7/28/2019 Tng hp cng thc lp 12
1/16
CHNG 0. C HC VT RN1. NG HC VT RN* Lin h gia tc gc v tc di
v = R
* Gia tc tip tuyn
an= R
2
* Gia tc php tuyna
t= R
* Gia tc ton phn
a = at
2+ a
n
2
* Tc gc trung bnh
=
t
* Gia tc gc trung bnh
=
t
* Gia tc gc tc thi = t+
0
* Gc quay c trong thi gian t
=1
2t
2+
0t
* Cng thc c lp vi thi gian
2
0
2= 2.
* Gc quay c trong giy th n
n= (n 0,5)+0
* Chu k quay u
T = 1f=2
* Tc quay
=n
30
n: svng quay/pht
2. MOMEN QUN TNH* Cng thc tng qut
I = (mR2)
* Momen qun tnh ca vnh trn ng chtI = mR
2
* Momen qun tnh ca a trn ng cht
I =1
2mR
2
* Momen qun tnh ca thanh quay quanh 1 uI =
1
3ml
2
* Momen qun tnh ca thanh quay quanh trc gia
I =1
12ml
2
* Momen qun tnh ca hnh cu ng cht
I =2
5mR
2
CNG THC LP 12
gocriengtrenban.com6 1
7/28/2019 Tng hp cng thc lp 12
2/16
* Momen qun tnh ca mt cu ng cht
I =2
3mR
2
* Cng thc nh l Huyghens
IM= I
O+md
2
d: khong cch MO
3. NG LC HC VT R N* Momen lc
MF/O = F.dd: cnh tayn
* Phng trnh ng lc hc vt rn
=M
I
=M
FM
can
I
* o hm ca momen ng lng
M =dL
dt4. NNG LNG* Momen ng lng
L = I
* Bo ton momen ng lng cho 2 a
I11 + I22 = (I1 + I2 )* ng nng quay
Wd=
1
2I
2
* ng nng va quay va tnh tin
W =1
2I
2+1
2mv
2
* Lin h gia ng nng quay v momen ng lng
Wd=
L2
2I
CNG THC LP 12
2
7/28/2019 Tng hp cng thc lp 12
3/16
7/28/2019 Tng hp cng thc lp 12
4/16
* Lc ko vF
kv= kx
* Lc n hiF
dh= k | l
0+ x |
* Lc n hi ln nhtFdhmax
= k(l0+ A)
* Lc n hi nh nhtFdhmin
= k | l0
A |
* Chiu di l xo v tr cn bng
lVTCB
=lmax
+ lmin
2
* Bin con lc
A = lmax
lmin
2
* L xo ghp song song
k//= k
1+ k
2
1
T2=
1
T12
+1
T22
* L xo ghp ni tip1
k//
=1
k1
+1
k2
T
2
= T12
+T22
k1l
1= k
2l
2
* Ghp hai vt nng m = m1 + m2
T2= T
1
2+T
2
2
* L xo b ct k1
k2
=
l2
l1
* iu kin vt m khng trt trn vt MkA
(m +M) g
: hsma st gia m v Mk:cng ca l xo (N/m)
3. CON L C N* Tn s gc, chu k con lc n
=gl
T = 2 lg
* Lin h gia n v di v n v gcs = l v = l '
v
2=
2gl(coscos0 )
* Tc cc iv0= l
0
* Lin h chu k v chiu di dy
T1
2
T22
=
l1
l2
* Lc cng dyT = mg(3cos 2cos
0)
CNG THC LP 12
4
7/28/2019 Tng hp cng thc lp 12
5/16
* Lc cng dy ln nhtTmax =
mg(3 2cos0 )
* Lc cng dy nh nhtT = mgcos
0
* Lc ko vFkv = mgsin
* Thay i nhit
T2
2
T1
2= 1+t
: hsndi (K-1)t:bin thin nhit
* Thay i caoT2
T1
=
R + h
R
R: bn knh trit
* Chu tc dng ca lc ngoi
T2
2
T12=
g
g '
g ' = g a g ' = g
2+ a
2
a: gia tc ngoi (m/s2)
* Lc in
a = q Em
E: cngin trng (V/m)
* Chnh lch thi gian ca ng h
t= t.(1T
2
T1
)
t: thi gian xem xtT1: chu kng
t > 0:ng hchy nhanh
* Chu k con lc b vng inh
T = (l1
g+
l2
g)
4. NNG LNG CON L C* ng nng con lc
Wd=
1
2
mv2
* Th nng con lc l xo
Wt=
1
2kx
2
* Th nng con lc n
Wt = mgl(1 cos) =1
2m
2l2
2
* C nng tng qutW =W
t+W
d
* C nng con lc l xo
W =1
2m
2A
2=
1
2kA
2
CNG THC LP 12
5
7/28/2019 Tng hp cng thc lp 12
6/16
* C nng con lc n
W =1
2m
2l2
0
2
* Lin h li v nng lngW
t
W= (
x
A)2= (
0
)2=
1 cos
1 cos0
* Lin h vn tc v nng lngW
d
W= (
v
vmax
)2
5. DAO NG T T DN - CNG HNG* Bin con lc gim sau N chu k
A =4NFc
k=
4Nmg
k* S dao ng thc hin cho n khi tt hn
N=A
A=
kA
4F=
kA
4mg
* Qung ng i c cho n lc tt hn
s =kA
2
2F=
kA2
2mg
* Chu k ngoi lc xe chy trn rnh
T =s
v6. T NG HP DAO NG* Cng pha
A = A1 + A2 = Amax
* Ngc qua
A = |A1 - A2| = Amin* Vung pha
A = A1
2+ A
2
2
* Cng bin
A = 2A1cos(
2)
* Bin tng qut
A2= A
1
2+ A
2
2+ 2A
1A
2cos
* Pha tng hp
tan=A
1sin
1+ A
2sin
2
A1cos
1+ A
2cos
2
7. CON L C TRNG PHNG* S ln con lc gp nhau cho n khi gp li
N1T1= N
2T2
CNG THC LP 12
6
7/28/2019 Tng hp cng thc lp 12
7/16
CHNG 2. SNG C HC1. SNG C HC TNG QUT* Bc sng ca sng c hc
=v
f
* lch pha gia 2 im trn phng truyn
= 2d
* Phng trnh lan truyn sng
u = Acos(2(t
T
x
))
* Tc lan truyn sng
v =FT
: khi lng trn mtn vchiu di dyFT: lc cng dy
2. GIAO THOA SNG
* Bc giao thoa
k=d
* Bc giao thoa ca hai bin b rng giao thoa L
k=L
* Phng trnh giao thoa
u = 2A cos(d
2 d
1)
cos(t
(d2 d
1)
)
3. SNG DNG* S b sng dng
k=2l
* Bc sng di nht trn dy 2 u cnh= 2l
* Bc sng di nht trn dy 1 u cnh= 4l
* Bin mt im trong b/4 - 2a; /12 - a
/8 - a2/2; /6 - a3/2
4. SNG M* Cng m
I =P
4R2
* T s cng m
I1
I2
=
R2
2
R1
2
* Mc cng m
L = log(I
I0
) LA L
B= log(
IA
IB
)
* Trung im
dM=dA+ d
B
2 d
M=
dA d
B
2
CNG THC LP 12
7
7/28/2019 Tng hp cng thc lp 12
8/16
CHNG 3. DNG IN XOAY CHIU1. I CNG IN XOAY CHIU* Gi tr hiu dng
I =I0
2 U=
U0
2
* Cng tc thi
i =U
R
R
i
I0
2
+uL
U0L
2
= 1 i
I0
2
+uC
U0C
2
= 1
* in p tc thiu = u
R
+ uL
+ uC
* Cm khngZL = L = 2 fL
* Dung khng
ZC =1
C=
1
2 fC* Tng tr
Z= R2+ Z
L Z
C( )2= (R + r)
2+ Z
L Z
C( )2
* in p ton mch
U= UR2
+ ULUC( )
2
* lch pha
tan=Z
L Z
C
R=
ULU
C
UR
* Cng dng in hiu dngI =
U
Z=
UR
R=
UL
ZL
=
UC
ZC
R :ZL
:ZC=U
R:U
L:U
C
* H s cng sut
cos= R
Z=UR
U
* Cng sut in
P =UIcos= RI2=
U2
Rcos
2
P = (R + r)I2
* Cng hng in: I max, P maxZ
L= Z
C Z= R
UL=U
CU=U
R
=0=
1
LC f = f0 =
1
2 LC
u, i cng phacos= 1 P =UI
0
=
f0
f
=
ZC
ZL
* u1 vung pha vi u2tan
1tan
2= 1
* iu kin URL khng ph thuc vo RZ
C= 2Z
L
CNG THC LP 12
8
7/28/2019 Tng hp cng thc lp 12
9/16
* iu kin URC khng ph thuc vo RZ
L= 2Z
C
2. KHI R THAY I* P max khi ch c R
R = ZL Z
C
Pmax
=
U2
2R cos=
2
2
* P max khi c R, r
R + r = ZL Z
C
Pmax
=U
2
2(R + r) cos=
2
2
* PR max khi c R, r
R = ZL Z
C( )2
+ r2
Pmax =U
2
2(R + r)
* R = R1; R = R2 cho cng P
P = U
2
R1+ R
2
R1R
2= Z
L Z
C( )2
(R2
0= R
1R
2 P
max=
U2
2 R1R
2
)
3. KHI L THAY I* P, I, UC max: cng hng
L =1
C2
* UL max
ZL=R2 + Z2C
ZC
; uRC v u vung pha
ULmax
=
U R2+ Z
2
C
R= U
2+U
2
L+U
2
R
(U2Lmax
UCU
Lmax U2 = 0 )
* URL max
ZL=
ZC+ Z
2
C+ 4R
2
2
URLmax
=2RU
4R2+ Z
2
C Z
C
* L = L1; L = L2 cho cng P
ZC= Z
L0=
ZL1
+ ZL2
2* L = L1; L = L2 cho cng UL
1
ZL1
+1
ZL2
=2Z
C
R2+ Z
2
C
1
ZL
=
1
2 (
1
ZL1
+
1
ZL2
) L =2L
1L
2
L1+ L
2
(UL max)
4. KHI C THAY I
CNG THC LP 12
9
7/28/2019 Tng hp cng thc lp 12
10/16
* P, I, UL max: cng hng
C=1
L2
* UC max
ZC=R
2+ Z
2
L
ZL
; uLR v u vung pha
UCmax =
U R2+ Z
2
L
R= U
2
+U
2
C +U
2
R
(U2
CmaxU
LU
CmaxU
2= 0 )
* URC max
ZC=
ZL+ Z
2
L+ 4R
2
2
URCmax
=2RU
4R2+ Z
2
L Z
L
* C = C1; C = C2 cho cng P
ZL= Z
C0=ZC1 + ZC2
2
* C = C1; C = C2 cho cng UC
1
ZC1
+1
ZC2
=2Z
L
R2+ Z
2
L
1
ZC
=1
2(1
ZC1
+1
ZC2
)C=C1 +C2
2(UC max)
5. KHI , f THAY I* P, I max: cng hng
=1
LC
* UL max
=1
CL
C
R2
2
ULmax
=
2LU
R 4LC R2C
2
* UC max
=1
L
L
C
R2
2 U
Cmax=
2CU
R 4LC R2C
2
* = 1; = 2 cho cng P
1
2=
0=
1
LC
* = 1; = 2 cho cng UL
1
1
2+
1
2
2= LC C
2
R2
2
* = 1; = 2 cho cng UC
1
1
2+
1
2
2=
L2
L
R
R2
2
6. MY BI N P
CNG THC LP 12
10
7/28/2019 Tng hp cng thc lp 12
11/16
* My l tngU
1
U2
=
N1
N2
=
I2
I1
* My khng l tng
H=Ptt
Ptp=
RI2
2
U1I1
H: hiu sut my bin p
7. TRUY N TI IN NNG I XA* in tr dy
R = l
S
:in trsut (m)
* Cng sut hao ph
P =RP
2
U2cos
2
* Hiu sut truyn ti
H =100%P
P
* gim thU = I.R
8. MY PHT IN XOAY CHI U* Tn s dng in
f =np
60
n: scp ccp: svng quay/pht
* T thng cc i gi qua khung dy
0 = NBSN: svng
B: cmng t(T)S: din tch khung dy (m2)
* Sut in ng cc i sinh raE
0=
0
* H thc c lp gia t thng v sut in ng
0
2
+E
E0
2
= 1
* Mch mc hnh sao
Id = Ip
Ud = 3Up
P = 3UpIp cos
* Mch mc hnh tam gic
Id = 3Ip
Ud =Up
P = 3UdIdcos
CNG THC LP 12
11
7/28/2019 Tng hp cng thc lp 12
12/16
CHNG 4. DAO NG IN T1. DAO NG IN T* Tn s dao ng
=1
LC f =
1
2 LC
* Cng dng in cc i
I0=Q
0 =U
0
C
L* Cng thc c lp vi thi gian
q = Cu
i
I0
2
+u
U0
2
= 1 i
I0
2
+q
Q0
2
= 1
* Nng lng in t trng
WL =1
2Li
2
WC=1
2Cu
2=1
2 qu=1
2
q2
C
W =WL +WC
* Nng lng hao ph
P = RI2= R
I0
2
2
2. SNG IN T* Bc sng in t
= 2c LC
* Ghp tC
//= C
1+C
2
1
Cnt
=1
C1
+1
C2
* T xoay
C : gc quayc
* Lin h gia bc sng v t
1
2
2
=
C1
C2
* Ghp cun dy Lnt= L
1+ L
2
1
L//
=1
L1
+1
L2
* Lin h gia bc sng v L
1
2
2
=
L1
L2
CNG THC LP 12
12
7/28/2019 Tng hp cng thc lp 12
13/16
CHNG 5. SNG NH SNG1. TN SC, KHC X NH SNG* Lin h gia chit sut v bc sng
1
2
=
n2
n1
(f khng i)
* nh lut khc x nh sngn1sini = n
2sinr
n1i = n2r
r = rdo r
tim
* Cng thc lng knhA = r
1+ r
2
D = i1+ i
2 A
D = Dtim D
do
i = i ' =D
min+ A
2 r = r ' =
A
2
* Cng thc thu knh
1f= (n 1)( 1
R1
+ 1R
2
)
2. GIAO THOA NH SNG* Hiu quang l
r=
xa
D
* Khong vn giao thoa
i =D
a
* Bc giao thoa
k=x
i=
r
* Mi trng trong sut chit sut n
in=
i
n
* t bn mng trn ng i nh sng
x0= n 1( )
eD
a
* Dch chuyn ngun sng ln xung
x0 =D
D1
d
D1: khong cch tngunn 2 khed:di ca ngun sng
x0:di ca hvn/VSTT
* B rng quang phx
1= x
1d x
1t
* Quang ph chng lnh = x
1d x
2t
* Vn sng trng nhau
k1
1
= k2
2
CNG THC LP 12
13
7/28/2019 Tng hp cng thc lp 12
14/16
CHNG 6. LNG T NH SNG1. HIN TNG QUANG IN* Nng lng ca photon
= hf =hc
=
1,9875.1025
* Cng thot
A =hc
0
* ng nng ban u cc i ca electron
Wd0max
=
1
2mv
0max
2
* Cng thc quang in
= A +Wd0max
* Hiu in th hm/in th cc i
eUh=W
d0max
* Cng dng quang in
I =n
e.e
t
* Cng sut bc x chiu ti
P =np.
t
* Hiu sut quang inH=
ne
np
* ng nng electron p vo anotW
dAW
d0max
= eUAK
* Bn knh electron bay trong t trng u
R =mv
eB
B: cmng t(T)m: khi lng electron
* ng nng electron p vo i catot (ng tia X)W
d=Q +
2. QUANG PH HIDRO* Nng lng electron mc n
En
=
13,6
n2 (eV
)
* Bc sng bc x khi chuyn di
Em En = hfmn =hc
mn
1
31
=
1
32
+
1
21
* Bn knh mc nng lng
r= n2r0
* Vn tc electron mc n
v =
9.109e2
r.n
* Nng lng ion ho t mc n: En
CNG THC LP 12
14
7/28/2019 Tng hp cng thc lp 12
15/16
CHNG 7. VT L HT NHN1. TNG QUT V HT NHN* ht khi
m = Zmp + (A Z)mn mX
* Nng lng lin kt/Nng lng phn r
W = m.c2= m.931(MeV)
* Nng lng lin kt ring
w =W
A
* S mol
n =N
NA
=
m
A
2. PHNG X* Hng s phng x
=ln2
T=
0,693
T
* Lng cn li
N = N02
t/T= N
0e
T
m = m02t/T
= m0eT
n = n02t/T
= n0eT
H = H02t/T
= H0eT
* S ht sinh ra
N = N0 (1 2t/T
)
* T s ht mi v c sau thi gian t
= 2t/T1
* Khi lng cht mi
mY=
AY
AX
m0(1 2
t/T)
* phng x
H = N1 Ci = 3,7.1010Bq1 Bq = 1 phn r/s
* Tc phn r/chiu x
t=N
H
N: sphn r
3. PHN NG HT NHN* Bo ton s khi/in tch
At= A
s
Zt= Z
s
* Nng lng to ra/thu vo
W = mtm
s( )c2= m
s m
t( )c2= W
s W
t
* ng nng - ng lng
p = mv = 2mK
pt = ps* ng lng trong phng x
pY = p
Y: ht nhn con
CNG THC LP 12
15
7/28/2019 Tng hp cng thc lp 12
16/16
PH LCMT S CNG THC THNG GP KHC* Mi quan h gia 2 i lng cng pha, ngcpha
x1
x01
=
x2
x02
* Mi quan h gia 2 i lng cng pha, ngc
phax1
x01
2
+x2
x02
2
= 1
x1,x2: cc gi trtc thix01,x02: cc gi trcci
* Cng thc log, mx
ax
b= x
a+b
(xa)b= x
ab
elnx
= x
CNG THC LP 12