tóm tắt lý thuyết vật lý lớp 12

7/31/2019 tm tt l thuyt vt l lp 12

1/91

Tm tt l thuyt & cc dng ton - Vt l 12 - 1 -

GV : NG NGC TON Mobi : 090 9894 590 Email :[email protected]

sin

3

4

6

6

4

3

23

2

43

65

65

2

32

43

2

3A

2

2A

2

1A

22A

21

A

23A

22A-

21

A-

23A-

2

3A

22A-

21A- A

0

-A

0

W =3W t

W =3W t

W =W t

W t=3W

W =W t

2/2vv max

23vv max

2/vv max2/vv max

22vv max

v < 0

23vv max

x

V > 0

W t=3W

+

cos

Tha s Tn tin t K hiu Tha s Tn tin t K hiu10 Tera T 10- dexi d109 Giga G 10-2 centi c10 Mega M 10- mili m10 Kilo K 10- micro 10 Hecto H 10-9 nano n10 Deca D 10- pico p
mailto:[email protected]:[email protected]


2/91



CH 1: C HC VT RN

VN 1.NG HC VT RN QUAY QUANH MT TRC C NHi lng vt l K hiu (n v) Quay u Quay bin i u Ghi ch1. Gia tc gc (rad/s2,vng/s2) 0 const

2. Tc gc (rad/s, vng/s)

2

2 f const T 0 t Phng trnhvn tc

3. Ta gc (rad) t 0 20 012

t t Phng trnhchuyn ng

4. Gc quay (rad)

0

0

t

t t

2 20

0 2Thng chn

t 0 = 0

Xt mt im M trn vt rn cch trc quay mt khongR 5. Tc di v (m/s) const Rv t av Rv t 0

6. Gia tc hngtm an (m/s2) Rv Ran2

2 Rv Ran

22 Gia tc phptuyn

7. Gia tc tiptuyn at (m/s

2) 0t a . Rat

8. Gia tc ton phn a (m/s

2) naa2 2

2 4

n t a a a

r

t n aa

Ch :o Mi im ca vt rn u chuyn ng trn trong mt phng vung gc vi trc quay, t

trn trc quay, bn knh bng khong cch t im xt n trc quay.o Cc i lng, , c gi tr i s, ph thuc vo chiu dng c chn

(thng chn chiu dng l chiu quay ca vt ).o i n v: 1 vng = 3600 = 2 rado >0: Chuyn ng quay nhanh dn.o


3/91



a. Thanh mnh 2121 mL I Cc vt ng ch t, c dng hnh hc i xng. L: Chiu di thanh.

b. Vnh trn ( hnh tr rng) 2mR I

c. a trn( hnh tr c) 221 mR I

d. Hnh cu c 2

5

2 mR I

2. Mmen nglng L (kg.m

2.s-1) mrv I L

3. Mmen lc M (N.m) Fd M d: Khong cch t trc quay n gi ca lc(cnh tay n ca lc)

2M mr I Phng trnh LH ca vt rn quay quanhmt trc c nh (dng khc ca L II Newton)

4. Dng khcdt dL M

Ch :o Cng thc Huyghen-Steiner:

2

md I I GO dng khi i trc quay.d = OG : Khong cch gia hai trc quay.o 0 F M : nu F c gi ct hoc song song vi trc quay.o nh l bin thin mmen ng lng:

2 1 2 2 1 10M M L L L M t I I

VN 3.NH LUT BO TON MMEN NG LNGNi dung: 1 1 2 20 M L const I I

I1, 1: Mmen qun tnh v tc gc ca vt lc u.I2, 2: Mmen qun tnh v tc gc ca vt lc sau.

Ch :o p dng nh lut cho h vt rn c cng trc quay:const L i vi trc quay .o Khi I = const = 0 : Vt rn khng quay.

Hoc = const: Vt rn quay u.o Vt c mmen qun tnh i vi trc quay thay i :- Nu I vt quay chm dn v dng li- Nu I vt quay nhanh dn

VN 4. KHI TM. NG NNG CA VT RN

1. Ta khi tm:i

iiC

m

xm x

i

iiC

m

ym y

i

iiC

m

z m z

2. Chuyn ng ca khi tm : F am c( F : Tng hnh hc cc vect lc tc dng ln vt rn.)

3. ng nng: ( J )Chuyn ng tnh tin Chuyn ng quay Chuyn ng song ph ng

2 2

1W C mv

2 2

1W I 22 2

121

W I mv C

R


4/91



Ch :o Xem khi tm trng vi trng tm G. Khi mt trng lng, trng tm khng cn nhng kh

tm lun tn ti.o Vt rn ln khng trt: R v C o Mi lc tc dng vo vt :

+) C gi i qua trng tm lm vt chuyn ng tnh tin.+) C gi khng i qua trng tm lm vt va quay va chuyn ng tnh tin.

o nh l ng nng: 12 ngoailuc W W W Ao Th nng trng trng: t W mgh

h: cao tnh t mc khng th nng.o nh lut bo ton c nng: Khi vt ch chu tc dng ca lc th:

tW=W W onstc * S tng t gia cc i lng gc v i lng di trong chuyn ng quay v chuyn ngthng

Chuy n ng quay(trc quay c nh, chiu quay khng i)

Chuy n ng th ng(chiu chuyn ng khng i)

To gcTc gcGia tc gcMmen lc MMmen qun tnh IMmen ng lng L = I

ng nng quay 21W2 I

rad To xTc vGia tc aLc FKhi lng mng lng p = mv

ng nng 21W2

mv

mrad/s m/srad/s m/s Nm Nkgm2 kg

kgm /s kgm/s

J J

Chuyn ng quay u:= const; = 0; = 0 + t

Chuyn ng quay bin i u:= const= 0 + t

20 12

t t 2 2

0 02 ( ) Phng trnh ng lc hc

M I

o Dng khc dL M dt

nh lut bo ton mmen ng lng1 1 2 2 i I I hay L const

nh l v ng nng2 2

2 11 1W2 2 I I A (cng ca ngoi lc)

Chuyn ng thng u:v = const; a = 0; x = x0 + at

Chuyn ng thng bin i u:a = constv = v0 + at

x = x0 + v0t + 212at

2 20 02 ( )v v a x x

Phng trnh ng lc hc F am

o Dng khc dp F dt

nh lut bo ton ng lngi i i p m v const

nh l v ng nng2 2 2 1

1 1W2 2

mv mv A (cng ca ngoi lc)

Cc cng thc lin h gia cc i lng gc v i lng di : . R s ; . Rv ; . Rat ; 2. Ran

CH 2: DAO NG C HC

VN 1.DAO NG IU HACc nh ngha


5/91



1. Dao ng L mt chuy n ng qua li v c gii hn quanh mt v tr cn b ng (v trm vt ng yn).2. Dao ng tun hon L dao ng m trng thi chuy n ng ca vt c lp li nh c saunhng khong thi gian bng nhau.3. Mt dao ng ton phn (chu trnh) L giai on nh nht c lp li trong dao ng tun hon.

4. Chu k Thi gian thc hin mt dao ng ton phn (khong thi gian ngn nht

hai ln vt i qua mt v tr xc nh vi cng chiu chuyn ng).5. Tn s S dao ng ton phn thc hin trong mt giy.

6. Dao ng iu ha L dao ng tun hon c m t bng mt nh lut dng cosin (hay sitheo thi gian.Trong A , , l nhng hng s )cos( t A x7. Dao ng t do (daong ring)

L dao ng ca h xy ra ch di tc dng ca ni lc, mi h dao ndo u c mt tn s gc ring0 nht nh.

8.Dao ng tt dn-L dao ng c bin gim d n theo thi gian; dao ng t t d n khngc tnh tun hon; s tt dn cng nhanh nu lc cn ca mi trng cng-Khi ma st nh, dao ng tt dn c th coi gn ng l tun hon vi tgc bng tn s gc ring0 ca h.

9.Dao ng duy trL dao ng c c khi cung cp thm nng lng b li s tiu hao do mst m khng lm thay i tn s gc ring ca h.

o ng dng : duy tr dao ng trong con lc ng h (ng h c dct)

10.Dao ng cng bc

-L dao ng c to ra di tc dng ca mt ngoi lc bin thin itheo thi gian c dng )cos(0 t F F ; f .2 (f tn s ngoi lc)-Dao ng cng bc l iu ha; c tn s gc bng tn s gcca ngoilc; bin t l vi F0 v ph thuc vo-Khi = 0 th bin ca dao ng cng bc t gi tr cc i: ta c htng cng hng. iu kin xy ra :0 hay 0 khi 00; T T f f

o c im:Vi cng mt ngoi lc tc dng nu ma st gim th gi tr cci ca bin tngLc cn cng nh (Amax) cng ln cng hng r cnghng nhnLc cn cng ln (Amax) cng nh cng hng khng rcng hng t

o ng dng : ch to tn s k , ln dy n 11. Phn bit dao ng cng bc vi dao ng duy tr

Dao ng cng bc Dao ng duy trGi ng nhau - u xy ra di tc dng ca ngoi lc.

- Dao ng cng bc khi cng hng cng c tn s bng tn s ring ca vt.

Khc nhau

- Ngoi lc l bt k, c lp vi vt- Sau giai on chuyn tip th dao ngcng bc c tn s bng tn s f ca ngoilc- Bin ca h ph thuc vo F0 v |f f 0|

- Lc c iu khin bi chnh daong y qua mt c cu no - Dao ng vi tn s ng bng tn sdao ng ring f 0 ca vt- Bin khng thay i

12. Phn bit cng hng vi dao ng duy trCng hng Dao ng duy tr

Ging nhau C hai u c iu chnh tn s ngoi lc bng vi tn s dao ng t do

Khc nhau

- Ngoi lc c lp bn ngoi.- Nng lng h nhn c trong mi chu kdao ng do cng ngoi lc truyn cho ln

- Ngoi lc c iu khin bi chnhdao ng y qua mt c cu no .- Nng lng h nhn c trong mi


6/91



hn nng lng m h tiu hao do ma sttrong chu k .

chu k dao ng do cng ngoi lctruyn cho ng bng nng lng m htiu hao do ma st trong chu k .

i lng vt l K hiu (n v) Cng thc Ghi ch1.Li ( lch khiVTCB)

x (m; cm)cos( )

sin2

x A t

A t

Phng trnh dao ng iu ha A, , l hng s

a. Bin daong A (m; cm) A = xmax

A>0, ph thuc vo cch kchthch dao ng

b. Pha ca daong (t) (rad) =( )t Xc nh trng thi dao ng c. Pha ban u(t=0) (rad)

C gi tr ty theo iu kin banu

d. Tn s gc (rad/s)2

2 f T

T: chu k (s) ; f: tn s (s-1; Hz)

2.Vn tc v (m/s) '( ) Asin t+

os t+ +2

v x t

Ac

Vn tc sm pha hn li gc2

3. Gia tc: a (m/s2) 2 2'( ) "( )

os t+

a v t x t

Ac x

Gia tc ngc pha vi li

4. Chu k T (s)2 1 t

T f N

N: S dao ng thc hin trong khong thi giant

5. Tc trung bnh v (m/s)

sv

t s: Qung ng vt i ctrong khong thi giant

6. Vn tc trung bnh vtb (m/s)

2 1tb

x x x v

t t

x: di vt thc hin c

trong khong thi giant Ch :

o Ti v tr cn bng:x = 0v = vmax= A (hoc bng -A)a = 0

o Ti hai bin:x = Av = 0a = amax= 2A (hoc bng -2A)

o Vn tc trung bnh ca vt dao ng iu hatrong mt chu k bng 0.

VN 2.TNG HP DAO NG

1. Biu din dao ng iu ha bng vect quay

Mi dao ng iu ha: x=Acos t+

c biu din bng mt vect quayOM (tm quay O):OM = ATc gc = Tn s gc

thi im t =0:( , )OM ox

Ox

M

t

x, v, a

A

-A

A

-A

2A

-2A

O TT/2

T

ng biu d in x(t), v(t) v a(t) v trong cng mth trc to , ng v i = 0

a(t)

v(t)

x(t)


7/91



2. Tng hp hai dao ng cng phng, cng tn s:

1 1 1

2 2 2

os t+

os t+

x Ac

x A c

*Dao ng tng hp: 1 2 os x x x Ac t cng phng, cng tn s vi hai dao ng thnh pha.Bin dao ng 2 2

1 2 1 22 os A A A A A c

b. lch pha 2 1

c. Pha ban u 1 1 2 21 1 2 2

sin sintan

os os A A A c A c

Ch :

o 2 10 : : x2 sm pha hn x1 mt gc (x1 tr pha hn x2 mt gc ).o

2 10 : : x2 tr pha hn x1 mt gc (x1 sm pha hn x2 mt gc ).o 2 10 : : hai dao ng cng pha (hoc 2k ): ax 1 2m A A A A o : hai dao ng ngc pha {hoc )12( k }: min 1 2 A A A A

o

2 : hai dao ng vung pha {hoc 2)12( k } :2221 A A A

o21 A A : 2

cos2 1 A A Vi 12

o0120

32

21 A A A

o 1 2 1 2 A A A A A so snh pha dao ng, phi chuyn cc phng trnh dao ng v cng mt hm s l

gic : cos sin2

x x

v sin os x-2

x c

VN 3.MT S H DAO NGi lng vt l Con lc l xo Con lc n Con lc vt l

1.Cu trc

Vt c khi lng m (kg), gn

vo l xo c cng k (N m )

Vt c khi lng m (kg)

treo u si dy nh,khng dn, chiu di l(m)

Vt rn khi lng m

(kg),quay quanh mttrc nm ngang khngqua trng tm

O

t

1 A

2 Ax

O

t

x

Cng pha Ngc pha Vung pha


8/91



2.Phng trnhng lc hc

2x"+ 0 x x: li thng

2s"+ 0s s: li cong

2"+ 0 : li gc

3.Phng trnhdao ng

x=Acos t+ 00

cos( )cos( )

s s t t

0= cos t+ )(1;100 rad

4.Tn s gc

ringl

g mk

g l

d

Img

5.Chu k g l

k mT 22 2 l T

g 2

dI

T mg

6.Tn s l g

mk f

21

21

l g

f 2

1 I

mgd f

21

7. Lc gy raDDH

- Lc ko v:F = - kx

* L xo treo thng ng : F = k( 0l x)

- Lc ko v:

t mg P s mg l

(vi nh)

- Mmen lc ca conlc vt l:

M mgd (vi nh)

8. Cng thcc lp vi thigian

1222

2

2

Av

A x

22

22 Av x

1220

2

20

2

S v

S s

202

22 S v s

9.Nng lng

a.ng nng W21=

2mv W 21=

2mv

Bin thin tun hon

vi chu k T=2T ; tn

s gc =2 ; tn s f=2f b.Th nng

Wh 21

x2

k Wt zmg

c.C nng t W W W

222

21

21 AmkAW

t W W W 20

20

2

21

21 mgl S mW

Ch :o Ti v tr cn bng: axmv v : Wt = 0; W = (W)maxo Ti hai bin:W = 0; W = (Wt)maxo d : Khong cch t trc quay n trng tm vt rn (m)

I: Momen qun tnh ca vt rn i vi trc quay (kg.m2)

VN 4.MT S DNG TON

Chn h quy chiu:+ Trc Ox...+ Gc to ti VTCB+ Chiu dng...+ Gc thi gian (t=0): thng chn lc vt bt u dao ng hoc lc vt qua VTCB thchiu (+)

Phng trnh dao ng c dng: x = Acos(t + )Phng trnh vn tc: v = -Asin( t + )

Dng 1Vit phng trnh dao ng diu ho.Xc nh cc c trng ca mt dao ng iu ho


9/91



1. Xc nh tn s gc: ( >0)

Khi cho dn ca l xo VTCB0 : 00

k g k mg m 0

g

2 2

v A x

2. Xc nh bin dao ng A:(A>0)

cho Cng thcChiu di qu o d ca vt dao ng2d A

Chiu di ln nht v nh nht ca l xo2

minmax l l A

Li x v vn tc v ti cng mt thi im2

22

v x A (nu bung nh v = 0)

Vn tc v gia tc ti cng mt thi im 2 22 4

v aA

Vn tc cc i vmax

maxv A

Gia tc cc i amax 2max

a A

Lc hi phc cc i Fmaxk

F A max

Nng lng ca dao ngk W A 2

Mt s ch v iu kin ca bin

Hnh 1 Hnh 2 Hnh 3Vt m1 c t trn vt m2 dao ng iu ho theo phng thng ng. (Hnh 1)

m1 lun nm yn trn m2 trong qu trnh dao ng th: k g mm g A )( 212max

Vt m1 v m2 c gn vo hai u l xo t thng ng, m1 dao ng iu ho.(Hnh 2)

m2 lun nm yn trn mt sn trong qu trnh m1 dao ng th: k g mm g A )( 212max

Vt m1 t trn vt m2 dao ng iu ho theo phng ngang. H s ma st gia m1 v m2 l , b qua ma st gia m2 v mt sn. (Hnh 3)

m1 khng trt trn m2 trong qu trnh dao ng th: k g mm g A )( 212max

3. Xc nh pha ban u : ( )Da vo cch chn gc thi gian xc nh

Khi t=0 : 0 00 0

o s = x

A s i n = v

x x A c

v v


10/91



Nu lc vt i qua VTCB :

00

os =00v 0sin

c Acos A A sin v A

Nu lc bung nh vt:

0

0

Acos x

A sin

0 0cos

sin 0

x A A

Ch :- Khi th nh, bung nh vt v0=0 , A=x0- Khi vt i theo chiu dng th v>0, theo chiu m th v


11/91



Vn t c v = x = x0; gia t c a = v = x = x0H thc c lp: a = -2x0 v A2 =Khi x = a Asin2( t + ) th ta h bc.

Cng thc lng gic :* cos2 =1 cos22

v sin2 =1 cos22

* cosa + cosb= 2cosa b2

cosa b2

Bin A/2; tn s gc 2, pha ban u 2.

Phng trnh dao ng c dng: x = Acos(t + )Phng trnh vn tc: v = -Asin( t + )

1.Khi vt i qua li x0:x0= Acos( t + ) cos( t + ) = 0 x

A= cos ( ) 2 t n

2

nt nT (s)

Vi n N Khi >0n N* Khi 0 )

Cch 1 :Tnh s chu k dao ng t thi im t1 n t2 : 2 1t t m N n

T T , vi 2T

Trong mt chu k:* Vt i c qung ng sT = 4A* Vt i qua li bt k 2 ln* Nu m= 0 th:

Qung ng i c: s = n.sT = n.4A

Dng 3 Xc nh thi im vt i qua li x 0 , vn tc vt t gi tr v0

Dng 4

Xc nh qung ng v s ln vt i qua li x 0 t thi im t 1 n t 2


12/91



S l n vt i qua x0 l m = n.mT = 2n* Nu m 0 th:

Khi t = t1 ta tnh x1 = Acos( t1 + ) v v1 dng hay m (khng tnh v1)Khi t = t2 ta tnh x2 = Acos( t2 + ) v v2 dng hay m (khng tnh v2)

Sau v hnh ca vt trong phn lmT

chu k ri da vo hnh v tnh slv s ln

mlvt i qua x0 tngng.

Khi : +Qung ng vt i c l: s = n.4A + sl+S ln vt i qua x0 l: m = 2n + ml* V d:

1 0 2

1 20, 0 x x xv v

Ta c hnh v:

Khi : + S ln vt i qua x0 l ml= 1+ Qung ng i c: sl 1 2 1 22 4 A Ax A x Ax x

Cch 2 :

Bc 1 : Xc nh : 1 1 2 21 1 2 2

x Acos( t ) x Acos( t )v

v Asin( t ) v Asin( t )

(v1 v v2 ch cn xc nh du)

Bc 2 : Phn tch : t2 t1 = nT + t (n N; 0 t < T) . (Nu 2T

2 At S2

)

Qung ng i c trong thi gian nT l S1 = 4nA, trong thi giant l S2.Qung ng tng cng l S = S1 + S2 :

* Nu v1v2 02 2 1

2

2 2 1

Tt S x x2T 2At S2Tt S 4A x x2

* Nu v1v2 < 0 1 2 1 21 2 1 2

v 0 S 2A x x

v 0 S 2A x x

bin dng ca l xo thng ng khi vt VTCB:mg l k

2 l T g

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

nghing : sinmg l k

2sin

l T g

+ Chiu di l xo ti VTCB:l CB= l 0 + l (l 0 l chiu di t nhin)

+ Chiu di cc tiu (khi vt v tr cao nht):l Min= l 0 + l A+ Chiu di cc i (khi vt v tr thp nht):

l Max= l 0 + l + A l CB= (l Min+ l Max )/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 ;(t =

vi cos = )- Thi gian l xo dn1 ln l thi gian ngn nht vt i t v tr x1 = - l n x2 = A ; (T/2t)

Lu :Trong mt dao ng (mt chu k) l xo nn 2 ln v gin 2 ln

-A AOx2 x1x0 x

Dng 5 Tnh thi gian l xo dn v nn trong mt chu k

l

dnO

-A

Ann

(A > l )

O

x


13/91



1. Lc hi phc (lc tc dng ln vt):F kx ma : Lun hng v v tr cn bng

ln: F = k|x| = m2|x|

Lc hi phc t gi tr cc i F max= kA khi vt i qua cc v tr bin (x = A) Lc hi phc c gi tr cc tiu F min= 0 khi vt i qua v tr cn bng (x = 0)2. Lc n hi v lc tc dng ln i m treo l xo:

Lc tc dng ln im treo l xo l lc n hi:o

0F k | x | Khi chn chiu dng hng xung.o

0F k | x | Khi chn chiu dng hng ln.+ Khi con lc l xo nm ngang: 0 = 0

+ Khi con lc l xo treo thng ng:0 2k

mg g l

+ Khi con lc nm nghing 1 gc : 0sink

mg l

Lc cc i tc dng ln im treo l:max 0F k( A) Lc cc tiu tc dng ln im treo l:

+Kkhi con lc nm ngang: F min= 0+ Khi con lc treo thng ng hoc nm nghing 1 gc

Nu 0l A th min 0F k( A) Nu 0 A th Fmin = 0

3. Chiu di l xo:l0 : l chiu di t nhin ca l xo:Khi con lc l xo nm ngang:

+ Chiu di cc i ca l xo :ax 0ml l A + Chiu di cc tiu ca l xo:min 0l l A

Khi con lc l xo treo thng ng hoc nm nghing 1 gc:+ Chiu di l xo khi vt VTCB: 0 0cbl l l + Chiu di cc i ca l xo: ax 0 0ml l l A + Chiu di cc tiu ca l xo: ax 0 0ml l l A + Chiu di li x: 0 0l l l x

1. Th nng Wt =2

1 kx2 =2

1 k A2cos2( t + ) = 2 21 1 os 24 4

kA kA c t 2. ng nng W =

21 mv2 =

21 m 2A2sin2( t + ) = 2 21 1 os 2

4 4kA kA c t Vi k = m

2

3. C nng W = Wt + W =21 k A2 =

21 m 2A2 = const

Ch :Khi tnh nng lng phi i khi lng v (kg) , vn tc v (m/s) , li v (m) .Khi W = nWt hoc Wt = nW

Ti v tr c W = nWt ta c :

Dng 6 Xc nh lc tc dng cc i v cc tiu tc dng ln vt v im treo l xo - Chiu di l xo

khi vt dao ng

Dng 7 Xc nh nng lng ca dao ng iu ho


14/91



o Ta :12

121)1( 2222

n A x Am xmn

o Vn tc :12

121

.1 222

nn

Av Ammvn

n

Ti v tr c Wt = nW ta c :

o Ta :12

1

2

1.

1 22

n

n A xkAkx

n

n

o Vn tc :12

121)1( 222

n Av Ammvn

Trng thi Ta Vn tcng nng bng th nng

2 A

2 A

ng nng bng hai ln th nng3

A32

A

ng nng bng ba ln th nng2 A

23 A

Th nng bng hai ln ng nng32 A 3

A

Th nng bng ba ln ng nng2

3 A2 A

Th nng v ng nng ca vt bin thin tun hon vi cng tn s gc = 2 , tn s daong f =2f v chu k T =

2T

- nh lut bo ton ng lng : const p p pconst p n....21 .(iu kin p dng l h kn)- nh lut bo ton c nng : E = constE + Et = const (iu kin p dng l h kn , khng ma st)

- nh l bin thin ng nng : ngoailucngoailuc ngoailuc Amvmv A E E A E 212212 21

21

- Ch : i vi va chm n hi ta c : 21'22'2122 21

21

21

21 mvmvmvmv

- Qung ng S m gi i c k t khi bt u chuyn ng n khi vt ri khi gi btng bin dng ca l xo trong khong thi gian . Khong thi gian t lc gi bt u ch

ng n khi vt ri khi gi c xc nh theo cng thc:21 22

S S at t a

(1)

(a l gia tc ca gi )- Vn tc ca vt khi ri khi gi l :2 .v a S (2)- Gi 0l l bin dng ca l xo khi vt VTCB ( khng cn gi ),l l bin dng ca l xokhi vt ri gi . Li x ca vt thi im ri khi gi l0 x l l

- Ta c2

2 22

v x A

Dng 8 Bi ton v va chm

Dng 9 Bi ton v dao ng ca vt sau khi ri khi gi


15/91



Ta dng mi lin h gia dao ng iu ho v chuy n ng trn u tnh.Vt chuyn ng trn u t M n N, hnh chiu ca vt ln trc Ox dao ng iu ho 1

n x2 .Thi gian ngn nht vt dao ng i t x1 n x2 bng thi gian vt chuyn ng trn u t M

n N.2 1t

vi1

1

22

s

s

xco A xco A

v ( 1 20 , )

Vt xut pht t VTCB: (x=0)

+ Khi vt i t: x = 02 A x th

12T t : S = A/2

+ Khi vt i t: x=0 22

A x th8T t : S = 2

2 A

+ Khi vt i t: x=0 32 A x th 6 T t : S = 32 A

+ Khi vt i t: x=0 x A th4

T t : S = A

Vt xut pht t v tr bin: ( x A)

+ Khi vt i t: x=A 32

A x th12

T t : S = A - 32

A

+ Khi vt i t: x=A 22

A x th8T t : S = A- 2

2 A

+ Khi vt i t: x =A2

A x th6

T t : S = A/2

+ Khi vt i t: x=A x= 0 th4

T t : S = A

1. L xo ghp ni tip: cng k ca h :

Hai l xo c cng k 1 v k 2 ghp ni tip c th xem nh mt l xo c cng k tho mn

biu thc:21

111k k k

Chu k dao ng: 2 2 21 2T T T , Tn s dao ng : 22

21

2111

f f f 2. L xo ghp song song:

cng k ca h :Hai l xo c cng k 1 v k 2 ghp song song c th xem nh mt l xoc cng k tho mn biu thc: k = k 1 + k 2

Chu k dao ng: 2 2 21 2

1 1 1T T T

, Tn s dao ng : 22

21

2 f f f

3. Khi ghp xung i cng thc ging ghp song song

Dng 11 H l xo ghp ni tip - ghp song song v xung i

M N

xO Ax1x2-A

Dng 10 Xc nh thi gian ngn nht vt i qua li x 1 n x 2

ml1,k 1 l2,k 2

l1, k 1

l2, k 2

m


16/91



Lu :Khi gii cc bi ton dng ny, nu gp trng hp mt l xo c di t nhin l0 ( cng k 0) c

ct thnh hai l xo c chiu di ln lt l l1 ( cng k 1) v l2 ( cng k 2) th ta c: k 0l0 = k 1l1 = k 2l2Trong : k 0 =

0

ES ; E: sut Young (N/m2); S: tit din ngang (m2)

Gn l xo k vo vt khi lng m1 c chu k T1; vo vt khi lng m2 c chu k T2; vovt khi lng (m1+m2) c chu k T3; vo vt khi lng (m1 m2) (m1 > m2) c chu k T4.Th ta c: 2 2 23 1 2T T T v 2 2 24 1 2T T T

Vt c vn t c ln nh t khi qua VTCB, nh nh t khi qua v tr bin nn trong cng mt khothi gian t qung ng i c cng ln khi vt cng gn VTCB v cng nh khi cng gv 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 (Hnh 1) ax 2Asin2

mS

Qung ng nh nht khi vt i t M1 n M2 i xng qua trc

cos (Hnh 2) 2 (1 os )2min

S A c

Lu :o Trong trng hpt > T/2

Tch '2T t n t trong *;0 '

2T n N t

Trong thi gian2T n qung ng lun l 2nA

Trong thi giant th qung ng ln nht, nh nht tnh nh trn.o Tc trung bnh ln nht v nh nht ca trong khong thi giant:

axax

mtbm

S vt

v mintbminS v

t (vi Smax; Smin tnh nh trn)

1) Phng trnh dao ng.Chn: + Trc OX trng tip tuyn vi qu o

+ Gc to ti v tr cn bng+ Chiu dng l chiu lch vt+ Gc thi gian .....

Phng trnh ly di: s = S0cos( t + ) hoc = 0cos( t + ) vi s = l , S0 =A= 0l v = - Asin( t + )

Tm >0:

+ = 2 f =T

2 , Vi N

t T , N: Tng s dao ng

+

g

, ( l:chiu di dy treo , g: gia tc trng trng ti ni ta xt: m/s2)

+mgd

I

, Vi d = OG: khong cch t trng tm n trc quay.

Dng 12 Bi ton tnh qung ng ln nht v nh nht vt i c trong khong thi gian 0


17/91



I: mmen qun tnh ca vt r n.

+ 2 2v

A s

Tm A>0:

+

22 2

2vA s

vi s .

+ Khi cho chiu di qu o l mt cung trnMN:2

MN A

+ 0A . vi 0 : ly gc (rad)Tm ( )Da vo cch chn gc thi gian xc nh ra

Khi t = 0 th0

0

x x

v v0

0

x Acos

v A sin

0

0

os

sin

xc Av A

= ?

Phng trnh ly gic: l s = 0 cos( t + ) rad. vi l Al S

00 rad

2) Chu k dao ng nh.

+ Con lc n:2T

g

2

2

2

2

44

T g

g T

+ Con lc vt l: 2 I

T mgd

2

2

2

2

44

T mgd I

I g T md

1. Nng lng con lc nChn gc th nng ti v tr cn bng O+ Th nng hp dn ly : Wt = )cos1( mgl mgh

+ C nng:W=W+Wt= )cos1(21

21

0020

220 mgl mghS mmv

+ ng nng: W = W-Wt = )cos(cos

2

10

2 mgl mv

*Khi gc nh:p dng cng thc gn ng2

1cos;2

1cos20

0

2

+ Th nng : Wt = 222 21

21 smmgl

+ C nng : W = 20220 21

21 S mmgl

+ ng nng : W = )(21)(

21 2

0222

02 S smmgl

Dng 14 Nng lng con lc n -Xc nh vn tc ca vt -Lc cng dy treo khi vt i qua li gc

N

OA

0

P


18/91



Ch :

- Nu W=nWt ta c :W=Wt+W =(n+1)Wt 121)1(

21 022

0 nmgl nmgl

- Nu Wt=nW ta c :W=Wt+W=(n+1)W0

20

220 .1

2)(21)1(

21

nnmgl nmgl

2. Vn tc ca vt khi i qua li (i qua A)p dng nh lut bo ton c nng ta c : C nng ti bin = c nng ti v tr ta xt

WA=W N WtA+WA=WtN+WN mg (1 cos ) +2

A1 mv2 = 0mg (1 cos ) +0

2A 0v 2g (cos cos )

A 0v 2g (cos cos ) 3. Lc cng dy (phn lc ca dy treo) treo khi i qua li (i qua A)

Theo nh lut II Newtn:P + =ma chiu ln ta c2A

htvmgcos ma m

2A

0

vm mgcos m2g(cos cos ) mgcos

Vy: 0 = mg(3cos - 2cos )

4. Khi gc nh 010

2

sin

cos 12

Khi

2 2 2A 0

2 20

v g ( )1 mg(1 2 3 )2

Ch :Ti VTCB Ti hai bin

= 0 ax 02 1 osmv v gl c

gl vv 0max

ax 03 2cosm mg )1( 20max mg

v = 0 ; = 0min 0osmgc

)211( 20min mg

GHI NH : Mt s cng thc gn ng

1 )(1;100 rad nn 1)1(

2121 1)1)(1(

212

1 111

21cos

2

tansin

Khi thay i cao, su v nhit thay i th chu k ca con lc n cng thay i

Gia tc trng trng mt t: 2 RGM g (R=6400km Bn knh Tri t)

1. Gia tc trng trng cao h

Gia tc trng trng cao h:

h 22

GM gg h(R h) (1 )R

.

Dng 15 S thay i chu k ca con lc n theo cao, su v nhit


19/91



Chu k con lc dao ngng mt t:1T 2 g

(1)

Chu h con lc dao ngsai cao h:2

h

T 2g

(2)

1 h2

T gT g m

hg 1

hg 1 R

1

2

T 1

hT 1 R 2 1 hT = T (1 + )R Khi a ln cao chu k dao ng tng ln.

2. Gia tc trng trng su d

su d:ddg = g(1- )R

Chng minh: Pd = Fhd

3

d 2

4m( (R d) .D)

3mg G(R d)

D: Khi lng ring Tri t33

3

d 2 3 2 3 2

4( .D)(R d)R

M(R d) GM d3g G G .(1 )(R d) .R (R d) .R R R

d dg = g(1- )R

Chu kcon lc dao ng su d:2

d

T 2g

(3)

d1

2

gTT g m

dg d1g R

12

12 1

T dT = T (1+ )R d1-

R Khi a xung su chu k dao ng tng ln nhng tng t hn a ln cao

3. Chiu di ca dy kim loi nhit tKhi nhit thay i: Chiu di bin i theo nhit := 0 (1 + t).

: L h s n di v nhit ca kim loi lm dy treo con lc.0 : Chiu di 00C

Chu k con lc dao ngng nhit t1(0C):1

1T 2 g

(1)

Chu k con lc dao ngsai nhit t2(0C):2

2T 2 g

(2)1 1

2 2

TT

Ta c:1 0 1 1 1

2 12 0 2 2 2

(1 t ) 1 t 11 (t t )

(1 t ) 1 t 2

V 1 1 1

2 1 2 1 2 12

2 1

T T1 11 (t t ) T T (1 (t t ))1T 2 21 (t t )2

Vy2 1 2 1

1(1 ( ))2

T T t t

+ Khi nhit tng th chu k dao ng tng ln+ Khi nhit gim th chu k dao ng gim xung

Ch : + Khi a ln cao m nhit thay i th:1

2 12

11 ( )2

T ht t T R


20/91



+ Khi a ln xung su d m nhit thay i th:1

2 12

11 ( )2 2

T d t t T R

Vit cng thc tnh chu k ca con lc ng h trong trng hp chy ng (T1) v chy sai (T2)

Lp t s 12

T

T (ri dng cng thc gn ng nu cn) hoc lp hiu2 1T T T

1

2

1T T

: ng h chy chm ( 0T )

1

2

1T T

: ng h chy nhanh ( 0T )

S dao ng con lc ng h chy sai trong khong thi gian t :2

t N

T

Thi gian ng h chy sai ch : 112

'T

t NT t T

Thi gian ng h chy sai: 12 2

' 1 T t t t t t T N T T T

Ch :o Ch c l thay i:

1 1 1

2 12 2 1 2 1

T l l 11 t tT l 2l 1 t t

o Ch c g thay i:12 0

hg T R T g R h

o Khi c l v g thay i:1 12 2 0

hg T l T l g

Ti cng mt ni con lc n chiu dil 1 c chu k T1; con lc n chiu dil 2 c chu k T2; conlc n chiu di (l 1 + l 2 ) c chu k T3; con lc n chiu di (l 1 - l 2 ) (l 1>l 2) c chu k T4. Th tac: 2 2 23 1 2T T T v 2 2 24 1 2T T T

Mt ngy m: t = 24h = 24.3600 = 86400s.Chu k dao ngng l: T1Chu k dao ngsai l T2

+ S dao ng con lc dao ngng thc hin trong mt ngy m:1

1

t NT

+ S dao ng con lc dao ngsai thc hin trong mt ngy m:2

2

t NT

+ S dao ngsai trong mt ngy m:1 1

2 1

1 1 N | N N | t | |T T

+ Thi gian chysai trong mt ngy m l:1

12

TT . N t | 1|T

Nu chu k tng con lc dao ng chm li Nu chu k gim con lc dao ng nhanh ln

Dng 16 Thi gian con lc ng h chy sai trong khong thi gian t

Dng 17 Xc nh thi gian dao ng nhanh chm trong mt ngy m.


21/91



* Khi a ln cao h con lc dao ng chm trong mt ngy l:ht.R

* Khi a xung su h con lc dao ng chm trong mt ngy l:d = t.

2R

* Thi gian chy nhanh chm khi nhit thay i trong mt ngy m l:|2 1

1 = t | t - t2

* Thi gian chy nhanh chm tng qut:) |2 1h 1 = t | (t - tR 2

1) Chu k con lc:

* Chu k cn lc trc khi vp inh:1

1T 2 g

, 1: Chiu di con lc trc khi vp inh

* Chu k con lc sau khi vp inh:2

2T 2 g

, 2 : Chiu di conlc sau khi vp inh

* Chu k ca con lc: 1 21T (T T )2

2) Bin gc sau khi vp inh0 :Chn mc th nng ti O. Ta c: WA=W N

WtA=WtN 2 0 1 0mg (1 cos ) mg (1 cos )

2 0 1 0(1 cos ) (1 cos ) v gc nh nn

2 22 0 1 0

1 1(1 (1 )) (1 (1 )2 2

1

0 02

= : Bin gc sau khi vp inh.

Bin dao ng sau khi vp inh:'

0 2. A l

Cho hai con lc n: Con lc 1 chu k1T bitCon lc 2 chu k2T cha bit 2 1T T

Cho hai con lc dao ng trong mt phng thng ng song song trc mt mt ngi qua Ngi quan st ghi li nhng ln chng i qua v tr cn bng cng lc cng chiu(trng phGi l thi gian hai ln trng phng lin tip nhau

a) Nu 1T

> 2T

: con lc 2T

thc hin nhiu hn con lc1T

mt dao ng

Ta c 1 2( 1)nT n T

1

2 1

T n

nT

11

2

T

T

111

1

2

T

T

11112 T T

b) Nu 1T < 2T : con lc 1T thc hin nhiu hn con lc2T mt dao ng

N

O

0

A0

Dng 19 Xc nh chu k con lc bn hn h trn hn

Dng 18 Xc nh chu k con lc vp (vng) inh bin sau khi vp inh


22/91



Ta c 2 1( 1)nT n T 1

1

2

T n

nT

11

2

T

T

111

1

2

T

T

11112 T T

Khi con lc chu tc dng thm ca ngoi lc khng inF :

Trng lc hiu dng (trng lc biu kin):hd nP P F

nhd n hd

Fmg mg F g gm

(*)

Khi con lc n s dao ng quanh v tr cn bng mi vi chu k:2hg

l T

g

Khi nF P

: nhdFg gm

Khi nF P

: nhdFg gm

Khi nF P

:

22 n

hd

Fg g m

Khi ( nF ,P ) = :2

2 n nhd

F Fg g 2g cosm m

V tr cn bng mi : n0FtanP

Cc loi lc thng gp:o Lc ht ca nam chm :Chiu (*)/xx :

m F g g x'

Nam chm t pha di : Fx > 0 F hng xung m F g g '

Nam chm t pha trn : Fx < 0 F hng ln m F g g '

o Lc tnh in: 9 1 22| q q |F 9.10

r (r: Khong cch gia hai in tch.)

Hai in tch cng du th y nhau; hai in tch tri du th ht nhau.o Lc in trng:F=|q|E vi UE

d(V/m)

Trng lc biu kin lc ny : m E q g g E q P P '' (*)

Chiu (*)/xx : )1('mg

qE g m

qE g g x x

F E

khi q>0;F E

khi q


23/91



V: Th tch ch t lng hoc ch t kh m vt chi m ch .o Lc qun tnh: am F qt

Trng lc biu kin lc ny : a g g F P P qt '' iu kin cn bng : T P F T P qt '0a : Gia tc ca h qui chiu gn con lc i vi h qui chiu qun tnh.

- Trng hpa hng xung : (*) g=g a

- Trng hpa hng ln : (*) g=g + a

- Trng hpa nm ngang p dng nh l Pitago : 22' a g g hoc cos

'g

g vi ( ' POP )

1) Bi ton t dy:Khi con lc t dy vt bay theo phng tip tuyn vi qu o tiim t.+ Khi vt i qua v tr cn bng th t dy lc vt chuyn ng

nn ngang vi vn tc u l vn tc lc t dy.Vn tc lc t dy:0 0v 2g (1 cos )

Phng trnh theo cc trc to :

0

2

theo ox : x v .t1theo oy : y gt2

phng trnh qu o:

22

20 0

1 x 1y g x2 v 4 (1 cos )

+ Khi vt t ly th vt s chuyn ng nm xin vi vn tc ban u l vn tc lc t dy.

Vn tc vt lc t dy:0 0v 2g (cos cos )

Phng trnh theo cc trc to :

0

20

theo ox : x (v cos ).t1theo oy : y (v sin ).t gt2

Khi phng trnh qu o l:2

20

1 gy (tan ).x x2 (v .cos )

Hay:2 2

20

1 gy (tan ).x (1 tan )x2 v

Ch : Khi vt t dy v tr bin th vt s ri t do theo phng trnh :21y gt

2

2) Bi ton va chm:+ Trng hp va chm mm: sau khi va chm h chuyn ng cng vn tc

Theo LBT ng lng:A B AB A A B B A BP P P m v m v (m m )V

Chiu phng trnh ny suy ra vn tc sau va chm V+ Trng hp va chm n hi: sau va chm hai vt chuyn ng vi cc vn tc k

nhau A2v v B2v .Theo nh lut bo ton ng lng v ng nng ta c

Dng 21 Bi ton con lc t dy - va chm

N

O

0vX

Y

N

O

0v

X

Y


24/91



A B A2 B2

dA dB dA2 dB2

P P P PW W =W +W

A A B B A A2 B A2

2 2 2 2A A B B A A2 B B2

m v m v m v m v1 1 1 1m v m v m v m v2 2 2 2

T y suy ra cc gi tr vn tc sau khi va chmA2v v B2v .

- Khi con lc gn vo h chuyn ng tnh tin vi gia tca th vt chu tc dng thm ca lc

qun tnh pt F m a

(ngc chiu via )

Trng lc hiu dng(trng lc biu kin):hd qt P P F

hd hd m g m g m a g g a

+ Khi h chuyn ng nhanh dn u tha cng chiu viv (chiu chuyn ng) khi qt F ngc chiu chuyn ng

+ Khi h chuyn ng chm dn u tha ngc chiu viv (chiu chuyn ng) khi qt F

cng chiu chuyn ng

1) Khi qt F P

(cng hng) thhdg g a khi T2 < T1:chu k gim

2) Khi qt F P

(ngc hng) thhdg g a khi T2 > T1:chu k tng

3) Khi qt F P

(vung gc) th2 2

hdg g a khi T2 < T1:chu k gim

V tr cn bng miqt

0F

tanP

4) Khi qt F hp vi P mt gc th:2 2 2

hdg g a 2ga.cos

cho h dao ng vi bin cc i hoc rung mnh hoc nc sng snh mnh nht xy ra cng hng dao ng.

Khi 0 0( ) f f T=T0

Vn tc khi xy ra cng hng l:svT

Con lc l xo Con lc n Con lc vt l

0

k

m

0

g

0

mgd

I

o Con lc vt l :mgd

I T 2 Con lc n ton hc :

g l

T 2'

o Con lc n ton hc ng b vi con lc vt l khi chng c cng chu k : T = T Vymd I l

o Nu con lc vt l l vt rn c dng i xng th p dng nh l Huyghen Steiner :

Dng 23 Bi ton v s cng hng dao ng

Dng 22

Xc nh chu k con lc khi n vo h chu n n t nh tin vi ia tca

Dng 24 Xc nh chiu di ca con lc n ton hc ng b vi con lc vt l


25/91



2md I I G Vy : md I d l G

o Theo nh lut bo ton v chuyn ha nng lng : (max)tBt E E E E

Hay I hhmg

vmghmghv I B

B)(2

.21 22

vi )cos(cos 0 d hh B

Vy I

mgd v )cos(cos2 0

o Khi i qua VTCB : I

mgd v )cos1(20 0max

nu dao ng ca con lc vt l l dao

ng b : 00max I mgd v

o Ch : Trong trng hp con lc vt l gm nhiu cht im th cht im ny c cng vgc

o gim nng lng ca con lc sauT 21 chnh l cng ca lc masat . Do ta c

)(.)(21 2

210

2

21

20

210 A Amg A Ak E E E

1. Bin ca con lc sauT 21 l :

k mg A A .20

21

2. Bin ca con lc sau 1T l :k mg A

k mg A A .4.2 0

211

3. Bin ca con lc sau nT l : k mg n A An .40 o Khi con lc l xo dng li th ton b c nng ban u ca n chuyn ha thnh cng ca l

masat . Do ta c : g A

mg kA s smg kA

.2.

.2..

21 2222

1. gim bin sau 1T l : 2.4.4

g k mg A

2. gim bin sau nT l : 2.4.4

g nk mg n A A A nn

3. S dao ng thc c l : g

A

A

A N

.4

.2

4. Thi gian thc hin dao ng n lc dng li : g AT N t .2...

o Vn tc ca qu nng t cc i khi lc masat bng lc hi phc :

k mg x xk mg F F hpms... 00

o Vn tc cc i khi vt dao ng ti v tr c ta 0 x . Lc ta c :

).()()(.21

21

21

00020

22 x Amk x Av x Amg kxmvkA

Dng 25 Xc nh v n tc ca con lc v t l

Dng 26 Bi ton v dao n tt dn


26/91



C nng ban u (cung cp cho dao ng) : 21(max)0 21 kA E E t (1)

Cng ca lc masat (ti lc dng li) : mgs s F A msms .. (2)

p dng nh lut bo ton v chuyn ha nng lng : mg kA

s E Ams .2

2

10

Cng bi q : v bin gim dn theo cp s nhn li v hng nn :

11

12

31212

3

1

2 .;...;.;.... Aq A Aq A Aq A A A

A A

A Aq nn

n

n (vi q < 1)

o ng i tng cng ti lc dng li : S Aqqq A A A A s nn 112121 2)...1(22...22

viq

qqqS n1

1...1 12 Vyq

A s12 1

Cng bi q : v bin gc gim dn theo cp s nhn li v hng nn :

11

12

31212

3

1

2 .;...;.;....

n

nn

n qqqq (vi q < 1) Vy 11

nnq

Nng lng cung cp (nh ln dy ct) trong thi gian t duy tr dao ng :

- C nng chu k 1 : 1max1 1 mgh E E tB hay211 2

1 mgl E

- C nng chu k 12 : 2max2 2 mgh E E tB hay222 2

1 mgl E

- gim c nng sau 1 chu k : )(21 2

2

2

1 mgl E Hay )1(

21 22

1qmgl E y chnh l

nng lng cn cung cp duy tr dao ng trong 1 chu k

- Trong thi gian t , s dao ngT t n . Nng lng cn cung cp duy tr dao ng sau n dao

ng E n E .

- Cng sut ca ng h :t

E P

1. T ng hp dao ng iu hoa. C s l thuyt: Nh ta bit mt dao ng iu hox = Acos( t + )

+ C th c biu din bng mt vect quayA c di t l vi bin A v to vi trc honh mgc bng gc pha ban u.+ Mt khc cng c th c biu din bng s phc di dng: z = a + bi+Trong ta cc:z =A(sin +i cos ) (vi mun:A= 2 2a b ) Hay Z = Ae j( t + ).+V cc dao ng cng tn s gcc tr s xc nh nn ngi ta thng vit vi quy cz = AeJ ,trong my tnhCASIO fx- 570ESk hiu di dng l: r (ta hiu l: A ) .+ c bit gic s c hin th trong phm vi : -1800< < 1800 hay - < < rt ph hp vi bi

Dng 29 Dng my tnh CASIO fx 570ES hoc CASIO fx 570MS gii mt s bi ton c dng

hm dao n iu ha

Dng 27 Con lc l xo dao ng tt dn .Bin gim dn theo cp s nhn li v hng .Tm cng bi

Dng 28Con lc n dao ng tt dn .Bin gim dn theo cp s nhn li v hng .Tm cng bi

nn l n cun c du tr dao n


27/91



ton t ng hp dao ng i u ho.Vy tng hp cc dao ng iu ho cng phng, cng tn s bng phng php Frexnen ngngha vi vic cng cc s phc biu din ca cc dao ng .b.Chn ch mc nh ca my tnh:CASIO fx 570ESMy CASIO fx570ES bmSHIFT MODE 1hin th1 dng (MthIO) Mn hnh xut hinMath.+ thc hin php tnh v s phc th bm my : MODE 2 mn hnh xut hin chCMPLX+ tnh dng to cc :A , Bm my :SHIFT MODE 3 2

+ tnh dng to cc:a + ib. Bm my :SHIFT MODE 3 1+ ci t n v o gc (Deg, Rad ):-Chn n v o gc l (D) ta bm my : SHIFT MODE 3 trn mn hnh hin th chD-Chn n v o gc l Rad (R ) ta bm my: SHIFT MODE 4 trn mn hnh hin th chR

+ nhp k hiu gc ca s phc ta n SHIFT (-).V d: Cch nhp:My tnh CASIO fx 570ESCho:x= 8cos( t+ /3) s c biu din vi s phc8 600 hay 8 /3 ta lm nh sau:-Chn mode: Bm my: MODE 2 mn hnh xut hin chCMPLX-Chn n v o gc l (D) ta bm: SHIFT MODE 3 trn mn hnh hin th chD-Nhp my: 8 SHIFT (-) 60 s hin th l:8 60-Chn n v o gc l Rad (R ) ta bm: SHIFT MODE 4 trn mn hnh hin th chR

-Nhp my: 8 SHIFT (-) (:3 s hin th l:8 1 3Kinh nghim cho thy: nhp vi n vnhanh hn n vrad nhng kt qu sau cng cn phichuyn sang n vrad cho nhng bi ton theo n vrad. (v nhp theo n v rad phi c du ngocn ( ) nn thao tc nhp lu hn,v d: nhp 90 th nhanh hn nhp (/2)c.Lu :Khi thc hin php tnh kt qu c hin th dng i s: a +bi (hoc dng cc: A ).

-Chuyn t dng :a + bi sang dngA , ta bmSHIFT 2 3 =V d:Nhp: 8 SHIFT (-) (:3 ->Nu hin th: 4+ 4 3 i , mun chuyn sang dng ccA :- Bm phmSHIFT 2 3 = kt qu: 8 /3

-Chuyn t dngA sang dng :a + bi : bmSHIFT 2 4 =V d: Nhp: 8 SHIFT (-) (:3 -> Nu hin th:8 /3,mun chuyn sang dng phca+bi :

- Bm phmSHIFT 2 4 = kt qu:4+4 3 i d. Xc nh A v bng cch bm my tnh:

+Vi my FX570ES :Bm chn MODE 2 trn mn hnh xut hin ch:CMPLX.-Nhp A1, bm SHIFT (-) nhp 1; bm+ , Nhp A2 , bm SHIFT (-) nhp 2 nhn = hin th kt qu.

(Nu hin th s phc dng:a+bi th bm SHIFT 2 3= hin th kt qu l: A )+Gi tr ca dng( nu my ci ch lD:)+Gi tr ca dngrad ( nu my ci ch lR: Radian)

+Vi my FX570MS :Bm chn MODE 2 trn mn hnh xut hin ch:CMPLX. Nhp A1, bm SHIFT (-) nhp 1 ;bm+ ,Nhp A2 , bm SHIFT (-) nhp 2 nhn =

Sau bm SHIFT + = hin th kt qu l: A. SHIFT = hin th kt qu l:

+Lu Ch hin th mn hnh kt qu:Sau khi nhp ta n du = c th hin th kt qu di dng sv t, mun kt qu di dngthp phnta nSHIFT = ( hoc dng phm SD) chuyn i kt quHin th.e. Nu cho x 1 = A1cos( t + 1 ) v x = x 1 + x 2 = Acos( t + ) .Tm dao ng thnh phn x2 : x2 = x - x1 vi: x2 = A2cos( t + 2)Xc nh A2 v 2 nh bm my tnh:*Vi my FX570ES :Bm chn MODE 2 Nhp A , bm SHIFT (-) nhp ; bm - (tr); Nhp A1 , bm SHIFT (-) nhp 1 nhn =kt qu.(Nu hin th s phc th bmSHIFT 2 3 = hin th kt qu trn mn hnh l: A2 2+Ta c s u lA2 v sau du l gi tr ca2 dng ( nu my ci n v l D:)


28/91



+Ta c s u lA2 v sau du l gi tr ca2 dng rad ( nu my ci n v l R: Radian)*Vi my FX570MS :Bm chn MODE 2 Nhp A , bm SHIFT (-) nhp ;bm - (tr); Nhp A1 , bm SHIFT (-) nhp 1 nhn =Sau bm SHIFT + = hin th kt qu l: A2. bm SHIFT = hin th kt qu l: 2

2. Tm li v vn tc mt thi imDng bi tp ny thng thng c thgii bng cch tnh ton i s thng thng hoc dng my tnV d : Mt cht im thc hin dao ng iu ho dc theo trc Ox xung quanh v tr cn bng O

chu k T = 2 s. Ti thi im t1 cht im c to x1 = 2 cm v vn tc v1 = 4 cm/s. Hy xc nh to v vn tc ca cht im ti thi im t2 = t1 + 3

1 s.

Cch gii Hng dn bm my v kt quGi s

. A.sin x Acos t v t Khi t=t1: 1 1. 2 x Acos t cm v

1 1A.sin 4v t cm/sKhi t=t2 = t1 + 3

1 s :

2 2 1 3. . x Acos t Acos t

2 1 1. . . .3 3 x A cos t cos A sin t sin

12 1

1 4 3. . 2. .

3 3 2 2v

x x cos sin

2

2 31 2,1027 x cm

Tnh v2: 2 1. 3v A sin t

2 1 1. . . .3 3v A sin t cos A cos t sin

2 1 1. . 3,4414 /3 3v v cos x sin cm s

2cos 60 ) + (4 shift 10x

) sin 60 ) =KQ: 2,1026577914cos 60 ) - -2shift 10x ) sin 60 ) =KQ: -3,441398093

Bi ton ny n v o gc bng bm my snhanh hn

3. Tm nhanh mt i lng cha bit trong biu thc vt l :S dng SOLVE ( Ch dng trong COMP : MODE 1 )SHIFT MODE 1 Mn hnh: Math

V d: Tnh khi lng m ca con lc l xo dao ng, khi bit chu k T =0,1(s) v cng k=100N/m.

Ta dng biu thck mT 2

Ch : Phm gnbin X: ALPHA ) ; SOLVE: SHIFT CALC; Nhp du= l phm ALPHACALC Phng php truyn thng Phng php dng SOLVE


29/91



Ta c :k mT

k mT 22 42

Suy ra: 22

4 kT m

Th s:2

2

4

)1,0(100

m =0,25kg

Vy :khi lng m ca con lc 0,25kg

-Vi my FX570ES:Bm: MODE 1

-Bm: 0.1 SHIFT X10 X ALPHACALC =2

SHIFT X10 X ALPHA ) X 100

Mn hnh xut hin :100

21.0 X

-Tip tc bm:SHIFT CALC SOLVE = ( ch khong 6s )

Mn hnh hin th:X l i lng m

Vy :m= 0,25 kg

CH 3: SNG C HC

VN 1.SNG C HC1. Sng c L nhng dao ng c hc lan truyn theo thi gian trong mi trng vt cht ltc (rn, lng, kh).

a. Sng ngang-Cc phn t c phng dao ng vung gc vi phng truyn sng.-Truyn c trong mi trng xut hin lc n hi khi c bin dng lch: mlng, cht rn.(v d : sng trn mt cht lng)

b. Sng dc-Cc phn t c phng dao ng trng vi phng truyn sng.-Truyn c trong mi trng xut hin lc n hi khi c bin dng nn-dn

lng, kh.(v d : sng m truyn trong khng kh)Cc i lng c trng cho sngi lng vt l Cng thc Ghi ch

1. Chu k, tn s f T 1 (s) Bng chu k, tn s ca ngun to ra sng.

2. Bc sng v vT f

(m)Qung ng sng truyn i trong mt chu k dao ng (khong cch gn nhau nht ca hai im trn phng truyn sng dao ng cng pha).

3. Tc sng v f T

(m/s) L tc truyn mt pha dao ng nht nh.

4.Nng lngsng )(21

22 J AmW sng Qu trnh truyn sng l qu trnh truyn nng lng a.Sng th ng b.Sng ph ng c.Sng cu

Sng truy n theomt phng (v d: sng truyn trnsi dy n hi ltng)

W=constA=const

- Sng truyn theo mt phng (v d:sng truyn mt nc)- Gn sng l nhng vng trn ngtm nng lng sng t ngun triu trn ton b vng trn .-Ta c :W0=2R M.WM=2R N.W N

-Sng truyn trong khng gian (v d:sng m pht ra t mt ngun im)-Mt sng c dng l mt cu nnglng sng t ngun tri u trn ton bmt cu-Ta c:W0=4R 2M.WM=4R 2 N.W N

10021.0 X

X = 0.25

LR = 0


30/91



2

2

N

M

M

N

N

M

A A

R R

W W Vy

R A

RW 1~;1~ 2

2

2

2

N

M

M

N

N

M

A A

R R

W W Vy

R A

RW 1~;1~ 2

5.Li ca mtim bt k trn phng truyn

sng

0

0

0

( ) cos

cos 2

2cos 2

M M

M

M

x u t A t

v

x t A

T

x A ft

x M :Ta ca M trn phng truyn sng . Daong ti im chn lm gc:

0cosOu A t

iu kin ti M c dao ng :M x

t v

6. lch pha 1 22 2d d d

d : Khong cch gia hai im

7. Bin casng mt im A : L bin dao ng ca phn t vt cht ti im .

Ch :- Ch c pha dao ng truyn i, cc phn t ca mi trng dao ng ti ch quanh v tr cbng. Cc phn t xa tm pht sng dao ng tr pha hn.- Sng c khng truyn c trong chn khng.

-2

22 k k d k (k = 1,2,3, ..) hai im dao ng cng pha

(khong cch gia hai im trn phng truyn sng bng mt s nguyn ln bc sng hocmt s chn ln na bc sng)

- )21(

2)12()12( k k d k (k = 0,1,2,3, ..) hai im dao ng ngc pha

(khong cch gia hai im trn phng truyn sng bng mt s bn nguyn ln bc sng hbng mt s l ln na bc sng)

-2

)21(

4)12(

2)12( k k d k (k = 0,1,2,3, ..) hai im dao ng vung pha

(khong cch gia hai im trn phng truyn sng bng mt s bn nguyn ln na bc s

hoc bng mt s l ln4

1 bc sng)

- Qu trnh truyn sng l qu trnh truyn nng lng, cng ra xa tm pht sng nng lng c gim lm bin sng cng gim.- Quan st hnh nh sng cn ngn sng lin tip th cn - 1 bc sng . Hoc quan st thy t

ngn sng th n n ngn sng th m (m > n) c chiu dil th bc sngnm

l

- S ln nh ln trn mt nc l N trong khong thi gian t giy th1 N

t T Hayt

N f 1

VN 2.GIAO THOA SNG

1. Giao thoa

-L s t ng hp ca hai (hay nhiu) sng kt hp trong khng gian.

-Trong vng giao thoa xut hin nhng vn giao thoa cc i v cc tiu xen k cu nhau.*Sng kt hp Do hai ngun kt hp pht ra: hai ngun dao ng c cng tn s, cng phng dng v hiu s pha khng i theo thi gian.2. lch pha ca hai sng thnh phn ti mt im

a. Hai ngun S1, S2 cng pha : 0 hoc 2k b. Hai ngun S1, S2 ngc pha : hoc

)12( k

d d d 22 12

d d d 22 12


31/91



o

222 k k d k

Dao ng ti im xt c bin cc i.o

)

21(

2)12()12( k k d k

Dao ng ti im xt c bin cc tiu.

o

)21(

2)12(2 k k d k

Dao ng ti im xt c bin cc i.o

22)12( k k d k

Dao ng ti im xt c bin cc tiu.S vn giao thoa cc i gia hai ngun

S 1S 2:

2121 S S k

S S k Z

S vn giao thoa cc tiu gia hai ngunS 1S 2:

21

21 2121

S S

k S S k Z

* S vn cc i l, s vn cc tiu chn.* ng trung trc ca S 1S 2 l vn cc i.

S vn giao thoa cc i gia hai ngunS 1S 2:

21

21 2121

S S

k S S k Z

S vn giao thoa cc tiu gia hai ngunS 1S 2:

2121 S S k

S S k Z

* S vn cc i chn, s vn cc tiu l.* ng trung trc ca S 1S 2 l vn cc tiu.

c. Hai ngun dao ng vung pha :2

hoc2

)12( k

22

22 12

d d d

o

)41(

2)

212(2 k k d k

Dao ng ti im xt c bin cc io

)

41(

2)

212()12( k k d k

Dao ng ti im xt c bin cc tiuS vn giao thoa cc i gia hai ngun S 1S 2

4

1

4

1 2121

S S k

S S k Z

S vn giao thoa cc tiu gia hai ngun S 1S 2

41

41 2121

S S k

S S k Z

*S vn dao ng cc i , cc tiu khng tnh hai ngun21S S Phng trnh sng ti M do hai sng t hai ngun truyn ti:

)2.2cos( 11

1 d

ft Au M v )2.2cos( 222 d

ft Au M

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

2.2cos

2cos2 212112

d d ft d d Au M

Bin dao ng ti M: )2

cos(2 12

d d A A M Vi 21

Ch :Tm s im dao ng cc i , cc tiu :

Cch 1 :o S cc i:

22l k l )( Z k

o S cc tiu:

221

221 l k l )( Z k


32/91



Cch 2 :o Ta ly pmS S ,21

(m nguyn dng , p phn l sau du phy)

Hai ngu n cng pha Hai ngu n ngc phaS cc i lun l : 2m+1S cc tiu l :

Trng hp 1 : nu p


33/91



)2sin(2)2

2cos(2

d Ad A A M )2cos(2

d A A M

i m bng i m nt- Ti M l bng sng khi sng ti v sng phn

x ti dao ng cng pha- Bin : ( M A )max= 2A- V tr ca cc im bng so vi gc ta B :

4)12( k xb (k = 0,1,2,3,.)

- Ti M l bng sng khi sng ti v sng phnx ti dao ng ngc pha

- Bin : ( M A )min= 0- V tr ca cc im nt so vi gc ta B :

2 k xb (k = 1,2,3,.)

* iu kin c sng dnga. Hai u dy c nh(hai u l nt sng)

b. Mt u c nh, mt u t do(u t do l bng sng)

iu kin v chiu di ca dy :Chiu di si dy:

42

2 k k l

k = 1,2,3: s b sngS im bng: N bng= k S im nt: Nnt = k+1

iu kin v chiu di ca dy :Chiu di si dy:

4)12(

2)

21( k k l

k = 0,1,2,3: s b sng- N bng= Nnt = k + 1

iu kin v tn s c sng dng :

l vk v f 2

vi k = 1, 2, 3

+ Tn s nh nht ( c bn) ng vi k = 1:

=2

, f 1 =v2


l

vk v f 2

)21(

vi k = 0, 1, 2, 3


=4

, f 1 =v4

c. Hai u dy l t do(hai u l bng sng)

iu kin v chiu di ca dy :

Chiu di si dy: 422

k k l k = 1,2,3: s b sng

S im bng: N bng= k S im nt: Nnt = k - 1


l vk v f 2

vi k = 1, 2, 3


=2

, f 1 =v2

Ch :- Sng dng c phng trnh: t kx Au coscos2 (hoc t kx Au sincos2 hoc

t kx Au sinsin2 hoc t kx Au cossin2 ) th vn tc truyn sng bng:k

v

- Trong khi sng ti v sng phn x vn truyn theo hai chiu khc nhau , nhng sng tng dng ti ch , n khng truyn i trong khng gian . Gi l sng dng - Nu dy l kim loi (st) c kch thch bi nam chm in (nam chm c nui bi d xoay chiu c tn s f dd ) th tn s dao ng ca dy l : f = 2f dd - Nu dy dn cng thng mang dng in xoay chiu (tn s f) t trong t trng khng B ( dy) th tn s rung ca dy dn bng tn s ca dng in : f = f


34/91



VN 3 .MT S DNG TON GIAO THOA SNG C 1. Xc nh bin , lch pha ca giao thoa sng tng hp :

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

)2.2cos( 11

1 d

ft Au M v )2.2cos( 222 d

ft Au M

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

2.2cos

2cos2 212112

d d ft d d Au M

Bin dao ng ti M: )2cos(212

d d

A A M Vi 21

lch pha hai dao ng tai M :

122d d Vi 21

TH1 : Hai ngun A , B dao ng cng pha (hai ngun ng b) : 021 hoc 2k

Bin dao ng tng hp ti M : )cos(2)cos(2 12

d Ad d A A M


122d d

o Dao ng ti im xt c bin cc i : A A M 2 Hai sng thnh phn ti M cng pha

k d d k d d k 1212 222

S vn giao thoa cc i gia hai ngun S1S2 :

2121 S S k S S k Z

H qu :

- Dao ng ti im xt c bin :6

3 12

d d A A M


2 12

d d A A M


12

d d A A M

S vn giao thoa trong ba trng hp ny bng 2 ln s vn giao thoa cc io Dao ng ti im xt c bin cc tiu :0 M A Hai sng thnh phn ti M ngc pha

)21

(2

)12()12(2)12( 1212 k k d d k

d d k

S vn giao thoa cc tiu gia hai ngun S1S2 :21

21 2121

S S k

S S k Z

Ch :o S vn cc i l, s vn cc tiu chn.o ng trung trc ca S1S2 l vn cc i.

- mt thi i m nh t nh mi i m trn dy dao ng cng pha vi nhau- Khong cch gia hai im bng k nhau hoc hai im nt k nhau bng

2 .

- Khong cch gia mt im nt v mt im bng k nhau bng 4

.

- Khong thi gian ngn nht gia hai ln si dy dui thng l2T

- B rng mt bng sng l L = 4A


35/91



o S vn dao ng cc i , cc tiu khng tnh hai ngun21S S o Nu im O l trung im ca on AB th ti O hoc cc im nm trn ng trung tr

on AB s dao ng vi bin cc i v bngA A M 2 (v lc ny 21 d d )TH2 : Hai ngun A , B dao ng ngc pha : 21 hoc )12( k

Bin dao ng tng hp ti M : )2

cos(2)2

cos(2 12

d Ad d A A M


122 d d


)21

(2

)12(222 1212 k k d d k

d d k

S vn giao thoa cc i gia hai ngun S1S2 :21

21 2121

S S

k S S k Z

H qu :


3 12

d d A A M

- Dao ng ti im xt c bin : 42 12 d d A A M


12

d d A A M

S vn giao thoa trong ba trng hp ny bng 2 ln s vn giao thoa cc io Dao ng ti im xt c bin cc tiu :0 M A Hai sng thnh phn ti M ngc pha

)12(2)12( 12 k d d

k

)1(2

)22(12

12

k k d d

k d d

S vn giao thoa cc tiu gia hai ngun S1S2 :

2121 S S k S S k Z

Ch :o S vn cc i chn, s vn cc tiu l.o ng trung trc ca S1S2 l vn cc tiu.o S vn dao ng cc i , cc tiu khng tnh hai ngun21S S o Nu im O l trung im ca on AB th ti O hoc cc im nm trn ng trung tr

on AB s dao ng vi bin cc tiu v bng0 M A (v lc ny 21 d d )

TH3 : Hai ngun A , B dao ng vung pha :221

hoc2

)12( k

Bin dao ng tng hp ti M : )4

cos(2)4

cos(2 12

d Ad d A A M

lch pha hai dao ng tai M :2

2 12

d d


)

41

(2

)21

2(22

22 1212 k k d d k

d d k

S vn giao thoa cc i gia hai ngun S1S2 :41

41 2121

S S k

S S k Z

H qu :


36/91




3 12

d d A A M

- Dao ng ti im xt c bin : 02 12 d d A A M - Dao ng ti im xt c bin :

1212

d d A A M

S vn giao thoa trong ba trng hp ny bng 2 ln s vn giao thoa cc io

Dao ng ti im xt c bin cc tiu :0 M A Hai sng thnh phn ti M ngc pha

)12(

22)12( 12 k

d d k

)43(

2)

232(

)41(

2)

212(

12

12

k k d d

k k d d

S vn giao thoa cc tiu gia hai ngun S1S2 :41

41 2121

S S k

S S k Z

Ch :o S vn dao ng cc i , cc tiu khng tnh hai ngun21S S o Nu im O l trung im ca on AB th ti O hoc cc im nm trn ng trung tr

on AB s dao ng vi bin v bng 2 A A M

(v lc ny 21 d d )2. Xc nh s im cc i , cc tiu trn on thng AB

TH1 : Hai ngun A , B dao ng cng pha (hai ngun ng b) : 021 hoc 2k o S im dao ng vi bin cc i trn on AB :

Hiu khong cch gia chng phi l : k d d 12 (1)Mc khc tng khong cch gia chng l : ABd d 12 (2)

Ly (1) + (2) v theo v ta c : 222 ABk d

Do M thuc on AB nn : ABd 20 Thay vo ta c :

ABk AB AB ABk 22

0

o S im dao ng vi bin cc tiu trn on AB :

Hiu khong cch gia chng phi l : )21(12 k d d (3)

Mc khc tng khong cch gia chng l : ABd d 12 (4)

Lm tng t nh trn ta c :21

21

ABk AB

TH2 : Hai ngun A , B dao ng ngc pha : 21 hoc )12( k o S im dao ng vi bin cc i trn on AB :


Mc khc tng khong cch gia chng l : ABd d 12 (2)

Lm tng t nh trn ta c :21

21

ABk AB

o S im dao ng vi bin cc tiu trn on AB :Hiu khong cch gia chng phi l : k d d 12 (3)Mc khc tng khong cch gia chng l : ABd d 12 (4)

A BM1d 2d


37/91



Lm tng t nh trn ta c :

ABk AB


hoc2

)12( k

o S im dao ng vi bin cc i , cc tiu trn on AB :


Mc khc tng khong cch gia chng l : ABd d 12 (2)Lm tng t nh trn ta c :

41

41

ABk AB

3. Xc nh s i m cc i , cc ti u trn on thng CD to vi AB mt hnh vung hoc hch nht :

TH1 : Hai ngun A , B dao ng cng pha (hai ngun ng b) : 021 hoc 2k o S im cc i trn on CD tha mn : k d d 12

& BC AC d d BD AD 12Suy ra : BC AC k BD AD Hay

BC AC k BD AD

o S im cc tiu trn on CD tha mn : )21(12 k d d

& BC AC d d BD AD 12Suy ra : BC AC k BD AD )

21( Hay

21

21

BC AC k BD AD

TH2 : Hai ngun A , B dao ng ngc pha : 21 hoc )12( k

o S im cc i trn on CD tha mn : )21(12 k d d & BC AC d d BD AD 12

Suy ra : BC AC k BD AD )21( Hay

21

21

BC AC k BD AD

o

S im cc tiu trn on CD tha mn : k d d 12 & BC AC d d BD AD 12Suy ra : BC AC k BD AD Hay

BC AC k BD AD


hoc2

)12( k

o S im cc i , cc tiu trn on CD tha mn : )41(12 k d d &

BC AC d d BD AD 12

Suy ra : BC AC k BD AD )41( Hay

41

41

BC AC k BD AD

4. Xc nh s im cc i , cc tiu trn on thng l ng cho ca mt hnh vung hohnh ch nht :

TH1 : Hai ngun A , B dao ng cng pha (hai ngun ng b) : 021 hoc 2k o S im cc i trn on BD tha mn : k d d 12 & 012 ABd d BD AD

(v im DB nn v phi AC thnh AB cn BCBB=0)

Suy ra : ABk BD AD Hay

ABk BD AD

o S im cc tiu trn on BD tha mn : )21(12 k d d

& 012 ABd d BD AD

A B

D C

O

I

A B

D C

O

I


38/91



Suy ra : ABk BD AD )21( Hay

21

21

ABk BD AD


o S im cc i trn on BD tha mn : )21(12 k d d & 012 ABd d BD AD


21

21

ABk BD AD

o S im cc tiu trn on BD tha mn : k d d 12 & 012 ABd d BD AD

Suy ra : ABk BD AD Hay

ABk BD AD


hoc2

)12( k

o S im cc i , cc tiu trn on BD tha mn : )41(12 k d d &

012 ABd d BD AD


41

41

ABk BD AD

5. Xc nh s im cc i , cc tiu trn on thng l ng trung trc ca AB cch AB on x :

TH1 : Hai ngun A , B dao ng cng pha (hai ngun ng b) : 021 hoc 2k o S im cc i trn ng trung trc tha mn :

Do 21 d d Nn lch gia M,A hoc B :

2

.2.2 12 k d d

Hay k d d 21 M AC d AO 1 AC k AO 22)2(

2OC

ABk

AB

22)2

(1

2OC

ABk

AB

(Do2

AB AO v 22)2

( OC AB

AC )

o S im cc tiu trn ng trung trc tha mn : lch gia M,A hoc B :

)12(.2.2 12 k d d Hay )

21(21 k d d

M AC d AO 1 AC k AO )21( 22)

2()

21

(2

OC AB

k AB

21

)2

(1

21

222 OC

ABk

AB

TH2 : Hai ngun A , B dao ng ngc pha : 21 hoc )12( k o S im cc i trn ng trung trc tha mn :

lch gia M,A hoc B :

2.2 1

k d

Hay )21

(21 k d d

M AC d AO 1 AC k AO )21( 22)

2()

21

(2

OC AB

k AB

21

)2

(1

21

222 OC

ABk

AB

o S im cc tiu trn ng trung trc tha mn :

lch gia M,A hoc B :

)12(

.2 1 k d Hay k d d 21

C

A BO

M1d


39/91



M AC d AO 1 AC k AO 22)2(

2OC

ABk

AB 22)

2(

12

OC AB

k AB


hoc2

)12( k

o S im cc i , cc tiu trn ng trung trc tha mn :

lch gia M,A hoc B :

2

2

.2 1 k d hoc )12( k Hay )

4

1(21 k d d

M AC d AO 1 AC k AO )41( 22)

2()

41

(2

OC AB

k AB

41

)2

(1

41

222 OC

ABk

AB

6. Xc nh s im cc i , cc tiu trn ng trn tm O l trung im ca AB :TH1 : Hai ngun A , B dao ng cng pha (hai ngun ng b) : 021 hoc 2k

o S im cc i trn ng trn tm O tha mn :

ABk AB

KL : Trn on AB c k im dao ng vi bin cc ith trn ng trn tm O c 2k im dao ng vi bin cc io S im cc tiu trn ng trn tm O tha mn :

21

21

ABk AB

KL : Trn on AB c k im dao ng vi bin cc tiuth trn ng trn tm O c 2k im dao ng vi bin cc tiu


o S im cc i trn ng trn tm O tha mn :21

21

ABk AB

KL : Trn on AB c k im dao ng vi bin cc i th trn ng trn tm O c 2k idao ng vi bin cc i

o S im cc tiu trn ng trn tm O tha mn : AB

k AB

KL : Trn on AB c k im dao ng vi bin cc tiu th trn ng trn tm O c 2k dao ng vi bin cc tiu


hoc2

)12( k

o S im cc i , cc tiu trn ng trn tm O tha mn :41

41

ABk AB

KL : Trn on AB c k im dao ng vi bin cc i th trn ng trn tm O c 2k idao ng vi bin cc i

VN 4.GII TON SNG CNHMY TNH FX-570ES1. Phng php s dngTABLE (MODE 7) gii bi ton sng c :

Bm:SHIFT 9 3 = = Reset allBm:SHIFT MODE 2 Line IOBm:MODE 7 : TABLE

V dTa c hm s f(x)=212 x

Bc 1: (MODE 7) TABLE

A BO

D

f(x)=


40/91



Bc 2: Nhp hm s vo my tnh

Bc 3: bm = nhp 1

Bc 4: bm = nhp 5

Bc 5: bm = nhp 1

Bc 6: bm =Ta c bng bin thin: f(X)

2. S dng my tnh cm tay gii cc bi tp sng c V d 1: Si dy di l = 1m c treo l lng ln mt cn rung. Cn rung theo phng ngang vithay i t 100Hz n 120Hz. Tc truyn sng trn dy l 8m/s. Trong qu trnh thay i tn th s ln quan st c sng dng trn dy l:

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

- l = (2k+1)4

= (2k+1) f v

4

f=(2k+1)l

v4

=(2k+1)2

Do 100Hz f 120Hz . Cho k=0,1,2..k=24 f =98Hzk=25 f =102Hzk=26 f =106Hzk=27 f =110Hzk=28 f =114Hzk=29 f =118Hzk=30 f =122Hz chn A

SHIFT MODE 2 : Line IOMODE 7 : TABLE.

148)( x

tuso f x f = tuso x 2 =(2X +1)x2

Vi tuso = (2 x X + 1).Nhp my:( 2 x ALPHA ) X + 1 ) x 2

= START 20 = END 30 = STEP 1 =K t qu x=k f(x)=f

24252627282930

98102106110114118122

V d 2:Cu 50 - thi tuyn sinh i hc khi A nm 2011 - M 817 Mt sng hnh sin truyn theo phng Ox t ngun O vi tn s 20 Hz, c tc truyn sng nkhong t 0,7 m/s n 1 m/s. Gi A v B l hai im nm trn Ox, cng mt pha so vi O v cnhau 10 cm. Hai phn t mi trng ti A v B lun dao ng ngc pha vi nhau. Tc truyl :

A. 100 cm/s B. 80 cm/s C. 85 cm/s D. 90 cm/sCch gii Hng dn bm my v kt qu

D

f(x)=x2+1 2

D

Start?1

D

End? 5

D

Step?1

Dx f(x)

123

123

1.54.59.5 1


41/91



- d = (2k+1)2

=(2k+1) f v

2

Do 0,7 m/s v 1 m/s.12

2k df v

Cho k=0,1,2.. v = 80 cm/schn B. vi k=2

Mode 7

mauso x xv x f 20102)( ; Mauso=2x ALPHA ) +1

Nhp my:...tng t nh trn....(400 : ( ALPHA ) X + 1 )

= START 0 = END 10 = STEP 1 =

Kt qu:

x=k f(x)=v

0123

400133.338057.142

Ch : Cch chn Start? End? V Step?-Chn Start?: Thng thng l bt u t 0 hoc ty theo bi-Chn End? : Ty thuc vo bi ton m cho nhng khng qu 30 (ngh thut ca tng ng bi)-Chn Step : 1(v k nguyn)

VN 4.SNG M(Sng m l nhng sng c lan truyn trong mi trng rn, lng, kh.)

1. Cc c trng ca m

a. cao

Ph thuc vo tn s ca m. m cng cao th tn s cng ln.

b. m sc Ph thuc vo dng th dao ng ca m.c. to Cm gic m nghe to hay nh, ph thuc vo cng m v tn s m.

d.Vn tc

Vn tc truyn m ph thuc vo tnh n hi, mt v nhit ca mi trBiu thc vn tc trong khng kh ph thuc nhit : t vv 10

v0 l vn tc truyn m C 00 ; v l vn tc truyn m t 0C;1

273 K -1

e. Cng m Nng lng sng m truy n qua mt n v din tch t vung gc vi ph

truyn sng trong mt n v thi gian . 2.4. r P

S P

t S W I

f. Mc cng m1

212

100

00

0lg10;10lg10)(;10lg)(

I I L L L I I

I I dB L I I

I I B L

L L

Ch :- I : Cng m (W/m2 ).- 120 10I W/m

2: Cng m chun (cng m nh nht m tai ngi c th nghe cvi L= 0dB)

- Cng m cc i m tai ngi nghe c: I max=10W/m2

(ngng au,ng vi L=130dB)- Ngng nghe l mc cng m nh nht gy c cm gic m cho tai ngi, thaytheo tn s ca m.- Gii hn nghe ca tai ngi: t ngng nghe n ngng au.- Khi cng m tng 10n ln th cm gic v to tng n ln (L tng 10n dB).

- Sng m trong khng kh c dng hnh cu: 2.4 r P

S P I

(P: Cng sut ca ngun pht m)

- Tc truyn m ph thuc vo tnh n hi v khi lng ring ca mi trng (mt trng) : vcht rn> vcht lng > vcht kh

2.Cc bi ton v cng sut ca ngun m

f(Hz)16 20 000

Tai ngicm nhn c Siu mH m


42/91



- Cng sut ca ngun m ng hng: I r IS P 2.4 (S l din tch ca mt cu c bn knhr bng khong cch gia tm ngun m n v tr ta ang x

I l cng m ti im ta xt)+ Nu m truyn i theo hnh nn c gc nh l th:

2cos1.2.2 2 I r rhI IS P ; h l cao ca chm cu

+ Din tch ca chm cu c gc nh l bng:

2cos1.2..2 2 r hr S

- B A I I , l cng m ca cc im A, B cch ngun m nhng khong r A, r B th: 22

A

B

B

A

r r

I I

- Mi lin h gia cng m v bin ca sng m:22

21

2

1

A A

I I

- Khi cng m tng (gim) k ln th mc cng m tng (gim)k N lg (B) v k N lg10 (dB)+ Trng hp nk 10 n N (B) hoc n N 10 (dB)

- Khi mc cng m tng hay gim N (B) th cng m tng hay gim N 10 ln.

- Ti mt im cch ngun m 1 khong x, mc cng m l L(B). Ngng nghe ca tai ng B L0 , th khong cch ti a m ngi ny cn cm gic c m thanh l: )(max 010 L L x x3. Ngun nhc ma. Dy n c hai u c nh

*Tc truyn sng:

v

: Lc cng (N);l

m0 : Mt di (khi lng trn mt n v chiu di kg/m)

*Khi xy ra sng dng:

l vk v f 2

k=1: 1 2v

f l

: Ha m c bn (ha m bc 1)

k=2: f 2=2f 1: Ha m bc 2b.ng so mt u kn, mt u h

*Khi xy ra sng dng :l

vk f .4

)12( Ch c th pht ra nhng ha m bc l.

ng vi k = 0 m pht ra m c bn c tn s1 4v f l

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

c. Hp cng hng-Hp rng c mt u h, c tc dng khuch i m.-Hp n c tc dng va khuch i m, va to m sc ring cho mi lonhc c.

4. Hiu ng p-ple

Ngun m ng yn, my thu chuyn ng vi vn tc vM.My thu chuyn ng li gn ngun m th thu c m c tn s:' M v v f f

v

My thu chuyn ng ra xa ngun m th thu c m c tn s:" M v v f f v

Ngun m chuyn ng vi vn tc vS , my thu ng yn.

My thu chuyn ng li gn ngun m vi vn tc vM th thu c m c tn s:'S

v f f v v


43/91



My thu chuyn ng ra xa ngun m th thu c m c tn s:"S

v f f v v

Ch : C th dng cng thc tng qut: ' M S

v v f f v v

Vi v : L vn tc truyn m, f l tn s ca m.vM: Tc ca my thu i vi mi trng.vS : Tc ca ngun pht i vi mi trng.

o My thu chuyn ng li gn ngun th ly du + trc vM, ra xa th ly du -o Ngun pht chuyn ng li gn ngun th ly du - trc vS, ra xa th ly du +

CH IV: DAO NG V SNG IN T

VN 1.MCH DAO NG LC (MCH DAO NG IN T)i lng vt l K hiu (n v) Cng thc Ghi ch

1. in tch trnhai bn t in q (C)

0

0

cos( )

sin2

q q t

q t

0 0q CU =

2. Dng introng mch

i (A) 00

'( ) q sin t+

os t+ +2

i q t

I c

- i sm pha hn q gc2

- LC

qq I 000

3. in p trnhai bn t

u (V) 0 os t +'( ) "

qu U c

c Li t Lq

- u tr pha hn i gc2

- u cng pha vi q ; 00qU C

=

4. Cm ng t B(T) 0 cos( )2 B B t - B cng pha vi i

-7

0 04 .10 B nI ; N nl

=

5. Tn s gc (rad/s)1LC

- t cm L (H)- in dung C (F)

6. Nng lng in ta. Nng lngin trng (tptrung t in)

WC (J)2

2 20C

1W os ( )

2 2q

Cu c t C

Bin thin tun hon vi chuk T=

2T ; tn s gc =2 ;

tn s f=2f b. Nng lng ttrng (tp trung cun cm)

WL(J)2

2 20L

1W sin ( )

2 2q

Li t C

c. Nng lngin t ton phn W (J)W = W + Wt =

2

2

oq

C = const

= C(max) L(max)W W =2 20 0q LI

2C 2

Trong qu trnh dao ng camch, nng lng t trng v nng lng in trng lun chuyn ha cho nhau

Cc nh ngha :

7.Dao ng int tt dn

- V trong mch dao ng lun c in tr R nng lng dao ng gim dn bin q0 U0 I0 B0 gim dn theo thi gian gi l dao ng in t tt dn- c im: Nu in tr R cng ln th dao ng in t tt dn cng nhanh vngc li

8. Dao ng in- Mu n duy tr dao ng ta phi b v ng ph n nng lng b tiu hao tromi chu k


44/91



t duy tr . H tdao ng

- lm vic ny ngi ta dng tranzito i u khi n vic b nng lng cho phhp- Mch dao ng iu ha c s dng tranzito to thnh h t dao ng

9. Dao ng in

t cng bc . Scng hng

a) Dao ng in t cng bc: M c mch dao ng LC vi t n s gc ring 0 n itip vi mt ngun in ngoi l ngun in xoay chiu c in p u = U0cost . lcny dng in trong mch LC bin thin theo tn s gc ca ngun in xoaychiu ch khng th dao ng theo tn s gc ring 0 qu trnh ny gi l dao

ng in t cng bc b)S cng hng :- Gi nguyn bin ca u , iu chnh khi = 0 th bin dao ng in I0trong khung t cc i hin tng ny gi l s cng hng- Gi tr cc i ca bin cng hng ph thuc vo in tr thun R

+ Nu R nh (I0)max cng hng nhn+ Nu R ln (I0)min cng hng t

Ch :- 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 2C U U RC P I 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 in tch dng th i>0ng vi dng in chy n b

xt

- Mi lin h gia cc gi tr u, i, U 0 v I 0 :2 2 2

0

2 2 20

Lu + i = U

CC

u + i = IL

- in dung ca t in phng : 9.SC

4 .9.10 .d

- Khong thi gian ngn nht gia hai ln m W L = W C l t min= 42

' T T

10. S tng t gia dao ng in v dao ng c i lng c i lng in Dao ng c Dao ng inx q x + 2x = 0 q + 2q = 0

v i k m

1

LC

m L x = Acos( t + ) q = q0cos( t + )k 1

C v = x

= - Asin( t + )i = q

= - q0sin( t + )

F u 2 2 2( )v A x

2 2 20 ( )

iq q

R F = -kx = -m

2

x2qu L q

C

W Wt (WL) W =12

mv2 Wt =12

Li2

Wt W (WC) Wt =12

kx2 W =2

2qC

VN 2.SNG IN T 1.nh ngha:Sng in t l qu trnh lan truyn ca in t trng trong khng gian.2.c im:

- Tc lan truyn trong khng gian v = c = 3.108m/s


45/91



- Sng in t l sng ngang:, E B

phng truyn sng (E, B u bin thin tun hon vlun cng pha vi nhau)- Sng in t truyn trong mi mi trng, k c chn khng (khc bit vi sng c

- Trong chn khng: Bc sng ca sng in t: ccT f

- Trong qu trnh lan truyn c mang theo nng lng

- Tun theo cc quy lut truyn thng, phn x, khc x, giao thoa, nhiu x3. Ngun pht:- Chn t (thng bng kim loi, bn trong c dng in bin thin)

Bt c vt th no to ra in trng hoc t trng bin thin: tia la in, dy dn in xoaydao ng ngt mch in

VN 3 :MT S DNG TON1. Xc nh in p cc i, cng dng in cc i

00 0

q I q LC

, 0 0C

I UL

(1)

0 0

0 0

q I LU I C C C

(2)

2. Tnh in p tc thi, cng dng in tc thi

2 20Lu I iC (3)

2 20Ci U uL = 2 20

1Q q

LC= 2 20Q q (4)

3. Mch LC c C thay i :C1 nt C2 v C1 // C2- Mch LC1 c tn s f 1, chu k T1. Mch LC2 c tn s f 2, chu k T2.- Mch L v C1 ni tip C2 c tn s f , chu k T.

1 2

1 1 1

C C C , 2 21 2f f f , 2 2 2

1 2

1 1 1

T T T 1 2

2 21 2

TT T T T

(5)

- Mch L v C1 song song C2 c tn s f , chu k T.

C=C1+C2 , 2 2 21 2

1 1 1f f f

, 2 2 21 2T T T 1 22 21 2

f f f f f

(6)

VN 4 :TRUYN THNG BNG SNG IN T Vic pht v thu sng in t Vic pht v thu sng in t

A. Pht sngPht sng::- Dao ng in t trong my pht dao ng s cm ng qua anten (mch dao ng h) ri ra khng gian.- Tn s cng cao th nng lng sng cng ln v sng lan truyn cng xa.

B. Thu sng:- iu chnh sao cho f

0= f th trong mch chn sng s c cng hng, sng

cn thu s c bin cc i.My pht hoc my thu sng in t s dng mch dao ng LC th tn ssng in t pht hoc thu c bng tn s ring ca mch.

Bc sng ca sng in t: 83.10 .2 ( )c LC m f

f

f 0 = f

L A L C


46/91



My pht My thu

(1): Micr.(2): Mch pht sng in t cao tn.(3): Mch bin iu.(4): Mch khuych i(5): Anten pht.

(1): Anten thu.(2): Mch khuych i dao ng in t cao tn.(3): Mch tch sng.(4): Mch khuych i dao ng in t m tn.(5): Loa.

Ch :o Mch dao ng c L bin i t Lmin Lmaxv C bin i t Cmin Cmax th bc sng casng in t pht (hoc thu)

+ min tng ng vi Lmin v Cmin+ max tngng vi Lmax v Cmax

o Gc quay ca t xoay:+ Khi t quay tmin n ( in dung t Cmin n C) th gc xoay ca t l:

minmin max min

max minC C .( )

C C

+ Khi t quay t v trmax v v tr ( in dung t C n Cmax) th gc xoay ca t l:max

max max minmax min

C C .( )C C

Tn sng Bc sng ng dng Tnh chtSng di > 3000m Thng tin di nc B tng in li phn x

vi mc khc nhauSng trung 3000 m 200 m Thng tin, truyn thanh,truyn hnh trn mt t

Sng ngn 1 200 m 50 mSng ngn 2 50 m 10 m

Sng cc ngn 10 m 0.01 m Truyn thng qua v tinhi xuyn qua tng in li

CH V: DNG IN XOAY CHIU

VN 1.DNG IN XOAY CHIUi lng vt l K hiu (n v) Cng thc Ghi ch1. Cng dng in

i (A) )cos(0 it I i - Mi giy i chiu 2f ln.- Nu pha ban u i= 2

2. in p u (V) )cos(0 ut U u

Tchsng

Chnsng

Khuchi m

tn

L o a

ngni Bin iu

Dao ngcao tn

Khuchi cao

tn

2

1

3 4 5 1 2 3 45


47/91



3.Cm ng t B(T) )cos(0 Bt B B hoc i =2 th ch giy u

tin i chiu (2f - 1) ln.- Trong 1 chu k dng in ichiu 2 ln- i ,u , B , , e : gi tr tc thi- I 0 ,U 0 ,B0 ,0 ,E 0 :gi tr cc i

4.T thng (Wb) t t NBS

coscos

0

5. Sut inng cm ng

e(V))cos(

sin

0 et E e

t NBS dt d e

6. lch phagiau v i u-i (rad) u i u i ( 2 2u i

)

+ u-i>0: u sm pha hn i+ u-i>0: u tr pha hn i+ u-i=0:u vi i cng pha

7. Cc gi tr hiu dngCng dng in in p Sut in ng

0

2

I I = 0

2U U = 0

2

E E =

Ch :- Cc thit b o lng in (ampe k, vn k..) ch gi tr hiu dng ca i lng cn o.cng dng in, mc ampe k ni tip vi mch in; mc vn k song song vi mch in p gia hai u on mch.- Dng in xoay chiu c gi tr thay i theo thi gian- Dng in xoay chiu c chiu thay i theo thi gian

VN 2.NH LUT M CHO CC ON MCH IN1. on mch RLC khng phn nhnh (mc ni tip)

Xt on mch gm c in tr thun R,cun cm thun L (c r=0) v t in C

0 cos R L C i i i i I t 0 cos R L C u iu u u u U t

0 0 0 0 R L C U U U U

i lng vt l Cng thc Ghi ch

in p cc i 22

0 0 0 0 R L C U U U U - in p hiu dng

2202 R L C

U U U U U

Tng tr 22 L C Z R Z Z - n v - Cm khng: 2 L Z L fL ( )

- Dung khng: 1 12C

Z C fC

( )

nh lut m U I Z

00

U I Z

tan u-i0 0

0

tan L C u i

L C L C

R R

Z Z R

U U U U U U

+ ZL>ZC: >0: u sm pha hn i (on mch c tnhcm khng)+ ZL


48/91



o

LC LC Z Z Z Z C L

112min

o u, i cng pha ( = u i = 0 u= i)o Zmin= R L C U U ; RU U

o Vi mt gi tr U xc nh : maxUI = I =R

RU R

o Cng sut cc i : Pmax= UI =2

UR o H s cng sut : 1cos

Z R

o in tr ca on mch nh, im cc i cng hng cao hn (cng hng nhn)

sin L C L C u i Z Z U U

Z U

Nu on mch cho cun dy c in tr R 0 th xem in tr ca ton mch l (R+ R 0) mc nitip vi cun thun cm L v t in C.

Nu trong on mch khng c phn t no th cho cc i lng lin quan n phn t ) bng 0 thay vo cc cng thc trn khi tnh ton.

2. on mch ch c in tr thun R S l R ; U I

Rv 00

U I R

uR , i cng pha ( = u i = 0 u= i)

Lu :in tr R cho dng in khng i i qua v cU I R

Nhit lng ta ra trn in tr R trong thi gian t:2

2 U Q RI t UIt t R

(J)

3. on mch ch c cun thun cm L

V l

N L .)(10.4 27 ; L

U I Z

; 00 L

U I Z

v2 2 2 2

2 2 2 20 0L L

i u i u1 1I U 2I 2U

Cm khng: ZL = L = 2 fLu L nhanh pha hni gc /2 ( u-i = u i = /2)

Lu :Cun thun cm khng cn tr dng in 1 chiu nhng c tc dng

tóm tắt lý thuyết vật lý lớp 12

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