bản chính bộ công thức vật lý 12

7/31/2019 bn chnh b cng thc vt l 12

1/51

Nguyn Vn Dn Long An - 0975733056

1

Chng I: DAO NG CHC

I - ICNGV DAO NGIU HO

T: chu k; f: tn s; x: li ; v: vn tc; a: gia tc; g: gia tc trngtrng; A: bin dao ng; ( t + ): pha dao ng; : pha ban u; : tc gc;1. Phng trnh dao ng

tAcosx

- Chu k:2

T (s) - Tn s:2

1

Tf (Hz)

- Nu vt thc hin c N dao ng trong thi gian t th:

t N

T v fN t

.

2. Phng trnh vn tctAxv sin'

- x = 0 (VTCB) th vn tc cc i: Avmax

- x A (bin) th 0v 3. Phng trnh gia tc

2 2' cosa v A t x

- x = A th 2maxa A

- x = 0 th 0a

Ghi ch: Lin h v pha: v sm pha2

hn x;

a sm pha 2 hn v;a ngc pha vi x.

4. H thc c lp thi gian gia x, v v a

- Gia x v v:2

222 v

xA

- Gia v v a:

222 2

2max

a

v A v

- Gia a v x: 2a x


2/51


2

5. Cc lin h khc

- Tc gc:max

max

v

a

- Tnh bin

2

222

2

22

max

2max

2

maxmax 242

avvxk

W

a

vav

n

SLA

6. Tm pha ban u

2

A 22

A

3

2A

3

A2

A O A

2

A 22

A 32

A

v < 0= + /2

v < 0= + /4

v < 0= + /6

v = 0

= 0

v < 0= + /3

v > 0= - /6

v < 0= + 2 /3

v > 0= - /2 v > 0= - /3

v > 0= - /4

v < 0= + 3 /4

v < 0= + 5 /6

v > 0

= -5 /6

v > 0= - 3 /4

v > 0= - 2 /3

v = 0=


3/51


3

6. Thi gian ngn nht vt i t:+ x1n x2 (gi s 21 xx ):

12t vi

Ax

A

x

2

2

1

1

cos

cos

21,0 .

+ x1n x2 (gi s 1 2x x ):

12t vi

A

x

A

x

2

2

1

1

cos

cos

1 2, 0

7. Vn tc trung bnh - tc trung bnh

- Tc trung bnh vS

t

- di x trong n chu k bng 0;qung ng vt i c trong n chu k bng nAS 4 .

- Vn tc trung bnhx

v t .

8. Tnh qung ng vt i c trong thi gian t

cos-A A

2

0 A2

A 2

2 A 3

2

+A

T/4 T/12 T/6T/8 T/8

T/6 T/12* Phng php chung tm qung ng itrong khong thi gian no

ta cn xc nh:- V tr vt lc t = 0 v chiu chuyn ng ca vt lc ;

- Chia thi gian t thnh cc khong nh: nT; nT/2; nT/4; nT/8; nT/6;T/12 vi n l s nguyn;


4/51


4

- Tm qung ng s1; s2; s3; tng ng vi cc qung thi gian nutrn v cngliTnh qung ng ngn nht v b nhtvt i c trong khong thi

gian t vi2

0T

t

Nguyn tc:+ Vt i c qung ng -A - x0 O x0 +A

di nhtkhi li im u v imcuic gi tr i nhau smax

Qung ngdi nht: max 2 sin2

tS A

+ Vt i c qung ng -A - x0 O x0 +Angnnhtkhi li im u v imcui c gi tr bng nhau

smin Smin

Qung ng ngnnht: min 2 1 cos

2

tS A

Trng hp2

Tt th ta tch t

Tnt

2 * 0

2

Tn N v t :

+ Qung ng ln nht: max 2 2 sin2

tS nA A

+ Qung ng nh nht: min 2 2 1 cos2

tS nA A

+ Tc trung bnh ln nhttrong thi gian t: maxaxtbmS

vt

+ Tc trung bnh nh nhttrong thi gian t: minmintbS

vt


5/51


5

II - CON LC L XO

l : bin dng ca l xo khi vt cn bng;k: cng ca l xo (N/m);

0l : chiu di t nhin ca l xo

1. Cng thc c bn- Tn s gc:

k g

m l;

+ Con lc l xo treo thng ng:2

mg gl

k;

+ t con lc trn mt phng nghing gc khng ma st:sinmg

l k - p dng cng thc v chu k v tn s:

2. Chiu di cc i v cc tiu ca l xo+ dao ng thng ng:

AlllAlll

0max

0min 2

minmax llA

+ dao ng phng ngang:

min 0

max 0

A

l A

l l

l

3.Ghp l xo.

- Ghp ni tip:nkkkk

1...11121

- Ghp song song: nkkkk ...21

- Gi T1 v T2l chu k khi treo m vo ln lt 2 l xo k1 v k2 th:

+ Khi ghp k1ni tip k2:2

2

2

1

2

2

2

2

1

111

fff

TTT

22 2

1 1 1

2 2

m lT

k g

k gf

T m l


6/51


6

+ Khi ghp k1 song song k2:2

2

2

1

2

2

2

2

1

111

TTT

fff

- Gi T1 v T2l chu k khi treo m1 v m2ln lt vo l xo k th:

+ Khi treo vt 21 mmm th:2

2

2

1 TTT

+ Khi treo vt 21 mmm th:2

2

2

1 TTT 21 mm

4. Ct l xo- Ct l xo c cng k, chiu di

0l thnh nhiu on c

chiu di nlll ...,,, 21 c cng

t ng ng nkkk ...,,, 21 lin h nhau theo h thc:

nnlklklkkl ...22110 .

-Nu ct l xo thnh n on bng nhau (cc l xo ccng cng k):

nkk' hay:

nff

n

TT

'

'

5. Lc n hi - lc hi phc

Nidung

Lc hi phucLc n hi

L xo nmngang

L xo thng ngA A

Gc ti V tr cn bng V tr l xo cha bin dng

Bn chthp dh

F P F Fh= k . ( bin dng)

nghav tcdng

- Gy ra chuyn ngca vt- Gip vt tr vVTCB

- Gip l xo phc hi hnh dng c- Cn gi l lc ko (hay lc y) ca lxo ln vt (hoc im treo)

Cc i Fmax = kAFmax = kA Fmax = k(l + A)

Cc tiu

Fmin = 0

Fmin = 0 Fmin = 0Fmin = k(l

A)V trbt k

F= k x F= k x F = k(l + x)


7/51


7

III - CON LCN

1. Cng thc c bnDi y l bng so snh cc c trng chnh ca hai h dao ng.

H dao ng Con lc l xo Con lc n

Cu trcHn bi m gn vo l xo (k). Hn bi (m) treo vo u si

dy (l).

VTCB

- Con lc l xo ngang: lxo khng gin- Con lc l xo thng ng

n dnk

mgl

Dy treo thng ng

Lc tc dng

Lc n hi ca l xo:F = - kxx l li di

Trng lc ca hn bi v lccng ca dy treo:

sl

gmF s l li cung

Tn s gcm

k=

g

l

l

g

Phng trnhdao ng.

x = Acos(t + ) s = s0cos(t + )Hoc = 0cos(t + )

C nng2 2 21 1

2 2W kA m A 0

(1 cos )W mgl

20s

l

gm

2

1

- Chu k dao ng ca con lc n c chiu di l1 v l2ln lt l T1 v T2

th:+ Chu k ca con lc c chiu di 21 lll :

2

2

2

1 TTT

+ Chu k ca con lc c chiu di 21 lll :2

2

2

1 TTT 21 ll .

- Lin h gia li di v li gc: s l - H thc c lpthi gianca con lc n:

a = - 2s = - 2

l; 2 2 2

0

( )v

S s 2

2 2

0

v

gl

2. Lc hi phc


8/51


8

2sins

F mg mg mg m sl

3. Vn tc - lc cng+ Khi con lc v tr li gc vn tc v lc cng tng ng ca vt:

0

0

2 cos cos

3cos 2cosc

v gl

T mgKhi 0 nh:

2 20

2 2

0

31

2c

v gl

T mg

+ Khi vt bin:0

0

cosc

v

T mg; khi 0 nh:

2

0

0

12

c

v

T mg

+ Khi vt qua VTCB: 0

0

2 1 cos

3 2cosc

v gl

T mg; khi 0 nh:

0

2

01c

v gl

T mg

4. Bin thin chu k ca con lc n ph thuc: nhit , su v cao. Thi gian nhanh chm ca ng h vn hnh bng con lc n

a.Cng thc c bn* Gi chu k ban u ca con lc l 0T (chu k chy ng), Chu k sau

khi thay i l T(chu k chy sai).

0TTT : bin thin chu k.+ 0T ng h chy chm li;+ 0T ng h chy nhanh ln.

* Thi gian nhanh chm trong thi gian N (1 ngy m24 86400N h s ) s bng:

0

TNT N

T T

b. Cc trng hp thng gp

Khi nhit thay it1

t n 2t :0

1

2

1

2

Tt

T

N t

( 2 1t t t )


9/51


9

Khi a con lc t cao1

h n cao2

h : 0

T h

T R

hN

R

( 2 1h h h )

Khi em vt ln cao 0h , khi em vt xung cao thp hn0h . Ban u vt mt t th 01h v hh

Khi a con lc t su1

h n su2

h : 02

2

T h

T R

N h

R

( 2 1h h h )

Khi em vt xung su 012 hhh , khi em vt ln cao hn banu 0h . Ban u vt mt t th 01h v hh

c. Cc trng hp c bit- Khi a con lc mt t (nhit

1t ) ln cao h (nhit

2t ):

0

1

2

T ht

T R

Nu ng h vn chy ng so vi di mt t th:

0

10

2

T ht

T R

- Khi a con lc t tri t ln mt trng (coi chiu di l khng i) th:

T

MT

MT

T

MT

T

M

M

R

R

T

T

5. Con lc n chu tc dng ca lc ph khng i

* Lc ph f

gp trong nhiu bi ton l:

+Lc qun tnh amFq

, ln: maFq , (a l gia tc ca h quychiu)

+Lc in trng F qE , ln: EqF ,

q l in tch ca vt, E l cng in trng ni t con lc ( /V m )

+Lc y Acsimet gV

AF , ln: VgFA .


10/51


10

l khi lng ring ca mi trung vt dao ng, V l th tch vtchim ch

Chu k dao ng trong trng hp ny s l:g

lT 2 ,

'g l gia tc trngtrng hiu dng.* Tnh g':

+ Trng hp Pf

:m

fgg '

Lc qun tnh: agg'

Lc in trng:m

Eqgg '

+ Trng hp Pf

:m

fgg '

Lc qun tnh: agg'

Lc in trng:m

Eqgg '

Lc y Acsimt: m

Vg

gg '

+ Trng hp Pf

:2

2'

m

fgg

Lc qun tnh: 22' agg

Lc in trng:2

2'

m

qEgg

Ch :+ Trng hp Pf

th gc lch ca si dy so viphng

thng ng c tnh:P

ftan

+ Khi con lc n gn trn xe v chuyn ng trn mt phngnghing gc khng ma st th VTCB mi ca con lc l si dy lchgc (si dy vung gc vi mt phng nghing)so vi phngthng ng v chu k dao ng ca n l:


11/51


11

cos2'

g

lT

V - NNGLNG DAO NG

-ng nng: tAmmvWd2222 sin

2

1

2

1

- Th nng: tAmkxWt2222 cos

2

1

2

1

- ng nng v th nng bin thin tun hon vi chu k bng 1/2 chu kdao ng iu ho (T = T/2).

- Khong thi gian gia 2 ln ng nng v th nng bng nhau lin tip lT/4.

cos-A A

2

0 A2

A 2

2 A 3

2

+A

T/4 T/12 T/6

Vi T/8 T/8

T/6 T/12

1. Con lc l xo (Chn gc th nng ti VTCB)

- ng nng: 22

1mvW

; Th nng: 2

2

1kxWt

- C nng: t WWW222

2

1

2

1AmkA

+ V tr ca vt khi t nWW :1n

Ax

W = 3 Wt WmaxWt = 0

Wt = 3 WW = WtW = 0Wtmax

W = Wtmax = Wmax = 1/2kA2


12/51


12

+ Vn tc ca vt lct

nWW :11

max

n

A

n

vv

+ ng nng khi vt li x: 222

1xAkW

+ T s ng nng v th nng:2

22

x

xA

W

W

t

2. Con lc n(Chn gc th nng ti VTCB)

- ng nng: 22

1mvW

; Th nng: cos1mglWt

- C nng: t WWW 0cos-1mgl

Khi gc 0 b th:21

2tW mgl ;

20

1W mgl2

+ V tr ca vt khi

t nWW :1

0

n

SS v

1

0

n

+ Vn tc ca vt lc

t

nWW :1

max

n

vv

1

0

n

S

+ ng nng ca vt khi n li :

220222

02

1

2

1SSmmglW

+ T s ng nng v th nng:2

22

0

2

22

0

S

SS

W

W

t

VI - TNGHP DAO NG

1. Phng php gin Frexnel- Bi ton:Tng hp 2 dao ng iu ho cng phng:

1 1 1

2 2 2

cos

cos

x A t

x A t cosx A t


13/51


13

Vi

2211

2211

2121

2

2

2

1

coscos

sinsintan

cos2

AA

AA

AAAAA

-Nu bit mt dao ng thnh phn 111 cos tAx v dao ngtng hp tAx cos th dao ng thnh phn cn li l

222 cos tAx c xc nh:

11

112

11

2

1

22

2

coscos

sinsintan

cos2

AA

AA

AAAAA

(vi 21 )-Nu 2 dao ng thnh phn vung pha th: 22

2

1 AAA

2. Tm dao ng tng hp xc nh A v bng cch dng mytnh thc hin php cng:

+ Vi my FX570ES: Bm chn MODE 2 mn hnh xut hin ch:CMPLX.

-Chn n v o gc l bm: SHIFT MODE3 mn hnh hin th ch D(hoc Chn n v gc l Rad bm: SHIFT MODE4 mn hnh hin th

ch R )-Nhp A1 SHIFT (-) 1, +Nhp A2 SHIFT (-) 2 nhn = hin th kt

qu.

(Nu hin th s phc dng:a+bi

th bmSHIFT 2 3 = hin th kt

qu: A )

+ Vi my FX570MS: Bm chn MODE 2 mn hnh xut hin ch:CMPLX.

Nhp A1 SHIFT (-) 1 + Nhp A2 SHIFT (-) 2 =Sau bm SHIFT + = hin th kt qu l: A. SHIFT = hin th kt

qu l:

+ Lu Ch hin th mn hnh kt qu:


14/51


14

Sau khi nhp ta n du = c th hin th kt qu di dng s v t, munkt qu di dng thp phnta n SHIFT = (hoc dng phm SD ) chuyn i kt qu Hin th.

VII - DAO NGTTDN- Tm tng qung ngS m vt i c cho n khi dng li:

SFkA C2

2

1

- gim bin sau 1 dao ng:2

4 CFAm k

FC4 , CF l lc cn

Nu Fcl lc ma st th :kNA 4

- S dao ng thc hin c:CF

Ak

A

AN

4

.' 11

Nu Fcl lc ma st th:N

kAN

4' 1

- Thi gian t lc b ma st n khi dng lit =N. T

- Da vo s dao ng thc hin xc nh c s ln qua VTCB cavt: khi 25,' nNn (n nguyn) th s ln qua VTCB s l 2n; khi

75,'25, nNn th s ln qua VTCB ca vt l 2n+1; khi1'75, nNn th s ln qua VTCB ca vt l 2n+2.

- V tr ca vt c vn tc cc i:

Fc = Fhp => .m.g = K.x0 => 0mg

xk

- Vn tc cc i khi dao ng t c ti v tr x0 :

0 0v (A x ).

VIII - DAO NGCNGBC.CNGHNG

-Khi vt dao ng cng bc th tn s (chu k) dao ng ca vt bngtn s (chu k) ca ngoi lc.


15/51


15

- Hin tng cng hng xy ra khi tn s (chu k) ca ngoi lc bng tns (chu k) dao ng ring ca h.

Ch :Chu k kch thchv

lT ; l l khong cch ngn nht gia 2 mi

ray tu ha hoc 2 g trn ng ; Vn tc ca xe con lc t trnxe c cng hng:

0

0

lfT

lv

IXCON LCTRNG PHNG

- xc nh chu k ca 1 con lc l xo (hoc con lc n) ngi ta sosnh vi chu k T0( bit) ca 1 con lc khc 0TT .

- Hai con lc ny gi l trng phng khi chng ng thi i qua 1 v trxc nh theo cng mt chiu

- Thi gian gia hai ln trng phng:0

0

TT

TT

Ch :+ Nu 0TT nTTn 01

+ Nu 0TT 01 nTTn (vi *Nn )

CHNG II: SNG CHC

I - ICNGV SNG CHC

T: chu k sng; v: vn tc truyn sng; :bc sng

1. Cc cng thc c bn

-Lin h gia , v v T (f): v fT

- Qung ng sngtruyn i c trong thi gian t: tT

vtS


16/51


16

- Vn tc truyn sngbit qung ng sng truyn c trong thi gian t

l S:t

Sv

-Khong cch gia n gn li lin tip l d th:

1n

d

- n ngn sngi qua trc mt trong thi gian t th:1n

tT

-Phao nh cao n lntrong thi gian t th:1n

tT

2. Phng trnh sng

- Sng truyn tN qua O v n M, gi s biu thc Sng ti O c dng:)cos(0 tAu , th:

)2

cos(x

tAuM

)'2

cos(x

tAuN

- lch phaca 2 im trn phng truyn sng cch nhau mt on d:

d2

2k hay kd 2 im dao ng cng pha

12k hay2

12kd 2 im dao ng ngc pha

- lch pha ca cng mt imti cc thi im khc nhau:

12 tt

- Cho phng trnh sngl )cos( kxtAu sng ny truyn vi vn

tc:k

v

Ch :C nhng bi ton cn lp phng trnh sng ti 1 im theo iukin ban u m h chn th ta lp phng trnh sng ging nh phn lp

phng trnh dao ng iu ha.


17/51


17

IIGIAO THOA SNG

1. Phng trnh sng tng hp ti mt im

*Trng hp tng qut:Phng trnh sng ti 2 ngun

1 1Acos(2 )u ft v 2 2Acos(2 )u ft

Phng trnh sng ti M do hai sng t hai ngun truyn ti:1

1 1Acos(2 2 )Mdu ft v 2

2 2Acos(2 2 )Mdu ft

Phng trnh sng ti M:

2 1 1 2 1 22 cos[ ] cos 22 2

M

d d d d u A ft

Bin dao ng ti M:

]2

cos[2 12dd

AAM vi = 2 - 1

C bc 0(k=0)

AB

CT th 1(k=0) C bc 1

k=1CT th 2( k=1)

O

/2

Gn lmGn li

d1d2

M


18/51


18

2.Tm s im dao ng cc i, s im dao ng cc tiu gia haingun:

Cng thc tng qut

* S cc i: (k Z)2 2

l lk

* S cc tiu: (1 1

2 2 2 2

k Z)l l

k

Ta xt cc trng hp sau y:a. Hai ngun dao ng cng pha: = =2k

* S Cc i:(k Z)

l l

k

* S Cc tiu: 1 1- (k Z)2 2

l lk

Hay 0,5 (k Z)l l

k

b. Hai ngun dao ng ngc pha: ==(2k+1)

* S Cc i: 1 1 (k Z)2 2

l lk

Hay 0,5 (k Z)l l

k

* S Cc tiu: (k Z)l l

k

c. Hai ngun dao ng vung pha: =(2k+1) /2(S cc i= S cc tiu)

* S Cc i: 1 1 (k Z)4 4

l lk

* S Cc tiu:1 1

(k Z)4 4

l lk

Hay 0,25 (k Z)l l

k

Nhn xt: s im cc i v cc tiu trn on AB l bng nhau nn cth dng 1 cng thc l


19/51


19

3. Tm s cc i , cc tiu ngoi on thng ni 2 ngun

12

'

1

'

2 ddkdd

(gi s'

1

'

212 dddd )- Xc nh s im (s ng) cctiu trn on AB (cng pha so vi ngthng 0102) l s nghim k nguyn thamn biu thc:

2

1

2

1 12'

1

'

2 ddkdd

(gi s'

1

'

212 dddd )Ch :Vi bi ton tm s ng dao ng cci v khng dao ng gia hai im M, N cchhai ngun ln lt l d1M, d2M, d1N, d2N.

t dM = d1M - d2M ; dN = d1N - d2Nv gi s dM < dN.+ Hai ngun dao ng cng pha:

Cc i: dM < k < dNCc tiu: dM < (k+0,5) < dN

+ Hai ngun dao ng ngc pha:Cc i: dM < (k+0,5) < dNCc tiu: dM < k < dN

S gi tr nguyn ca k tho mn cc biu thc trn l s ng cn tm.+ Hai ngun dao ng vung pha:

IIISNG DNG

- Bin ca sng ti v sng phn xl A th bin dao ng cabng sng a =2A.

- B rng ca bng sng l: L = 4A.- Vn tc cc i ca mt im bng sng trn dy: vmax = 2A- Phng trnh sng ti v sng phn x ti M cch B mt khong d l:

os(2 2 )M

du Ac ft v ' os(2 2 )M

du Ac ft

- Phngtrnh sng dng ti M: 'M M Mu u u


20/51


20

2 os(2 ) os(2 ) 2 sin(2 ) os(2 )2 2 2

M

d du Ac c ft A c ft

Ch : Khong thi gian gia 2 ln lin tipsi dy dui thng l T/2.Khong cch gia 2 nt lin k bng khong cch 2 bng lin k

v bng 2 .Khong cch gia 2 nt hoc 2 bng

2k .

- iu kin c sng dng trn si dy n hi:

+ C 2 u c nh:2

kl ( *Nk )

S nt trn dy l 1k ; s bng trn dy l k

+ C mt u c nh, mt u t do:4

12kl ( Nk )

S nt trn dy l 1k ; s bng trn dy l 1k

IVSNG M

1. i cng v sng m

- V sng m cng l sng c nn cc cng thc ca sng c c th pdng cho sng m.- Vn tc truyn m ph thuc vo tnh n hi, mt v nhit ca mi

trng. Biu thc vn tc trong khng kh ph thuc nhit :tvv 10

v0l vn tc truyn m C0

0 ; v l vn tc truyn m t0C;1

273K-1

2. Cc bi ton v to ca m

- Mc cng m k hiu l L, n v l ben (B) :0

lgI

L BI

-Nu dng n v xiben th :0

10lgI

L dBI

; 1 10B dB

Vi I l cng m (n v 2W/m , I0l cng m chun,2-12

0 W/m10I .3. Cc bi ton v cng sut ca ngun m


21/51


21

-Cng sut ca ngun m ng hng: IIP .r4S 2 (S l din tch ca mt cu c bn knh rbng khong cch gia tm

ngun m n v tr ta ang xt, I l cng m ti im ta xt)- BA II , l cng m ca cc im A, B cch ngun m nhng khong

rA, rB th:2

2

A B

B A

I rI r

- Mi lin h gia cng m v bin ca sng m:2

2

2

1

2

1

A

A

I

I

- Khi cng m tng (gim) k ln th mc cng m tng (gim)kN lg (B) v kN lg10 (dB).

+ Trng hpn

k 10 nN (B) hoc nN 10 (dB)

4. Giao thoa sng mGiao thoa sngsng dng p dng cho:

+ Dy n c 2 u c nh:

m c bn:l

vf

20 (cn gi l ha m bc 1)

ho mbc 2 l : f2 = 2f0;

ha m bc 3 l : f3 = 3f0 bc n:l

vnfn

2.

+ ng so:

H mt u: m c bnl

vf

40 (cn gi l ha m bc 1);

ho mbc 3 l f3 = 3f0; f5 = 5f0 bc n:

l

vnfn

4

12 .

H 2 u: m c bnl

vf

20 ;

ho m f1 = 2f0; f1 = 3f0 ; f bc n:l

vnfn

2. .

Ch : i vi ng so h 1 u, u kn s l 1 nt, u h s l bngsng nu m nghe tonht v s l nt nu m nghe b nht


22/51


22

CHNG III: DNG IN XOAY CHIU

I. ICNG VIN XOAY CHIU

1. Sut in ng xoay chiu- Chu k v tn s quay ca khung:

2

1;

2

TfT

-Biu thc ca t thng qua khung dy:ttNBS coscos

0

0NBS: T thng cc i gi qua khung dy.

-Biu thc ca sut in ngxut hin trong khung dy dn:

tEtNBSt

e sinsin 0

vi 00 NBSE : Sutin ng cc i xut hin trong khung.2. in p (hiu in th) xoay chiu. Dng in xoay chiu

- Hiu in th xoay chiu: 0 cos( )uu U t (V)

- Dng inxoay chiu:0

cos( ) (A)i

i I t

i lng u i gi l lch pha ca u so vi i.+ Nu 0 th u sm pha so vi imt gc+ Nu 0 th u tr pha so vi i

+ Nu 0 th u cng (ng) pha so vi i.Ch :

+ Nu c mt in p xoay chiu (in p cc i l0

U ) c t

vo hai u bng n non m n ch sng ln mi khi in p u ln hnmt gi tr no 01 Uuu th trong mt chu k n sng ln 2 ln v

tt i 2 ln. Trong mt giy n sng ln hoc tt i 2f ln.+ Cc my o ch cc gi tr hiu dng ca cc i lng

3. Cc gi tr hiu dng 02

II ; 0 0,

2 2

U EU E

4. Cc cng thc khc- Tnh nhit l-ngta ra trn in tr thun theo cng thc: RtIQ 2


23/51


23

(I l gi tr hiu dng ca dng inchy qua R trong thi gian t).-in trca on dy dn ng cht, tit din u c chiu di l l ,

in tr sut , din tch tit din l S:S

lR ;

-Mt khi chtc khi lng m, nhit dung ring l KkgJc . nhn nhit

lng Q tng nhit t1

t n 2t , th: 12 ttmcQ -in lng chuyn qua tit dinca dy dn trong khong thi gian tt

1t n 2t :

2

1

t

t

dqq2

1

t

t

idt

5. Dng in xoay chiu trong mch ch c in tr thun R; ch c cundy thun cm L v ch c t in C

Ch c R Ch c L Ch c Cnh lutm

RIU R 00 ,

IRUR LL ZIU 00 ,

LL IZU CC ZIU 00 ,

CC IZU

Tr khng R LZ L 1C

ZC

lch pha(u v i)

ui = 0 ui = + /2 ui = - /2

Gin vct

Lin h giau v i: 0

00 I

i

U

u 1

2

0

2

2

0

2

I

i

U

u 1

2

0

2

2

0

2

I

i

U

u

II. MCH R, L, C MCNITIP.CNGHNGIN

1. Cc cng thc c bn


24/51


24

Ccmt

Mch RLC Mch RL Mch RC Mch LC

Dngmch

Vectquay

Tng

trZ = 2

L

2)(ZR CZ Z =

2

L

2R Z Z =2

C

2R Z

Z =

CL ZZ

Gclchpha

L CZ - Z

tan

R0L 0C

0R

U - Utan

U

L C

R

U - Utan

U

+

ZL >ZC:tnh cmkhng

+ ZL< ZC:tnh dung

khng.+ ZL=ZC:cnghng

LZ

tan

R

0L L

0R R

U Utan

U U

Mch c tnh cmkhng: > 0

CZ

tan -

R

0C C

0R R

U Utan - -

U U

Mch c tnhdung khng:

< 0

tg

nhlutm

Z

UI;

Z

UI 00

Z

UI;

Z

UI 00

Z

UI;

Z

UI 00

Z

UI;

Z

UI 00

Cngsut

P = UIcos

P = RI2

P = UIcos

P = RI2

P = UIcos

P = RI2P = 0

innng

W = P t W = P t W = P t W = 0


25/51


25

2. Cng hng in.Nu gi nguyn gi tr ca in p hiu dng U gia hai u mch v

thay i tn s gc sao cho CL ZZ hayC

L1

, th trong mch xy

ra hin tng c bit, l hin tng cng hng. Khi :+ Tng tr ca mch t gi tr nh nht RZmin .

+ Cng dng in qua mch t gi tr cc iR

UImax .

+ Cc in p tc thi hai u t in v hai u cun cm c bin bng nhau nhng ngc pha nn trit tiu ln nhau, in p hai u intr bng in p hai u on mch.

iu kin xy ra cng hng l :

01

CL

LC

1 .

3. iu kin hai i lng tha mn h thc v pha+Khi hiu in th cng pha vi dng in (cng h ng):

0tanR

ZZ CL hay CL ZZ

+Khi hai hiu in th u1 v u2 cng pha: 2121 tantan .Sau lp biu thc ca 1tan v 2tan th vo v cn bng biu

thc ta s tm c mi lin h.+ Hai hiu in th c pha vung gc:

1tan.tan2

2121 .

Sau lp biu thc ca1

tan v2

tan th vo v cn bng biu thcta cng s tm c mi lin h.

Tr-ng hp tng qut hai i l ng tho mn mt h thc no ta sdngphng phpgin vectl tt nht hoc dng cng thc hm stan gii ton:

21

21

21tan.tan1

tantantan


26/51


26

4. MT S CNG THC P DNG NHANH CHO DNG CUHI TRC NGHIM (dng hi p)

Cc dng sau y p dng cho on mch xoay chiu L RC mcni tip

Dng 1: Hi iu kin c cng hng in mch RLC v cc h qup: iu kin ZL = Zc LC2 = 1

Khi Z = Zmin = R ; I = Imax=U

R

cos = 1 ; P = Pmax =2

U

R

Dng 2:Cho R bin iHiR Pmax, tnh Pmax, h s cng sut coslc ?

p : R = ZL - ZC,2

Max

U 2P = , cos =

2R 2

Dng 3: Cho R bin i ni tip cun dy c rHiR cng sut trn R cc i

p : R2

= r2

+ (ZL - ZC)2

Dng 4:Cho R bin i, nu vi 2 gi tr R1 , R2 m P1 = P2HiR PMax

p R = ZL - ZC= 1 2R R Dng 5: Cho C1, C2 m I1 = I2 (P1 = P2)

HiC PMax(cng hng in)

p C1 C2c L

Z + ZZ = Z =

2

Dng 6: Cho L1, L2 m I1 = I2 (P1 = P2)HiL PMax (cng hng in)

p L1 L2L C

Z + ZZ = Z =

2

Dng 7: Hi vi gi tr no ca C th in p hiu dng trn t in UCmax

p ZC =2 2

L

L

R + Z

Z

, Khi

R CLM N BA


27/51


27

2 2

ax

L

CM

U R ZU

Rv

2 2 2 2 2 2

ax ax ax; 0

CM R L CM L CM U U U U U U U U

Dng 8: Hi vi gi tr no ca Lth in p hiu dng trn t in ULmax

p ZL =2 2

C

C

R + Z

Z

, Khi

2 2

ax

C

LM

U R ZU

Rv 2 2 2 2 2 2ax ax ax; 0LM R C LM C LM U U U U U U U U

Dng 9: Hi iu kin 1, 2lch pha nhau

2(vung pha nhau)

p p dng cng thc tan 1.tan2 = -1Dng 10: Hi khi cho dng in khng i trong mch RLC th tc dng

ca R, ZL, ZC?p : I = U/R ZL = 0 ZC =

Dng 11: Hi Vi = 1hoc = 2th I hoc P hoc UR c cngmt gi tr th IMax hoc PMaxhoc URMax

p khi : 1 2 tn s 1 2f f f

Dng 12: Gi tr = ? th IMax URmax; PMax cn ULCMin

p : khi

1

LC (cng hng)Dng 13: Hi: Hai gi tr ca :

1 2P P

p 21 2 0

Dng 14: HiHai gi tr ca L :1 2L L

P P

p1 2 2

0

2L L

C

Dng 15: HiHai gi tr ca C :1 2C C

P P

p2

1 2 0

1 1 2

C C L

Dng 16: HiHai gi tr ca R :1 2R R

P P

p R1R2 =2

( )L CZ Z v R1 + R2=2

U

P

Dng 17: Hi khi iu chinh L URCkhng ph thuc vo R thp: Khi ZL = 2 ZC


28/51


28

5. Cng sut ca mch in xoay chiu. H s cng sut.- Cng thc tnh cng sut ca mch in xoay chiu bt k:

cosUIP ; cos l h s cng sut.

- Ring vi mch ni tip RLC:2

2 R

R

UP I R U I

R

.

- H s cng sut ca on mch ni tip RLC:Z

R

U

URcos

- i vi ng c in: 2cosco

P UI P I R ;

trong R l in tr thun ca ng c, cos l h s cng sutca ng c, I l cng dng in chy qua ng c, U l in pt vo hai u ng c v

ciP l cng sut c ch ca ng c.

- Hiu sut ca ng c in:os

ciP

HUIc

Ch : + tm cng sut hoc h s cng sut ca mt on mch no th cc i l ng trong biu thc tnh phi c trong on mch .

+ Trong mch in xoay chiu cng sut ch c tiu th trnin tr thun.

III. MY PHT IN XOAY CHIU

1. My pht in xoay chiu mt pha- Tn s dng in xoay chiu do my pht pht xoay chiu mt pha pht ra:f np trong : p s cp cct, n s vng quay ca roto trong mt giy.

2. My pht in xoay chiu ba phaa. Ngun mc theo kiu:

+ hnh sao: 0 0

3

d p

d p

I I

I

U U

+ hnh tam gic:3d p

d p

I I

U U

b. Phi hp mc ngun v ti

Ngun v ti u mc hnh sao:p dng cho ngun A:


29/51


29

3pd

dp

UU

II

p dng cho ti B:

3'

'

pd

dp

UUII

Ngun v ti u mc tam gic:p dng cho ngun A:

pd

pd

UU

II 3

p dng cho ti B:

'

'3

pd

pd

UU

II

Ngun mc hnh sao v ti mc tam gic:p dng cho ngun A:

3pd

dp

UU

II

p dng cho ti B:

'

'3

pd

pd

UU

II

Ngun mc tam gic v

ti mc hnhsao:

p dng cho ngun A:

pd

pd

UU

II 3

p dng cho ti B:

3'

'

pd

pd

UU

II


30/51


30

IV. MY BIN P V TRUYNTIINNNG

1. My bin pa. Gi

1N v

2N l s vng dy ca cun s cp v th

cp; 1 2,i i v 1 2,e e l c ng v sut in ngtc thica mch s cp v th cp; 1 2,r r v 1 2,u u lin tr ca

cun dy s cp v th cp v hiu inth tc thi hai u mch s cp vth cp. Ta c cc lin h:

kN

N

e

e

2

1

2

1

(k gi l h s ca MBA)

cun s cp: 1 1 1 1u e i r cun th cp: 2 2 2 2u e i r

b. Nu in tr cc cun dy khng ng k:Gi

1U v

2U l in p hiu dng xut hin hai u ca cun s

cp v th cp;1

I v2

I l c ng hiu dng dng in ca mch s

cp v th cp khi mch kn. H l hiu sut ca MBA.

Ta c cc lin h: 1 1

2 2

U NU N

+ 2 1N N th 2 1U U , ta gi MBA l my tng th.

+ 2 1N N th 2 1U U , ta gi MBA l my h th.

- Hiu sut ca my bin p :111

222

cos

cos

IU

IUH

vi 1cos v 2cos l h s cng sut ca mch s cp v th cp.-Nu mch s cp v th cp c u v i cng pha th:

2 2

1 1

U IH

U Ihay

HI

I

U

U

.1

2

2

1 ;

Khi %100H th 2 2 1

1 1 2

U N I

U N I

c. Nu in tr cun s cp v th cp ln lt l r1 v r2, v mchin hai u cun th cp c in tr R:


31/51


31

Quy c: 12

N= k

N

- in p hai u cun th cp U2 = 1 22 1

k.RU

k (R + r ) + r

- Hiu sut my bin p: H =2

2

2 1

k .Rk (R + r ) + r

2. Truyn ti in nngUP, : l cng sut v in p

ni truyn i, ',' UP : l cngsut v in p nhn c ni

tiu th; I: l cng dngin trn dy, R: l in trtng cng ca dy dntruynti.

+ gim th trn dy dn: IRUUU ' + Cng sut hao ph trn -ng dy:

RU

PRIPPP .

cos'

22

22

+Hiu sut ti in:P

PP

P

PH

'' ,

Ch :+ Ch phn bit hiu sut ca MBA H v hiu sut ti in 'H .+ Khi cn truyn ti in khong cch l th ta phi cn si dy dn

c chiu di l2 .


32/51


32

CHNG IV: DAO NG SNG INT

i, I0: cng tc thi v cng cc i trong mch; q, Q0: in tchtc thi v in tch cc i trn t in; u, U0: in p tc thi v inpcc i trn t in.

1. i cng : Chu k, tn s ca mch dao ng

- Tn s gc:LC

1;

0

0

Q

I

- Chu k dao ng ring:0

0222IQLCT

- Tn s ring:LC

f2

1

2

Ch : Nu mch dao ng c t 2 t tr ln th ta coi b t l mt t cin dung C t ng ng c tnh nh sau:

+ Ghp ni tip:n

i in CCCCC 121

11...

111

n

CCCC ,...,,21

+ Ghp song song:n

i

in CCCCC1

21 ... nCCCC ,...,, 21

- Gi T1 v T2l chu k dao ng in t khi mc cun thun cm L lnlt vi t C1 v C2 th:

+ Khi mc L vi C1 ni tip C2:

2

2

2

1

2

2

2

2

1

2

111

TTT

fff

+ Khi mc L vi C1 song song C2:

2

2

2

1

2

2

2

2

1

2

111

fff

TTT


33/51


33

2. Nng lng ca mch dao ngNng l-ng in tr-ng:

222 201 1 cos

2 2 2tt

QqW Cu t

C C

22

02

1iIL

Nng l-ng t tr-ng:2 2 2

0

1 1sin

2 2dt

W Li LI t 2202

1uUC

Nng l-ng in t:

2 2

0 0

1 1

2 2dt tt W W W CU LI

2

2 2 01 1 1

2 2 2

QLi Cu

C

- Lin h gia in tch cc i v in pcc i: 00 CUQ - Lin h gia in tch cc i v dng in cc i: 00 QI

- Biu thc c lp thi gian gia in tch v dng in:2

222

0

iqQ

3. Qu trnh bin i nng lng mch dao ng

Nu mch dao ng c chu k T v tn s f thNng lng in trngv v nng lng t trng (

td WW , ) dao ng vi tn s f= 2f, chu k

T=2

T

u-U0 0U

2

0 0U2

0U 2

2 0

U 3

2

+U0T/4 T/12 T/6

T/8 T/8

T/6 T/12

Wtt = 3 Wt WtmaxW = 0

Wt = 3 WttWt = WttWtmin = 0Wmax


34/51


34

Ghi ch:- Hai ln lin tip Wt = Wtt l T/4- Khi q cc i th u cc i cn khi i cc tiu (bng 0) v ngc li.

4. Thu v pht sng in t-Khung dao ng c th pht v thu cc sng in t c b c sng:

LCcTc 2. ;c l tc truyn sng in t trong chn khng ( smc /10.3 8 )

-Nu mch dao ng c L thay it21 LL 21 LL th mch chn

sng c th chn c sng c bc sng:CLcCLc 21 22

-Nu mch dao ng c C thay it 21 CC 21 CC th mch chn

sng c th chn c sng c bc sng:21 22 LCcLCc

-Nu mch dao ng c L thay it 21 LL 21 LL v c C thayi t 21 CC 21 CC th mch chn sng cth chn c sng c

bc sng:

2211 22 CLcCLc

Gi 1 v 2 lbc sng mch dao ng hot ngkhi dng cun thun cm L mc vi C1 v C2th bc sngmch dao ng hot ng khi mc L vi:

+ C1 // C2 :2

2

2

1

2

2

2

2

1

2

111

fff

+ C1ni tip C2 :2

2

2

1

2

2

2

2

1

2

111fff

-Nu mch dao ng c C thay it 21 CC

21 CC th mch hot ng vi bc sng trong khong

21 21 th:

1

22

2

2

2

22

2

1

44 CcL

Cc


35/51


35

-Nu mch dao ng c L thay it 21 LL 21 LL th mch hotng vi bc sng trong khong 21 21 th:

1

22

2

2

2

22

2

1

44 LcC

Lc

Ch :Hai cng thc cui vn p c cho trng hp L v C l hng scn bc sng bin thin 21 .

5. Mch dao ng tt dn- Khung dy c in trhot ng nnc s

2 2 2 22 0 0

2 2

C U U RC I R R

L

cng l cng sut to nhit ca in tr.-Nng l-ng cn cung cp trong khong thi gian t: RtIQA 2

6. Di sng in t

Nidung

SNGDI

SNGTRUNG

SNG NGNSNG CC

NGNBc

sng

> 1000 m1.000 m

100 m

100 m10 m 10 m0,01m

cim

- C nnglng nh,- khngtruyn ci xa trnmt t.- t b nchp th

- C nnglng kh ln,- Truyn i

c trn mtt.- B tng in

ly hp th voban ngy v

phn x voban m

- C nng lngln,- Truyn i cmi a im trnmt t- C kh nng

phn x nhiu lngia tng in lyv mt t

- C nng lng rtln ln- Truyn c ic trn mt t- Khng btngin ly hp th hoc

phn x v c khnng truyn i rt xatheo mt ngthng

ngdng

Dng thng tindi nc

Dng thngtin vo banm

Dng thngtin trn mt t

Dng thng tintrong v tr.


36/51


36

CHNG V: SNG NH SNG

1. Tn sc nh sng

a. i vi lng knh

- Cng thc lng knh:

AiiD

Arr

rni

rni

'

'

'sin'sin

sinsin

vi i, i l gc ti v gc l; A l gc chit

quang; D l gc lch to bi tia ti v tia l.- Tr-ng hp gc nh: D = (n - 1)A- Gc lch cc tiu.

+ Khi c gc lch cc tiu, ng i ca tia sng i xng qua mt phngic ca gc chit quang.

+ K hiu gc lch cc tiu l mD , gc ti ng vi gc lch cc tiu l

mi , ta c:

2sin

2sin

2

2'

An

AD

AiD

Arr

m

mm

- Gc lch gia 2 tia sng n scqua lng knh (chit sut i vi lngknh ln lt l n1 v n2 21 nn ): AnnD 21

-B rng quang ph lin tctrn mn chn t pha sau lng knh cchlng knh mt khong l:lAnnl

t

(vi nt v nl chit sut ca nh sng tm v nh sng i vilngknh v A tnh bng radian)

b. Tn sc t mi trng ny sang mi trng khc

*Nu dngnh sng n sc th:+ Mu n sc khng thay i (v f khng i)


37/51


37

+ Bc sng n sc thay iVn tc v bc sng ca nh sng trong mi trng c chit sut n:

n

cv ;

n' ;

trong c v l vn tc v bc sng ca nh sng trong chn khng.

+ Dng nh lut khc x tm gc khc x 211

2

sin

sinn

n

n

r

i

+ Nu nh sng t mi trng chit quang ln sang mi trng chit

quang nh phi xc nh ghi :1

2sin

n

ni gh

*Nu dngnh sng trngth:+ C hin tng tn sc v xut hin chum quang ph lin tc.+ Cc tia n sc u b lch

- Tia lch t so vi tia ti;- Tia tm lch nhiu so vi tia ti.

c. Thang sng in t

10-11

m 10-8

m 0,001m (m)f(Hz)

2. Giao thoa nh sngGi khong cch gia hai khe S1S2 la, khong cch t mt phng cha 2khe v mn chn l D, l b csng ca nh sng.a. Cc cng thc c bn

-Hiu ng ica mt im c

ta x trn mn:

Sngvtuyn

Tiahngngoi

nhsngtrng

Tia tngoi

TiaX

Tiagama

0,4m

0,75m


38/51


38

D

axdd 12

- V tr vn sng: kia

Dkx

Vn sng bc n ng vi nk - V tr vn ti:

212

212

ik

a

Dkx hoc

1

2

sk s k

tk

x xx

0k vn ti bc n ng vi 1nk ; 0k vn ti th n ng vi nk ;

v d: vn ti th 5 ng vi 5k hoc k=4.

-Khong vn:D

i

a

-Bc sng ca nh sng:D

ia

- Tn s ca bc x:c

f

-Khong cch gia n vn sng lintip l d th:1n

di

-Khong cch gia 2 vn sng bc kbng: ki2

b. S vn sng, ti trn mnTnh s vn sng, ti trn trng giao thoa L, c 2 cch:Cch 1:+ S vn sng trn L l s nghim k (nguyn) tha mn h thc:

i

Lk

i

L

22

+ S vn ti trn L l s nghim k (nguyn) tha mn h thc:

2

1

22

1

2 i

Lk

i

L

Cch 2:

+ S vn sng: 12

2i

LN ,

i

L

2l phn nguyn ca biu thc

i

L

2.


39/51


39

Tnh s vn sng ti trn on AB c ta xA v xBbt k xA< xB+ S vn sng trn on AB l s nghim k (nguyn) tha mn h thc:

i

xk

i

x BA

+ S vn ti trn on AB l s nghim k nguyn tha mn h thc:

2

1

2

1

i

xk

i

x BA Zk

Dch chuyn ca h vnGi: D l khong cch t 2 khe ti mn

D1 l khong cch t ngun sng ti 2 khe

+ Khi ngun sng S di chuyntheo phng song song vi S1S2th hvn di chuyn ngc chiu, khong vn i vn khng i v di ca hvn l:

0

1

Dx d

D, d l dch

chuyn ca ngun sng.

+Khi ngun S dng yn vhai khe dch chuyn theo phngsong song vi mn th h vn dch chuyn cng chiu, khong vn i vnkhng i v di ca h vn l:

dD

Dx

1

0 1 , d l dch chuyn ca hai khe S1 v S2.

e. Giao thoa vi nh sng trng hoc nh sng nhiu thnh phni vi nh sng gm nhiu thnh phn n sc (nh sng a sc).- V tr chng chp (trng) ca nhiu bc x:

nnkkkx ...2211

l

k

k

k

1

2

2

1 (*) (l

kl dng ti gin ca

1

2 , k, l nguyn dng)

T suy ra:nlknkk

..

2

1 ....,2,1,0n


40/51


40

Suy ra v tr ca bc x trng: 111

1 knia

Dkn

a

Dkx

- Cc bc x trngmu vn trung tm cch nhau mt khong u n

a

Dkd 1 , k l hng s bit (*)

i vi nh sng trng 0,38 0,76m m .

- B rng vn sng (quang ph) bc k:

ttk iika

kDx .

-nh sng n sc c vn sng ti im ang xt:.k D xa

x a kD ,

k c xc nh t bt ph ng trnh: 0,38 0,76xa

m mkD

- nh sng n sc c vn ti ti im ang xt:2

2 12 2 1

D xax k

a k D,

k c xc nh t bt ph ng trnh 20,38 0,762 1xam m

k D

Lu : V tr c mu cng mu vi vn sng trung tm l v tr trng nhauca tt c cc vn sng ca cc bcx thnh phn c trong ngun sng.

CHNG VI: LNGTNH SNG

0 : gii hn quang in, 0f : tn s gii hn quang in, : bc sng

nh sng, f : tn s ca nh sng, A: Cng thot, max0v : vn tc ban u

cc i, bhI : cng dng quang in bo ha, hU : in p(hiu in th)

hm, h : hng s Flng ( Jsh 3410.625,6 ) , c : vn tc nh sng trong

chn khng ( smc /10.3 8 ), e : in tch ca electron ( Ce 1910.6,1 )

1. Cc cng thcvhintng quang in


41/51


41

+Nng lng ca photon: 2cmhc

hf ph

+ng lng ca photon:c

hcmp ph ,

mphl khi lng tng i tnh ca photon.+ Gii hn quang in: 0

hc

A

+Ph-ng trnh Anhxtanh: 201

2maxhf A mv ,

khi lng ca electron 319,1.10m kg .+Bc x n sc(bc sng ) c pht ra v nng lng ca mi

xung l E th s photon pht ra trong mi giy bng:hc

E

hf

EEn

+ Vn tc ban u cc i:m

hc

v0

max0

112

+ Vt dn c chiu sng: max2

max0

2

1Vemv

( maxV l in th cc i ca vt dn khi b chiu sng)+Nu in trng cnl u c cng E v electron bay dc theo

ng sc in th:

max

2

max02

1Edemv

( maxd l qung ng ti a m electron c th ri xa c Catot).

Ch :+Nu chiu vo Catt ng thi 2 bc x 1, 2 th hin tng

quang in xy ra i vi bc x c bc sng b hn 0 0ff . Nu c2 bc x cng gy ra hin tng quang in th ta tnh ton vi bc x c

bc sng b hn.+Ban nng cao

- in p hmtrit tiu dng quang in21

2omax h

mv eU


42/51


42

- Cng dng quang in bo ha:I = ne (n: s electron v anot trong 1 s)

- Tc electron khi v n anodDng nh l ng nng WA - Womax= eUAK

2. Chuynngca electron trong trngint

a. Chuyn ng ca electron trong in trng

+ in p U tng tc cho electron: 22

1vmeU e

2

02

1vme

( 0v v v ln lt l vn tc u v vn tc sau khi tng tc ca e).

+ Trong in tr ng u: dF e E

ln: EeF C 3 trng hp:

-Nu0

v E: chuyn ng chm dn u vi gia tceE

am

-Nu 0v E: chuyn ng nhanh dn u vi gia tc

eE

a m

-Nu0

v E: chuyn ng cong qu o Parabol+ Theo phng xx: thng u: x = v0t

+ Theo phng yy nhanh dn u vi gia tceE

am

b. Chuyn ng ca electron trong t trng

+ Trong t tr ng u: B qua trng lc ta ch xt lc Lorenx:sinvBef = ma =

2v

mR

Bv

,

+Nu vn tc ban u vung gc vi cm ng t: lectron chuyn ngtrn u vi bn knh

.m vR

e B; bn knh cc i:

Be

mvR max0

max


43/51


43

+Nu vn tc ban u xin gc vi cm ng t: lectron chuyn

ng theo vng xon c vi bn knh ng vng c:sin

max0

Be

mvR

3. Cng sutcangun sng - dng quang in bo ho

- hiusutlngt

a. Cng sut ca ngun sng.

.P n IS eH

I

hc

PPn bh

n l s photon ca ngun sng pht ra trong mi giy; l l ng t

nng l ng (photon); (I l cng ca chm sng, H l hiu sutlng t)

b. C-ng dng inbo ha.

eHnent

qI ebh

t

N

e

In bhe

N l selectron n c Ant trong thi gian t giy, en l slectron n Ant trong mi giy.

e l in tch nguyn t

19

1,6.10e Cc.Hiu sut l-ng t.' bhInH

n P e

'n l s lectron bt ra khi Katt kim loi trong mi giy.n l s photon p vo Katt trong mi giy.

Ch : Khi dng quang in bo ho th n = ne4. Chu k,tns,bc sng tia X ngRn Ghen pht ra

- Gi nng l ng ca 1 electron trong chm tia Catot c c khi n im cc l

W , khi chm ny p vo i m cc n s chia lm 2 phn:

+ Nhit l ng ta ra (Qi) lm nng i m cc v

+ phn cn li c gii phng d i dng nng lng photon ca tia X(bc x Rn-ghen).


44/51


44

i QW

Trong : c

hhf

(l nng lng photon ca tia Rnghen)

22

2

0

2

mvUemvW

l ng nng ca electron khi p vo i catt (i m cc)U l hiu in th gia ant v cattv l vn tc electron khi p vo i cattv0l vn tc ca electron khi ri catt (thng v0 = 0)m = 9,1.10-31kg l khi lng electron

- Cng dng in qua ng Rn-ghen: enI ,(n l s e p vo i Catot trong 1s).Trng hp b qua nhit lng ta ra trn i m cc:

Ta c:

W ngha l

Wc

h

Hay

W

hc

- ng Rn Ghen s pht bc x c b c sng nh nht nu ton b nngl ng ca chm tia Katot chuyn hon ton thnh nng l ng ca bc xRn Ghen. B-c sng nh nht -c tnh bng biu thc trn khi du = xyra:

W

hcmin

Trng hp ton b nng lng ca electron bin thnh nhit

lng:-Nhit lng ta ra trn i Catot trong thi gian t:W = Q RI

2t = mctt: tng nhit ca i m cc (anot)c: nhit dung ring ca kim loi anot.

m: khi lng anotTrng hp tng qut:

- Hiu sut ca ng Rnghen: i

W

QW

WH


45/51


45

5. Mu nguyn tBo

+Khi nguyn tang mc nng l ng cao chuyn xung mc nngl ng thp th pht ra photon, ng c li chuyn t mc nng l ng thpchuyn ln mc nng l ng cao nguyn t s hp thu photon

hfEE thpcao

+Bn knh qu o dngth n ca electron trong nguyn t hir: 0

2rnrn

Vi mr 110 10.3,5 : l bn knh Bo ( qu o K)+Mi lin h gia cc bc sng v tn sca cc vch quang ph

ca nguyn t hir:

Th d 31 = 32 + 21

213231

111v 213231 fff

+Nng lng electron trong nguyn t hir:

2

13,6( )nE eV

n Vi n N*: lng t s.

+Nng lng ion ha hydro (t trng thi c bn)

Wcung cp = E - E1

Ch : Khi nguyn t trng thi kch thch n (trng thi th n)c th pht ra s bc x in t ti a cho bi cng thc:

2

12 nnCN n ; trong 2

nC l t hp chp 2 ca n.

+ Cc dy quang ph (ban nng cao)

L mann

K

M

N

O

L

P

Balmer

Paschen

H H H H

n = 1

n = 2

n = 3

n = 4n = 5n = 6


46/51


46

. 1 21; 2, 3, 4, ...n n dy Laiman (t ngoi)

1 22; 3, 4, 5, ...n n dy Banme (nhn thy)

1 23; 4, 5, 6,...n n dy Pasen (hng ngoi).

CHNG VII: HT NHN NGUYN TI - ICNG VHT NHN NGUYN T

1. Cu to ca ht nhn nguyn t- Ht nhn nguyn t l phn cn li ca nguyn t sau khi loi b electron,

ht nhn nguyn t X k hiu l: AZ X , XA , X

A .

Trong : Z l nguyn t s hay s proton trong ht nhn.N: S ntronA Z N : S khi.

- Kch thc (bn knh) ca ht nhn: 31

15.10.2,1 AR m ; vi A l skhi ca ht nhn.

2. n v khi lng nguyn t

- n v khi lng nguyn t l n v Cacbon (k hiu l u)kgu

2710.66055,11

-Ngoi ra theo h thc gia nng lng v khi lng ca Anhxtanh, khi

lng cn c th o bng n v2

c

eVhoc

2c

MeV;

2/5,9311 cMeVu

3. Nng lng linkt nng lng lin kt ring

Ht nhn AZ X c khi lng m c cu to bi Z proton v N

notron. Cc php o chnh xc cho thy khi lng m ca ht nhn AZ X bao

gi cng b hn tng khi lng ca cc nuclon to thnh ht nhn AZ X :

p mm Zm Nm m . m c gi l ht khi ca ht nhn.

-Nng lng lin kt v nng lng lin kt ring:A

lk

lkr

2

lk

WW

m.cW


47/51


47

Nng lng lin kt ring cng ln th ht nhn cng bn vng.-Nng lng ngh: 2mcE , vi m l khi lng ngh ca ht nhn

4. Cng thc Einstein gia nng lng v khi lng

Nng lng ht = Nng lng ngh + ng nng ca htE = E0 + W = mc

2 + mv2

II - PHNG X

- S ht nhn cn li: 0 0.2 .t

tTN N N e

-Khi lng cn li: tTt

emmm 00 2

Vi T l chu k phng x, l hng s phng xT

2ln

- S ht nhn b phn r: teNN 10 Tt

N 210

khi Tt : tNN 0

-Phn trm s nguyn t b phn r: tTt

eN

N121

0

-Khi lng b phn r: tTt

emmm 121 00

-Phn trm khi lng b phn r: tTt

em

m121

0

- S ht sinh ra bng s ht phng x mt i

- Tnh tui ca mu cht phngx:H

H

N

Nt 00 ln

1ln

1

-Khi c cn bng phng x: 2211 NN

-

Khi lng:A

N

Nm

A

.

- Cho phng trnh phng x: ZYX AZA

Z

'

'


48/51


48

+Nu ban u ht X ng yn th t s ng nng:Y

Z

Z

Y

m

m

W

W

+Nu cc ht sinh ra c cng vn tc th:Z

Y

Z

Y

m

m

W

W

- Cho 1 lng ng v phng x Xc chu k phng x l T, phng xban u l H0vo th tch V ca cht lng, sau thi gian 0t ly ra th tchv cht lng th phng x l H. Th tch cht lng bng:

T

tt

H

vH

He

vHV

00

2.

00

- Gi Nl s xung phng x pht ratrong thi gian1

t , 'N l s xung

phng x pht ra trong thi gian 2t k t thi im sau thi im ban umt khong thi gian

0t , th:

2

1

0

1

1.

' t

tt

e

ee

N

N

+ Nu 21 tt :0

'

te

N

N

+ Nu Ttt 21, :2

1.'

0

tte

NN t

Ch : Tui ca ming g c xc nh t thi im cht (cht) nthi im taxt.

Nu khong thi gian kho st rt nh so vi chu k bn r( Tt ) th ta vn dng h thc gn ng xex 1 (khi 1x ). y

ta c: tet

1 v Tt nn teNN 10 tN0

Phn ring ban nng cao+ phng x thi im t (n v Becren Bq):

0.tH N N e tT

t

eHH 00 2 00 NH

+ Lin h gia khi lng v phng x:AN

AHm


49/51


49

+ Lu : Khi tnh phng x H, H0th chu k phng x T tnh bngnv giy(s).

III - PHNNGHT NHN

Phng trnh phn ng: 31 2 41 2 3 4

AA A A

Z Z Z ZA B C D

1. Cc nh lut bo ton.+ nh lut bo ton s khi: 4321 AAAA .

+ Bo ton in tch: 4321 ZZZZ

+ nh lut bo ton ng lng: A B C DP P P P + nh lut bo ton nng lng ton phn: Nng lng tng cng

trong phn ng ht nhn l khng i.

Ch :Trong phn ng ht nhn khng c nh lut bo ton khi lng .

2. Xc nh nng lng, to hay thu bao nhiu?

Trong phn ng ht nhn31 2 4

1 2 3 4

AA A A

Z Z Z ZA B C D Cc ht nhn A, B, C, D c:

Nng lng lin kt ring tng ng l 1, 2, 3, 4.Nng lng lin kt tng ng l E1, E2, E3, E4 ht khi tng ng l m1, m2, m3, m4

a. ht khi: m m3 + m4 - m1 - m2

b.Cng thc tnh nng lng ca phn ng ht nhn:

Bit cc khi lng W = (MtrcMsau) c2

Nu Bit nng lng lin kt W = Esau - Etrc

Bit ht khi cc ht W = ( msau - mtrc)c2

Ch : p, n v electron c ht khi bng 0.


50/51


50

c. bit phn ng ta hay thu nng lng:Gi tng khi lng ca cc ht nhn v phi l m0, v to thnh l m.Nu:

0m m Phn ng to nng lng

Nng lng ta ra ca 1 phn ng: 20' cmmW

Nng lng ta ra thng dng ng nng cc ht.Cc ht sinh ra khi bn hon cc ht ban u

0m m Phn ng thu nng lng

+ Nng lng cn cung cp ti thiu phnng xy ra(chnh l nng lng thu vo ca phn ng): 20min cmmW

Nng lng thu vo thng di dngng nng cc ht

hoc bc xCc ht sinh ra khi khng bn hon cc ht ban u+ Nu ng nng ca cc ht ban u l minWW th:

'2

0 WcmmW

( 'W l ng nng ca cc ht sinh ra)3. Tnh ng nng v vn tc cc htca phn ht nhn, s dng

cc cch sau:

Dng nh lut bo ton nng lng ton phn:'

2

0 WWcmm

(S dng ht khi ca cc ht nhn: 20 cmm )Kt hp vi nh lut bo ton ng lng:

A B C DP P P P

2 2

A B C DP P P P

Dng phng php gii ton vecto v hnh hoc

T suy ra i lng cn tm v d gc hp bi chiu chuyn ng cacc ht so vi mt phng no Cc trng hp c bitkhi so snh ng nng cc ht sinh ra:

-Nu cc ht nhn ban u ng yn th:'

'

'

'

X

Y

Y

X

m

m

W

W

-Nu cc ht sinh ra c cng vn tc th:'

'

'

'

Y

X

Y

X

m

m

W

W

Ch :Cng thc gia ng lng v ng nng:


51/51


p2 = 2m WNhit ta ra khi t m kg cht t c nng sut ta nhit l L bng:

LmQ , L: nng sut to nhit (J/kg). JKWh 000.600.31

Nc nng OD22

1 dng chit sut ng v tri (2

1D ) dng lmnguyn liu phn ng nhit hch c trong nc chim 0,015% v khi lng.* Cc trng hp c bit thng gp

+ Trc ht ta c nh lut bo ton nng lngA + B C + D

WC + WD = (mtrc - msau)c2 + WA (gi s ht B ng yn) (1)

+ Hai ht sinh ra c vn tc vung gc nhau2 2 2

C D A C Dp p p p p

mCWC+ mDWD = mAWA (2)T (1) v (2) ta gii v tm c WC v WD

+ Mt trong hai ht sinh ra vung gc vi ht A2 2 2

C A D A C p p p p p

mDWD - mCWC= mAWA (2)T (1) v (2) ta gii v tm c WC v WD

+ Hai ht sinh ra ging ht nhau v vec t p cc ht i xngv hp

Ap vi cc gc bng nhau

Ta c cos2

A

C

pp

2cos2

A A

C C

m Wm W

Nh ta tm WC v WD.

+ Phng x sinh ra hai ht chuyn ng ngc chiu nhau

Cp

Ap

Dp

Ap

Cp

Dp

Cp

Ap

Dp

bản chính bộ công thức vật lý 12

Documents

Transcript of bản chính bộ công thức vật lý 12