Tom Tat Cong Thuc Vat Li 12

Tm tt cng thc vt l 12

Page 1

CHNG I: DAO NG C I. DAO NG IU HO

1. P.trnh dao ng : x = Acos(t + )

2. Vn tc tc thi : v = -Asin(t + )

3. Gia tc tc thi : a = -2Acos(t + ) = -2x

a

lun hng v v tr cn bng

4. Vt VTCB : x = 0; vMax = A; aMin = 0

Vt bin : x = A; vMin = 0; aMax = 2A

5. H thc c lp: 2 2 2( )v

A x

; 2

2 2 2

2

av A

6. C nng: 2 2

1W W W

2t m A

2 2 2 2 2

1 1W sin ( ) Wsin ( )

2 2mv m A t t 2 2 2 2 2 2

1 1W ( ) W s ( )

2 2t m x m A cos t co t

7. Dao ng iu ho c tn s gc l , tn s f, chu k T. Th ng nng v th nng bin

thin vi tn s gc 2, tn s 2f, chu k T/2.

8. T s gia ng nng v th nng : 2

1d

t

E A

E x

9. Vn tc, v tr ca vt ti :

+.nng= n ln th nng : 1 1

n Av A x

n n

+Th nng= n ln .nng : 11

A nv x A

nn

10. Khong thi gian ngn nht vt i t v tr c li x1 n x2

2 1t

vi

11

22

s

s

xco

A

xco

A

v 1 20 , )

11. Chiu di qu o: 2A 12. Qung ng i trong 1 chu k lun l 4A; trong 1/2 chu k lun l 2A 13. Qung ng vt i c t thi im t1 n t2.

Phn tch: t2 t1 = nT + t (n N; 0 t < T) -Qung ng i c trong thi gian nT l S1 = 4nA

-Trong thi gian t l S2. Qung ng tng cng l S = S1 + S2 Lu :

+ Nu t = T/2 th S2 = 2A + Tnh S2 bng cch nh v tr x1, x2 v v vng trn mi quan h

+ Tc trung bnh ca vt i t thi im t1 n t2: 2 1

tb

Sv

t t

14. Bi ton tnh qung ng ln nht v nh nht vt i c trong khong thi gian 0 < t < T/2.

-A A

x1 x2

O


Page 2

- Vt c vn tc ln nht khi qua VTCB, nh nht khi qua v tr bin nn trong cng mt khong thi gian qung ng i c cng ln khi vt cng gn VTCB v cng nh khi cng gn v tr bin. - S dng mi lin h gia dao ng iu ho v chuyn ng trn u.

+ Gc qut = t.

+ Qung ng ln nht khi vt i t M1 n M2 i xng qua trc sin ax 2Asin2

MS

+ Qung ng nh nht khi vt i t M1 n M2 i xng qua trc cos 2 (1 os )2

MinS A c

Lu : + Trong trng hp t > T/2

Tch '2

Tt n t (trong *;0 '

2

Tn N t )

Trong thi gian 2

Tn qung ng lun l 2nA

Trong thi gian t th qung ng ln nht, nh nht tnh nh trn.

+ Tc trung bnh ln nht v nh nht ca trong khong thi gian t: axaxM

tbM

Sv

t

v

MintbMin

Sv

t

vi SMax; SMin tnh nh trn. 14. Cc bc lp phng trnh dao ng iu ho:

* Tnh * Tnh A da vo phng trnh c lp

* Tnh da vo /k u v v vng trn:

thng t0=0 0

0

Acos( )

sin( )

x t

v A t

Lu : + Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0

+ Trc khi tnh cn xc nh r thuc gc phn t th my ca ng trn lng gic

(thng ly - < )

15. Cc bc gii bi ton tnh thi im vt i qua v tr bit x (hoc v, a, Wt, W, F) ln th n * Xc nh M0 da vo pha ban u * Xc nh M da vo x (hoc v, a, Wt, W, F)

* p dng cng thc

t (vi OMM 0 )

* Gii phng trnh lng gic ly cc nghim ca t (Vi t > 0 phm vi gi tr ca k ) * Lit k n nghim u tin (thng n nh) * Thi im th n chnh l gi tr ln th n

A -A

M M 1 2

O

P

x x O

2

1

M

M

-A A P 2 1 P

P 2

2


Page 3

Lu : ra thng cho gi tr n nh, cn nu n ln th tm quy lut suy ra nghim th n + C th gii bi ton bng cch s dng mi lin h gia dao ng iu ho v chuyn ng trn u 16. Cc bc gii bi ton tm li , vn tc dao ng sau (trc) thi im t mt khong thi

gian t.

* Xc nh gc qut trong khong thi gian t : t .

* T v tr ban u (OM1) qut bn knh mt gc li (tin) mt gc , t xc nh M2

ri chiu ln Ox xc nh x 17. Cc bc gii bi ton tm s ln vt i qua v tr bit x (hoc v, a, Wt, W, F) t thi im t1 n t2. * Gii phng trnh lng gic c cc nghim

* T t1 < t t2 Phm vi gi tr ca (Vi k Z) * Tng s gi tr ca k chnh l s ln vt i qua v tr . Lu : + C th gii bi ton bng cch s dng mi lin h gia dao ng iu ho v chuyn ng trn u. + Trong mi chu k (mi dao ng) vt qua mi v tr bin 1 ln cn cc v tr khc 2 ln. 18. Cc bc gii bi ton tm li , vn tc dao ng sau (trc) thi im t mt khong thi

gian t. Bit ti thi im t vt c li x = x0.

* T phng trnh dao ng iu ho: x = Acos(t + ) cho x = x0

Ly nghim t + = vi 0 ng vi x ang gim (vt chuyn ng theo chiu m v v < 0)

hoc t + = - ng vi x ang tng (vt chuyn ng theo chiu dng)

* Li v vn tc dao ng sau (trc) thi im t giy l

x Acos( )

Asin( )

t

v t

hoc

x Acos( )

Asin( )

t

v t

19. Dao ng c phng trnh c bit:

* x = a Acos(t + ) vi a = const

Bin l A, tn s gc l , pha ban u

x l to , x0 = Acos(t + ) l li . To v tr cn bng x = a, to v tr bin

x = a A Vn tc v = x = x0, gia tc a = v = x = x0

H thc c lp: a = -2x0 2 2 20 ( )

vA x

* x = a Acos2(t + ) (ta h bc)

Bin A/2; tn s gc 2, pha ban u 2. II. CON LC L XO

+ Phng trnh dao ng: cos( )x A t

Phng trnh vn tc: '; sin( ) cos( )2

dxv x v A t A t

dt

+ Phng trnh gia tc: 2

2 2

2'; ''; cos( );

dv d xa v a x a A t a x

dt dt

Hay 2 cos( )a A t

+ Tn s gc, chu k, tn s v pha dao ng, pha ban u:


Page 4

a. Tn s gc: 2

2 ( / ); k g

f rad sT m l

; ( )

mgl m

k

b. Tn s: 1 1

( ); 2 2

N kf Hz f

T t m

c. Chu k: 1 2

( ); 2t m

T s Tf N k

d. Pha dao ng: ( )t

e. Pha ban u:

Ch : Tm , ta da vo h phng trnh 0

0

cos

sin

x A

v A

lc 0 0t

MT S TRNG HP THNG GP

Chn gc thi gian 0 0t l lc vt qua v tr cn bng 0 0x theo chiu

dng 0 0v : Pha ban u 2

Chn gc thi gian 0 0t l lc vt qua v tr cn bng 0 0x theo chiu m

0 0v : Pha ban u 2

Chn gc thi gian 0 0t l lc vt qua bin dng 0x A : Pha ban u 0

Chn gc thi gian 0 0t l lc vt qua bin m 0x A : Pha ban u

Chn gc thi gian 0 0t l lc vt qua v tr 02

Ax theo chiu dng

0 0v : Pha ban u 3


Ax theo chiu dng

0 0v : Pha ban u

2

3


Ax theo chiu m 0 0v :

Pha ban u 3



Pha ban u 2

3


2

Ax theo chiu dng

0 0v : Pha ban u 4


Page 5


2

Ax theo chiu dng

0 0v : Pha ban u

3

4


2


Pha ban u 4


2

Ax theo chiu m

0 0v : Pha ban u 3

4


2

Ax theo chiu dng

0 0v : Pha ban u 6


2

Ax theo chiu dng

0 0v : Pha ban u

5

6


2


Pha ban u 6


2

Ax theo chiu m

0 0v : Pha ban u 5

6

cos sin( )2

; sin cos( )

2


Page 6

Gi tr hm s ca cc cung (gc) lng gic c bit.

5. Phng trnh c lp vi thi gian:

22 2

2

vA x ;

2 22

4 2

a vA

Ch : 2

: Vat qua v tr can bang

: Vat bien

M M

MM

v A a

va A

6. Lc n hi, lc hi phc:

a. Lc n hi:

( )

( ) ( ) neu

0 neu l A

hM

h hm

hm

F k l A

F k l x F k l A l A

F

b. Lc hi phc: 0

hpM

hp

hpm

F kAF kx

F

hay

2

0

hpM

hp

hpm

F m AF ma

F

lc hi phc

lun hng vo v tr cn bng. Ch : Khi h dao ng theo phng nm ngang th lc n hi v lc hi phc l nh nhau

h hpF F .

7. Thi gian, qung ng, tc trung bnh

a.Thi gian:Gii phng trnh cos( )i ix A t tm it

Ch :

Gi O l trung im ca qu o CD v M l trung im ca OD; thi gian i t O n M l

12OM

Tt , thi gian i t M n D l

6MD

Tt .

T v tr cn bng 0x ra v tr 2

2x A mt khong thi gian

8

Tt .

T v tr cn bng 0x ra v tr 3

2x A mt khong thi gian

6

Tt .

Goc

Hslg

00

300

450

600

900

1200

1350

1500

1800

3600

0

6

4

3

2

3

2

4

3

6

5

2

sin 0

2

1

2

2

2

3

1

2

3

2

2

2

1

0 0

cos 1

2

3

2

2

2

1

0

2

1

2

2

2

3

-1 1

tg 0

3

3

1 3 kx 3 -1

3

3

0 0

cotg kx 3 1

3

3

0

3

3

-1 3 kx kx


Page 7

Chuyn ng t O n D l chuyn ng chm dn ( 0; av a v

), chuyn ng t D n

O l chuyn ng nhanh dn ( 0; av a v

)

Vn tc cc i khi qua v tr cn bng (li bng khng), bng khng khi bin (li cc i).

b. Qung ng:

Neu th 4

Neu th 22

Neu th 4

Tt s A

Tt s A

t T s A

suy ra

Neu th 4

Neu th 44

Neu th 4 22

t nT s n A

Tt nT s n A A

Tt nT s n A A

Ch :

2 22 neu vat i t

2 2

neu vat i t 4

Ms A x A x A

Tt s A x O x A

2 2 2 2 neu vat i t

2 2

2 2 neu vat i t 0

2 2

8 2 21 neu vat i t

2 2

m

M

m

s A x A x A x A

s A x x AT

t

s A x A x A

3 3 neu vat i t 0

2 2

neu vat i t 6 2 2

3 3 2 3 neu vat i t

2 2

M

m

s A x x A

T A At s x x A

s A x A x A x A

neu vat i t 0

2 2

3 3121 neu vat i t

2 2

M

m

A As x x

Tt

s A x A x A

1.

2

2

2

2

42

4

kTm

mT

k mk

T

m = m1 + m2 ----> T2 = (T1)

2 + (T2)

2

m = m1 - m2 ----> T2 = (T1)

2 - (T2)

2

* Ghp ni tip cc l xo 1 2

1 1 1...

k k k cng treo mt vt khi lng n

h nhau th: T2 = T12 + T2

2

m t l thun vi T2

k t l nghch vi T2


Page 8

* Ghp song song cc l xo: k = k1 + k2 + cng treo mt vt khi lng nh nhau

th:2 2 2

1 2

1 1 1...

T T T

* Tn s gc: k

m ; chu k:

22

mT

k

;

tn s: 1 1

2 2

kf

T m

iu kin dao ng iu ho: B qua ma st, lc cn v vt dao ng trong gii hn n hi

2. C nng: 2 2 21 1

W2 2

m A kA

3. * bin dng khi l xo nm ngang : l = 0 * bin dng ca l xo thng ng khi vt VTCB:

mg

lk

2l

Tg

* bin dng ca l xo khi vt VTCB vi con lc l xo nm trn mt phng nghing

c gc nghing :

sinmg

lk

2

sin

lT

g

+ Chiu di l xo ti VTCB: lCB = l0 + l (l0 l chiu di t nhin) + Chiu di cc tiu (khi vt v tr cao nht):

lMin = l0 + l A + Chiu di cc i (khi vt v tr thp nht):

lMax = l0 + l + A lCB = (lMin + lMax)/2 + Khi A >l (Vi Ox hng xung):

- Thi gian l xo nn 1 ln l thi gian ngn nht vt i t v tr x1 = -l n x2 = -A. - Thi gian l xo gin 1 ln l thi gian ngn nht vt i t v tr x1 = -l n x2 = A, Trong mt dao ng (mt chu k) l xo nn 2 ln v gin 2 ln!

4. Lc ko v hay lc hi phc F = -kx = -m2x c im: * L lc gy dao ng cho vt. * Lun hng v VTCB * Bin thin iu ho cng tn s vi li 5. Lc n hi l lc a vt v v tr l xo khng bin dng. C ln Fh = kx

* (x

* l

bin dng ca l xo) * Vi con lc l xo nm ngang th lc ko v v lc n hi l mt (v ti VTCB l xo khng bin dng) * Vi con lc l xo thng ng hoc t trn mt phng nghing. ln lc n hi c biu thc:

* Fh = kl + x vi chiu dng hng xung

* Fh = kl - x vi chiu dng hng ln

+ Lc n hi cc i (lc ko): FMax = k(l + A) = FKmax (lc vt v tr thp nht)

+ Lc n hi cc tiu:

* Nu A < l FMin = k(l - A) = FKMin * Nu A l FMin = 0 (lc vt i qua v tr l xo khng bin dng)


Page 9

6. Mt l xo c cng k, chiu di l c ct thnh cc l xo c cng k1, k2, v chiu di tng ng l l1, l2, th c: kl = k1l1 = k2l2 = 7. o chu k bng phng php trng phng xc nh chu k T ca mt con lc l xo (con lc n) ngi ta so snh vi chu k T0

( bit) ca mt con lc khc (T T0). Hai con lc gi l trng phng khi chng ng thi i qua mt v tr xc nh theo cng mt chiu.

Thi gian gia hai ln trng phng 0

0

TT

T T

Nu T > T0 = (n+1)T = nT0.

Nu T < T0 = nT = (n+1)T0. vi n N*

Trong mt chu k, cht im qua v tr 0x x l 4 ln, nn

2

t k

8. Nng lng trong dao ng iu ha: tE E E

a. ng nng: 2 2 2 2 21 1

sin ( ) sin ( )2 2

E mv m A t E t

b. Th nng: 2 2 2 2 21 1

cos ( ) cos ( ); 2 2

tE kx kA t E t k m

Ch :

2 2 2

2 2 2

2

1 1

2 2

1 1: Vat qua v tr can bang

2 2

1: Vat bien

2

M M

tM

E m A kA

E mv m A

E kA

Th nng v ng nng ca vt bin thin tun hon vi

' 2

'2

' 2

f f

TT

ca dao ng.

Trong mt chu k, cht im qua v tr 0x x l 4 ln, nn

2

t k

III. CON LC N 1. Con lc dao ng vi li gc b ( T2 = (T1)

2 + (T2)

2

Tn s gc: g

l ; chu k:

22

lT

g

;

tn s: 1 1

2 2

gf

T l

2.Lc hi phc

2sins

F mg mg mg m sl

+ Vi con lc n lc hi phc t l thun vi khi lng.


Page 10

+ Vi con lc l xo lc hi phc khng ph thuc vo khi lng.

0ss

0

hpM

hp

hpm

gF mg

F m ll

F

3.1 Phng trnh dao ng:

a. Phng trnh li gc: 0 cos( )t (rad)

b. Phng trnh li di: 0 cos( )s s t

vi s = l, S0 = 0l

c. Phng trnh vn tc di: 0'; sin( )ds

v s v s tdt

v = s = -S0sin(t + ) = -l0sin(t + ) d. Phng trnh gia tc tip tuyn:

22 2

02'; ''; cos( ); t t t t

dv d sa v a s a s t a s

dt dt Ch : 00;

ss

l l

e. Tn s gc, chu k, tn s v pha dao ng, pha ban u:

3.2 a. Tn s gc: 2

2 ( / ); g mgd

f rad sT l I

b. Tn s: 1 1

( ); 2 2

N gf Hz f

T t l

c. Chu k: 1 2

( ); 2t l

T s Tf N g

d. Pha dao ng: ( )t

e. Pha ban u:

Ch : Tm , ta da vo h phng trnh 0

0

cos

sin

s s

v s

lc 0 0t

Lu : S0 ng vai tr nh A cn s ng vai tr nh x

4. H thc c lp: a = -2s = -2l

2 2 20 ( )

vS s

22 2

0

v

gl

Ch : 0

2

0

: Vat qua v tr can bang

: Vat bien

M M

MM

v s a

va s

5. Cnng:

2 2 2 2 2 2 2

0 0 0 0

1 1 1 1W

2 2 2 2

mgm S S mgl m l

l

6. Khi con lc n dao ng vi 0 bt k.

C nng W = mgl(1-cos0); Tc v2 = 2gl(cos cos0) Lc cng T = mg(3cos 2cos0)

- Khi con lc n dao ng iu ho (0


Page 11

2 20(1 1,5 )CT mg

7. Nng lng trong dao ng iu ha: tE E E

a. ng nng: 2 2 2 2 201 1

sin ( ) sin ( )2 2

E mv m s t E t

b. Th nng:

2 2 2 2 2

0

1 1(1 cos ) cos ( ) cos ( );

2 2t

g g gE mgl m s m s t E t

l l l Ch :

2 2 2

0 0 0

2 2 2

0

2

0 0

1 1(1 cos )

2 2

1 1: Vat qua v tr can bang

2 2

1(1 cos ): Vat bien

2

M M

tM

gE m s m s mgl

l

E mv m s

gE m s mgl

l

Th nng v ng nng ca vt dao ng iu ha vi

' 2

'2

' 2

f f

TT

Vn tc: 20 02 (1 cos ) 2 (cos cos )v v gl gl

Lc cng dy: 0(3cos 2cos )mg

8. Cng thc tnh gn ng v s thay i chu k tng qut ca con lc n (ch

l ch p dng cho s thay i cc yu t l nh):

g

g

l

l

T

T

T

TT

T

T '.

'1

'1

'

'

'

0

' 2 2 2 2

cao sauh hT t g l

T R R g L

vi : R = 6400km, ' , ' , 'T T T g g g l l l

Nu bi ton cho thay i yu t no th dng yu t tnh cn cc yu cn li coi

nh- bng khng

S sai lch ng h trong mt ngy m s l : 86400'

T

T

+ Ti cng mt ni con lc n chiu di l1 c chu k T1, con lc n chiu di l2 c chu k T2, con lc n chiu di l1 + l2 c chu k T2,con lc n chiu di l1 - l2 (l1>l2) c chu k T4.

Th ta c: 2 2 23 1 2T T T v

2 2 2

4 1 2T T T

9. Con lc n c chu k ng T cao h1, nhit t1. Khi a ti cao h2, nhit t2 th ta c:

2

T h t

T R

Vi R = 6400km l bn knh Tri t, cn l h s n di ca thanh con lc. 10. Con lc n c chu k ng T su d1, nhit t1. Khi a ti su d2, nhit t2 th ta c:

2 2

T d t

T R


Page 12

Lu : * Nu T > 0 th ng h chy chm (ng h m giy s dng con lc n)

* Nu T < 0 th ng h chy nhanh

* Nu T = 0 th ng h chy ng * Thi gian chy sai mi ngy

(24h = 86400s): 86400( )T

sT

11. Khi con lc n chu thm tc dng ca lc khng i: Lc ph khng i thng l:

* Lc qun tnh: F ma

, ln F = ma ( F a

)

* Lc in trng: F qE

, ln F = qE (Nu q > 0 F E

; cn nu q < 0

F E

)

* Lc y csimt: F = DgV ( F

lung thng ng hng ln) Trong : D l khi lng ring ca cht lng hay cht kh. g l gia tc ri t do. V l th tch ca phn vt chm trong cht lng hay cht kh .

Khi : 'P P F

gi l trng lc hiu dng hay trng lc biu kin (c vai tr nh trng

lc P

)

'F

g gm

gi l gia tc trng trng hiu dng hay gia tc trng trng biu kin.

Chu k dao ng ca con lc n khi : ' 2'

lT

g

Cc trng hp c bit:

* F

c phng ngang:

+ Ti VTCB dy treo lch vi phng thng ng mt gc c: tanF

P th

2 2' ( )F

g gm

* F

c phng thng ng th 'F

g gm

+ Nu F

hng xung th 'F

g gm

+ Nu F

hng ln th 'F

g gm

12. S thay i chu k dao ng ca con lc n:

a. Theo cao (v tr a l):

2

0h

Rg g

R h

nn 2h

h

l R hT T

g R

b. Theo chiu di dy treo (nhit ): 00(1 )l l t nn

0

0

2 ( 1)2t

l tT T

g

Thi gian con lc chy nhanh (chm trong 1s): 2 1

1 1

T TT

T T

lch trong mt ngy m: 1

86400T

T


Page 13

c. Nu 1 2l l l th 2 2

1 2T T T ; nu 1 2l l l th 2 2

1 2T T T

d. Theo lc l lF

:

2 2

hay

hay 2

hay cos

l hd

l hd hd

hd

l hd

F P a g g g al

F P a g g g a Tg

gF P a g g g a

Ch : Lc l c th l lc in, lc t, lc y Acsimet, lc qun tnh ( qta a

)

Gia tc php tuyn: 2

; : ban knh quy aonv

a ll

Lc qun tnh: F ma

, ln F = ma

( F a

)

Chuyn ng nhanh dn u a v

( v

c hng chuyn ng)

Chuyn ng chm dn u a v

Lc in trng: F qE

, ln F = qE;

Nu q > 0 F E

;

Nu q < 0 F E

Lc y csimt: F = DgV ( F

lun thng ng hng ln)

Trong : D l khi lng ring ca cht lng hay cht kh.

g l gia tc ri t do.

V l th tch ca phn vt chm trong cht lng hay cht kh .

Khi : hdP P F

gi l trng lc hiu dng hay trong lc biu kin (c vai tr nh trng

lc P

v hdF

g gm

gi l gia tc trng trng hiu dng hay gia tc trng trng biu

kin).

IV. TNG HP DAO NG

A. 1. Tng hp hai dao ng iu ho cng phng cng tn s x1 = A1cos(t + 1) v x2 =

A2cos(t + 2) c mt dao ng iu ho cng phng cng tn s

x = Acos(t + ).

Trong : 2 2 21 2 1 2 2 12 os( )A A A A A c

1 1 2 2

1 1 2 2

sin sintan

os os

A A

A c A c

vi 1 2 (nu 1 2 )

* Nu = 2k (x1, x2 cng pha) AMax = A1 + A2

`* Nu = (2k+1) (x1, x2 ngc pha)

AMin = A1 - A2 A1 - A2 A A1 + A2

2. Thng thng ta gp cc trng hp c bit sau: +

12 =00 th A =A1+A2 21


Page 14

+ 12 =90

0 th

2

2

2

1 AAA

+ 12 =120

0 v A1=A2 th A=A1=A2

+ 12 =180

0 th 21 AAA

3. Khi bit mt dao ng thnh phn x1 = A1cos(t + 1) v dao ng tng hp x = Acos(t

+ ) th dao ng thnh phn cn li l x2 = A2cos(t + 2).

Trong : 2 2 22 1 1 12 os( )A A A AAc

1 121 1

sin sintan

os os

A A

Ac A c

vi 1 2 ( nu 1 2 )

4. Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cng tn s x1 =

A1cos(t + 1;

x2 = A2cos(t + 2) th dao ng tng hp cng l dao ng iu ho cng phng cng tn s

x = Acos(t + ).

Chiu ln trc Ox v trc Oy Ox .

Ta c: 1 1 2 2os os os ...xA Ac Ac A c

1 1 2 2sin sin sin ...yA A A A

2 2

x yA A A v tany

x

A

A vi [Min;Max]

B. 1.

2. Phng php lng gic:

a. Cng bin : 1 1 2 2cos( ) va cos( )x A t x A t . Dao ng tng hp

1 2 cos( )x x x t A c bin v pha c xc nh:

1 2 1 22 cos cos ( )2 2

x A t

; t 1 22 cos2

A

A v 1 22

nn

cos( )x t A .

b. Cng pha dao ng: 1 1 0 2 2 0sin( ) va cos( )x A t x A t . Dao ng tng hp

1 2 cos( )x x x t A c bin v pha c xc nh: 1 0cos ( )cos

Ax t

;

t 1 22 2 2

2 1 2

1tan cos

1 tan

A A

A A A

Trong : 2

cos

A

A ; 0

VI. DAO NG TT DN-DAO NG CNG BC-CNG HNG A. 1. Dao ng tt dn ca con lc l xo

+ gim c nng sau mt chu k bng cng ca lc ma st cn tr trong chu k , nn : gim bin sau mi chu k l:

k

FA ms

4 ; 2

4 4mg gA

k

+ S dao ng thc hin -c: 2

4 4

A Ak AN

A mg g

+ Thi gian k t lc bt u dao ng cho n khi dng hn: -

k

mNNTN

2.

2.


Page 15

- .4 2

AkT At N T

mg g

(Nu coi dao ng tt dn c tnh tun hon vi chu k 2

T

)

+ Gi maxS l qung -ng i -c k t lc chuyn ng cho n khi dng hn. C nng

ban u bng tng cng ca lc ma st trn ton b qung -ng , tc l:

m s

m sF

kASSFkA

2.

2

1 2

maxmax

2 ; 2 2 2

2 2

kA AS

mg g

2. Dao ng tt dn ca con lc n

+ Suy ra, gim bin di sau mt chu k: 2

4

m

FS m s

+ S dao ng thc hin -c: S

SN

0

+ Thi gian k t lc chuyn ng cho n khi dng hn:

g

lNTN 2..

+ Gi maxS l qung -ng i -c k t lc chuyn ng cho n khi dng hn. C nng

ban u bng tng cng ca lc ma st trn ton b qung -ng , tc l:

?.2

1maxmax

2

0

2 SSFSm ms

3. Hin tng cng hng xy ra khi: f = f0 hay = 0 hay T = T0

Vi f, , T v f0, 0, T0 l tn s, tn s gc, chu k ca lc cng bc v ca h dao ng. B. 1. Dao ng tt dn:

a. Phng trnh ng lc hc: ckx F ma

b. Phng trnh vi phn: '' ( )cFk

x xm k

t cF

X xk

suy ra 2''k

X X Xm

c. Chu k dao ng: 2m

Tk

d. bin thin bin : 4 cFAk

e. S dao ng thc hin c: 1 14 c

A kAN

A F

Do ma st nn bin gim dn theo thi gian nn nng lng dao ng cng gim

2. Dao ng cng bc: cng bc ngoai lcf f . C bin ph thuc vo bin ca ngoi lc

cng bc, lc cn ca h, v s chnh lch tn s gia dao ng cng bc v dao ng ring.

3. Dao ng duy tr: C tn s bng tn s dao ng ring, c bin khng i. 4. S cng hng c:

0

0 Max

0

ieu kien lam A A lc can cua moi trng

f f

T T

C. I. Dao ng tt dn :

1. Th no l dao ng tt dn : Bin dao ng gim dn 2. Gii thch : Do lc cn ca mi trng (lc ma st) lm tiu hao c nng ca con lc 3. ng dng : Thit b ng ca t ng hay gim xc.


Page 16

II. Dao ng duy tr : Gi bin dao ng ca con lc khng i m khng lm thay i chu k dao ng ring bng cch cung cp cho h mt phn nng lng ng bng phn nng lng tiu hao do ma st sau mi chu k. III. Dao ng cng bc : 1. Th no l dao ng cng bc : Gi bin dao ng ca con lc khng i bng cch tc dng vo h mt ngoi lc cng bc tun hon 2. c im : - Tn s dao ng ca h bng tn s ca lc cng bc. - Bin ca dao ng cng bc ph thuc bin lc cng bc v chnh lch gia tn s ca lc cng bc v tn s ring ca h dao ng. IV. Hin tng cng hng : 1. nh ngha : Hin tng bin ca dao ng cng bc tng n gi tr cc i khi tn

s f ca lc cng bc tin n bng tn s ring f0 (hay =o) ca h dao ng gi l hin tng cng hng. 2. Tm quan trng ca hin tng cng hng : Hin tng cng hng khng ch c hi m cn c li

CHNG II: SNG C I. SNG C HC

1. 1. Bc sng: = vT = v/f

Trong : : Bc sng; T (s): Chu k ca sng; f (Hz): Tn s ca sng

v: Tc truyn sng (c n v tng ng vi n v ca ) 2. Phng trnh sng

Ti im O: uO = Acos(t + )

Ti im M1 : uM1 = Acos(t + -

12d )

Ti im M2 : uM2 = Acos(t + +

22d )

3. lch pha gia hai im trn cng mt phng truyn cch nhau mt khong d l :

d2

Nu 2 im nm trn mt phng truyn sng v cch nhau mt khong x th:

2x x

v

Lu : n v ca x, d, v v phi tng ng vi nhau 4. Trong hin tng truyn sng trn si dy, dy c kch thch dao ng bi nam chm in vi tn s dng in l f th tn s dao ng ca dy l 2f. II. SNG DNG 1. Mt s ch * u c nh hoc m thoa hoc u dao ng nh l nt sng. * u t do l bng sng * 2 im i xng vi nhau qua nt sng lun dao ng ngc pha. * 2 im i xng vi nhau qua bng sng lun dao ng cng pha.

* Cc im trn dy u dao ng vi bin khng i nng lng khng truyn i * Khong thi gian gia hai ln si dy cng ngang (cc phn t i qua VTCB) l na chu k. 2. iu kin c sng dng trn si dy di l:

* Hai u l nt sng: * ( )2

l k k N


Page 17

S bng sng = s b sng = k S nt sng = k + 1

* Mt u l nt sng cn mt u l bng sng: ( 1;3;5;7...)2

l m k

(2 1) ( )4

l k k N

S b sng nguyn = k S bng sng = s nt sng = k + 1 3. Phng trnh sng dng trn si dy CB (vi u C c nh hoc dao ng nh l nt sng)

* u B c nh (nt sng):

Phng trnh sng ti v sng phn x ti B: os2Bu Ac ft v

' os2 os(2 )Bu Ac ft Ac ft

Phng trnh sng ti v sng phn x ti M cch B mt khong d l:

os(2 2 )Md

u Ac ft

v ' os(2 2 )Md

u Ac ft

Phng trnh sng dng ti M: 'M M Mu u u

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

M

d du Ac c ft A c ft

Bin dao ng ca phn t ti M: 2 os(2 ) 2 sin(2 )2

M

d dA A c A

* u B t do (bng sng):

Phng trnh sng ti v sng phn x ti B: ' os2B Bu u Ac ft

Phng trnh sng ti v sng phn x ti M cch B mt khong d l:

os(2 2 )Md

u Ac ft

v ' os(2 2 )Md

u Ac ft

Phng trnh sng dng ti M: 'M M Mu u u

2 os(2 ) os(2 )Md

u Ac c ft

Bin dao ng ca phn t ti M: 2 cos(2 )Md

A A

Lu : * Vi x l khong cch t M n u nt sng th bin : 2 sin(2 )Mx

A A

* Vi x l khong cch t M n u bng sng th bin : 2 cos(2 )Md

A A

III. GIAO THOA SNG

Giao thoa ca hai sng pht ra t hai ngun sng kt hp S1, S2 cch nhau mt khong l: Xt im M cch hai ngun ln lt d1, d2

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:

11 1Acos(2 2 )M

du ft

v 22 2Acos(2 2 )M

du ft

Phng trnh giao thoa sng ti M: uM = u1M + u2M


Page 18

1 2 1 2 1 22 os os 22 2

M

d d d du Ac c ft

Bin dao ng ti M: 1 22 os2

M

d dA A c

vi 1 2

Ch : * S cc i: (k Z)2 2

l lk

* S cc tiu: 1 1

(k Z)2 2 2 2

l lk

1. Hai ngun dao ng cng pha ( 1 2 0 )

* im dao ng cc i: d1 d2 = k (kZ)

S ng hoc s im (khng tnh hai ngun): l l

k

* im dao ng cc tiu (khng dao ng):

d1 d2 = (2k+1)2

S ng hoc s im (khng tnh hai ngun): 1 12 2

l lk

2. Hai ngun dao ng ngc pha:( 1 2 )

* im dao ng cc i: d1 d2 = (2k+1)2

(kZ)

S ng hoc s im (khng tnh hai ngun): 1 12 2

l lk

* im dao ng cc tiu (khng dao ng):

d1 d2 = k (kZ)

S ng hoc s im (khng tnh hai ngun): l l

k

Ch : Vi bi ton tm s ng dao ng cc i v khng dao ng gia hai im M, N cch hai ngun ln lt l d1M, d2M, d1N, d2N.

t dM = d1M - d2M; dN = d1N - d2N v gi s:

dM< dN. + Hai ngun dao ng cng pha:

Cc i: dM < k < dN

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

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

Cc tiu: dM < k < dN S gi tr nguyn ca k tho mn cc biu thc trn l s ng cn tm. IV. SNG M

1. Cng m: W P

I= =tS S

ViW (J), P(W) l N.lng, cng sut pht m ca ngun S (m

2) l din tch mt vung gc vi phng truyn m (vi sng cu th S l din tch mt cu S=4R2)

2. Mc cng m 0

( ) lgI

L BI

Hoc 0

( ) 10.lgI

L dBI

Vi I0 = 10-12

W/m2 f = 1000Hz: cng m chun.


Page 19

3. * Tn s do n pht ra (hai u dy c nh hai u l nt sng) ( k N*)2

vf k

l

ng vi k = 1 m pht ra m c bn c tn s 1

2

vf

l

k = 2,3,4 c cc ho m bc 2 (tn s 2f1), bc 3 (tn s 3f1)

* Tn s do ng so pht ra (mt u bt kn, mt u h mt u l nt sng, mt u l bng sng)

(2 1) ( k N)4

vf k

l

ng vi k = 0 m pht ra m c bn c tn s 1

4

vf

l

k = 1,2,3 c cc ho m bc 3 (tn s 3f1), bc 5 (tn s 5f1)

CHNG III: DAO NG V SNG IN T 1. Dao ng in t

* in tch tc thi q = q0cos(t + ) * Hiu in th (in p) tc thi

00os( ) os( )

qqu c t U c t

C C

* Dng in tc thi

i = q = -q0sin(t + ) = I0cos(t + +2

)

* Cm ng t: 0 os( )2

B B c t

Trong : 1

LC l tn s gc ring

2T LC l chu k ring

1

2f

LC l tn s ring

00 0q

I qLC

0 00 0 0q I L

U LI IC C C

* Nng lng in trng: 2

2

1 1W

2 2 2

qCu qu

C

220

W os ( )2

qc t

C

* Nng lng t trng: 2

2 201W sin ( )2 2

t

qLi t

C

* Nng lng in t: W=W Wt

2

2 200 0 0 0

1 1 1W

2 2 2 2

qCU q U LI

C

Ch : + Mch dao ng c tn s gc , tn s f v chu k T th W v Wt bin thin vi

tn s gc 2, tn s 2f v chu k T/2


Page 20

+ Mch dao ng c in tr thun R 0 th dao ng s tt dn. duy tr dao ng cn

cung cp cho mch mt nng lng c cng sut: 2 2 2 2

2 0 0

2 2

C U U RCI R R

L

+ Khi t phng in th q v u gim v ngc li + Quy c: q > 0 ng vi bn t ta xt tch in dng th i > 0 ng vi dng in chy n bn t m ta xt.

2. Sng in t Vn tc lan truyn trong khng gian v = c = 3.108m/s My pht hoc my thu sng in t s dng mch dao ng LC th tn s sng in t pht hoc thu c bng tn s ring ca mch.

Bc sng ca sng in t 2v

v LCf

Lu :

* Mch dao ng c L bin i t LMin LMax v C bin i t CMin CMax th bc sng

ca sng in t pht (hoc thu) Min tng ng vi LMin v CMin

Max tng ng vi LMax v CMax * Cho mch dao ng vi L c nh. Mc L vi C1 c tn s dao ng l f1, mc L vi C2 c tn s l f2.

+ Khi mc ni tip C1 vi C2 ri mc vi L ta c tn s f tha : 2

2

2

1

2 fff

+ Khi mc song song C1 vi C2 ri mc vi L ta c tn s f tha : 22

2

1

2

111

fff

CHNG IV: IN XOAY CHIU 1. Biu thc in p tc thi v dng in tc thi:

u = U0cos(t + u) v i = I0cos(t + i)

Vi = u i l lch pha ca u so vi i,

c 2 2

2. Dng in xoay chiu i = I0cos(2ft + i) * Mi giy i chiu 2f ln

* Nu pha ban u i = 2

hoc i =

2

th ch giy u tin i chiu 2f-1 ln.

Lu : Cng thc tnh thi gian n hunh quang sng trong mt chu k

Khi t in p u = U0cos(t + u) vo hai u bng n, bit n ch sng ln khi u U1.

4

t

Vi 1

0

osU

cU

, (0 < < /2)

3. Dng in xoay chiu trong on mch R,L,C

* on mch ch c in tr thun R: uR cng pha vi i, ( = u i = 0) U

IR

v 00

UI

R

Lu : in tr R cho dng in khng i i qua v c U

IR

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

L

UI

Z v 00

L

UI

Z vi ZL = L l cm khng

Lu : Cun thun cm L cho dng in khng i i qua hon ton (khng cn tr).

* on mch ch c t in C: uC chm pha hn i l /2, ( = u i = -/2)


Page 21

C

UI

Z v 00

C

UI

Z vi

1CZ

C l dung khng

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

2 2 2 2 2 2

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

tan ;sin ; osL C L CZ Z Z Z R

cR Z Z

vi 2 2

+ Khi ZL > ZC th u nhanh pha hn i + Khi ZL < ZC th u chm pha hn i + Khi ZL = ZC th u cng pha vi i.

Lc Max

UI =

R gi l hin tng cng hng dng in

4. Cng sut to nhit trn on mch RLC:

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

* Cng sut trung bnh: P = UIcos = I2R.

4.1 6. in p u = U1 + U0cos(t + ) c coi gm mt in p khng i U1 v mt in

p xoay chiu u=U0cos(t + ) ng thi t vo on mch. 5. Tn s dng in do my pht in xoay chiu mt pha c p cp cc, rto quay vi vn tc n vng/giy pht ra: f = pn (Hz)

T thng gi qua khung dy ca my pht in :

= NBScos(t +) = 0cos(t + )

Vi 0 = NBS l t thng cc i gi qua N vng dy, B l cm ng t ca t trng, S l

din tch ca vng dy, = 2f Sut in ng trong khung dy:

e = NSBcos(t + - 2

) = E0cos(t + -

2

)

Vi E0 = NSB l sut in ng cc i. 6. Dng in xoay chiu 3 pha l h thng ba dng in xoay chiu, gy bi ba sut in ng xoay chiu cng tn s, cng bin

nhng lch pha tng i mt l 2

3

1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

e E c t

e E c t

e E c t

1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

i I c t

i I c t

i I c t

(tii xng)

My pht mc hnh sao: Ud = 3 Up My pht mc hnh tam gic: Ud = Up Ti tiu th mc hnh sao: Id = Ip

Ti tiu th mc hnh tam gic: Id = 3 Ip


Page 22

7. Cng thc my bin p l tng: 1 1 2 1

2 2 1 2

U E I N

U E I N

Lu : my pht v ti tiu th thng chn cch mc tng ng vi nhau. 10. Cng sut hao ph trong qu trnh truyn ti in nng:

2

cos

i

i

U

PRP

l

RS

l in tr tng cng ca dy ti in

(lu : dn in bng 2 dy) Trong : P l cng sut truyn i ni cung cp U l in p ni cung cp

cos l h s cng sut ca dy ti in

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

Hiu sut ti in: i

i

i

n

P

PP

P

PH

; .100%H

8. on mch RLC c R thay i:

* Khi R=ZL-ZC th 2 2

ax2 2

M

L C

U U

Z Z R

* Khi R=R1 hoc R=R2 th P c cng gi tr. Ta c

R1, R2 th.mn phng trnh bc 2 0222 CL ZZPRUPR

22

1 2 1 2; ( )L CU

R R R R Z Z

V khi 1 2R R R th

2

ax

1 22M

U

R R

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

Khi 2 2

0 ax

02 2( )L C M

L C

U UR Z Z R

Z Z R R

Khi 2 2

2 2

0 ax2 2

00 0

( )2( )2 ( ) 2

L C RM

L C

U UR R Z Z

R RR Z Z R

9. on mch RLC c L thay i: * Vi L = L1 hoc L = L2 th UL c cng gi tr th ULmax khi

1 2

1 2

1 2

21 1 1 1( )

2L L L

L LL

Z Z Z L L

* Khi ZL=ZC ( 21

LC

)th IMax

URmax; PMax cn ULCMin

* Khi 2 2

CL

C

R ZZ

Z

th

2 2

ax

C

LM

U R ZU

R

v

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

* Vi

2

1

LL

LL th UL c cng gi tr th ULmax khi

21

212

LL

LLL

ZZ

ZZZ


Page 23

* Khi 2 24

2

C C

L

Z R ZZ

th

ax2 2

2 R

4RLM

C C

UU

R Z Z

Lu : R v L mc lin tip nhau

10. on mch RLC c C thay i:

* Khi 2

1C

L (Khi ZL=ZC) th IMax URmax; PMax cn ULCMin

*Khi 2 2

LC

L

R ZZ

Z

th

2 2

ax

L

CM

U R ZU

R

v

2 2 2 2 2 2

ax ax ax; 0CM R L CM L CMU U U U U U U U

*Vi

2

1

CC

CC th UC c cng gi tr th UCmax khi21

212

CC

CCC

ZZ

ZZZ

;

1 2

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

C CC

Z Z Z

* Khi 2 24

2

L L

C

Z R ZZ

th

ax2 2

2 R

4RCM

L L

UU

R Z Z

11. Mch RLC c thay i:

* Khi 1

LC th IMax URmax; PMax cn ULCMin

* Khi 2

1 1

2

C L R

C

th ax2 2

2 .

4LM

U LU

R LC R C

* Khi 21

2

L R

L C th ax

2 2

2 .

4CM

U LU

R LC R C

* Vi = 1 hoc = 2 th I hoc P hoc UR c cng mt gi tr th IMax hoc PMax hoc

URMax khi 1 2 tn s 1 2f f f

12. Hai on mch AM gm R1L1C1 ni tip v on mch MB gm R2L2C2 ni tip mc ni

tip vi nhau c UAB = UAM + UMB uAB; uAM v uMB cng pha tanuAB = tanuAM = tanuMB

13. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau

Vi 1 111

tanL CZ Z

R

v 2 22

2

tanL CZ Z

R

(gi s 1 > 2)

C 1 2 = 1 2

1 2

tan tantan

1 tan tan

Trng hp c bit = /2 (vung pha nhau) th tan1tan2 = -1. VD: * Mch in hnh 1 c uAB v uAM lch pha nhau

R L C M A B

Hnh 1


Page 24

y 2 on mch AB v AM c cng i v uAB chm pha hn uAM

AM AB = tan tan

tan1 tan tan

AM AB

AM AB

Nu uAB vung pha vi uAM th tan tan =-1 1L CL

AM AB

Z ZZ

R R

* Mch in hnh 2: Khi C = C1 v C = C2 (gi s C1 > C2) th i1 v i2 lch pha nhau

y hai on mch RLC1 v RLC2 c cng uAB Gi 1 v 2 l lch pha ca uAB so vi i1 v i2

th c 1 > 2 1 - 2 =

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

Nu I1 I2 th tnh 1 2

1 2

tan tantan

1 tan tan

CHNG V: SNG NH SNG 1. Hin tng tn sc nh sng. * /n: L hin tng nh sng b tch thnh nhiu mu khc nhau khi i qua mt phn cch ca hai mi trng trong sut. * nh sng n sc l nh sng khng b tn sc nh sng n sc c tn s xc nh, ch c mt mu.

Bc sng ca nh sng n sc f

v , truyn trong chn khng

f

c0

* Chit sut ca mi trng trong sut ph thuc vo mu sc nh sng. i vi nh sng mu l nh nht, mu tm l ln nht. * nh sng trng l tp hp ca v s nh sng n sc c mu bin thin lin tc t n tm.

Bc sng ca nh sng trng: 0,38 m 0,76 m. 2. Hin tng giao thoa nh sng (ch xt giao thoa nh sng trong th nghim Ing). * /n: L s tng hp ca hai hay nhiu sng nh sng kt hp trong khng gian trong xut hin nhng vch sng v nhng vch ti xen k nhau. Cc vch sng (vn sng) v cc vch ti (vn ti) gi l vn giao thoa.

* Hiu ng i ca nh sng (hiu quang trnh) : D

axddd 12

* Khong vn i l khong cch gia hai vn sng hoc hai vn ti lin tip:: a

Di

* V tr (to ) vn sng: xs=ki ( Zk ) k = 0: Vn sng trung tm

k = 1: Vn sng bc (th) 1

* V tr (to ) vn ti: xt=ki+2

i( Zk )

k = 0, k = -1: Vn ti th (bc) nht k = 1, k = -2: Vn ti th (bc) hai * Nu th nghim c tin hnh trong mi trng trong sut c chit sut n th bc sng v

khong vn u gim n ln : n

ii

n ';'

R L C M A B

Hnh 2


Page 25

* Khi ngun sng S di chuyn theo phng song song vi S1S2 th h vn di chuyn ngc chiu v khong vn i vn khng i.

di ca h vn l: 01

Dx d

D=

Trong : D l khong cch t 2 khe ti mn D1 l khong cch t ngun sng ti 2 khe d l dch chuyn ca ngun sng * Khi trn ng truyn ca nh sng t khe S1 (hoc S2) c t mt bn mng dy e, chit

sut n th h vn s dch chuyn v pha S1 (hoc S2) mt on: 0( 1)n eD

xa

-=

* Xc nh s vn sng, vn ti trong vng giao thoa (trng giao thoa) c b rng L (i xng qua vn trung tm)

+ S vn sng (l s l): 12

2

i

LNS

+ S vn ti (l s chn):

2

1

22

i

LN t

* Xc nh s vn sng, vn ti gia hai im M, N c to x1, x2 (gi s x1 < x2) + Vn sng: x1 < ki < x2

+ Vn ti: x1 < (k+0,5)i < x2

S gi tr k Z l s vn sng (vn ti) cn tm Lu : M v N cng pha vi vn trung tm th x1 v x2 cng du. M v N khc pha vi vn trung tm th x1 v x2 khc du. * Xc nh khong vn i trong khong c b rng L. Bit trong khong L c n vn sng.

+ Nu 2 u l hai vn sng th: 1

Li

n=

-

+ Nu 2 u l hai vn ti th: L

in

=

+ Nu mt u l vn sng cn mt u l vn ti th: 0,5

Li

n=

-

* S trng nhau ca cc bc x 1, 2 ... (khong vn tng ng l i1, i2 ...)

+ Trng nhau ca vn sng: xs = k1i1 = k2i2 = ... k11 = k22 = ...

+ Trng nhau ca vn ti: xt = (k1 + 0,5)i1 = (k2 + 0,5)i2 = ... (k1 + 0,5)1 = (k2 + 0,5)2 = ...

Lu : V tr c mu cng mu vi vn sng trung tm l v tr trng nhau ca tt c cc vn sng ca cc bc x.

* Trong hin tng giao thoa nh sng trng (0,4m 0,76m)

- B rng quang ph bc k: tk iik vi v t l bc sng nh sng v tm - Xc nh s vn sng, s vn ti v cc bc x tng ng ti mt v tr xc nh ( bit x)

+ Vn sng: ax

, k ZD

x ka kD

ll= =

Vi 0,4 m 0,76 m cc gi tr ca k

+ Vn ti: ax

( 0,5) , k Z( 0,5)

Dx k

a k D

ll= + =

+

Vi 0,4 m 0,76 m cc gi tr ca k


Page 26

- Khong cch di nht v ngn nht gia vn sng v vn ti cng bc k:

[k ( 0,5) ]Min tD

x ka

ax [k ( 0,5) ]M tD

x ka

Khi vn sng v vn ti nm khc pha i vi vn trung tm.

ax [k ( 0,5) ]M tD

x ka

Khi vn sng v vn ti nm cng pha i vi vn trung tm.

CHNG VII: LNG T NH SNG 1. Nng lng mt lng t nh sng (ht phtn)

2hchf mcel

= = =

Trong h = 6,625.10-34 Js l hng s Plng. c = 3.10

8m/s l vn tc nh sng trong chn khng.

f, l tn s, bc sng ca nh sng (ca bc x). m l khi lng ca phtn 2. Tia Rnghen (tia X)

Bc sng nh nht ca tia Rnghen E

hcmin

Trong 22

0

2 2

mvmvE e U= = + l ng nng ca electron khi p vo i catt (i m

cc) U l hiu in th gia ant v catt v l vn tc electron khi p vo i catt v0 l vn tc ca electron khi ri catt (thng v0 = 0) m = 9,1.10

-31 kg l khi lng electron

3. Hin tng quang in

*Cng thc Anhxtanh : 2

2

max0mvAhc

hf

Trong 0

hcA l cng thot ca kim loi dng lm catt

0 l gii hn quang in ca kim loi dng lm catt

* dng quang in trit tiu th UAK Uh (Uh < 0), Uh gi l hiu in th hm:

2

0 ax

2

Mh

mveU =

Lu : Trong mt s bi ton ngi ta ly Uh > 0 th l ln. * Xt vt c lp v in, c in th cc i VMax v khong cch cc i dMax m electron chuyn ng trong in trng cn c cng E c tnh theo cng thc:

2ax 0 ax ax

1

2M M Me V mv e Ed= =

* Vi U l hiu in th gia ant v catt, vA l tc cc i ca electron khi p vo ant,

vK = v0Max l tc ban u cc i ca electron khi ri catt th: 2 21 1

2 2A Ke U mv mv= -

* Hiu sut lng t (hiu sut quang in) 0

nH

n=

Vi n v n0 l s electron quang in bt khi catt v s phtn p vo catt trong cng mt khong thi gian t.


Page 27

Cng sut ca ngun bc x: 0 0 0n n hf n hc

pt t t

e

l= = =

Cng dng quang in bo ho: bh

n eqI

t t= =

bh bh bhI I hf I hc

Hp e p e p e

e

l = = =

* Bn knh qu o ca electron khi chuyn ng vi vn tc v trong t trng u B :

sinBe

mvR ( Bv, )

Xt electron va ri khi catt th v = v0Max

Khi sin 1mv

v B Re B

a^ = =r ur

Lu : Hin tng quang in xy ra khi c chiu ng thi nhiu bc x th khi tnh cc i lng: Tc ban u cc i v0Max, hiu in th hm Uh, in th cc i

VMax, u c tnh ng vi bc x c Min (hoc fMax) 4. Tin Bo - Quang ph nguyn t Hir

* Tin Bo mn m nmn

hchf E Ee

l= = = -

* Bn knh qu o dng th n ca electron trong nguyn t hir: rn = n

2r0 Vi r0 =5,3.10

-11m l bn knh Bo ( qu oK) * Nng lng electron trong nguyn t hir:

2

13,6( )nE eV

n= - Vi n N

*.

Nng lng ion ha l nng lng ti thiu a e t qu o K ra xa v cng (lm ion ha nguyn t Hir): Eion=13,6eV * S mc nng lng - Dy Laiman: Nm trong vng t ngoi:ng vi e chuyn t qu o bn ngoi v qu o K

Lu : Vch di nht LK khi e

chuyn t L K

Vch ngn nht K khi e

chuyn t K. - Dy Banme: Mt phn nm trong vng t ngoi, mt phn nm trong vng nh sng nhn thy ng vi e chuyn t qu o bn ngoi v qu o L Vng nh sng nhn thy c 4 vch:

Vch H ng vi e: M L

Vch lam H ng vi e: N L

Vch chm H ng vi e: O L

Vch tm H ng vi e: P L

Laiman

K

M

N

O

L

P

Banme

Pasen

H H H H

n=1

n=2

n=3

n=4

n=5

n=6


Page 28

Lu : Vch di nht ML (Vch H )

Vch ngn nht L khi e chuyn t L. - Dy Pasen: Nm trong vng hng ngoi ng vi e chuyn t qu o bn ngoi v qu o M

Lu : Vch di nht NM khi e chuyn t N M.

Vch ngn nht M khi e chuyn t M. Mi lin h gia cc bc sng v tn s ca cc vch quang ph ca nguyn t hir:

13 12 23

1 1 1

CHNG IX. VT L HT NHN 1. Hin tng phng x

* S n.t cht phng x cn li sau thi gian t t

T

teN

NN 0

0

2

* S ht nguyn t b phn r bng s ht nhn con c to thnh v bng s ht ( hoc e-

hoc e+) c to thnh: NNN 0 0 0(1 )tN N N N e l-D = - = -

* Khi lng cht phng x cn li sau thi gian t: t

T

tem

mm 0

0

2

Trong : N0, m0 l s nguyn t, khi lng cht phng x ban u T l chu k bn r

2 0,693ln

T Tl = = l hng s phng x

v T khng ph thuc vo cc tc ng bn ngoi m ch ph thuc bn cht bn trong ca cht phng x.

* Khi lng cht b phng x sau thi gian t : mmm 0

0 0(1 )tm m m m e l-D = - = -

* Phn trm cht phng x b phn r: t

T

te

m

m

1

2

11

0

* Phn trm cht phng x cn li: t

T

te

m

m

2

1

0

* Lin h gia khi lng v s nguyn t : AN

A

mN

NA = 6,022.10-23

mol-1

l s Avgar (s ht trong mt mol) * Khi lng cht mi c to thnh sau thi gian t

12

23

13

1

2

3


Page 29

1 0 11 1 0(1 ) (1 )t t

A A

A N ANm A e m e

N N A

l l- -D= = - = -

Trong : A, A1 l s khi ca cht phng x ban u v ca cht mi c to thnh

Lu : Trng hp phng x +, - th A = A1

m1 = m * phng x H:L i lng c trng cho tnh phng x mnh hay yu ca mt lng

cht phng x, o bng s phn r trong 1 giy : NHeHH

H t

T

t ;

2

00

H0 = N0 l phng x ban u. n v: Becren (Bq); 1Bq = 1 phn r/giy Curi (Ci); 1 Ci = 3,7.10

10 Bq

Lu : Khi tnh phng x H, H0 (Bq) th chu k phng x T phi i ra n v giy(s). 2. H thc Anhxtanh, ht khi, nng lng lin kt * H thc Anhxtanh gia khi lng v nng lng Vt c khi lng m th c nng lng ngh E = m.c2

Vi c = 3.108 m/s l vn tc nh sng trong chn khng.

* ht khi ca ht nhn AZ X : m = m0 m

Vi: m0 = Zmp + Nmn = Zmp + (A-Z)mn l khi lng cc nucln. m l khi lng ht nhn X.

* Nng lng lin kt : E = m.c2 = (m0-m)c2

* Nng lng lin kt ring (l nng lng lin kt tnh cho 1 nucln): A

E

Lu : Nng lng lin kt ring cng ln th ht nhn cng bn vng. 3. Phn ng ht nhn

* Phng trnh phn ng: 43214

4

3

3

2

2

1

1XXXX

A

Z

A

Z

A

Z

A

Z

Trong s cc ht ny c th l ht s cp nh nucln, e, phtn ...

Trng hp c bit l s phng x: X1 X2 + X3

X1 l ht nhn m, X2 l ht nhn con, X3 l ht hoc * Cc nh lut bo ton + Bo ton s nucln (s khi): A1 + A2 = A3 + A4 + Bo ton in tch (nguyn t s): Z1 + Z2 = Z3 + Z4 Hai nh lut ny dng vit phng trnh phn ng ht nhn + Bo ton nng lng

ts

ts

st

EE

cmm

cmmQ

2

2

Q>0 phn ng ta nng lng; Q


Page 30

Lu : - Khng c nh lut bo ton khi lng.

- Mi quan h gia ng lng pX v ng nng KX ca ht X l: 2 2X X Xp m K=

* Nng lng phn ng ht nhn

E = (M0 - M)c2

Trong : 1 20 X X

M m m= + l tng khi lng cc ht nhn trc phn ng.

3 4X XM m m= + l tng khi lng cc ht nhn sau phn ng.

Lu : - Nu M0 > M th phn ng to nng lng E di dng ng nng ca cc ht X3,

X4 hoc phtn . Cc ht sinh ra c ht khi ln hn nn bn vng hn.

- Nu M0 < M th phn ng thu nng lng E di dng ng nng ca cc ht X1,

X2 hoc phtn . Cc ht sinh ra c ht khi nh hn nn km bn vng.

* Trong phn ng ht nhn 31 2 41 2 3 41 2 3 4

AA A A

Z Z Z ZX X X X+ +

Cc ht nhn X1, X2, X3, X4 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 Nng lng ca phn ng ht nhn

E = A33 +A44 - A11 - A22

E = E3 + E4 E1 E2

E = (m3 + m4 - m1 - m2)c2

* Quy tc dch chuyn ca s phng x

+ Phng x ( 42 He ):

4 4

2 2

A A

Z ZX He Y-

- +

So vi ht nhn m, ht nhn con li 2 trong bng tun hon v c s khi gim 4 n v.

+ Phng x - ( 10e

- ): 01 1

A A

Z ZX e Y- + +

So vi ht nhn m, ht nhn con tin 1 trong bng tun hon v c cng s khi.

Thc cht ca phng x - l mt ht ntrn bin thnh mt ht prtn, mt ht electrn v mt ht ntrin:

n p e v- + +

Lu : - Bn cht (thc cht) ca tia phng x - l ht electrn (e-) - Ht ntrin (v) khng mang in, khng khi lng (hoc rt nh) chuyn ng vi vn tc ca nh sng v hu nh khng tng tc vi vt cht.

+ Phng x + ( 10e

+ ): 01 1

A A

Z ZX e Y+ - +

So vi ht nhn m, ht nhn con li 1 trong bng tun hon v c cng s khi.

Thc cht ca phng x + l mt ht prtn bin thnh mt ht ntrn, mt ht pzitrn v mt ht ntrin:

p n e v+ + +

Lu : Bn cht (thc cht) ca tia phng x + l ht pzitrn (e+)

+ Phng x (ht phtn) Ht nhn con sinh ra trng thi kch thch c mc nng lng E1 chuyn xung mc nng lng E2 ng thi phng ra mt phtn c nng lng

1 2

hchf E Ee

l= = = -

4. Cc hng s v n v thng s dng


Page 31

* S Avgar: NA = 6,022.1023

mol-1

* n v nng lng: 1eV = 1,6.10-19 J; 1MeV = 1,6.10-13 J * n v khi lng nguyn t (n v Cacbon): 1u = 1,66055.10

-27kg = 931 MeV/c

2

* in tch nguyn t: e = 1,6.10-19 C * Khi lng prtn: mp = 1,0073u * Khi lng ntrn: mn = 1,0087u * Khi lng electrn: me = 9,1.10

-31kg = 0,0005u

CHNG X. T VI M N V M. 1. HT S CP - Hat s cap la nhng hat co kch thc va khoi lng nho hn hat nhan

nguyen t. ac trng chnh cua cac hat s cap la:

+ Khoi lng ngh m0 ht nang lng ngh E0 = m0c2.

+ So lng t ien tch q cua hat s cap co the la +1, -1, 0 (tnh theo ien tch nguyen to e).

+ So lng spin s la ai lng ac trng cho chuyen ong noi tai cua hat s cap.

+ Thi gian song trung bnh. Ch co 4 hat s cap khong phan ra thanh cac hat

khac, o la proton, electron, photon, ntrino; con lai la cac hat khong ben

co thi gian song rat ngan, c t 10-24

s en 10-6

s, tr ntron co thi gian song

la 932s.

+ Phan ln cac hat s cap eu tao thanh cap: hat va phan hat.

Phan hat co cung khoi lng ngh, cung spin, ien tch co cung o ln nhng

trai dau.

- Cac hat s cap c phan thanh 4 loai: photon, lepton, mezon va barion.

Mezon va barion c goi chung la haron.

Co 4 loai tng tac c ban oi vi hat s cap la: tng tac hap dan, tng tac

ien t, tng tac yeu, tng tac manh. - Tat ca cac haron eu co cau tao t hat quac.

Co 6 loai quac la u, d, s, c, b va t.

ien tch cac hat quac la 3

e,

2

3

e.

Cac barion la to hp cua ba quac.

Quan niem hien nay ve cac hat thc s la s cap gom cac quac, cac lepton va

cac hat truyen tng tac la gluon, photon, W

, Z0 va graviton.

2. H MT TRI - He Mat Tri gom Mat Tri trung tam he; 8 hanh tinh ln va cac ve tinh

cua no gom Thuy tinh, Kim tinh, Trai at, Hoa tinh, Moc tinh, Tho tinh, Thien

Vng tinh va Hai Vng tinh. Cac hanh tinh nay chuyen ong quanh Mat Tri

theo cung mot chieu va gan nh trong cung mat phang. Mat Tri va cac hanh

tinh con t quay quanh mnh no.

Khoi lng Mat Tri bang 1,99.1030

kg, gap 333000 lan khoi lng Trai at.

Khoang cach t Trai at en Mat Tri xap x 150 trieu km, bang 1 n v

thien van.

- Mat Tri gom quang cau va kh quyen Mat Tri.

Mat Tri luon bc xa nang lng ra xung quanh. Hang so Mat Tri la H=

1360W/m2. Cong suat bc xa nang lng cua Mat Tri la P = 3,9.10

26W. Nguon

nang lng cua Mat Tri chnh la cac phan ng nhiet hach. thi k hoat


Page 32

ong cua Mat Tri, tren Mat Tri xuat hien cac vet en, bung sang nhieu hn

luc bnh thng.

- Trai at co dang phong cau co ban knh xch ao bang 6378km, co khoi

lng la 5,98.1024

kg. Mat Trang la ve tinh cua Trai at co ban knh 1738km va

khoi lng la 7,35.1022

kg. Gia toc trong trng tren Mat Trang la 1,63m/s2.

3. SAO. THIN H

- Sao la mot khoi kh nong sang giong nh Mat Tri nhng rat xa Trai at.

a so sao trang thai on nh. Ngoai ra co mot so sao ac biet nh sao bien

quang, sao mi, sao ntron.

Khi nhien lieu trong sao can kiet, sao tr thanh sao lun, sao ntron hoac lo en.

- Thien ha la he thong gomnhieu loai sao va tinh van.

Ba loai thien ha chnh la thien ha xoan oc, thien ha elip, va thien ha khong

nh hnh.

Thien Ha cua chung ta la thien ha xoan oc co ng knh khoang 100 ngan

nam anh sang, day khoang 330 nam anh sang, khoi lng bang 150 t lan khoi

lng Mat Tri. He Mat Tri nam ra Thien Ha, cach trung tam khoang 30

000 nam anh sang va quay vi toc o khoang 250km/s.

4. THUYT BIG BANG Theo Thuyet Big Bang, vu tru c tao ra bi mot vu no cc ln, manh cach ay khoang 14 t nam, hien ang dan n va loang dan. Hai hien tng thien

van quan trong la vu tru dan n va bc xa nen vu tru la minh chng cua thuyet Big Bang.

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