bản chính bộ công thức vật lý 12
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Chng I: DAO NG CHC
I - ICNGV DAO NGIU HO
T: chu k; f: tn s; x: li ; v: vn tc; a: gia tc; g: gia tc trngtrng; A: bin dao ng; ( t + ): pha dao ng; : pha ban u; : tc gc;1. Phng trnh dao ng
tAcosx
- Chu k:2
T (s) - Tn s:2
1
Tf (Hz)
- Nu vt thc hin c N dao ng trong thi gian t th:
t N
T v fN t
.
2. Phng trnh vn tctAxv sin'
- x = 0 (VTCB) th vn tc cc i: Avmax
- x A (bin) th 0v 3. Phng trnh gia tc
2 2' cosa v A t x
- x = A th 2maxa A
- x = 0 th 0a
Ghi ch: Lin h v pha: v sm pha2
hn x;
a sm pha 2 hn v;a ngc pha vi x.
4. H thc c lp thi gian gia x, v v a
- Gia x v v:2
222 v
xA
- Gia v v a:
222 2
2max
a
v A v
- Gia a v x: 2a x
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5. Cc lin h khc
- Tc gc:max
max
v
a
- Tnh bin
2
222
2
22
max
2max
2
maxmax 242
avvxk
W
a
vav
n
SLA
6. Tm pha ban u
2
A 22
A
3
2A
3
A2
A O A
2
A 22
A 32
A
v < 0= + /2
v < 0= + /4
v < 0= + /6
v = 0
= 0
v < 0= + /3
v > 0= - /6
v < 0= + 2 /3
v > 0= - /2 v > 0= - /3
v > 0= - /4
v < 0= + 3 /4
v < 0= + 5 /6
v > 0
= -5 /6
v > 0= - 3 /4
v > 0= - 2 /3
v = 0=
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6. Thi gian ngn nht vt i t:+ x1n x2 (gi s 21 xx ):
12t vi
Ax
A
x
2
2
1
1
cos
cos
21,0 .
+ x1n x2 (gi s 1 2x x ):
12t vi
A
x
A
x
2
2
1
1
cos
cos
1 2, 0
7. Vn tc trung bnh - tc trung bnh
- Tc trung bnh vS
t
- di x trong n chu k bng 0;qung ng vt i c trong n chu k bng nAS 4 .
- Vn tc trung bnhx
v t .
8. Tnh qung ng vt i c trong thi gian t
cos-A A
2
0 A2
A 2
2 A 3
2
+A
T/4 T/12 T/6T/8 T/8
T/6 T/12* Phng php chung tm qung ng itrong khong thi gian no
ta cn xc nh:- V tr vt lc t = 0 v chiu chuyn ng ca vt lc ;
- Chia thi gian t thnh cc khong nh: nT; nT/2; nT/4; nT/8; nT/6;T/12 vi n l s nguyn;
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- Tm qung ng s1; s2; s3; tng ng vi cc qung thi gian nutrn v cngliTnh qung ng ngn nht v b nhtvt i c trong khong thi
gian t vi2
0T
t
Nguyn tc:+ Vt i c qung ng -A - x0 O x0 +A
di nhtkhi li im u v imcuic gi tr i nhau smax
Qung ngdi nht: max 2 sin2
tS A
+ Vt i c qung ng -A - x0 O x0 +Angnnhtkhi li im u v imcui c gi tr bng nhau
smin Smin
Qung ng ngnnht: min 2 1 cos
2
tS A
Trng hp2
Tt th ta tch t
Tnt
2 * 0
2
Tn N v t :
+ Qung ng ln nht: max 2 2 sin2
tS nA A
+ Qung ng nh nht: min 2 2 1 cos2
tS nA A
+ Tc trung bnh ln nhttrong thi gian t: maxaxtbmS
vt
+ Tc trung bnh nh nhttrong thi gian t: minmintbS
vt
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II - CON LC L XO
l : bin dng ca l xo khi vt cn bng;k: cng ca l xo (N/m);
0l : chiu di t nhin ca l xo
1. Cng thc c bn- Tn s gc:
k g
m l;
+ Con lc l xo treo thng ng:2
mg gl
k;
+ t con lc trn mt phng nghing gc khng ma st:sinmg
l k - p dng cng thc v chu k v tn s:
2. Chiu di cc i v cc tiu ca l xo+ dao ng thng ng:
AlllAlll
0max
0min 2
minmax llA
+ dao ng phng ngang:
min 0
max 0
A
l A
l l
l
3.Ghp l xo.
- Ghp ni tip:nkkkk
1...11121
- Ghp song song: nkkkk ...21
- Gi T1 v T2l chu k khi treo m vo ln lt 2 l xo k1 v k2 th:
+ Khi ghp k1ni tip k2:2
2
2
1
2
2
2
2
1
111
fff
TTT
22 2
1 1 1
2 2
m lT
k g
k gf
T m l
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+ Khi ghp k1 song song k2:2
2
2
1
2
2
2
2
1
111
TTT
fff
- Gi T1 v T2l chu k khi treo m1 v m2ln lt vo l xo k th:
+ Khi treo vt 21 mmm th:2
2
2
1 TTT
+ Khi treo vt 21 mmm th:2
2
2
1 TTT 21 mm
4. Ct l xo- Ct l xo c cng k, chiu di
0l thnh nhiu on c
chiu di nlll ...,,, 21 c cng
t ng ng nkkk ...,,, 21 lin h nhau theo h thc:
nnlklklkkl ...22110 .
-Nu ct l xo thnh n on bng nhau (cc l xo ccng cng k):
nkk' hay:
nff
n
TT
'
'
5. Lc n hi - lc hi phc
Nidung
Lc hi phucLc n hi
L xo nmngang
L xo thng ngA A
Gc ti V tr cn bng V tr l xo cha bin dng
Bn chthp dh
F P F Fh= k . ( bin dng)
nghav tcdng
- Gy ra chuyn ngca vt- Gip vt tr vVTCB
- Gip l xo phc hi hnh dng c- Cn gi l lc ko (hay lc y) ca lxo ln vt (hoc im treo)
Cc i Fmax = kAFmax = kA Fmax = k(l + A)
Cc tiu
Fmin = 0
Fmin = 0 Fmin = 0Fmin = k(l
A)V trbt k
F= k x F= k x F = k(l + x)
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III - CON LCN
1. Cng thc c bnDi y l bng so snh cc c trng chnh ca hai h dao ng.
H dao ng Con lc l xo Con lc n
Cu trcHn bi m gn vo l xo (k). Hn bi (m) treo vo u si
dy (l).
VTCB
- Con lc l xo ngang: lxo khng gin- Con lc l xo thng ng
n dnk
mgl
Dy treo thng ng
Lc tc dng
Lc n hi ca l xo:F = - kxx l li di
Trng lc ca hn bi v lccng ca dy treo:
sl
gmF s l li cung
Tn s gcm
k=
g
l
l
g
Phng trnhdao ng.
x = Acos(t + ) s = s0cos(t + )Hoc = 0cos(t + )
C nng2 2 21 1
2 2W kA m A 0
(1 cos )W mgl
20s
l
gm
2
1
- Chu k dao ng ca con lc n c chiu di l1 v l2ln lt l T1 v T2
th:+ Chu k ca con lc c chiu di 21 lll :
2
2
2
1 TTT
+ Chu k ca con lc c chiu di 21 lll :2
2
2
1 TTT 21 ll .
- Lin h gia li di v li gc: s l - H thc c lpthi gianca con lc n:
a = - 2s = - 2
l; 2 2 2
0
( )v
S s 2
2 2
0
v
gl
2. Lc hi phc
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2sins
F mg mg mg m sl
3. Vn tc - lc cng+ Khi con lc v tr li gc vn tc v lc cng tng ng ca vt:
0
0
2 cos cos
3cos 2cosc
v gl
T mgKhi 0 nh:
2 20
2 2
0
31
2c
v gl
T mg
+ Khi vt bin:0
0
cosc
v
T mg; khi 0 nh:
2
0
0
12
c
v
T mg
+ Khi vt qua VTCB: 0
0
2 1 cos
3 2cosc
v gl
T mg; khi 0 nh:
0
2
01c
v gl
T mg
4. Bin thin chu k ca con lc n ph thuc: nhit , su v cao. Thi gian nhanh chm ca ng h vn hnh bng con lc n
a.Cng thc c bn* Gi chu k ban u ca con lc l 0T (chu k chy ng), Chu k sau
khi thay i l T(chu k chy sai).
0TTT : bin thin chu k.+ 0T ng h chy chm li;+ 0T ng h chy nhanh ln.
* Thi gian nhanh chm trong thi gian N (1 ngy m24 86400N h s ) s bng:
0
TNT N
T T
b. Cc trng hp thng gp
Khi nhit thay it1
t n 2t :0
1
2
1
2
Tt
T
N t
( 2 1t t t )
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Khi a con lc t cao1
h n cao2
h : 0
T h
T R
hN
R
( 2 1h h h )
Khi em vt ln cao 0h , khi em vt xung cao thp hn0h . Ban u vt mt t th 01h v hh
Khi a con lc t su1
h n su2
h : 02
2
T h
T R
N h
R
( 2 1h h h )
Khi em vt xung su 012 hhh , khi em vt ln cao hn banu 0h . Ban u vt mt t th 01h v hh
c. Cc trng hp c bit- Khi a con lc mt t (nhit
1t ) ln cao h (nhit
2t ):
0
1
2
T ht
T R
Nu ng h vn chy ng so vi di mt t th:
0
10
2
T ht
T R
- Khi a con lc t tri t ln mt trng (coi chiu di l khng i) th:
T
MT
MT
T
MT
T
M
M
R
R
T
T
5. Con lc n chu tc dng ca lc ph khng i
* Lc ph f
gp trong nhiu bi ton l:
+Lc qun tnh amFq
, ln: maFq , (a l gia tc ca h quychiu)
+Lc in trng F qE , ln: EqF ,
q l in tch ca vt, E l cng in trng ni t con lc ( /V m )
+Lc y Acsimet gV
AF , ln: VgFA .
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l khi lng ring ca mi trung vt dao ng, V l th tch vtchim ch
Chu k dao ng trong trng hp ny s l:g
lT 2 ,
'g l gia tc trngtrng hiu dng.* Tnh g':
+ Trng hp Pf
:m
fgg '
Lc qun tnh: agg'
Lc in trng:m
Eqgg '
+ Trng hp Pf
:m
fgg '
Lc qun tnh: agg'
Lc in trng:m
Eqgg '
Lc y Acsimt: m
Vg
gg '
+ Trng hp Pf
:2
2'
m
fgg
Lc qun tnh: 22' agg
Lc in trng:2
2'
m
qEgg
Ch :+ Trng hp Pf
th gc lch ca si dy so viphng
thng ng c tnh:P
ftan
+ Khi con lc n gn trn xe v chuyn ng trn mt phngnghing gc khng ma st th VTCB mi ca con lc l si dy lchgc (si dy vung gc vi mt phng nghing)so vi phngthng ng v chu k dao ng ca n l:
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cos2'
g
lT
V - NNGLNG DAO NG
-ng nng: tAmmvWd2222 sin
2
1
2
1
- Th nng: tAmkxWt2222 cos
2
1
2
1
- ng nng v th nng bin thin tun hon vi chu k bng 1/2 chu kdao ng iu ho (T = T/2).
- Khong thi gian gia 2 ln ng nng v th nng bng nhau lin tip lT/4.
cos-A A
2
0 A2
A 2
2 A 3
2
+A
T/4 T/12 T/6
Vi T/8 T/8
T/6 T/12
1. Con lc l xo (Chn gc th nng ti VTCB)
- ng nng: 22
1mvW
; Th nng: 2
2
1kxWt
- C nng: t WWW222
2
1
2
1AmkA
+ V tr ca vt khi t nWW :1n
Ax
W = 3 Wt WmaxWt = 0
Wt = 3 WW = WtW = 0Wtmax
W = Wtmax = Wmax = 1/2kA2
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+ Vn tc ca vt lct
nWW :11
max
n
A
n
vv
+ ng nng khi vt li x: 222
1xAkW
+ T s ng nng v th nng:2
22
x
xA
W
W
t
2. Con lc n(Chn gc th nng ti VTCB)
- ng nng: 22
1mvW
; Th nng: cos1mglWt
- C nng: t WWW 0cos-1mgl
Khi gc 0 b th:21
2tW mgl ;
20
1W mgl2
+ V tr ca vt khi
t nWW :1
0
n
SS v
1
0
n
+ Vn tc ca vt lc
t
nWW :1
max
n
vv
1
0
n
S
+ ng nng ca vt khi n li :
220222
02
1
2
1SSmmglW
+ T s ng nng v th nng:2
22
0
2
22
0
S
SS
W
W
t
VI - TNGHP DAO NG
1. Phng php gin Frexnel- Bi ton:Tng hp 2 dao ng iu ho cng phng:
1 1 1
2 2 2
cos
cos
x A t
x A t cosx A t
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Vi
2211
2211
2121
2
2
2
1
coscos
sinsintan
cos2
AA
AA
AAAAA
-Nu bit mt dao ng thnh phn 111 cos tAx v dao ngtng hp tAx cos th dao ng thnh phn cn li l
222 cos tAx c xc nh:
11
112
11
2
1
22
2
coscos
sinsintan
cos2
AA
AA
AAAAA
(vi 21 )-Nu 2 dao ng thnh phn vung pha th: 22
2
1 AAA
2. Tm dao ng tng hp xc nh A v bng cch dng mytnh thc hin php cng:
+ Vi my FX570ES: Bm chn MODE 2 mn hnh xut hin ch:CMPLX.
-Chn n v o gc l bm: SHIFT MODE3 mn hnh hin th ch D(hoc Chn n v gc l Rad bm: SHIFT MODE4 mn hnh hin th
ch R )-Nhp A1 SHIFT (-) 1, +Nhp A2 SHIFT (-) 2 nhn = hin th kt
qu.
(Nu hin th s phc dng:a+bi
th bmSHIFT 2 3 = hin th kt
qu: A )
+ Vi my FX570MS: Bm chn MODE 2 mn hnh xut hin ch:CMPLX.
Nhp A1 SHIFT (-) 1 + Nhp A2 SHIFT (-) 2 =Sau bm SHIFT + = hin th kt qu l: A. SHIFT = hin th kt
qu l:
+ Lu Ch hin th mn hnh kt qu:
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Sau khi nhp ta n du = c th hin th kt qu di dng s v t, munkt qu di dng thp phnta n SHIFT = (hoc dng phm SD ) chuyn i kt qu Hin th.
VII - DAO NGTTDN- Tm tng qung ngS m vt i c cho n khi dng li:
SFkA C2
2
1
- gim bin sau 1 dao ng:2
4 CFAm k
FC4 , CF l lc cn
Nu Fcl lc ma st th :kNA 4
- S dao ng thc hin c:CF
Ak
A
AN
4
.' 11
Nu Fcl lc ma st th:N
kAN
4' 1
- Thi gian t lc b ma st n khi dng lit =N. T
- Da vo s dao ng thc hin xc nh c s ln qua VTCB cavt: khi 25,' nNn (n nguyn) th s ln qua VTCB s l 2n; khi
75,'25, nNn th s ln qua VTCB ca vt l 2n+1; khi1'75, nNn th s ln qua VTCB ca vt l 2n+2.
- V tr ca vt c vn tc cc i:
Fc = Fhp => .m.g = K.x0 => 0mg
xk
- Vn tc cc i khi dao ng t c ti v tr x0 :
0 0v (A x ).
VIII - DAO NGCNGBC.CNGHNG
-Khi vt dao ng cng bc th tn s (chu k) dao ng ca vt bngtn s (chu k) ca ngoi lc.
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- Hin tng cng hng xy ra khi tn s (chu k) ca ngoi lc bng tns (chu k) dao ng ring ca h.
Ch :Chu k kch thchv
lT ; l l khong cch ngn nht gia 2 mi
ray tu ha hoc 2 g trn ng ; Vn tc ca xe con lc t trnxe c cng hng:
0
0
lfT
lv
IXCON LCTRNG PHNG
- xc nh chu k ca 1 con lc l xo (hoc con lc n) ngi ta sosnh vi chu k T0( bit) ca 1 con lc khc 0TT .
- Hai con lc ny gi l trng phng khi chng ng thi i qua 1 v trxc nh theo cng mt chiu
- Thi gian gia hai ln trng phng:0
0
TT
TT
Ch :+ Nu 0TT nTTn 01
+ Nu 0TT 01 nTTn (vi *Nn )
CHNG II: SNG CHC
I - ICNGV SNG CHC
T: chu k sng; v: vn tc truyn sng; :bc sng
1. Cc cng thc c bn
-Lin h gia , v v T (f): v fT
- Qung ng sngtruyn i c trong thi gian t: tT
vtS
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- Vn tc truyn sngbit qung ng sng truyn c trong thi gian t
l S:t
Sv
-Khong cch gia n gn li lin tip l d th:
1n
d
- n ngn sngi qua trc mt trong thi gian t th:1n
tT
-Phao nh cao n lntrong thi gian t th:1n
tT
2. Phng trnh sng
- Sng truyn tN qua O v n M, gi s biu thc Sng ti O c dng:)cos(0 tAu , th:
)2
cos(x
tAuM
)'2
cos(x
tAuN
- lch phaca 2 im trn phng truyn sng cch nhau mt on d:
d2
2k hay kd 2 im dao ng cng pha
12k hay2
12kd 2 im dao ng ngc pha
- lch pha ca cng mt imti cc thi im khc nhau:
12 tt
- Cho phng trnh sngl )cos( kxtAu sng ny truyn vi vn
tc:k
v
Ch :C nhng bi ton cn lp phng trnh sng ti 1 im theo iukin ban u m h chn th ta lp phng trnh sng ging nh phn lp
phng trnh dao ng iu ha.
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IIGIAO THOA SNG
1. Phng trnh sng tng hp ti mt im
*Trng hp tng qut:Phng trnh sng ti 2 ngun
1 1Acos(2 )u ft v 2 2Acos(2 )u ft
Phng trnh sng ti M do hai sng t hai ngun truyn ti:1
1 1Acos(2 2 )Mdu ft v 2
2 2Acos(2 2 )Mdu ft
Phng trnh sng ti M:
2 1 1 2 1 22 cos[ ] cos 22 2
M
d d d d u A ft
Bin dao ng ti M:
]2
cos[2 12dd
AAM vi = 2 - 1
C bc 0(k=0)
AB
CT th 1(k=0) C bc 1
k=1CT th 2( k=1)
O
/2
Gn lmGn li
d1d2
M
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2.Tm s im dao ng cc i, s im dao ng cc tiu gia haingun:
Cng thc tng qut
* S cc i: (k Z)2 2
l lk
* S cc tiu: (1 1
2 2 2 2
k Z)l l
k
Ta xt cc trng hp sau y:a. Hai ngun dao ng cng pha: = =2k
* S Cc i:(k Z)
l l
k
* S Cc tiu: 1 1- (k Z)2 2
l lk
Hay 0,5 (k Z)l l
k
b. Hai ngun dao ng ngc pha: ==(2k+1)
* S Cc i: 1 1 (k Z)2 2
l lk
Hay 0,5 (k Z)l l
k
* S Cc tiu: (k Z)l l
k
c. Hai ngun dao ng vung pha: =(2k+1) /2(S cc i= S cc tiu)
* S Cc i: 1 1 (k Z)4 4
l lk
* S Cc tiu:1 1
(k Z)4 4
l lk
Hay 0,25 (k Z)l l
k
Nhn xt: s im cc i v cc tiu trn on AB l bng nhau nn cth dng 1 cng thc l
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3. Tm s cc i , cc tiu ngoi on thng ni 2 ngun
12
'
1
'
2 ddkdd
(gi s'
1
'
212 dddd )- Xc nh s im (s ng) cctiu trn on AB (cng pha so vi ngthng 0102) l s nghim k nguyn thamn biu thc:
2
1
2
1 12'
1
'
2 ddkdd
(gi s'
1
'
212 dddd )Ch :Vi bi ton tm s ng dao ng cci v khng dao ng gia hai im M, N cchhai ngun ln lt l d1M, d2M, d1N, d2N.
t dM = d1M - d2M ; dN = d1N - d2Nv gi s dM < dN.+ Hai ngun dao ng cng pha:
Cc i: dM < k < dNCc tiu: dM < (k+0,5) < dN
+ Hai ngun dao ng ngc pha:Cc i: dM < (k+0,5) < dNCc tiu: dM < k < dN
S gi tr nguyn ca k tho mn cc biu thc trn l s ng cn tm.+ Hai ngun dao ng vung pha:
IIISNG DNG
- Bin ca sng ti v sng phn xl A th bin dao ng cabng sng a =2A.
- B rng ca bng sng l: L = 4A.- Vn tc cc i ca mt im bng sng trn dy: vmax = 2A- Phng trnh sng ti v sng phn x ti M cch B mt khong d l:
os(2 2 )M
du Ac ft v ' os(2 2 )M
du Ac ft
- Phngtrnh sng dng ti M: 'M M Mu u u
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2 os(2 ) os(2 ) 2 sin(2 ) os(2 )2 2 2
M
d du Ac c ft A c ft
Ch : Khong thi gian gia 2 ln lin tipsi dy dui thng l T/2.Khong cch gia 2 nt lin k bng khong cch 2 bng lin k
v bng 2 .Khong cch gia 2 nt hoc 2 bng
2k .
- iu kin c sng dng trn si dy n hi:
+ C 2 u c nh:2
kl ( *Nk )
S nt trn dy l 1k ; s bng trn dy l k
+ C mt u c nh, mt u t do:4
12kl ( Nk )
S nt trn dy l 1k ; s bng trn dy l 1k
IVSNG M
1. i cng v sng m
- V sng m cng l sng c nn cc cng thc ca sng c c th pdng cho sng m.- Vn tc truyn m ph thuc vo tnh n hi, mt v nhit ca mi
trng. Biu thc vn tc trong khng kh ph thuc nhit :tvv 10
v0l vn tc truyn m C0
0 ; v l vn tc truyn m t0C;1
273K-1
2. Cc bi ton v to ca m
- Mc cng m k hiu l L, n v l ben (B) :0
lgI
L BI
-Nu dng n v xiben th :0
10lgI
L dBI
; 1 10B dB
Vi I l cng m (n v 2W/m , I0l cng m chun,2-12
0 W/m10I .3. Cc bi ton v cng sut ca ngun m
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-Cng sut ca ngun m ng hng: IIP .r4S 2 (S l din tch ca mt cu c bn knh rbng khong cch gia tm
ngun m n v tr ta ang xt, I l cng m ti im ta xt)- BA II , l cng m ca cc im A, B cch ngun m nhng khong
rA, rB th:2
2
A B
B A
I rI r
- Mi lin h gia cng m v bin ca sng m:2
2
2
1
2
1
A
A
I
I
- Khi cng m tng (gim) k ln th mc cng m tng (gim)kN lg (B) v kN lg10 (dB).
+ Trng hpn
k 10 nN (B) hoc nN 10 (dB)
4. Giao thoa sng mGiao thoa sngsng dng p dng cho:
+ Dy n c 2 u c nh:
m c bn:l
vf
20 (cn gi l ha m bc 1)
ho mbc 2 l : f2 = 2f0;
ha m bc 3 l : f3 = 3f0 bc n:l
vnfn
2.
+ ng so:
H mt u: m c bnl
vf
40 (cn gi l ha m bc 1);
ho mbc 3 l f3 = 3f0; f5 = 5f0 bc n:
l
vnfn
4
12 .
H 2 u: m c bnl
vf
20 ;
ho m f1 = 2f0; f1 = 3f0 ; f bc n:l
vnfn
2. .
Ch : i vi ng so h 1 u, u kn s l 1 nt, u h s l bngsng nu m nghe tonht v s l nt nu m nghe b nht
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CHNG III: DNG IN XOAY CHIU
I. ICNG VIN XOAY CHIU
1. Sut in ng xoay chiu- Chu k v tn s quay ca khung:
2
1;
2
TfT
-Biu thc ca t thng qua khung dy:ttNBS coscos
0
0NBS: T thng cc i gi qua khung dy.
-Biu thc ca sut in ngxut hin trong khung dy dn:
tEtNBSt
e sinsin 0
vi 00 NBSE : Sutin ng cc i xut hin trong khung.2. in p (hiu in th) xoay chiu. Dng in xoay chiu
- Hiu in th xoay chiu: 0 cos( )uu U t (V)
- Dng inxoay chiu:0
cos( ) (A)i
i I t
i lng u i gi l lch pha ca u so vi i.+ Nu 0 th u sm pha so vi imt gc+ Nu 0 th u tr pha so vi i
+ Nu 0 th u cng (ng) pha so vi i.Ch :
+ Nu c mt in p xoay chiu (in p cc i l0
U ) c t
vo hai u bng n non m n ch sng ln mi khi in p u ln hnmt gi tr no 01 Uuu th trong mt chu k n sng ln 2 ln v
tt i 2 ln. Trong mt giy n sng ln hoc tt i 2f ln.+ Cc my o ch cc gi tr hiu dng ca cc i lng
3. Cc gi tr hiu dng 02
II ; 0 0,
2 2
U EU E
4. Cc cng thc khc- Tnh nhit l-ngta ra trn in tr thun theo cng thc: RtIQ 2
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(I l gi tr hiu dng ca dng inchy qua R trong thi gian t).-in trca on dy dn ng cht, tit din u c chiu di l l ,
in tr sut , din tch tit din l S:S
lR ;
-Mt khi chtc khi lng m, nhit dung ring l KkgJc . nhn nhit
lng Q tng nhit t1
t n 2t , th: 12 ttmcQ -in lng chuyn qua tit dinca dy dn trong khong thi gian tt
1t n 2t :
2
1
t
t
dqq2
1
t
t
idt
5. Dng in xoay chiu trong mch ch c in tr thun R; ch c cundy thun cm L v ch c t in C
Ch c R Ch c L Ch c Cnh lutm
RIU R 00 ,
IRUR LL ZIU 00 ,
LL IZU CC ZIU 00 ,
CC IZU
Tr khng R LZ L 1C
ZC
lch pha(u v i)
ui = 0 ui = + /2 ui = - /2
Gin vct
Lin h giau v i: 0
00 I
i
U
u 1
2
0
2
2
0
2
I
i
U
u 1
2
0
2
2
0
2
I
i
U
u
II. MCH R, L, C MCNITIP.CNGHNGIN
1. Cc cng thc c bn
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Ccmt
Mch RLC Mch RL Mch RC Mch LC
Dngmch
Vectquay
Tng
trZ = 2
L
2)(ZR CZ Z =
2
L
2R Z Z =2
C
2R Z
Z =
CL ZZ
Gclchpha
L CZ - Z
tan
R0L 0C
0R
U - Utan
U
L C
R
U - Utan
U
+
ZL >ZC:tnh cmkhng
+ ZL< ZC:tnh dung
khng.+ ZL=ZC:cnghng
LZ
tan
R
0L L
0R R
U Utan
U U
Mch c tnh cmkhng: > 0
CZ
tan -
R
0C C
0R R
U Utan - -
U U
Mch c tnhdung khng:
< 0
tg
nhlutm
Z
UI;
Z
UI 00
Z
UI;
Z
UI 00
Z
UI;
Z
UI 00
Z
UI;
Z
UI 00
Cngsut
P = UIcos
P = RI2
P = UIcos
P = RI2
P = UIcos
P = RI2P = 0
innng
W = P t W = P t W = P t W = 0
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2. Cng hng in.Nu gi nguyn gi tr ca in p hiu dng U gia hai u mch v
thay i tn s gc sao cho CL ZZ hayC
L1
, th trong mch xy
ra hin tng c bit, l hin tng cng hng. Khi :+ Tng tr ca mch t gi tr nh nht RZmin .
+ Cng dng in qua mch t gi tr cc iR
UImax .
+ Cc in p tc thi hai u t in v hai u cun cm c bin bng nhau nhng ngc pha nn trit tiu ln nhau, in p hai u intr bng in p hai u on mch.
iu kin xy ra cng hng l :
01
CL
LC
1 .
3. iu kin hai i lng tha mn h thc v pha+Khi hiu in th cng pha vi dng in (cng h ng):
0tanR
ZZ CL hay CL ZZ
+Khi hai hiu in th u1 v u2 cng pha: 2121 tantan .Sau lp biu thc ca 1tan v 2tan th vo v cn bng biu
thc ta s tm c mi lin h.+ Hai hiu in th c pha vung gc:
1tan.tan2
2121 .
Sau lp biu thc ca1
tan v2
tan th vo v cn bng biu thcta cng s tm c mi lin h.
Tr-ng hp tng qut hai i l ng tho mn mt h thc no ta sdngphng phpgin vectl tt nht hoc dng cng thc hm stan gii ton:
21
21
21tan.tan1
tantantan
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4. MT S CNG THC P DNG NHANH CHO DNG CUHI TRC NGHIM (dng hi p)
Cc dng sau y p dng cho on mch xoay chiu L RC mcni tip
Dng 1: Hi iu kin c cng hng in mch RLC v cc h qup: iu kin ZL = Zc LC2 = 1
Khi Z = Zmin = R ; I = Imax=U
R
cos = 1 ; P = Pmax =2
U
R
Dng 2:Cho R bin iHiR Pmax, tnh Pmax, h s cng sut coslc ?
p : R = ZL - ZC,2
Max
U 2P = , cos =
2R 2
Dng 3: Cho R bin i ni tip cun dy c rHiR cng sut trn R cc i
p : R2
= r2
+ (ZL - ZC)2
Dng 4:Cho R bin i, nu vi 2 gi tr R1 , R2 m P1 = P2HiR PMax
p R = ZL - ZC= 1 2R R Dng 5: Cho C1, C2 m I1 = I2 (P1 = P2)
HiC PMax(cng hng in)
p C1 C2c L
Z + ZZ = Z =
2
Dng 6: Cho L1, L2 m I1 = I2 (P1 = P2)HiL PMax (cng hng in)
p L1 L2L C
Z + ZZ = Z =
2
Dng 7: Hi vi gi tr no ca C th in p hiu dng trn t in UCmax
p ZC =2 2
L
L
R + Z
Z
, Khi
R CLM N BA
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2 2
ax
L
CM
U R ZU
Rv
2 2 2 2 2 2
ax ax ax; 0
CM R L CM L CM U U U U U U U U
Dng 8: Hi vi gi tr no ca Lth in p hiu dng trn t in ULmax
p ZL =2 2
C
C
R + Z
Z
, Khi
2 2
ax
C
LM
U R ZU
Rv 2 2 2 2 2 2ax ax ax; 0LM R C LM C LM U U U U U U U U
Dng 9: Hi iu kin 1, 2lch pha nhau
2(vung pha nhau)
p p dng cng thc tan 1.tan2 = -1Dng 10: Hi khi cho dng in khng i trong mch RLC th tc dng
ca R, ZL, ZC?p : I = U/R ZL = 0 ZC =
Dng 11: Hi Vi = 1hoc = 2th I hoc P hoc UR c cngmt gi tr th IMax hoc PMaxhoc URMax
p khi : 1 2 tn s 1 2f f f
Dng 12: Gi tr = ? th IMax URmax; PMax cn ULCMin
p : khi
1
LC (cng hng)Dng 13: Hi: Hai gi tr ca :
1 2P P
p 21 2 0
Dng 14: HiHai gi tr ca L :1 2L L
P P
p1 2 2
0
2L L
C
Dng 15: HiHai gi tr ca C :1 2C C
P P
p2
1 2 0
1 1 2
C C L
Dng 16: HiHai gi tr ca R :1 2R R
P P
p R1R2 =2
( )L CZ Z v R1 + R2=2
U
P
Dng 17: Hi khi iu chinh L URCkhng ph thuc vo R thp: Khi ZL = 2 ZC
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5. Cng sut ca mch in xoay chiu. H s cng sut.- Cng thc tnh cng sut ca mch in xoay chiu bt k:
cosUIP ; cos l h s cng sut.
- Ring vi mch ni tip RLC:2
2 R
R
UP I R U I
R
.
- H s cng sut ca on mch ni tip RLC:Z
R
U
URcos
- i vi ng c in: 2cosco
P UI P I R ;
trong R l in tr thun ca ng c, cos l h s cng sutca ng c, I l cng dng in chy qua ng c, U l in pt vo hai u ng c v
ciP l cng sut c ch ca ng c.
- Hiu sut ca ng c in:os
ciP
HUIc
Ch : + tm cng sut hoc h s cng sut ca mt on mch no th cc i l ng trong biu thc tnh phi c trong on mch .
+ Trong mch in xoay chiu cng sut ch c tiu th trnin tr thun.
III. MY PHT IN XOAY CHIU
1. My pht in xoay chiu mt pha- Tn s dng in xoay chiu do my pht pht xoay chiu mt pha pht ra:f np trong : p s cp cct, n s vng quay ca roto trong mt giy.
2. My pht in xoay chiu ba phaa. Ngun mc theo kiu:
+ hnh sao: 0 0
3
d p
d p
I I
I
U U
+ hnh tam gic:3d p
d p
I I
U U
b. Phi hp mc ngun v ti
Ngun v ti u mc hnh sao:p dng cho ngun A:
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3pd
dp
UU
II
p dng cho ti B:
3'
'
pd
dp
UUII
Ngun v ti u mc tam gic:p dng cho ngun A:
pd
pd
UU
II 3
p dng cho ti B:
'
'3
pd
pd
UU
II
Ngun mc hnh sao v ti mc tam gic:p dng cho ngun A:
3pd
dp
UU
II
p dng cho ti B:
'
'3
pd
pd
UU
II
Ngun mc tam gic v
ti mc hnhsao:
p dng cho ngun A:
pd
pd
UU
II 3
p dng cho ti B:
3'
'
pd
pd
UU
II
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IV. MY BIN P V TRUYNTIINNNG
1. My bin pa. Gi
1N v
2N l s vng dy ca cun s cp v th
cp; 1 2,i i v 1 2,e e l c ng v sut in ngtc thica mch s cp v th cp; 1 2,r r v 1 2,u u lin tr ca
cun dy s cp v th cp v hiu inth tc thi hai u mch s cp vth cp. Ta c cc lin h:
kN
N
e
e
2
1
2
1
(k gi l h s ca MBA)
cun s cp: 1 1 1 1u e i r cun th cp: 2 2 2 2u e i r
b. Nu in tr cc cun dy khng ng k:Gi
1U v
2U l in p hiu dng xut hin hai u ca cun s
cp v th cp;1
I v2
I l c ng hiu dng dng in ca mch s
cp v th cp khi mch kn. H l hiu sut ca MBA.
Ta c cc lin h: 1 1
2 2
U NU N
+ 2 1N N th 2 1U U , ta gi MBA l my tng th.
+ 2 1N N th 2 1U U , ta gi MBA l my h th.
- Hiu sut ca my bin p :111
222
cos
cos
IU
IUH
vi 1cos v 2cos l h s cng sut ca mch s cp v th cp.-Nu mch s cp v th cp c u v i cng pha th:
2 2
1 1
U IH
U Ihay
HI
I
U
U
.1
2
2
1 ;
Khi %100H th 2 2 1
1 1 2
U N I
U N I
c. Nu in tr cun s cp v th cp ln lt l r1 v r2, v mchin hai u cun th cp c in tr R:
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Quy c: 12
N= k
N
- in p hai u cun th cp U2 = 1 22 1
k.RU
k (R + r ) + r
- Hiu sut my bin p: H =2
2
2 1
k .Rk (R + r ) + r
2. Truyn ti in nngUP, : l cng sut v in p
ni truyn i, ',' UP : l cngsut v in p nhn c ni
tiu th; I: l cng dngin trn dy, R: l in trtng cng ca dy dntruynti.
+ gim th trn dy dn: IRUUU ' + Cng sut hao ph trn -ng dy:
RU
PRIPPP .
cos'
22
22
+Hiu sut ti in:P
PP
P
PH
'' ,
Ch :+ Ch phn bit hiu sut ca MBA H v hiu sut ti in 'H .+ Khi cn truyn ti in khong cch l th ta phi cn si dy dn
c chiu di l2 .
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CHNG IV: DAO NG SNG INT
i, I0: cng tc thi v cng cc i trong mch; q, Q0: in tchtc thi v in tch cc i trn t in; u, U0: in p tc thi v inpcc i trn t in.
1. i cng : Chu k, tn s ca mch dao ng
- Tn s gc:LC
1;
0
0
Q
I
- Chu k dao ng ring:0
0222IQLCT
- Tn s ring:LC
f2
1
2
Ch : Nu mch dao ng c t 2 t tr ln th ta coi b t l mt t cin dung C t ng ng c tnh nh sau:
+ Ghp ni tip:n
i in CCCCC 121
11...
111
n
CCCC ,...,,21
+ Ghp song song:n
i
in CCCCC1
21 ... nCCCC ,...,, 21
- Gi T1 v T2l chu k dao ng in t khi mc cun thun cm L lnlt vi t C1 v C2 th:
+ Khi mc L vi C1 ni tip C2:
2
2
2
1
2
2
2
2
1
2
111
TTT
fff
+ Khi mc L vi C1 song song C2:
2
2
2
1
2
2
2
2
1
2
111
fff
TTT
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2. Nng lng ca mch dao ngNng l-ng in tr-ng:
222 201 1 cos
2 2 2tt
QqW Cu t
C C
22
02
1iIL
Nng l-ng t tr-ng:2 2 2
0
1 1sin
2 2dt
W Li LI t 2202
1uUC
Nng l-ng in t:
2 2
0 0
1 1
2 2dt tt W W W CU LI
2
2 2 01 1 1
2 2 2
QLi Cu
C
- Lin h gia in tch cc i v in pcc i: 00 CUQ - Lin h gia in tch cc i v dng in cc i: 00 QI
- Biu thc c lp thi gian gia in tch v dng in:2
222
0
iqQ
3. Qu trnh bin i nng lng mch dao ng
Nu mch dao ng c chu k T v tn s f thNng lng in trngv v nng lng t trng (
td WW , ) dao ng vi tn s f= 2f, chu k
T=2
T
u-U0 0U
2
0 0U2
0U 2
2 0
U 3
2
+U0T/4 T/12 T/6
T/8 T/8
T/6 T/12
Wtt = 3 Wt WtmaxW = 0
Wt = 3 WttWt = WttWtmin = 0Wmax
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Ghi ch:- Hai ln lin tip Wt = Wtt l T/4- Khi q cc i th u cc i cn khi i cc tiu (bng 0) v ngc li.
4. Thu v pht sng in t-Khung dao ng c th pht v thu cc sng in t c b c sng:
LCcTc 2. ;c l tc truyn sng in t trong chn khng ( smc /10.3 8 )
-Nu mch dao ng c L thay it21 LL 21 LL th mch chn
sng c th chn c sng c bc sng:CLcCLc 21 22
-Nu mch dao ng c C thay it 21 CC 21 CC th mch chn
sng c th chn c sng c bc sng:21 22 LCcLCc
-Nu mch dao ng c L thay it 21 LL 21 LL v c C thayi t 21 CC 21 CC th mch chn sng cth chn c sng c
bc sng:
2211 22 CLcCLc
Gi 1 v 2 lbc sng mch dao ng hot ngkhi dng cun thun cm L mc vi C1 v C2th bc sngmch dao ng hot ng khi mc L vi:
+ C1 // C2 :2
2
2
1
2
2
2
2
1
2
111
fff
+ C1ni tip C2 :2
2
2
1
2
2
2
2
1
2
111fff
-Nu mch dao ng c C thay it 21 CC
21 CC th mch hot ng vi bc sng trong khong
21 21 th:
1
22
2
2
2
22
2
1
44 CcL
Cc
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-Nu mch dao ng c L thay it 21 LL 21 LL th mch hotng vi bc sng trong khong 21 21 th:
1
22
2
2
2
22
2
1
44 LcC
Lc
Ch :Hai cng thc cui vn p c cho trng hp L v C l hng scn bc sng bin thin 21 .
5. Mch dao ng tt dn- Khung dy c in trhot ng nnc s
2 2 2 22 0 0
2 2
C U U RC I R R
L
cng l cng sut to nhit ca in tr.-Nng l-ng cn cung cp trong khong thi gian t: RtIQA 2
6. Di sng in t
Nidung
SNGDI
SNGTRUNG
SNG NGNSNG CC
NGNBc
sng
> 1000 m1.000 m
100 m
100 m10 m 10 m0,01m
cim
- C nnglng nh,- khngtruyn ci xa trnmt t.- t b nchp th
- C nnglng kh ln,- Truyn i
c trn mtt.- B tng in
ly hp th voban ngy v
phn x voban m
- C nng lngln,- Truyn i cmi a im trnmt t- C kh nng
phn x nhiu lngia tng in lyv mt t
- C nng lng rtln ln- Truyn c ic trn mt t- Khng btngin ly hp th hoc
phn x v c khnng truyn i rt xatheo mt ngthng
ngdng
Dng thng tindi nc
Dng thngtin vo banm
Dng thngtin trn mt t
Dng thng tintrong v tr.
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CHNG V: SNG NH SNG
1. Tn sc nh sng
a. i vi lng knh
- Cng thc lng knh:
AiiD
Arr
rni
rni
'
'
'sin'sin
sinsin
vi i, i l gc ti v gc l; A l gc chit
quang; D l gc lch to bi tia ti v tia l.- Tr-ng hp gc nh: D = (n - 1)A- Gc lch cc tiu.
+ Khi c gc lch cc tiu, ng i ca tia sng i xng qua mt phngic ca gc chit quang.
+ K hiu gc lch cc tiu l mD , gc ti ng vi gc lch cc tiu l
mi , ta c:
2sin
2sin
2
2'
An
AD
AiD
Arr
m
mm
- Gc lch gia 2 tia sng n scqua lng knh (chit sut i vi lngknh ln lt l n1 v n2 21 nn ): AnnD 21
-B rng quang ph lin tctrn mn chn t pha sau lng knh cchlng knh mt khong l:lAnnl
t
(vi nt v nl chit sut ca nh sng tm v nh sng i vilngknh v A tnh bng radian)
b. Tn sc t mi trng ny sang mi trng khc
*Nu dngnh sng n sc th:+ Mu n sc khng thay i (v f khng i)
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+ Bc sng n sc thay iVn tc v bc sng ca nh sng trong mi trng c chit sut n:
n
cv ;
n' ;
trong c v l vn tc v bc sng ca nh sng trong chn khng.
+ Dng nh lut khc x tm gc khc x 211
2
sin
sinn
n
n
r
i
+ Nu nh sng t mi trng chit quang ln sang mi trng chit
quang nh phi xc nh ghi :1
2sin
n
ni gh
*Nu dngnh sng trngth:+ C hin tng tn sc v xut hin chum quang ph lin tc.+ Cc tia n sc u b lch
- Tia lch t so vi tia ti;- Tia tm lch nhiu so vi tia ti.
c. Thang sng in t
10-11
m 10-8
m 0,001m (m)f(Hz)
2. Giao thoa nh sngGi khong cch gia hai khe S1S2 la, khong cch t mt phng cha 2khe v mn chn l D, l b csng ca nh sng.a. Cc cng thc c bn
-Hiu ng ica mt im c
ta x trn mn:
Sngvtuyn
Tiahngngoi
nhsngtrng
Tia tngoi
TiaX
Tiagama
0,4m
0,75m
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D
axdd 12
- V tr vn sng: kia
Dkx
Vn sng bc n ng vi nk - V tr vn ti:
212
212
ik
a
Dkx hoc
1
2
sk s k
tk
x xx
0k vn ti bc n ng vi 1nk ; 0k vn ti th n ng vi nk ;
v d: vn ti th 5 ng vi 5k hoc k=4.
-Khong vn:D
i
a
-Bc sng ca nh sng:D
ia
- Tn s ca bc x:c
f
-Khong cch gia n vn sng lintip l d th:1n
di
-Khong cch gia 2 vn sng bc kbng: ki2
b. S vn sng, ti trn mnTnh s vn sng, ti trn trng giao thoa L, c 2 cch:Cch 1:+ S vn sng trn L l s nghim k (nguyn) tha mn h thc:
i
Lk
i
L
22
+ S vn ti trn L l s nghim k (nguyn) tha mn h thc:
2
1
22
1
2 i
Lk
i
L
Cch 2:
+ S vn sng: 12
2i
LN ,
i
L
2l phn nguyn ca biu thc
i
L
2.
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Tnh s vn sng ti trn on AB c ta xA v xBbt k xA< xB+ S vn sng trn on AB l s nghim k (nguyn) tha mn h thc:
i
xk
i
x BA
+ S vn ti trn on AB l s nghim k nguyn tha mn h thc:
2
1
2
1
i
xk
i
x BA Zk
Dch chuyn ca h vnGi: D l khong cch t 2 khe ti mn
D1 l khong cch t ngun sng ti 2 khe
+ Khi ngun sng S di chuyntheo phng song song vi S1S2th hvn di chuyn ngc chiu, khong vn i vn khng i v di ca hvn l:
0
1
Dx d
D, d l dch
chuyn ca ngun sng.
+Khi ngun S dng yn vhai khe dch chuyn theo phngsong song vi mn th h vn dch chuyn cng chiu, khong vn i vnkhng i v di ca h vn l:
dD
Dx
1
0 1 , d l dch chuyn ca hai khe S1 v S2.
e. Giao thoa vi nh sng trng hoc nh sng nhiu thnh phni vi nh sng gm nhiu thnh phn n sc (nh sng a sc).- V tr chng chp (trng) ca nhiu bc x:
nnkkkx ...2211
l
k
k
k
1
2
2
1 (*) (l
kl dng ti gin ca
1
2 , k, l nguyn dng)
T suy ra:nlknkk
..
2
1 ....,2,1,0n
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Suy ra v tr ca bc x trng: 111
1 knia
Dkn
a
Dkx
- Cc bc x trngmu vn trung tm cch nhau mt khong u n
a
Dkd 1 , k l hng s bit (*)
i vi nh sng trng 0,38 0,76m m .
- B rng vn sng (quang ph) bc k:
ttk iika
kDx .
-nh sng n sc c vn sng ti im ang xt:.k D xa
x a kD ,
k c xc nh t bt ph ng trnh: 0,38 0,76xa
m mkD
- nh sng n sc c vn ti ti im ang xt:2
2 12 2 1
D xax k
a k D,
k c xc nh t bt ph ng trnh 20,38 0,762 1xam m
k D
Lu : V tr c mu cng mu vi vn sng trung tm l v tr trng nhauca tt c cc vn sng ca cc bcx thnh phn c trong ngun sng.
CHNG VI: LNGTNH SNG
0 : gii hn quang in, 0f : tn s gii hn quang in, : bc sng
nh sng, f : tn s ca nh sng, A: Cng thot, max0v : vn tc ban u
cc i, bhI : cng dng quang in bo ha, hU : in p(hiu in th)
hm, h : hng s Flng ( Jsh 3410.625,6 ) , c : vn tc nh sng trong
chn khng ( smc /10.3 8 ), e : in tch ca electron ( Ce 1910.6,1 )
1. Cc cng thcvhintng quang in
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+Nng lng ca photon: 2cmhc
hf ph
+ng lng ca photon:c
hcmp ph ,
mphl khi lng tng i tnh ca photon.+ Gii hn quang in: 0
hc
A
+Ph-ng trnh Anhxtanh: 201
2maxhf A mv ,
khi lng ca electron 319,1.10m kg .+Bc x n sc(bc sng ) c pht ra v nng lng ca mi
xung l E th s photon pht ra trong mi giy bng:hc
E
hf
EEn
+ Vn tc ban u cc i:m
hc
v0
max0
112
+ Vt dn c chiu sng: max2
max0
2
1Vemv
( maxV l in th cc i ca vt dn khi b chiu sng)+Nu in trng cnl u c cng E v electron bay dc theo
ng sc in th:
max
2
max02
1Edemv
( maxd l qung ng ti a m electron c th ri xa c Catot).
Ch :+Nu chiu vo Catt ng thi 2 bc x 1, 2 th hin tng
quang in xy ra i vi bc x c bc sng b hn 0 0ff . Nu c2 bc x cng gy ra hin tng quang in th ta tnh ton vi bc x c
bc sng b hn.+Ban nng cao
- in p hmtrit tiu dng quang in21
2omax h
mv eU
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- Cng dng quang in bo ha:I = ne (n: s electron v anot trong 1 s)
- Tc electron khi v n anodDng nh l ng nng WA - Womax= eUAK
2. Chuynngca electron trong trngint
a. Chuyn ng ca electron trong in trng
+ in p U tng tc cho electron: 22
1vmeU e
2
02
1vme
( 0v v v ln lt l vn tc u v vn tc sau khi tng tc ca e).
+ Trong in tr ng u: dF e E
ln: EeF C 3 trng hp:
-Nu0
v E: chuyn ng chm dn u vi gia tceE
am
-Nu 0v E: chuyn ng nhanh dn u vi gia tc
eE
a m
-Nu0
v E: chuyn ng cong qu o Parabol+ Theo phng xx: thng u: x = v0t
+ Theo phng yy nhanh dn u vi gia tceE
am
b. Chuyn ng ca electron trong t trng
+ Trong t tr ng u: B qua trng lc ta ch xt lc Lorenx:sinvBef = ma =
2v
mR
Bv
,
+Nu vn tc ban u vung gc vi cm ng t: lectron chuyn ngtrn u vi bn knh
.m vR
e B; bn knh cc i:
Be
mvR max0
max
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+Nu vn tc ban u xin gc vi cm ng t: lectron chuyn
ng theo vng xon c vi bn knh ng vng c:sin
max0
Be
mvR
3. Cng sutcangun sng - dng quang in bo ho
- hiusutlngt
a. Cng sut ca ngun sng.
.P n IS eH
I
hc
PPn bh
n l s photon ca ngun sng pht ra trong mi giy; l l ng t
nng l ng (photon); (I l cng ca chm sng, H l hiu sutlng t)
b. C-ng dng inbo ha.
eHnent
qI ebh
t
N
e
In bhe
N l selectron n c Ant trong thi gian t giy, en l slectron n Ant trong mi giy.
e l in tch nguyn t
19
1,6.10e Cc.Hiu sut l-ng t.' bhInH
n P e
'n l s lectron bt ra khi Katt kim loi trong mi giy.n l s photon p vo Katt trong mi giy.
Ch : Khi dng quang in bo ho th n = ne4. Chu k,tns,bc sng tia X ngRn Ghen pht ra
- Gi nng l ng ca 1 electron trong chm tia Catot c c khi n im cc l
W , khi chm ny p vo i m cc n s chia lm 2 phn:
+ Nhit l ng ta ra (Qi) lm nng i m cc v
+ phn cn li c gii phng d i dng nng lng photon ca tia X(bc x Rn-ghen).
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i QW
Trong : c
hhf
(l nng lng photon ca tia Rnghen)
22
2
0
2
mvUemvW
l ng nng ca electron khi p vo i catt (i m cc)U l hiu in th gia ant v cattv l vn tc electron khi p vo i cattv0l vn tc ca electron khi ri catt (thng v0 = 0)m = 9,1.10-31kg l khi lng electron
- Cng dng in qua ng Rn-ghen: enI ,(n l s e p vo i Catot trong 1s).Trng hp b qua nhit lng ta ra trn i m cc:
Ta c:
W ngha l
Wc
h
Hay
W
hc
- ng Rn Ghen s pht bc x c b c sng nh nht nu ton b nngl ng ca chm tia Katot chuyn hon ton thnh nng l ng ca bc xRn Ghen. B-c sng nh nht -c tnh bng biu thc trn khi du = xyra:
W
hcmin
Trng hp ton b nng lng ca electron bin thnh nhit
lng:-Nhit lng ta ra trn i Catot trong thi gian t:W = Q RI
2t = mctt: tng nhit ca i m cc (anot)c: nhit dung ring ca kim loi anot.
m: khi lng anotTrng hp tng qut:
- Hiu sut ca ng Rnghen: i
W
QW
WH
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5. Mu nguyn tBo
+Khi nguyn tang mc nng l ng cao chuyn xung mc nngl ng thp th pht ra photon, ng c li chuyn t mc nng l ng thpchuyn ln mc nng l ng cao nguyn t s hp thu photon
hfEE thpcao
+Bn knh qu o dngth n ca electron trong nguyn t hir: 0
2rnrn
Vi mr 110 10.3,5 : l bn knh Bo ( qu o K)+Mi lin h gia cc bc sng v tn sca cc vch quang ph
ca nguyn t hir:
Th d 31 = 32 + 21
213231
111v 213231 fff
+Nng lng electron trong nguyn t hir:
2
13,6( )nE eV
n Vi n N*: lng t s.
+Nng lng ion ha hydro (t trng thi c bn)
Wcung cp = E - E1
Ch : Khi nguyn t trng thi kch thch n (trng thi th n)c th pht ra s bc x in t ti a cho bi cng thc:
2
12 nnCN n ; trong 2
nC l t hp chp 2 ca n.
+ Cc dy quang ph (ban nng cao)
L mann
K
M
N
O
L
P
Balmer
Paschen
H H H H
n = 1
n = 2
n = 3
n = 4n = 5n = 6
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. 1 21; 2, 3, 4, ...n n dy Laiman (t ngoi)
1 22; 3, 4, 5, ...n n dy Banme (nhn thy)
1 23; 4, 5, 6,...n n dy Pasen (hng ngoi).
CHNG VII: HT NHN NGUYN TI - ICNG VHT NHN NGUYN T
1. Cu to ca ht nhn nguyn t- Ht nhn nguyn t l phn cn li ca nguyn t sau khi loi b electron,
ht nhn nguyn t X k hiu l: AZ X , XA , X
A .
Trong : Z l nguyn t s hay s proton trong ht nhn.N: S ntronA Z N : S khi.
- Kch thc (bn knh) ca ht nhn: 31
15.10.2,1 AR m ; vi A l skhi ca ht nhn.
2. n v khi lng nguyn t
- n v khi lng nguyn t l n v Cacbon (k hiu l u)kgu
2710.66055,11
-Ngoi ra theo h thc gia nng lng v khi lng ca Anhxtanh, khi
lng cn c th o bng n v2
c
eVhoc
2c
MeV;
2/5,9311 cMeVu
3. Nng lng linkt nng lng lin kt ring
Ht nhn AZ X c khi lng m c cu to bi Z proton v N
notron. Cc php o chnh xc cho thy khi lng m ca ht nhn AZ X bao
gi cng b hn tng khi lng ca cc nuclon to thnh ht nhn AZ X :
p mm Zm Nm m . m c gi l ht khi ca ht nhn.
-Nng lng lin kt v nng lng lin kt ring:A
lk
lkr
2
lk
WW
m.cW
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Nng lng lin kt ring cng ln th ht nhn cng bn vng.-Nng lng ngh: 2mcE , vi m l khi lng ngh ca ht nhn
4. Cng thc Einstein gia nng lng v khi lng
Nng lng ht = Nng lng ngh + ng nng ca htE = E0 + W = mc
2 + mv2
II - PHNG X
- S ht nhn cn li: 0 0.2 .t
tTN N N e
-Khi lng cn li: tTt
emmm 00 2
Vi T l chu k phng x, l hng s phng xT
2ln
- S ht nhn b phn r: teNN 10 Tt
N 210
khi Tt : tNN 0
-Phn trm s nguyn t b phn r: tTt
eN
N121
0
-Khi lng b phn r: tTt
emmm 121 00
-Phn trm khi lng b phn r: tTt
em
m121
0
- S ht sinh ra bng s ht phng x mt i
- Tnh tui ca mu cht phngx:H
H
N
Nt 00 ln
1ln
1
-Khi c cn bng phng x: 2211 NN
-
Khi lng:A
N
Nm
A
.
- Cho phng trnh phng x: ZYX AZA
Z
'
'
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+Nu ban u ht X ng yn th t s ng nng:Y
Z
Z
Y
m
m
W
W
+Nu cc ht sinh ra c cng vn tc th:Z
Y
Z
Y
m
m
W
W
- Cho 1 lng ng v phng x Xc chu k phng x l T, phng xban u l H0vo th tch V ca cht lng, sau thi gian 0t ly ra th tchv cht lng th phng x l H. Th tch cht lng bng:
T
tt
H
vH
He
vHV
00
2.
00
- Gi Nl s xung phng x pht ratrong thi gian1
t , 'N l s xung
phng x pht ra trong thi gian 2t k t thi im sau thi im ban umt khong thi gian
0t , th:
2
1
0
1
1.
' t
tt
e
ee
N
N
+ Nu 21 tt :0
'
te
N
N
+ Nu Ttt 21, :2
1.'
0
tte
NN t
Ch : Tui ca ming g c xc nh t thi im cht (cht) nthi im taxt.
Nu khong thi gian kho st rt nh so vi chu k bn r( Tt ) th ta vn dng h thc gn ng xex 1 (khi 1x ). y
ta c: tet
1 v Tt nn teNN 10 tN0
Phn ring ban nng cao+ phng x thi im t (n v Becren Bq):
0.tH N N e tT
t
eHH 00 2 00 NH
+ Lin h gia khi lng v phng x:AN
AHm
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+ Lu : Khi tnh phng x H, H0th chu k phng x T tnh bngnv giy(s).
III - PHNNGHT NHN
Phng trnh phn ng: 31 2 41 2 3 4
AA A A
Z Z Z ZA B C D
1. Cc nh lut bo ton.+ nh lut bo ton s khi: 4321 AAAA .
+ Bo ton in tch: 4321 ZZZZ
+ nh lut bo ton ng lng: A B C DP P P P + nh lut bo ton nng lng ton phn: Nng lng tng cng
trong phn ng ht nhn l khng i.
Ch :Trong phn ng ht nhn khng c nh lut bo ton khi lng .
2. Xc nh nng lng, to hay thu bao nhiu?
Trong phn ng ht nhn31 2 4
1 2 3 4
AA A A
Z Z Z ZA B C D Cc ht nhn A, B, C, D c:
Nng lng lin kt ring tng ng l 1, 2, 3, 4.Nng lng lin kt tng ng l E1, E2, E3, E4 ht khi tng ng l m1, m2, m3, m4
a. ht khi: m m3 + m4 - m1 - m2
b.Cng thc tnh nng lng ca phn ng ht nhn:
Bit cc khi lng W = (MtrcMsau) c2
Nu Bit nng lng lin kt W = Esau - Etrc
Bit ht khi cc ht W = ( msau - mtrc)c2
Ch : p, n v electron c ht khi bng 0.
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c. bit phn ng ta hay thu nng lng:Gi tng khi lng ca cc ht nhn v phi l m0, v to thnh l m.Nu:
0m m Phn ng to nng lng
Nng lng ta ra ca 1 phn ng: 20' cmmW
Nng lng ta ra thng dng ng nng cc ht.Cc ht sinh ra khi bn hon cc ht ban u
0m m Phn ng thu nng lng
+ Nng lng cn cung cp ti thiu phnng xy ra(chnh l nng lng thu vo ca phn ng): 20min cmmW
Nng lng thu vo thng di dngng nng cc ht
hoc bc xCc ht sinh ra khi khng bn hon cc ht ban u+ Nu ng nng ca cc ht ban u l minWW th:
'2
0 WcmmW
( 'W l ng nng ca cc ht sinh ra)3. Tnh ng nng v vn tc cc htca phn ht nhn, s dng
cc cch sau:
Dng nh lut bo ton nng lng ton phn:'
2
0 WWcmm
(S dng ht khi ca cc ht nhn: 20 cmm )Kt hp vi nh lut bo ton ng lng:
A B C DP P P P
2 2
A B C DP P P P
Dng phng php gii ton vecto v hnh hoc
T suy ra i lng cn tm v d gc hp bi chiu chuyn ng cacc ht so vi mt phng no Cc trng hp c bitkhi so snh ng nng cc ht sinh ra:
-Nu cc ht nhn ban u ng yn th:'
'
'
'
X
Y
Y
X
m
m
W
W
-Nu cc ht sinh ra c cng vn tc th:'
'
'
'
Y
X
Y
X
m
m
W
W
Ch :Cng thc gia ng lng v ng nng:
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p2 = 2m WNhit ta ra khi t m kg cht t c nng sut ta nhit l L bng:
LmQ , L: nng sut to nhit (J/kg). JKWh 000.600.31
Nc nng OD22
1 dng chit sut ng v tri (2
1D ) dng lmnguyn liu phn ng nhit hch c trong nc chim 0,015% v khi lng.* Cc trng hp c bit thng gp
+ Trc ht ta c nh lut bo ton nng lngA + B C + D
WC + WD = (mtrc - msau)c2 + WA (gi s ht B ng yn) (1)
+ Hai ht sinh ra c vn tc vung gc nhau2 2 2
C D A C Dp p p p p
mCWC+ mDWD = mAWA (2)T (1) v (2) ta gii v tm c WC v WD
+ Mt trong hai ht sinh ra vung gc vi ht A2 2 2
C A D A C p p p p p
mDWD - mCWC= mAWA (2)T (1) v (2) ta gii v tm c WC v WD
+ Hai ht sinh ra ging ht nhau v vec t p cc ht i xngv hp
Ap vi cc gc bng nhau
Ta c cos2
A
C
pp
2cos2
A A
C C
m Wm W
Nh ta tm WC v WD.
+ Phng x sinh ra hai ht chuyn ng ngc chiu nhau
Cp
Ap
Dp
Ap
Cp
Dp
Cp
Ap
Dp
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