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Vt L 12

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• Trc nghim l 12- Luyn thi i hc Trang - 1 -

BNG TM TT CNG THC LNG GIC THNG DNG TRONG VT L 1. n v o Gi tr lng gic cc cung.

* 10 = 60 (pht), 1= 60 (giy); 10 =

180

()

* Gi l s o bng ca 1 gc, a l s o tnh bng radian tng ng vi khi ta c php bin i sau:

a =.180

()

* i n v: 1mF = 10-3F; 1F = 10-6F; 1nF = 10-9F; 1pF = 10-12F; 10

A = 10-10m. Cc n v khc cng i tng t. * Bng gi tr lng gic cung c bit.

Cung i nhau ( v -)

Cung b nhau v ( - )

Cung hn km ( v + )

Cung ph nhau ( v /2 -)

Cung hn km /2 ( v /2 +)

cos(-) = cos sin(-) = -sin tan(-) = -tan cot(-) = -cot

cos( - )= -cos sin( - ) = sin tan( - ) = -tan cot( - ) = -cotg

cos( + ) = -cos sin( + ) = -sin tan( + ) = tan cot( + ) = cotg

cos(/2 -)= sin sin(/2 -) = cos tan(/2 -) = cot cot(/2 -) = tan

cos(/2 +) = -sin sin(/2 +) = cos tan(/2+)= -cot cot(/2 +) = -tan

2. Cc i lng vt l Cc n v ca h SI di m Thi gian s Vn tc m/s Gia tc m/s2

Vn tc gc rad/s Gia tc gc rad/s2 Khi lng Kg Khi lng ring Kg/m3 Lc N p sut hoc ng sut Pa Xung lng Kg.m/s Momen ca lc N.m Nng lng, cng J

Cng sut W Momen xung lng Kg.m2/s Momen qun tnh Kg.m2 nht Pa.s Nhit K in lng C Cng in trng V/m in dung F Cng dng in A in tr in tr sut .m Cm ng t T T thng Wb Cng t trng A.m

• Trc nghim l 12- Luyn thi i hc Trang - 2 -

Momen t A.m2 Vecto t ha A/m t cm H Cng sng Cd Cch c tn mt s i lng vt l Anpha beta Gamma enta Epxilon Zeta T Fi ta Tta Nuy Muy Lamda Kxi Khi Omega Ipxilon Xicma R Pi o Omikron Kappa Ita

Cc hng s vt l c bn Vn tc nh sng trong chn khng

c = 3.108 m/s

Hng s hp dn G = 6,67.10-11 m3/(kg.s2)

Gia tc ri t do G = 9,8 m/s2 S Avogadro 6,02.1023 mol-1 Th tch kh tiu chun V0 = 2,24 m

3/kmol Hng s kh R = 8,314 J/kmol Hng s Bolzman k = 1,38,10-23

J/kmol S Faraday 0,965.108 C/kg i n v Chiu di 1A0 = 10-10 m

1 n v thin vn (a.e) = 1,49.1011 m 1 nm nh sng = 9,46.1015 m 1 inches = 2,54.10-2 m 1 fecmi = 10-15 m 1 dm = 1,61.103 m 1 hi l = 1,85.103 m

Din tch 1 ha = 104 m2 1 bac = 10-28 m2

Khi lng 1 tn = 10 t = 1000 kg 1 phun = 0,454 kg 1 a.e.m = 1,67.10-27 kg (Khi lng nguyn t) 1 cara = 2.10-4 kg

Cng v cng sut

1 erg/s = 10-7 W 1 m lc (HP) = 636 W 1 kcal/h = 1,16 W 1 calo (cal) = 4,19 J 1 W.h = 3,6.103 J

p sut 1 dyn/cm2 = 0,1 Pa 1 atm = 1,01.105 Pa 1 kG/m2 = 9,81 m2 1 mmHg = 133 Pa 1 at = 1 kG/cm2 = 9,18.104 Pa

3. Cc hng ng thc lng gic c bn:

sin2 + cos2 = 1; tan.cot = 1

22

cot1sin

1

2

2tan1

cos

1

4. Cng thc bin i a. Cng thc cng cos(a + b) = cosa.cosb - sina.sinb cos(a - b) = cosa.cosb + sina.sinb sin(a + b) = sina.cosb + sinb.cosa sin(a - b) = sina.cosb - sinb.cosa

• Trc nghim l 12- Luyn thi i hc Trang - 3 -

tan(a - b) = ba

ba

tan.tan1

tantan

tan(a + b) = ba

ba

tan.tan1

tantan

b. Cng thc nhn i, nhn ba cos2a = cos2a - sin2a = 2cos2a - 1 = 1 - 2sin2a; sin3a = 3sina 4sin3a sin2a = 2sina.cosa; cos3a = 4cos3a 3cosa;

tan2a = a

a2tan1

tan2

c. Cng thc h bc: cos2a = 1+cos2a

2 ; sin2a =

1-cos2a2

; tan2a = 1-cos2a1+cos2a

; cotan2a = 1+cos2a1-cos2a

d. Cng thc tnh sin, cos, tan theo t = tan 2

21

2sin

t

t

2

2

1

1cos

t

t

21

2tan

t

t

(

2 + k, k Z)

e. Cng thc bin i tch thnh tng

cosa.cosb = 12

[cos(a-b) + cos(a+b)] sina.sinb =12

[cos(a-b) - cos(a+b)]

sina.cosb = 12[sin(a-b) + sin(a+b)]

f. Cng thc bin i tng thnh tch

cosa + cosb = 2cosa+b

2 cos

a-b2

sina + sinb = 2sina+b

2cos

a-b2

cosa - cosb = -2sina+b

2sin

a-b2

sina - sinb = 2cosa+b

2sin

a-b2

tana + tanb =sin(a+b)

cosa.cosb tana - tanb =

sin(a-b)cosa.cosb

(a,b 2 +k )

5. PHNG TRNH V H PHNG TRNH LNG GIC a. Cc cng thc nghim pt c bn:

sinx = a = sin

2

2

kx

kx cosx = a = cos x = + k2

tanx = a = tan x = +k cotx = a = cot x = +k b. Phng trnh bc nht vi sin v cos:

Dng phng trnh: a.sinx + b.cosx = c (1) vi iu kin (a2 + b2 0 v c2 a2 + b2)

Cch gii: chia c 2 v ca (1) cho 22 ba ta c:22 ba

a

sinx +

22 ba

b

cosx =

22 ba

c

Ta t:

sin

cos

22

22

ba

b

ba

a

ta c pt:

)2()sin(

cos.sinsin.cos

22

22

ba

cx

ba

cxx

Gii (2) ta c nghim. c. Phng trnh i xng: Dng phng trnh: a.(sinx + cosx) + b.sinx. cosx = c (1) (a,b,c R)

Cch gii: t t = sinx + cosx = 2.cos(x - 4), iu kin - 2 t 2

t2 = 1+ 2sinx.cosx sinx.cosx = t2-12

th vo (1) ta c phng trnh:

a.t + b.t2-12

= c b.t2 + 2.a.t - (b + 2c) = 0

Gii v so snh vi iu kin t ta tm c nghim x. Ch : Vi dng phng trnh: a.(sinx - cosx) + b.sinx. cosx = c Ta cng lm tng t, vi cch t t = sinx - cosx = 2.cos(x +/4).

• Trc nghim l 12- Luyn thi i hc Trang - 4 -

d. phng trnh ng cp. Dng phng trnh: a.sin2x + b.cosx.sinx + c.cos2x = 0 (1) Cch gii: - b1 Xt trng hp cosx = 0

- b2 Vi cosx 0 (x = 2 + k) ta chia c 2 v ca (1) cho cos

2x ta c pt: a.tan2x + b.tanx + c =

0 t t = tanx ta gii phng trnh bc 2: a.t2 + b.t +c = 0. Ch : Ta c th xt trng hp sinx = 0 ri chia 2 v cho sin2x. 6. Mt s h thc trong tam gic: a. nh l hm s cos: a2 = b2 + c2 2bc.cosA;

b. nh l hm sin: a

sinA =

bsinB

= c

sinC

c. Vi tam gic vung ti A, c ng cao AH:

222

111

ABACAH ; AC2 = CH.CB; AH2 = CH.HB; AC.AB = AH.CB

• Trc nghim l 12- Luyn thi i hc Trang - 5 -

DAO NG C HC

I CNG DAO NG IU HA 1. Dao ng c, dao ng tun hon + Dao ng c l chuyn ng c gii hn, qua li ca vt quanh v tr cn bng. + Dao ng tun hon l dao ng m nhng khong thi gian bng nhau (gi l chu k T) vt tr li v tr c theo hng c 2. Dao ng iu ha + Dao ng iu ha l dao ng trong li ca vt l mt hm coossin (hay sin) theo thi gian. + Phng trnh dao ng: x = Acos(t + ) Trong : + A: Bin dao ng, l gi tr cc i ca li x; n v m, cm. A lun dng + (t + ): l pha ca dao ng ti thi im t; n v rad + l pha ban u ca dao ng, n v rad + : Tn s gc ca dao ng iu ha; n v rad/s + Cc i lng: bin A ph thuc vo cch kch thch ban u lm cho h dao ng; pha ban u ph thuc vo vic chn mc (ta v thi gian) xt dao ng, cn tn s gc (chu k T, tn s f) ch ph thuc cu to ca h dao ng. + Phng trnh dao ng iu ha x = Acos(t + ) l nghim ca phng trnh x + 2x = 0 l phng trnh ng lc hc ca dao ng iu ha + Hnh chiu ca mt chuyn ng trn u ln 1 trc c nh qua tm l mt dao ng iu ha. Mt dao ng iu ha c th biu din tng ng 1 chuyn ng trn u c bn knh R = A, tc v = vmax = A. 3. Cc i lng c trng ca dao ng iu ha + Chu k T ca dao ng iu ha l khong thi gian thc hin mt dao ng ton phn; n v: giy (s). + Tn s f ca dao ng iu ha l s dao ng ton phn thc hin c trong mt giy; n v hc (H).

+ Lin h gia , T v f:

f2

tgian_thoi

Ndong_dao_Sof

2

f

1T

Nhn xt: + Mi chu k vt qua v tr bin 1 ln, cc v tr khc 2 ln (1 ln theo chiu dng v 1 ln theo chiu m). + Mi chu k vt i c qung ng 4A, chu k vt i c 2A, chu k i c qung ng A (nu xut pht t VTCB hoc v tr bin). 4. Vn tc trong dao ng iu ha:

+ Vn tc l o hm bc nht ca li theo thi gian: v = x = -Asin(t+) = Acos(t + + 2

)

+ Vn tc ca vt dao ng iu ha bin thin iu ha cng tn s nhng sm pha 2

so vi li .

+ V tr bin: x = A v = 0 + V tr cn bng: x = 0 |v| = vmax = A 5. Gia tc trong dao ng iu ha + Gia tc l o hm bc nht ca vn tc (o hm bc 2 ca li ) theo thi gian: a = v = x = -2Acos(t+) = - 2x.

• Trc nghim l 12- Luyn thi i hc Trang - 6 -

Bin : A Ta VTCB: x = a Ta v tr bin: x = a A

+ Gia tc trong dao ng iu ha bin thin iu ha cng tn s nhng ngc pha vi li v sm

pha 2

so vi vn tc.

+ Vect gia tc ca vt dao ng iu ha lun hng v v tr cn bng, c ln t l vi ln ca li . + v tr bin: x = A gia tc c ln cc i: amax =

2A + v tr cn bng: x = 0 gia tc bng 0. Nhn xt: Dao ng iu ha l chuyn ng bin i nhng khng u. 6. Lc tc dng ln vt dao ng iu ha: F = ma = -k.x lun hng v v tr cn bng, gi l lc ko v. 7. Cng thc c lp:

A2 = x2 + 2

2v

v A2 =

2

2v

+

4

2a

8. Phng trnh c bit: x = a Acos(t + ) vi a = const

x = a Acos2(t + ) vi a = const Bin : 2

A; = 2; = 2

9. th dao ng: + th dao ng iu ha (li , vn tc, gia tc) l ng hnh sin, v th ngi ta cn gi dao ng iu ha l dao ng hnh sin. + th gia tc li : dng on thng nm gc phn t th 2 v th 4 + th li - vn tc; vn tc gia tc: dng elip. 10. Vit phng trnh dao ng: * Xc nh bin :

- Nu bit chiu di qu o ca vt L th A = 2

L.

- Nu vt c ko khi VTCB 1 on x0 v c th khng vn tc y th A = x0.

- Nu bit vmax v th A = maxv .

- Nu bit max v min l chiu di cc i v cc tiu ca l xo khi n dao ng th A = 2

minmax

- Bit gia tc cc i amax th A = 2maxa

* Xc nh tn s gc: =

T

22. =

gianthoi

* Xc nh pha ban u: lc t = 0 th x = x0 v du ca v (theo chiu (+): v >0, theo chiu (-): v < 0,

bin: v = 0.

0

0

tsinAv

tcosAx

Lu : + Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0. + Gc thi gian t = 0 ti v tr bin dng: = 0. + Gc thi gian t = 0 ti v tr bin m: = .

+ Gc thi gian t = 0 ti v tr cn bng theo chiu m: = 2

+ Gc thi gian t = 0 ti v tr cn bng theo chiu dng: = 2

11. Thi gian vt i t li x1 n li x2 (ho