt cht lng.c.Sng dc: l sng c trong cc phn t ca mi trng dao ng theo phng trng vi phng truyn sng. Sng

dc truyn c c trong mi trg rn, lng, kh.d.c trng ca sng hnh sin:- Bin sng:bin ca sng l bin dao ng ca mt phn t mi trng c sng truyn qua.- Chu k:l chu k ca mt phn t ca mi trng c sng truyn qua. (f = 1

T)

- Tc truyn sng:Tc truyn sng v l tc lan truyn dao ng trong mi trng. Vi mi mi trng c vkhng i.

- Bc sng:+ l qung ng m sng truyn trong mt chu k.+ Hoc l khong cch gn nht ca hai im cng pha trn

phng truyn sng. = v. T =vf( m, cm)

- Nng lng sngl nng lng dao ng ca cc phn t ca mi trng c sng truyn qua.2.PHNG TRNH SNG

Xt ti ngun O: c phng trnh sng l: u O = Uocost

Sng truyn t O n M: u M= Uocos( t - t) = U

ocos( t -

dv) = Uocos(t -

dv

)

= Uocos( t -2fdvf

) = Uocos( t -2d

) t dv.

lch pha dao ng ca hai im trn phng truyn sng: = 2 d = 2d2- d

1

Nu:

- = k2(hai im cng pha) k2= 2d

d = k

Nhng im cng pha trn phng truyn sng cch nhau nguyn ln bc sng.

- = ( 2k + 1)( hai im ngc pha) ( 2k + 1)= 2d

d = ( 2k + 1).2

Nhng im ngc pha trn phng truyn sng cch nhau mt s l ln na bc sng.

BI 2: GIAO THOA SNG C.I. PHNG PHP.1. NH NGHA GIAO THOA SNG

-Hin tng hai sng kt hp, khi gp nhau ti nhng im xc nh, lun lun hoc tng cng nhau to thnh cci hoc lm yu nhau ( to thnh cc tiu) gi l s giao thoasng.- Ngun kt hp l hai ngun c cng tn s v lch pha khng i theo thi gian.

2. GIAO THOA SNG.A.Hai ngun sng cng pha.

O Mu O = U

ocost


23




u1

M= Uocos( t -

2d1

)

u2

M= Uocos( t -

2d2

)

u M= u1

M + u

2

M= U

ocos( t -

2d1

) + Uocos( t -2d2

)

= 2. Uocos( d2- d

1)

.cos

t -

( d1+ d2)

= A M.cos

t -( d

2+ d

2)

Vi A M = |2. Uocos

( d2- d1)

|

Xt bin A = |2.Uocos( d2- d1)

|

A maxkhi cos

( d2- d1)

= 1. ( d2- d

1)

= k d = d2- d

1= k. vi k = 0, 1, 2,

KL: Bin ca sng giao thoa t cc i ti v tr c hiu ng i bng nguyn ln bc sng.

A min khi cos

( d2- d1)

= 0

( d2- d1)

= (k +

12). d = d2- d

1= ( k +

12). vi k = 0, 1, 2 .

KL: Bin ca sng giao thoa t cc tiu ti v tr c hiu ng i bng l ln na bc sng.

B. Hai ngun lch pha bt k.u1

M= U

ocos( t +

1 -

2d1

)

u2

M= Uocos( t +

2 -

2d2

)

u M= u1

M + u

2

M= U

ocos( t +

1 -

2d1

) + Uocos( t +2 -

2d2

)

= 2.Uocos

1- 2

2+

( d2- d1).

cos

t +

1+2

2-

( d2+ d1)

= A M.cos

t +

1+

2

2-

( d2+ d1)

Vi A M= |2.Uocos 1-

2

2 +( d2- d

1).

| = |2.U o.cos -2 +

( d2- d1)

| Trong : = 2- 1

Xt bin A = |2.Uo.cos

-

2

+( d2- d

1)

|

A maxkhi cos

-

2

+( d2- d

1)

= 1.

-

2

+( d2- d

1)

= k

A minkhi cos

-

2

+( d2- d

1)

= 0

-

2

+( d2- d

1)

= (k +12).

3. CC BI TON QUAN TRNGBi ton 1:xc nh s cc i - cc tiu gia hai im MN bt k vi lch pha bt k.


24




Ti M v N

d

M= d

2

M- d

1

M

d N= d2

N- d

1

N

gisd M


25




2d

= k.2 ( d1= d

2= d). k =

d

(3)

V ta c: d 2

k =d

2

k 2

( K l s nguyn). (4)

Thay ( 4)vo (2)v sau thay (2)vo (1 )ta c: u M = 2. Uo.cos( t - k.2)

*** Bi ton tm MI

min

Ta c: k k2

( k nguyn)

V MI min k

min d = k.

MI min = d2 - (

2)2 = (k. )

2 - (

2.)2

S1 S2

M

d 1d2

/2/2 I

***Bi ton xc nh s im dao ng cng pha vi ngun trong on MI

2 k

d Trong :d = MI2 + ( /2)

2

Tng kt:

Khong cch gia hai cc i lin tip l 2.

Khong cch gia hai cc tiu lin tip l 2

Khong cch gia mt cc i v mt cc tiu lin tip l 4.

k = 0 k = 1 k = 2 k = 3 k = 4

S1S2

k = -1k = -2k = -3k = -4

k = 0 k = 1 k = 2 k = 3k = -1k = -2k = -3k = -4

Ct1 Ct2 Ct3 Ct4Ct1Ct2Ct3Ct4

c 1 c 2 c 3 c 4c= 0c -1c -2c -3c -4

BI 3: SNG DNG

1.SNG PHN X.- Sng phn x c cng tn s v cng bc sng vi sng ti.- Nu u phn x c nh th sng phn x ngc pha vi sng ti- Nu u phn x t do th sng ti v sng phn x cng pha vi nhau.2.SNG DNG.

A. Th nghim:

Quan st th nghim nh hnh v:- Ban u khi my cha rung th si dy dui thng.- Khi my rung, iu chnh tn s ca si dy n mt gi tr no th trn si dy hnh thnh mt hnh nh xc nh

vi cc bng v cc nt nh hnh v.Hnh nh quan st trn c gi l sng dng.


26




My rung

Khi my cha rung Khi my rung

My rung

B. nh ngha sng dngSng dng l trng hp c bit ca giao thoa sng, trong c s giao thoa gia sng ti v sng phn x. Nhng imtng cng ln nhau gi l bng sng, nhng im trit tiu ln nhau gi l nt sng.

- Cc bng sng lin tip( cc nt lin tip) cch nhau 2

- Khong cch gia mt bng v mt nt lin tip l 4.

- Cc im trong cng mt bng th lun dao ng cng pha vi nhau.- Cc im bt k hai bng lin tip lun dao ng ngc pha vi nhau.- Bin cc i ca cc bng l 2A, b rng cc i ca bng l 4A.- Thi gian si dy dui thng lin tip l T

2.

b sng2

4

bng sngnt sng

3.IU KIN C SNG DNG.A.Sng dng trn si dy c hai u c nh

l = k.2

Vi k = ( 1,2,3 l min =2 khi k

= 1.

l = k.v2f

f = k.v2l

fmin =v2l

khi k = 1.

2

l

B.Sng dng trn si dy c mt du c nh - mt u t do.l = k.

2 +

4 = (2k + 1).

4= m.

4

Vi m = ( 1,3,5)

l min =4

. Khi k = 1.

l = m.v4f

f = m.v4l

vi k = ( 1,3,5)

f min =v4l

vi khi m = 1.2

4

l

4.PHNG TRNH SNG DNG.A.Trng hp sng dng c u phn x l u c nh.

Loi 1: Ti im M trn dy nh hnh v c phng trnhsng ti utM= U

0cos(t +). Hy xy dng phng trnh

sng dng ti M.M

O

ut

M= U

0cos( t + )

d

Hng dn:

u M = u

tM + u

pM Trong : u tMl sng titiMu pMl sng phnxtiM

Mun c u pMta cn c u

pO( sng phn x ti O) mun c u

pOta cn c u

tO( sng ti ti O).


27




u tO= U0cos( t +-

2d

). u pO= U0cos( t + -

2d

- ) ( v sng ti v sng phn x ngc pha).

u pM= U0cos( t + -

4d

- )

u M = u

tM + u

pM = U0cos(t +) + U0cos( t +- 4d - )

= 2 U0cos(2d + 2)cos( t +- 2d - 2).

Loi 2: Ti im O trn dy nh hnh v c phngtrnh sng ti ut

O = U

0cos(t + ). Hy xy dng

phng trnh sng dng ti M.

MO

d

utO= U

0cos( t + )

Hng dn:Phng trnh sng ti M: u M = u

tM + u

pM

- Xy dng u tM : ut

M= U

0cos( t ++

2d

).

-Xy dng u pM:

u

pO= U

0cos( t + - ) u

pM= U

0cos( t + - -

2d

)

u M = u

tM + u

pM = U0cos( t + +

2d

) + U0cos( t +- -2d

)

= 2U0cos(2d + 2)cos( t +- 2).

Nhn xt:Vi trng hp sng dng c u phn x l u c nh th bin ca sngA = 2U0cos(2d +

2

)

B.Phng trnh sng dng trong trng hp u phn x l u t do:

Loi 3: Ti im M trn dy nh hnh v cphng trnh sng ti ut M= U0cos(t +). Hy xydng phng trnh sng dng ti M.

MO

ut

M= U0cos( t + )

Hng dn:u M = u

tM+ u

pM

Xy dng u tM: ut

M= U

0cos(t +).

Xy dng u pM: u

tO = U0cos( t + -

2d

). u pO= U0cos( t +-

2d

) ( v sng ti v sng phn x cng pha)

u pM= U0cos( t +- 4d )u M= u

tM+ u

pM= ut

M= U

0cos( t +) +U

0cos( t +- 4d

) = 2U0cos( 2d )cos( t + - 2d )Nhn xt: Bin ca sng dng trong trng hp u phn x l t do l A = 2U0cos(

2d )

BI 4: SNG M

1.SNG M.- Sng m l nhng sng c hc truyn trong mi trng rn lng kh


28




- Mt vt dao ng pht ra m gi l ngun m- Sng m c th truyn trong mi trng n hi ( rn lng kh).- Sng m khng truyn c trong chn khng.- Tnh n hi ca mi trng cng cao th tc m cng ln. tc truyn m theo th ( kh, lng, rn).- Trong cht kh v cht lng sng m l sng dc, cn trong cht rn sng m l sng dc hoc sng ngang.2.C TRNG VT L CA SNG M.

A. Tn s m:l mt trong nhng c trng vt l quan trng nht ca m. - m c tn s nh hn 16Hz th tai ngi khng nghe c gi l h m.- m c tn s ln hn 20000Hz th tai ngi cng khng nghe c gi l sng siu m.-Nhng m m tai c th nghe c gi l m thanh. m thanh c tn s nm trong khong t ( 16Hz n 20000Hz)B. Cng m- I : ( W/ m2 )L i lng o bng lng nng lng m sng m ti qua mt n v din tch t ti im , vung gc vi

phng truyn sng trong mt n v thi gian.

I =PS

=Wt

=P

4R2 I A.R A2 = IB.R B2 Trong : P l cng sutngunm: WS: dintch sng m truynqua (m2)C. Mc cng m:L (B) = lg(

IIo

) (B) = 10lg(IIo

) ( dB) Trong :I:Cngm tiim nghin cu( W/ m2 )Io: Cngm chun( W/m

2 )

3.C TRNG SINH L CA SNG M.- cao: cao ca m l mt c trng sinh l ca m gn lin vi tn s n.- to: to ch l mt khi nim ni v c trng sinh l ca m gn lin vi c trng vt l mc cng m v tn

s.- m sc:m sc l mt c trng sinh l ca m, gip ta phn bit m do cc ngun khc nhau pht ra c cng tn s

v khc nhau v bin .4.NHC M- Nhc m l cc m do nhc c pht ra.- Nhc m c th l cc ng cong tun hon.Ha m:

A. Vi n c hai u dy c nh:l = k.

2 = k.

v2f

f = k.v2l

= k. f min

-Trong :f min=

v2l

k l ham bck vik = ( 0,1,2,3 ..)

- Vi v l vn tc truyn sng m trn dy: v =

l lccngcady (N)l mtdi ( kg)

B. Vi ng so c mt u kn - mt u h.l = m.

4 = m.

v4f

f = mv4l

= m.f min.

-Trong :f min=

v4l

m l ham bcm vim = (1;3;5;7 )

5. CC CNG THC LOGARIT C BN:1.logab = xb = ax 3. lgb = x b = 10x2.lg( a.b) = lg a + lgb 4. log a

b = lga - lgb

2

4

2


29




CHNG III: SNG IN TBI 1: MCH DAO NG LC

I. PHNG PHP1. Phng trnh in tch

q = Qo.cos( t + ) (C )2. Phng trnh dngin

i = q = .Qo.cos( t ++2) A.

= Io.cos( t ++ 2) ( A ) Trong : ( Io = .Qo)

3. Phng trnh hiu in th

u =qC

=QoC

.cos( t + ) ( V)

= Uo.cos( t +) ( V) Trong : ( Uo =

QoC

)

C

L

+-

S mch LC

Mch LC hot ng datrn hin tng t cm

4. Chu k - Tn s:A. Tn s gc:( rad/s)=

1LC

Trong :L giltcmcacundy ( H)C l indung catin( F) C =

.S4Kd Trong :

: l hng sinmi

S: dintch tipxc cahai bntK = 9.109d: khongcch giahai bnt

B. Chu k T(s)

T =2

= 2 LC

C. Tn s: f ( Hz)

f =2

=1

2 LC

5. Qui tc ghp t in - cun dyA. Ghp ni tip

- T in:1C

=1C1

+1C2

C =C1. C

2

C1+ C2; C2 =

C. C1C 1- C

; C1 =C.C2C2- C

- Cun dy: L = L

1+ L

2

C 1 C2

L 1 L2

B. Ghp song song

- Ghp t in: C = C1+ C2

- Ghp cun dy:1L

=1L1

+1L2

L1

L2

C 1

C 2

6. Bi ton lin quan n ghp t


30




L C 1 T 1

C 2 T2

C 1ntC 2 C 1// C 2 T2 = T12 + T22

Bi ton 1

T =T1.T

2

T 12 + T

22

L C1 f

1

C 2 f2

C 1ntC2 f2 = f12 + f22

C 1// C 2

Bi ton 2

f =f1.f2

f12 + f

2

2

7 . Bng qui i n vQui i nh ( c) Qui i ln ( bi)Stt

K hiu Qui i K hiu Qui i1 m ( mini) 10-3 K ( kilo) 10

3

2 ( micro) 10-6 M ( m ga) 106

3 N ( nano) 10-9 Gi ( giga) 109

4 A0 ( Axittrom) 10-10

5 P ( pico) 10-12 T ( tetra) 1012

6 f ( fecmi) 10

-15

8. Bi ton vit phng trnh (u - i - q)Loi 1: Gii s bi cho phng trnh : q = Qo.cos(t +) C

i = Io.cos( t ++2) A Trong : [ ]Io=.Qo

u = Uo.cos( t +) V Trong :

Uo=

QoC

Loi 2: Gii s bi cho phng trnh : i = Io.cos(t + ) Aq = Qo.cos( t +-

2) C Trong :

Qo=

Io

u = Uo.cos( t +-

2

) V Trong :

Uo= IoL

C

Loi 3: Gii s bi cho phng trnh : u = Uo.cos(t +) Vq = Qo.cos( t+ ) C Trong : [ ]Q

o= C.U

o

i = Io.cos( t ++2) A. Trong :

Io= Uo.

CL

BI 2: NNG LNG MCH LCI. PHNG PHP.

1.Nng lng ca mch LC.Nng lng mch LC: W = Wd+ W

t trong :

- W : Nng lng mch dao ng ( J)- Wd: Nng lng in trng ( J) tp trung t in- Wt: Nng lng t trng ( J) tp trung cun dy.Wd=

12Cu2 =

12qu = 1

2q

2

C =1

2Q

2

Ccos2 ( t ).

W dmax =12CUo

2 =

12Q2C

- Wt: Nng lng t trng ( J). wt=12Li2 = 12L2Q2sin2( t).W tmax=

12LIo

2.

C

L

+-

S mch LC

t(s)0

WW0

W0/2

W

Wt


31




Tng Kt

W = Wd+ Wt

=12Cu21 +

12Li1

2 =

12Cu22+

12Li22=

12qu +

12Li2 =

12q2C

+12Li2

= W dmax=12Qo

2

C =

12C.Uo

2

= W tmax=12LIo

2

Ta c mt s h thc sau:LIo

2 - Li

2 = Cu

2 L ( I

o2 - i

2 ) = C.u

2

LIo2 - Li

2 =

q2C

L(Io2 - i

2) =

q2C

I02 - i

2 =

2.q

Qo2

C =

q2C

+ Li2 Qo2 - q

2 = LC.i Q

o2 - q

2 =

i2

C( Uo2 - u

2 ) = Li

2

CL

(Uo2 - u

2) = i

2

Io= Uo

CL

; Uo= Io

LC

2.Cng thc xc nh nng lng ta( nng lng cn cung cp duy tr mch LC)P = I2.R =

Io2 .R

2

Mt s kt lun quan trng.

- Nng lng in trng v nng lng t trng bin thin tun hon vi chu k l T2

- Nng lng in trng v nng lng t trng bin thin tun hon vi tn s l 2f.- Thi gian lin tip ng nng v th nng bng nhau l t = T

4.

BI 3: SNG IN T V TRUYN THNG BNG SNG V TUYN1.IN T TRNG

Mi bin thin theo thi gian ca t trng u sinh ra trong khng gian xung quanh mt in trng xoy bin thi n theothi gian, v ngc li, mi bin thin theo thi gian ca in trng cng sinh ra mt t trng bin thin theo thi gian trong

khng gian xung quanh.2.SNG IN TA. nh nghaSng in t l qu trnh lan truyn in t trng trong khng gianB. c im ca sng in t- Lan truyn vi vn tc 3.108 m/s trong chn khng-Sng in t l sng ngang, trong qu trnh lan truyn in trng v t trng ln truyn cng pha v c phng vung

gc vi nhau-Sng in t c th lan truyn c trong chn khng, y l s khc bit gia sng in t v sng cC. Tnh cht sng in t- Trong qu trnh lan truyn n mang theo nng lng- Tun theo cc quy lut truyn thng, phn x, khc x.- Tun theo cc quy lut giao thoa, nhiuxNgun pht sng in t ( chn t) c th l bt k vt no pht ra in trng hoc t trng bin thin nh: tia la in,

cu dao ng ngt mch inD. Cng thc xc nh bc sng ca sng in t:

= c.T =cf Trong :

gilbcsng sdt

c = 3.108 m/sT: chu ksng int

3.TRUYN THNG BNG SNG V TUYNA.Cc khong sng v tuynMc Loi sng Bc sng c im/ng dng

1 Sng di > 1000 m - Khng b nc hp th- Thng tin lin lc di nc


32




2 Sng trung 100 1000 m - B tng in ly hp th ban ngy, phn x ban mln ban m nghe radio r hn ban ngy- Ch yu thng tin trong phm vi hp

3 Sng ngn 10 100 m - B tng in ly v mt t phn x- My pht sng ngn cng sut ln c th truynthng tin i rt xa trn mt t

4 Sng cc ngn 0,01 10 m - C th xuyn qua tng in ly-Dng thng tin lin lc ra v tr

B.S my thu pht sng v tuyn

1

2

3 4 5

1 32 4S mypht sng

S my thusng

5

Trong :B

phnMy pht B phn My thu

1 My pht sng cao tn 1 n ten thu2 Micro( ng ni) 2 Chn sng3 Bin iu 3 Tch sng4 Khuych i cao tn 4 Khuych i m tn5 Anten pht 5 Loa

C.Truyn thng bng sng in t.Nguyn tc thu pht f my = f

sng

f my =1

2 LC = fsng =

c.

Bc sng my thu c: = c.2 LC4. MT S BI TON THNG GP.Loi 1: Xc nh bc sng my c th thu c:

bi 1:Mch LC ca my thu c L = L1; C = C1, cho c = 3.10

8 m/s. Xc nh bc sng m my c th thu c:

= c.2 L1C1

bi 2:Mch LC ca my thu c t in c th thay i c t C 1n C2( C

1< C

2) v t cm L. Hy xc nh khong

sng m my c th thu c:

=[ ]

1

2

Vi1= c.2 L. C

1

2= c.2 L. C2

bi 3:Mch LC ca my thu c C c th iu chnh t [ ]C1C2 ; L iu chnh c t [ ]L1L2 . Xc nh khong

sng m my c th thu c.

=[ ]

1

2 Vi

1= c.2 L

1.C

1

2= c.2 L2.C

2

bi 4:


33




L C 1 1

C 2 2

C 1ntC 2 C 1// C 2 = 12 + 22

= 1.

2

12 +

22

L C1 f

1

C2 f

2

C 1ntC 2 f2 = f12 + f22

C 1// C 2 f = f1.f

2

f12 + f

22


34




CHNG IV: DNG IN XOAY CHIUBI 1: I CNG DNG IN XOAY CHIU.

I. PHNG PHP.1.GII THIU V DNG IN XOAY CHIU.

A.nh ngha:Dng in xoay chiu l dng din c cng bin thin iu ha theo thi gianB.Phng trnh

i = Io.cos( t + ) ( A)

Hocu = U

o.cos( t +) (V)

Trong :- i:gi l cng dng in tc thi ( A)- Io:gi l cng dng in cc i ( A)- u:gi l hiu in th tc thi (V)- Uo:gi l hiu in th cc i ( V)-:gi l tn s gc ca dng in ( rad/s)C. Cc gi tr hiu dng:

- Cng dng in hiu dng: I =Io2 (A)

- Hiu in th hiu dung: U =Uo

2 (V)

- Cc thng s ca cc thit b in thng l gi tr hiu dngCc bi ton ch cn ch :

Bi ton 1: Xc nh s ln dngin i chiu trong 1s:- Trong mt chu k dng in i chiu 2 ln- Xc nh s chu k dng in thc hin c trong mt giy ( tn s)S ln dng in i chiu trong mt giy: n = 2f

Ch : Nu bi yu cu xc nh s ln i chiu ca dng in trong 1s u tin th n = 2f.- Nhng vi trng hp c bit khi pha ban u ca dng in l= 0 hocth trong chu k u tin dng in ch

i chiu 1 ln:n = 2f - 1.Bi ton 2: Xc nhthi gian n sng - ti trong mt chu k

ts=s

Trong :

s= 4

cos =|u|Uo

tt= t = 2- s

= T - ts

Gi H l t l thi gian n sng v ti trong mt chu k: H =tstt =

st

2. GII THIU V CC LINH KIN IN.Nidung in tr T in Cun dy thun cmK hiu

Tngtr( )R =

.lS

Z C=1

C Z

L= L

c im - Cho c dng in mt chiuv xoay chiu qua n nhng

ta nhit

- Ch cho dng in xoay chiu iqua

- Ch cn ch dng in xoaychiu

Cng thcnh lut

I =UR

; Io=UoR

; i =uR

I =UZl

; Io=UoZl

I =UZ C

; Io=UoZ C

Cng sut P = I2.R 0 0 lch pha u

- iu v i cng pha vi nhau

u chm pha hn i gc 2 u nhanh pha hn i gc

2

Phng trnh u = Uo.cos( t + ) (V)i = I0.cos( t +) A

u = Uo.cos( t +) (V) u = Uo.cos( t + ) (V)

R C

L


35




i = I0.cos( t +2) A i = I0.cos( t -

2) A

Gin u -i

u

i

u

i

u

i

3.QUI TC GHP LINH KIN.

Mc R Z L ZC

Mc ni tipR = R1+ R

2 Z

L= Z

L

1+ Z

L

2 Z

C= Z

C

1+ Z

C

2

Mc songsong

1R

=1R1

+1R2

R =R1. R

2

R1+ R2

1Z L

=1

Z L1+

1Z L

2 ZL=

Z L1. Z

L

2

Z L1+ Z

L

2 Z C=

Z C1. Z

C

2

Z C1+ Z

C

2

4.CNG THC C LP THI GIAN:Vi on mch ch c C hoc ch c cun dy thun cm ( L ) ta c:

(iIo

)2 + (u

Uo)2 = 1

BI 2: MCH IN RLCI. PHNG PHP

1.GII THIU V MCH RLCCho mch RLC nh hnh v:Gi s trong mch dng in c dng: i = Iocos( t +) A

U R= U

oRcos( t) V; uL= U

oLcos( t +

2 ) V; uC= U

oCcos( t -

2)

VGi u l hiu in th tc thi hai u mch: u = u R+ u

L+ u

C

u =U

oRcos( t) + U

oLcos( t +

2 ) + U

oCcos( t -

2)(1)Uo

2 = U

oR

2 + ( U

o

L- U

oC)

2 ( Chia hai v ca (1) cho 2 )

U2 = UR

2 + ( U

L- U

C)

2

(2)Gi l lch pha gia u v i camch in tan =Uo

L- U

oC

U oR =

U L- UC

U R

(3)H s cng sut ( cos ):cos =U oRUo

=U RU

2. NH LUT

Io=U

o

Z

=U oR

R

=U oL

Z

L

=U oC

Z

C

I =UZ

=U RR

=ULZ L

=U CZ C

- V dng in trong mch l nh nhau ti mi im, ta chia hai v ca (1)cho I0

Z = R2 + ( ZL- Z

C)

2 Trong :

Z l Tngtrcamch( )R l intr()

Z Ll cmkhng ( ZL)

Z Cl dung khng( ZC)


tan =Z L- Z

C

R

U oR

U oCUo

L

U oL- U

oC

U o


36





cos =RZ

- Nu tan > 0 ZL> ZC( mch c tnh cm khng)

- Nu tan < 0 ZC> ZL( mchh c tnh dung khng)

- Tan = 0 Mch ang c hin tng cng hng in

3.CNG SUT MCH RLC - P(W)P = UI.cos = I2 .R

U l hiuinthhiudngcamch( V)

I l cngdng inhiudng( A)cos l hscngsut

4.CNG HNG INHin tng cng hng sy ra khi dngin =

ring =

1

LC

2 =1

LC L =

1C

ZL= Z CH qu ca cng hng:

Z

min= R ; I

max=

U

R; i =

u

Z; tan = 0; = 0; cos = 1; P

max = U.I;

5.DNG TON VIT PHNG TRNH HIU IN TH - DNG IN ( u - i)Loi 1: Vit phng trnh u khi bit i.

Cho mch RLC c phng trnh i c dng: i = Iocos(t).phng trnh on mch X bt k c dng: u X = Ucos(t +

X) Trong : tan

X =

Z LX- Z

CX

RX

Trng s trng hp c bit:

- Vit phng trnh uL. uL= UoL.cos( t + 2)(V) Trong : Uo L = Io. Z L- Vit phng trnh u C: u C= Uo C. cos( t + 2)(V) Trong : Uo C = Io. ZC- Vit phng trnh u R: u R= Uo R. cos( t )( V) Trong : Uo R= Io.R

Loi 2: Vit phng trnh i khi bit phng trnh u.Cho on mch RLC, bit phng trnh hiu in th on mch X c dng: u X = U

O.cos(t) (V)

Phng trnh i s c dng: i = I Ocos( t -X).(A) Trong : tan

X =

Z LX- Z

CX

RX

Mt s trng hp c bit:- Bit phng trnh u R= U ORcos( t +) i = IOcos(t +)- Bit phng trnh u L= U OLcos( t +) i = IOcos(t +- 2)- Bit phng trnh u C= U OCcos( t +) i = IOcos( t ++ 2)

Loi 3: Vit phng trnh u Y khi bit phng trnh u

X .

Mch in RLC c phng trnh u Y dng: uY = U

o

Y.cos(t +) (V).Hy vit phng trnh hiu in th hai u

on mch X:Bc 1: Xy dng phng trnh i

i = Io.cos( t +-Y) (A) Trong : tan

Y =

Z LY- Z

C

Y

RY; I0 =

U OYZ Y

Bc 2: Xy dng phng trnh hiu in th bi yu cu:

u X = Uo

X.cos( t + -

Y +

X) Trong : tan

X =

Z LX- Z

CX

RX ; U OX = I

0. Z

X


37




BI 3: CNG SUT V CC TR CNG SUT1.Cng sut:P = UIcos= I2.R.trong :- P l cng sut ( W )- U l hiu in th hiu dng ca mch ( V )-I l cng dng in hiu dng ( A )

- cos=RZ

gi l h s cng sut.

2. Cc tr cng sut

P = I2 .R = U2

. RR2 + ( ZL- Z

C)

2

a.Nguyn nhn do cng hng ( sy ra vi mch RLC)-Khi thay i (L, C,, f)lm cho cng sut tng n cc i kt lun y l hin tng cng hng.ZL = Z CL = 1Choc 2fL = 12fCH qu( Khi mch c hin tng cng hng)

= 0; tan = 0; cos = 1; R = Z; P max =U2R

= U.I; I max =UR

;

Mt s ch :Nu khi thay i = 1v khi =

2th cng sut trong mch ( cng dng in trong mch) nh nhau. Hi

thay i bng bao nhiu cng sut trong mch l cc i.

= 12Nu khi thay i f = f1v khi f = f2th cng sut trong mch ( cng dng in trong mch) nh nhau. Hi thayi f bng bao nhiu cng sut trong mch l cc i.f = f1f2b.Nguyn nhn do in tr thay i.TH1:Mch RLC mc ni tip, cun dy thun cm.

P = I2 .R =U2. R

R2 + ( ZL- Z

C)

2

=U

R +(ZL- Z

C)

2

R

=UY

P max khi Y

min

Xt hm Y = R +(ZL- Z

C)

2

R 2 (Z L- Z

C)

2 ( p dng bt ng thc Cosi)

V ZL- Z Cl hng s, nn du bng sy ra khi: R = (Z

L- Z

C)

2

R R2 = (ZL- Z C)2

R = |Z L- ZC|

H qu:

Tan =Z L- Z

C

R = 1; =

4; cos =

22

; Z = R 2; P =U22R

TH2:Mch RLC mc ni tip, cun dy c in tr trong (r).

Khi R thay i P max. R = | ZL- Z

C| + r P max= U22(R+r)

Khi R thay i cng sut ta nhit trn in tr l cc iP Rmax khi R = r2+(Z L-ZC)2Bi ton ch :

Mch RLC. Nu khi thay i R = R

1v khi R = R

2th cng sut trong mch nh nhau. Hi thay i R bng baonhiu cng sut trong mch l cc i, gi tr cc i l bao nhiu?

R = R1R2 = | ZL- Z C| ; P max = U22 R1R

2

Mch RLC. Nu khi thay i R = R1v khi R = R2th cng sut trong mch nh nhau. Hi cng sut l bao

nhiu:

P =U2

R1+ R2


38




BI 4: HIU IN TH V CC TR HIU IN THI. PHNG PHP1. T CM THAY I.Cho mch RLC c L thay i

A.L thay i U R maxU R= I.R =

U.RR2 + ( Z

L- Z

C)

2

L thay i khng nh hng n t; UR

max khi mu t gi tr nh nht. ZL= Z

C ( Hin tng cng hng)

B.L thay i U C maxU C= I. Z C= U. Z

C

R2 + (ZL- Z

C)

2

Tng t nh trn: U C

max khi mch c hin tng cng hng.C.Nu L thay i UL maxUL= I. Z

L=

U. ZLZ

=U. ZL

R2 + ( ZL- Z

C)

2

( Chia c t v mu cho Z L)

=U

R2Z L

2+

(ZL- ZC)

2

Z L2

=U

YUL

maxkhi Y

min

Y =R2Z L

2

+ 1 - 2Z CZ L

+Z C

2

Z L2

=R2 + Z

C

2

Z L2

- 2Z CZ L

+ 1. ( t x =1Z L

)

Y = ( R2 + ZC

2 ) .x

2 - 2. Z

C.x + 1

Cch 1: Phng php o hmY = 2( R2 + Z

C

2 ).x - 2. Z

C = 0

x =Z C

R2 + ZC

2

Y = 2.(R2 + ZC

2) >0 Khi x =

Z CR2 + Z

C

2th Y min

x =Z C

R2 + ZC

2=

1Z L

ZL =R2 + Z

C

2

Z C

Cch 2: Phng php thY = ( R2 + Z

C

2 ) .x

2 - 2. Z

C.x + 1

V ( R2 + ZL

2) > 0 th c dng nh hnh v

Y min khi x = -b2a = Z

CR2 + Z

C

2

= 1Z LZL=R

2

+ Z

C

2

Z C

Y min = -4a

=R2

R2 + ZC

2

UL

max =UY

U LMAX= U Z C2+R2R UL max= U UC

2 + U

R

2

U R

Cch 3: Dng gin :p dng nh l sin ta c:

U Lsin

= Usin

UL=U

sin.sin (1)

Ta li c: sin =U RU RC

=U R

U R2 + U

C

2

(2)

Thay (2)vo (1):UL= UU R

2 + U

C

2

U R .sin

ULt gi tr ln nht khi sin = 1.( tc=2)

U

U L

U C

U R

U RC

-4a

y

- b2a


39




UL

max= UU R

2 + U

C

2

U R Hoc U L

max = U

R2 + ZC

2

R

Mt s h qu:UL

2 = U

2 + U

R

2 + U

C

2 U

L. U

R= U

RC.U = U . U

R

2 + U

C

2 U

L.( U

L- U

C) = U

2

UL.UC= U

2RC = U

R

2 + U

C

2 U

C.( U

L- U

C) = U

R

2

D.BI TON PH: bi: Mch RLC c L thay i, khi L = L1 v L = L2th thy U L u nh nhau. Xcnh L hiu in th hai umch t cc i.

Hng dn: U Lmax khi12

1

Z L1+

1Z L

2

=1Z L

L =2L1.L

2

L1+ L2

2: IN DUNG THAY I.A. C thay i U R max; U L max ( Phn tch tng t nh trn)Z L= Z

C L = 1C C = 12L

B.C thay i U CmaxZ C= R2+ZL2Z L U CMAX= U ZL

2+R

2

R

C.BI TON PH: bi: Mch RLC c C thay i. Khi C = C1 v C = C

2th thy U

C u nh nhau. U

Ctrong mch t cc i th

in dung ca t phi l bao nhiu?Hng dn:

U C

max khi:12

1

Z C1+

1Z C

2 =

1Z C

C =C 1+ C

2

2

3: IN TR THAY I.A. R thay i U Rmax:

U R= I .R =U.R

R2 + (ZL- Z

C)

2

=U

1 +(ZL- Z

C)

2

R2

t Y = 1 +(ZL- Z

C)

2

R2

U R=UY

UR

maxkhi Y

min

Y min khi (ZL- Z C)2R2

= 0 R

B. R thay i U Lmax:

U L = I. ZL =

U. ZLR2 + ( Z

L- Z

C)

2

U L

max khi R = 0.

B. R thay i U C max:U C= I.Z

C =

U. Z CR2 + (Z

L- Z

C)

2

U C

maxkhi R = 0

4: THAY I TN S GC:

A.thay i U Rmax: UR= I.R = U.RR2 + ( Z L- Z C)2

UR

maxkhi ZL= Z

C( cng hng) = 1

LC f =

1

2 LCB.thay i U Cmax:U C = I. Z

C=

U

C R2 + ( L +1

C)2

=U

C 2R + 4.L

2 - 2.

2LC

+1C2

=U

C. Y

Vi Y = 4.L2 +

2( R

2 -

2LC

) +1C2

Vy U Ct gi tr cc i khi Y

min:t x = 2.


40




Y c dng: Y = L2. x2 + ( R

2 -

2LC

). x +1C2

( L2 > 0)

Y t gi tr nh nht kh: x =-b2a

=

2LC

- R2

2L2=

1LC

-R22L2

= 2.

Y min ( Tc U C

max) khi: =1

LC-

R22L2

Hoc f =12

1

LC-

R22L2

***Bi ton ph: Mch RLC c tn s gc thay i c, Khi = 1v khi = 2th U

Ctrong mch l nh nhau.

Xc nh gi tr ca UCtrong mch t gi tr ln nht:

2 =

12 [ ]

12 +

22

C. thay i UL max: ( Phn tch tng t)=

1

LC -C2R

2

2

f =12

1

LC -C2R

2

2

***Bi ton ph: Mch RLC c tn s gc thay i c, Khi = 1v khi = 2th U

Ltrong mch l nh nhau.

Xc nh gi tr ca U Ltrong mch t gi tr ln nht:12

=12

1

12+

1 2

2

6. MCH RLC C C THAY I U R C MAXU RC= I. Z

RC = U.

Z RCZ

= UR2 + Z

C

2

R2 + (ZL- Z

C)

2

= U. Y

URCt gi tr cc i khi Y t gi tr cc i( Y

max)

tU = R2 + Z

C

2

V = R2 + ( ZL- Z

C)

2

U Z C= 2. Z

C

V

Z C= - 2( Z L- Z

C)

Y =U.V - V.U

V2=

2.Z C[ ]R2 + ( Z

L- Z

C)

2 + 2( Z L- Z

C) ( R

2 + Z

C

2)

[ ]R2 + ( Z L- Z C)2 2= 0

2. Z C.R2 + 2Z

C. Z

L

2 - 4Z

L. Z

C

2 + 2. Z

C

3 + 2Z

L.R

2 + 2Z

L. Z

C

2 - 2Z

C.R

2 - 2. Z

C

3 = 0

- 2. ZL. ZC

2 + 2. Z

C. Z

L

2 + 2. Z

L.R

2 = 0

2ZL( Z C2 - ZL. Z C - R2 ) = 0 Z C2 - ZL. Z C - R2 = 0. Gii phng trnh bc 2 theo Z Cta c: Z C=Z L+ Z L2 + 4R22

7. MCH RLC C L THAY I U RL

MAX:Tng t nh phn trn ( C thay i U Cmax).ZL

2 - Z

C. Z

L- R

2 = 0

ZL=Z C+ Z

C

2 + 4R

2

2

BI 5: PHNG PHP GIN VEC T

1. C S L THUYT HNH HCa.Cc cng thc lung gic c bn trong tam gic vung

Sin =i

Huyn =

ca

Cos =K

Huyn =

ba

Tan =iK

=c

b

Cotan =Ki

=bc

A

B

Cb

a

c


41




b.Cc h thc trong tam gic vungnh l 1 :(Pitago) BC2= AB2+ AC2

nh l 2 :AB2 = BC.BHAC2 = BC. CH

nh l 3 : AH2= BH.CH

nh l 4 : AB.AC = BC.AHnh l 5 : 1/AH2= 1/AB2+ 1/AC2

A

B

Cb

a

c

H

c.nh l cos - sinnh l cos: a2 = b

2 + c

2 -2b.c.cos

nh l sin:a

sin

A

=b

sin

B

=c

sin

C

A

B Ca

bc

d.Cc kin thc khc:-Tng ba gc trong tam gic l 180o- Hai gc b nhau tng bng 180o- Hai gc ph nhau tng bng 90o-Nm kin thc v tam gic ng dng, gc i nh, sole, ng v

2.

C S KIN THC VT L:

- Z = R2 + ( ZL- Z

C)

2 ; U = U

2R+ (U

L- U

C)

2

- Cos =RZ

=U RU

; tan =Z L- Z

C

R

-nh lut : I =U RR

=U LZ L

=U CZ C

=UZ

- Cng thc tnh cng sut: P = U.I. cos = I2.R- Cc kin thc v cc linh kin R,L,C.

Mch ch c L:

+ u nhanh pha hn i gc2

+ Gin vc t

i

u L

Mch ch c C:


42




+ u chm pha hn i gc2

+ Gin vc ti

u L

Mch ch c R:+ u v i cng pha+ Gin vc t u

i

Ch :Hai ng thng vung gc: K1. K

2= -1. tan

1.tan

2 = -1.

Nu hai gc(1> 0;

2>0)

1 + 2= 90o

tan 1. tan 2 = 1

Hoc:1< 0;

2< 0

1+2= - 90

o

tan 1. tan 2 = 1

tan ( 1 + tan 2) =

tan 1+ tan 2

1 - tan 1. tan 2

3. CC PHNG PHP V GIN 3.1 V ni tip:

V d 1 :Mch RLCmc ni tip, trong : 2R = 2Z L= ZC; xc nh h s

gc ca mch trn?

Gii:

Ta c: ZL= R

Z C= 2R

Z

ZL

ZC

R

Z

C - ZL

V d 2:Mch RL ni tip c mc vo mng in xoay chiu c phng trnh hiu in th u = 200 2 cos( 100t +3) V, th thy trong mch c dng in i = 2 2 cos( 100t) A. Hy xc nh gi tr ca R v L?

Gii:

Z =UI =

2002

= 100

= 3 rad

R = Z.cos = 100 cos 3 = 100

12 = 50

Z L= Z. sin = R.tan = 50. tan3 = 50 3

L =Z L

=50 3100

=0,5 3

H

R

Z L

Z

V d 3:Mch RLC ni tip ( trong cun dy thun cm Z L= 50 3 ). c mc vo mng in xoay chiu c

phng trnh hiu in th u = 100 2 cos( 100t -6) V, th thy dng in trong mch c m t bng phng trnh i =

2 cos( 100 t +6) A. Hy xc nh gi tr ca R v C.

Gii:


43




Ta c: Z =UI =

1001

= 100

= -3( Z C> Z

L)

Ta c gin sau:

R = Z.cos = 100. cos3 = 50

(ZC- ZL) = R.tan

3 = 50 3

ZC= ZL+ 50 3 = 50 3 + 50 3 = 100 3

V d 4: Mch RlC mc ni tip, C c th iu chnh c, c mc vomng in xoay chiu c hiu in th U, Diu chnh t C u U C

max Xc

nh gi tr U C

max.Gii:

Theo nh l sin ta c:U Csin

=U

sin UC=

Usin

. sin

Trong : sin =U RU R

L =

U RU R

2 + U

L

2

UC=U. U R

2 + U

L

2

U R.sin

UC

max khi sin = 1

UC

max = U. UR2 + U L2

U R

U

U C

U L

U R

U RL

V d 5: Mch RlC mc ni tip, C c th iu chnh c, c mc vomng in xoay chiu c hiu in th U, Khi iu chnh C UC

maxth

thy U C

max = 2U. Hy tnh gi tr ca ZLtheo R.

Gii:

Ta c:

U C= 2U sin =UU C

=U2U

=12=

6

tan =U

RU L = RZL= 13 Z

L= 3 R U

U = 2U

U L

U R

U RL

V d 5: Mch gm cun dy c in tr thun ng k mc ni tip vi t C,C c th iu chnh c, hai u mch c mc vo mng in xoay chiuc hiu in th U = 80 V, iu chnh C UC

maxth thy U

C

max = 100 V.

Xc nh hiu in th hai u cun dy?Gii:

Theo nh l Pitago ta c:

Ucd = U C

max2 - U

2 = 100

2 - 80

2 = 60 V

U = 80V

U = 100V

UL

UR

U

cd

Cu 6 :Hai cun dy (R1, L1) v (R2, L2) mc ni tip ri mc vo ngunxoay chiu ht U. Gi U1v U2l ht 2 u mi cun. iu kin U = U1+ U2l:

A. L1/R1= L2/R2 B. L1/R2= L2/R1C. L1.L2= R1R2 D. L1+ L2= R1+ R2

L 1; R1 L

2; R

2

U

U1 U

2

R

Z C- ZL

Z

3


44




U = U1+ U2khi hiu in th hai u cun dy cng pha

tan 1= tan 2

Z L

1

R1 =

Z L2

R2

L 1R1

=L 2R

L1R1

=L2R2

Chn p n A U 1U

2

R1

R2

ZL

2

ZL

1

1

2

U

Cu 1: Mch in AB gm cun dy c in tr trong r v t cm L,mc ni tip vi t in C. Gi U AMl hiu in th hai u cun dy v c gitr U AM= 40 V, U

MB= 60V hiu in th u

AMv dngin i lch pha gc 30

o.

Hiu in th hiu dng U ABl:A.122,3V B.87,6VC.52,9V D.43,8V

Gii:

Theo nh l cos ta c:

U AB2 = U

AM

2 + U

MB

2 - 2.U

AM.U

MBcos

AMB

= 402 + 602 - 2.40.60. cos 60o = 2800U AB = 52,9V Chn p n C

L; R

U

ABM

A

M

B

30o

60o

40V

60V

Cu 2: Mt on mch in xoay chiu c dng nh hnh v.Bit hiu inth u

AEv u

EBlch pha nhau 900.Tm mi lin h gia R,r,L,.C

A BC r

R,LE

A.R = C.r.L B.r =C. R..L C.L = C.R.r D. C = L.R.rGii:

Gi 1l gc lch gia hiu in th on AE v cng dng in trong mch2l gc lch giahiu in th on EB v cng dng in trong mch

V u

AEvung pha u

EBtan 1. tan 2= - 1.

-Z Cr

.Z LR

= -1

1.L

C.r.R = 1

L = C.r.R Chn p n CCu 3: Cho mt mch in gm mt t in c in dung C mc ni tip vi bin tr R. Mc vo hai u mch in mthiu in th xoay chiu c tn s f. Khi R=R1th cng dng in lch pha so vi hiu in th gia hai u on mchmt gc 1. Khi R=R2th cng dng in lch pha so vi hiu in th gia hai u on mch mt gc 2. Bit tng ca1v 2l 90

o. Biu thc no sau y l ng?

A.212 RR

Cf

. B.C

RRf

2

21 . C.21

2

RRC

f

. D.212

1

RRC

f

.

Gii:V1 +

2 = 90

otan 1.tan 2 = 1

( -Z CR1

). ( -Z CR2

) = 1 1

C.R1 .

1CR2

= 1

2 =1

C2.R1.R

2

f =1

2C R1. R2


45




3.2 Phng php v chung gcV d 1 :Mch RLCmc ni tip, trong : 2R = 2Z L= Z

C; xc nh h s

gc ca mch trn?

Gii:

Ta c:Z L= RZ C= 2R

tan =Z

L- Z

CR =R - 2RR = - 1

= - 4

cos = cos ( -4) =

22

Z

Z L

Z C

R

ZC - Z

L

3.3 Phng php v hn hp ( kt hp chung gc v ni tip)Cu 4: Cho mch in nh hnh v 0 50 3R ,

50L CZ Z AMU v MBU lch pha 750. in tr R c gi tr l

B

L, R0R C

MA

A.25 3 B.50 C.25 D.50 3

Gii:

Ta c:u AMlch pha lch pha u

MB gc2

u MBlch pha so vi i gc6

u AMlch pha vi i gc4

tan

AM =

Z C

R = 1R = ZC= 50

p n B

ZL= 50

Zc = 50

Ro = 50 3

MB

AM

30o

45o

BI 6: BI TON HP EN

Cha kha 1: lch pha u v i.1.Hp en c 1 phn t:

- Nu = 2rad l L- Nu = 0 rad l R- Nu = -

2 l C

X

2.Hp en cha hai phn t:- Nu

2 > > 0 l RL

- Nu - 2 < < 0 l RC


46




- Nu = 2 l LC

Cha kha 2: Cn c vo hiu in th: ( Cho s nh hnh v, gi s trong X v Y ch cha mt phn t)

- Nu U = |UX - UY| l L v C

- Nu U = U X2 + U

Y

2 l

R v CR v L

- Nu U = U X+ UYX v Y cha cng mt loi phn t

( cng R, cng L hoc cng C)

X Y

U x Uy

U

BI 6: MY BIN P V TRUYNTI IN I XA.I. PHNG PHP1.MY BIN PA.nh ngha:L thit b dng bin i in p ca dng in xoay chiu.- My bin p khng lm thay i gi tr tn s ca dng in xoay chiu.- My bin p khng bin i in p ca dng in mt chiu.B.Cu to

Gm hai phn:Phn 1: Li thp.

- c ghp t cc tm st non - silic mng song song v cch in vinhau.( chng li dng Phuco)

Phn 2: Cun dy:- Gm hai cun l cun s cp v th cp:- Cun s cp( N1):oGm N1 cun dy qun quanh li thpoCun s cp c ni vi ngun in

- Cun th cp( N2):oGm N2cun dy qun quanh li thpoCho in ra cc ti tiu thoNuN2

N1 > 1y l my tng p.

oNuN2N1 < 1y l my h p.

N2N

1

C.Nguyn tc hot ng:- Da trn hin tng cm ng in t.- Dng in bin thin trong cun s cp T thng bin thin trong li thp Dng in bin thin cun th cp

D.Cng thc my bin p.-My bin p hiu sut H = 100 %( cos 1 = cos 2)

U1U2

=N1N2

=I2I1

- My bin p H 100%o H =P2

P 1x 100% =

U2. I2.cos

2

U1.I1.cos

1 x100%.

*** Nu coi cun s cp c in tr trong - cun th cp c in tr trong khng ng k

Ta c:U L1U2

=N1N2

Trong : U L12 + U

R1

2 = U

12

*** Nu coi cun th cp c in tr trong ( mch ngoi mc vi in tr R) - cun s cp c in tr trong khngng k:

Ta c:N1N2

=U1

U2+ I2.r

2


47




2.TRUYN IN I XA:TI SAO PHI TRUYN TI IN:

- Ngun in c sn xut ra tp trung ti cc nh my in nh: nhit in,thy in, in ht nhn nhng vic tiu th in li rng khp quc gia,tp trung hn ti cc khu dn c, nh my, t thnh th n nng thn cngu cn in.

- Cn ng truyn ti in chia s gia cc vng, phn phi li in nng,xut nhp khu in nng....

V th truyn ti in l nhu cu thc t v cng quan trng:

BI TON TRUYN IN:Trong qu trnh truyn ti in bi ton c quan tm nht l lm sao gim hao ph in nng xung thp nht.

- Cng thc xc nh hao ph truyn ti: P = I2. R =P2

U2cosR Trong :

P l cng suttruynti(W)

R =.lS

l intrngdy truyn

Ul hiuinthtruynticosl hscng sutngtruyn

- Gii php lm gim hao ph kh thi nht l tng hiu in th in trc khi truyn ti U tnga lnhao ph gima2 ln

Cng thc xc nh gim th trn ng truyn ti in: U = I.R

Cng thc xc nh hiu sut truyn ti in: H =P - P

P.100% = 100% - % P

CHNG IV: DNG IN XOAY CHIUBI 7: MY PHT IN -NG C IN

I. PHNG PHP.1.NGUYN TC TO RA DNG IN- My pht in xoay chiu hot ng da trn hin tng cm

ng in t.- Cho khung dy c in tch S quay quanh trc t vung gcvi t trng u

B, lm xut hin t thng bin thin theo

thi gian qua cun dy lm cho trong cun dy xut hindng in.

Ta c:Phng trnh t thng: = BScos( t +) Wb = ocos( t +) Wb {

o= BS }

Trong :o :l t thng tc thi qua cun dy ( Wb- V be)o o: t thng cc i qua cun dy ( Wb - V be)o B: cm ng t ( T - Tesla)o S:din tch khung dy ( m2)o :l gc lch gia vc t ca cm ng t

B v vc t php tuyn

n ca khung dy.

Phng trnh sut in ng:Xt cho 1 vng dy:

e = - e = .osin ( t +)


48




e = ocos( t +-2) V

e = Eocos( t +-2) V

o Eo: sut in ng cc i trong 1 vng dy ( V) Eo= . O = BSNu cun dy c N vng dy:

e = Eo. cos( t +-

2) V

Eo= NBS.

2.MY PHT IN XOAY CHIU MT PHAA. Cu to:

M hnh 1 M hnh 2Gm hai phn chnh:Phn 1: Phn ng ( to ra dng in)

- Vi m hnh 1 phn cm l phn ng yn ( stato)- M hnh 2, phn cm quay( ro to) v vy a c in ra ngoi cn thm mt b gpo B gp gm 2 vnh khuyn v hai chi quyet t ln 2 vnh khuyn a in ra ngoio Nhc im ca b gp l nu dng in c cng sut ln truyn qua s to ra cc tia la in phng ra thnh camy gy nguy him cho ngi s dng. ( v th ch thit k cho cc my c cng sut nh).

Phn 2: l phn cm( to ra t trng - nam chm).- Vi m hnh 1, phn cm l phn quay ( ro to)- Vi m hnh 2, phn cm l phn ng yn ( stato)B. Nguyn tc hot ng.- Ti thi im ban u cc bc ca nam chm hng thng cun dy, t thng qua khung dy l cc i- Khi ro to quay to ra t thng bin thin trong khung dy to ra sut in ng cm ng trong cun dy

Nguyn tc hot ng da trn hin tng cm ng in t.Cng thc xc nh tn s ca my pht in xoay chiu 1 pha:

f =n.p60

Trong:n: l svng quay car t trong 1phtp: l scpcccanam chm

f = n.pTrong :n: svng quay caro to trong 1sp: scpcccanam chm

3. MY PHT IN XOAY CHIU 3 PHAA. Cu to+) R t ( phn cm): l mt nam chm in c

nui bi dngin mt chiu, c th quay quanh trc tora t trng bin thin.

+) Stato ( phn ng): l 3 cun dy ging ht nhauc t lch nhau 120 o trn vng trn.

B.Nguyn tc hot ng:Nguyn tc hot ng:

- Ti t =0 cc bc ca nam chm hng thng cung dy s 1, t thng qua cun dy s 1 l cc i: 1 = o.cos( t)- Sau T3cc bc ca nam chm hng thng cun dy s 2, t thng qua cun 2 t cc i: 2 = o.cos( t + 23 )


49




- Tip sau T3cc bc ca nam chm hng thng cun dy s 3, t thng qua cun 3 t cc i: 3 = o.cos( t + 43 )- Sau T

3 na ccbc ca nam chm quay tr li cun s 1, c nh vy r t quay to ra t thng bin thin trong 3 cun dy ca

phn ng lch pha nhau23

v cng tn s:

T thng bin thin trong 3 cun dy to ra sut in ng cm ng ba cun dy c phng trnh ln lt nh sau:- 1 = Eosin( t) V- 2 = Eosin( t + 2

3

) V

- 3 = Eosin( t + 43 ) VC.Cch mc dng in ba pha:

Mc hnh sao

dy pha

dy

trung ha

dy pha

dy pha

UP

UP

UP

Ud

Ud

U d

IP

Id

Ud= 3Up; I

d= I

p

Mc tam gic

dy pha

dy pha

dy pha

U d

IP

Id

Ud= Up; I

d = 3 I

p

4. NG C KHNG NG B 3 PHA

A. Cu to ng c khng ng b 3 pha:Gm hai phn:

- Stato c cun dy ging ht nhau qun trn ba li st b trlch nhau 1/3 vng trn.

- R t l mt hnh tr to bi nhiu l thp mng ghp cchin vi nhau. Trong cc rnh x mt ngoi r t c t ccthanh kim loi. Hai u mi thanh c gn vi cc vnh tothnh mt chic lng, lng ny cch in vi li thp c tc dngnh nhiu khung dy ng trc t lch nhau. R t ni trn cgi l r t lng sc.

B. Hot ng:- Nguyn tc hot ng da trn hin tng cm ng in tv tc dng ca t trng quay.

- Khi mc cc cun dy stato vi ngun in ba pha, ttrng quay to thnh c tc gc bng tn s ca dng in.T trng quay tc dng ln dng in cm ng trong cc khungdy r t cc m men lc lm r t quay vi tc nh hn tc ca t trng quay. Chuyn ng quay ca r t s dng lmquay cc my khc.

- Cng sut ca ng c khng ng b 3 pha:P = 3.UIcos= P c+ P nhit


50




- Hiu sut ca ng c: H =P cP

.100%

Vi ng c khng ng b 1 pha:P = U.I.cosP = P c + P

nhitP c= P - P nhit = U.I.cos - I2.R


51




CHNG V: SNG NH SNGBI 1: HIN TNG TN SC NH SNG - CC LOI QUANG PH

I. L THUYT1.HIN TNG TN SC NH SNG

Th nghim: Chiu tia sng trng qua lng knh, pha sau lng knhta t mn hng M. Trn M ta quan st c di mu bin thin lintc t n tm.Kt lun: Hin tng tn sc nh sng l hin tng m khi mt

chm sng khi i qua lng knh th n b phn tch thnh nhiu nhsng n sc khc nhau.*nh sng n sc l nh sng khi i qua lng knh ch b

lch m khng b tn sC:*nh sng a sc l nh sng gm hai nh sng n sc tr

ln.Th nghim v hin tng tn sc nh sng.

- nh sng n sc l nh sng c mt tn s nht nh v khng b tn sc khi truyn qua lng knh.- nh sng trng l hn hp ca nhiu nh sng n sc c mu bin thin lin tc t n tm. ( 0,76m > > 0,38

m )- Chit sut ca cc cht trong sut bin thin theo tn s ca nh sng n sc v tng dn t n tm.- Cng thc xc nh bc sng nh sng: =c

f.Trong :

lbcsng nh sng ( m)cl vntcnh sng trong chn khng ( m/s)fl tnscanh sng. (Hz)

2. GII THCH V HIN TNG TN SC NH SNG.Hin tng tn sc nh sng c gii thch nh sau:- nh sng trng l hn hp ca nhiu nh sng n sc khc nhau, c mu lin tc t n tm.- Chit sut ca thy tinh ( v ca mi mi trng trong sut khc) c gi tr khc nhau i vi nh sng n sc c mu

khc nhau, gi tr nh nht i vi nh sng v ln nht i vi nh sng tm. Mc khc, ta bit gc lch camt tia sng n sc khc x qua lng knh ph thuc vo chit sut ca lng knh: chit sut lng knh cng ln thgc lch cng ln. V vy sau khi khc x qua lng knh, b lch cc gc khc nhau , tr thnh tch ri nhau. Kt qul, chm sng trng l ra khi lng knh b tri rng ra thnh nhiu chm n sc , to thnh quang ph ca nh sngtrng m ta quan st c trn mn.

3.NG DNG CA TN SC NH SNG- ng dng trong my quang ph phn tch chm sng a sc, do vt pht ra thnh cc thnh phn n sc- Gii thch v nhiu hin tng quang hc trong kh quyn, nh cu vng4. MY QUANG PH:

My quang ph cu to gm ba b phno B phn th nht l ng trun trc, ng chun trc l mt ci

ng mt u l mt thy knh hi t L1, y kia l khe hp c l nhsng i qua nm ti tiu im vt ca thu knh hi t. c tc dngto ra cc chm sng song song n lng knh.

o Lng knh P:l b phn chnh ca my quang ph nhm tnsc nh sng trng thnh cc di mu bin thin lin tc t ntm.

o Mn Mhay gi l bung nh dng hng nh trn mn- Nguyn tc hot ng da trn hin tng tn sc nh sng.

ng chun trc Lng knhM

S

5.CC LOI QUANG PH.Cc loiquang ph

nh ngha Ngun pht c im ng dng


52




Quang phlin tc

L mt di mu c mu t n tm ni lin nhau mtcch lin tc

Do cc cht rn, lng,kh c p sut lnpht ra khi b nngnng

Quang ph lin tcca cc cht khcnhau cng mtnhit th hon tonging nhau v chph thuc vo nhit ca chng

Dng o nhit ccvt c nhit cao, xa,nh cc ngi sao.

Quang phvch phtx

L mt h thng nhng vchsng ring l, ngn cch nhaubi nhng khong ti

Quang ph vch docht kh p sutthp pht ra khi bkch thch bng nhithay in.

Quang ph vch cacc nguyn t khcnhau th rt khc nhauv s lng vch, vv tr v sng t ica cc vch. Minguyn t ha hc cmt quang ph vachc trng.

Dng nhn bit, phntch nh lng v nhtnh thnh phn ha hcca cc cht

Quang phvch hpth

L nhng vach ti nm trnnm sng ca quang ph lintc

Quang ph vch docht kh p sutthp pht ra khi bkch thch bng nhithay in. v c tchn trn quang phlin tc

- thu c quangph hp th th iukin nhit cangun phi thp hnnhit ca quangph lin tc- Trong cng mtiu kin v nhit v p sut , Nguynt c th pht raquang ph pht xmu g th hp thmu .

Dng nhn bit, phntch nh lng v nhtnh thnh phn ha hcca cc cht

***Hin tng o vch quang ph:

Hin tng m vch sng ca quang ph lin tc, tr thnh vch ti ca quang ph hp th hoc ngc li gi l hin tngo vch quang ph.

BI 2: CC LAI BC X KHNG NHN THY.1. HNG NGOI

nh ngha L bc x sng in t c bc sng ln hn bc sng ca nh sng ( hn>

)

Ngun pht V lthuyt cc ngun c nhit ln hn 0o K s pht ra tia hng ngoi

Tnh cht - Tc dng c bn nht ca tia hng ngoi l tc dng nhit- C kh nng gy ra mt s phn ng ha hc, tc dng ln mt s loi phim nh- Tia hng ngoi cng c th bin iu c nh sng in t c

Documents

Hệ thống lý thuyết + công thức giải nhanh vật lý 12