CÁC CÔNG THỨC GIẢI NHANH TRẮC NGHIỆM VẬT LÍ - TRẦN NGUYÊN TƯỜNG

8/10/2019 CC CNG THC GII NHANH TRC NGHIM VT L - TRN NGUYN TNG

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TRN NGUYN TNG

CC CNG THC

GII NHANH TRC NGHIM

VT L N THI TT NGHIP THPT

THI TUYN SINH I HC V CAO ANG

( T i b n l n t h n h t) c s a c h a v b su n g )

B C C A

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W.DAYKEMQUYNHON.UCOZ.COM

ng gp PDF bi GV. Nguyn Thanh T


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16 Hng Chui - Hai B Trng - H Ni

in thoai: Bin to-Ch bn: (04) 39714896:

Hnh chnh: (04 39714899: Tng bin tp: (04> 39714897

Fax: 04 39714899

Chu trch nhim xu t bn

Gim c - Tng bin tp:

Bin tp:

Sa bi:

Trnh by ba:

i tc lin kt xut bn:

TS. PHM TH TRM

NGUYN THY

NH SCH HNG N

NH SCH HNG N

SCH LIN KT

CC CNG THC GII NHANH TRC NGHIM VT L

M s: 1L - 17H2013In 1.000 cun, kh 16 X 24cm, ti Cng c phn Vn ha Vn Lang - Tp. H Ch Minh.S xut bn: 152-2013/CXB/05M 3/HQGHN.Quyt nh xut bn s: 26LK-TN/Q-NXBHQGHN.n xong v np lu chiu qu I! nm 2013.



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C h n g I. CC CNG THC GII NHANHTRONG VT L 12

1. Cc cng thc nu mi ii h gia X, Vv a

. 2 2 VA = X + -(1)

Hayot

u1(1')

a = -0)2X (2)

Mt vt dao ng iu ha theo phng trnh:

X = lOcoslOt - j (cm).

Vn tc ca vt khi bt u dao ng bng bao nhiu ?

Li g i i+ Theo cng thc (1'): V2 = C02(A2 - X2)

+ t = 0: X = lOcos^- j = 10 X - - j = -5 (cm)

+ co = 10 (rad/gy)

+ A = 10 (cm)

^ V2 = 102[102 - (-5)2]

= > V = 1 0 X V = + V ( c m / s )

+ V = -lOOsini- I> 0 3J

Vy chn V = 5 0 V3 (cm/s).

V d 2. Mt con lc l xo nm theo phng ngang. Qu cu c khilng m = 200 (g) v l xo c cng k = 40 (N/m). Tvtr cn bng

ko vt raxa mt don 4 (cm) ri truyn cho n tc 30V2 (cm/s) vt dao ng iu ha. Xc nh bin dao ng ca con lc.

Li g i i

V2+ Theo cng thc (1): A2 = X2 + -

C2

n k 4-0 I+ OI2 = = = 200 => O= 10V2

m 0,2

+ X = 4 (cm)

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4/291

Vy A = 42 +30V2

I 0 V2= 1/4 2 + 32 = 5 (cm).

V d 3. Mt vt ao ng iu ha vi chu k T = 0,5 (giy). Bit tc

ca vt idao ng A.

T

ca vt ng vi pha dao ng l 2 (m/s). Hay xc inh bin 3

Li g i i

+ X= Acos(cot + )= Acos= 3 2

+ 0) = ^ = = 4n (ra/s)T 0,5

a 2 _ 2 X2 fA)2-' 200)2 _ A 107 , .

+ A = X + + => A = pr (cm).c I2 J U *J Vs V d 4. Mt vt dao ng iu ha vi lncc i ca vn tc v

gia tc tng ng l 62,8 (cm/s) v 4 (m/s2). Hy xcnh bin A vchu k dao dng T.

Li g i i

+ 1v 1max = = 62,8 (Clll/s) = 2071 (cm/s)

+ !a Imax = CO*A = 4 (xn/s2) = 400 (cin /s2)

Nmax ^ 2A _ _ 400 _ 20|v| ~ toA ~ 2(br :r\ Imax - '

271 27T

+ T = = l ( g i y )C 0

71

] , . . 20tt 20n , f w__ \+ !V1max = A = 20n => A = = = 5* = 10 (cm)

t) zU7

Vy T = 1 (giy) v A = 10 (cm).

2. Cng thc tnh chu k ca con lc i X v con lc n

a) Nu con lc l xo c khi lng 111= 111! + m2 hay con lc n c chiudi dy treo = li + l->th:

T = Vt ? + T| (3)

f = - r (4 )t f 7 i

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5/291

b) Nu con lc l xo c khi lng ra = m.1 - m21 hay C011 lc dII c chiudi dy treo - \ \ - h \ th:

V d 5, Mt C011 lc l xo treo thng ng.

- Nu vt c khi lng mi th chu k dao ng Ti = 0,3 (giy).

- Nu vt c khi lng m2 th chu k dao ng T2= 0,4 (giy).

Hi nu vt c khi lng m = mi + m2 th ch k dao ng ca con lcbng bao nhiu ?

Theo cng thc (3): T = y/l f + Tf = tJq,Z2 + 0,42 = 0,5 (giy)

Vy T = 0,5 (giy).

V d 6. Mt con lc n dao ng i ha.

- Nu chiu di dy treo lxth tn sdao ng fi = 6 (Hz).

- Nu chiu di dy treo 2th tn s' dao ng f2 = 8 (Hz).

i nu chiu di dy treo. / = I - 1%th tn s dao ng l bao nhiu ?

(31)

f= (4)

Li g i i

Li g i i' .''heo cng thc: f =

Vy f = 9 (Hz).

c) Ghp con lc l xo

+ Nu hai l xo mc ni tipth

(5)

f= (6)

+ Nu hai l xo mc song song hay hai im cnh A v B th

TToT -------- (7)



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6/291

V d 7. Hai l xo cng di t nhin v cng ln lt l k] v k2.Khi tr T2 = ! + AT = 2 (giy)+ 0,001 (giy)

Vy T2 = 2,001 (giy).

V d 10. Mt ng h thc hin bi con lc n c chu k dao ng= 2 (giy) nhit t i = 25c. Bit h s X - 2.10"5 1/d. Xc dinh

chu k ao ng ca con lc nhit 2= 20c.

Li g i i

AT = 2 txXT = X2 .10_5(20 25).2 (giy) = lo ' 4 (giy)z Z

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7/291

Tj R

Lu :

(10)

- Du (+) ng vi trng hp a con lc t v tr ngang mc nc binln cao h.

Du (-) ng vi trng h.p t cao h a con lc xung v tr ngangmc nc bin.

V d 11. Ti mt ni trn mt t con lc dao ng vi chu k Ti =2 (giy). a con lc ln cao h = 3,2 (km) so vi mc nc bin th chu k dao ng bng bao nhiu ? Cho R = 6400 (km).

=> AT = .Tx = -M - X 2 (giy) = 1CT3 (giy)R 6400

M AT = T2 - T\

=> T2 = Ti + AT = 2 + 1CT3 = 2,001 (giy).

V d 12. Mt con lc dn dao ng vi chu k Ti - 2 (giy) cao

h = 1,6 (km) so vi mc nc bin. a con lc xung v tr ngang mcnc bin th chu k dao ng bng bao nhiu ? Ch.0R = 6400 (km).

=> AT = T, = X2.(giy) = 0,0005 (giy)R 1 6400

=> T2 = -Ti + AT = 2 - 0,0005 = 1,9995 (giy)

Vy T2 = 1,9995 (giy).

c) Cng thc v dn g h thc hi n bi con lc n chy nhanhhay chm trong mt ngy

Gi 'Z?l thi gian dng h chy nhanh hay chm trong mt ngy.

Li g i i

_ AT ha con lc ln d cao h. nn: - =

T, R

Li g i i

AT

T, .

86400 Cgiy)(11)

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8/291

Lu :

- Du (+)khi AT > 0: ng h chy chm.

- Du (-) khi AT < 0: ng h chy nhanh.

= -X(t2 - t i ) - Ti 2 2 1 R

( 12)

Dng kL nhit v h thay i.

V d 13. Mt con lc n c chu k dao ng Ti = 2 (s) nhit tj = 15c. Bit h s' n di ca ytreo X = 5.10"5 1/. Nu ng hthc hin bi con lc trn ni nhit t2 = 35c, th trong mt ngy,ng h chy nhanh hay chm vi thi gian l bao nhiu ?

Li g i i

AT----X(t2 t-)

a t > 0 ng h chy chm:

AT

%

'86400 (giy)

|M t2 - t x) =

- X5.10"2

86400 (giy)

'(35-1 5) =

86400

=> rc - 43,2 (giy).

V d 14. Ti mt ni ngang mc nc bin, vi nhit ti = 10c,mt ng h qu lc trong mt ngy chy nhanh 6,48 (giy). Bit h s n di X= 2.10-5 1/. Hi v tr trn vi nhit no th ng hchy ng ? .

Li g i i

ng h chy ng th phi chy chm li hay AT > 0:

'

86400 (giy)

1 '~X(t2 -1^) = --------2 2 1 86400

X2.10_5(t2 -10C) =2

6,48

86400

2 10c = 7,5c=> ta = 17,5C.

V d 15. Mt ng h qu lc chy nhanh trong mt ngy 6,48 (giy)ti ni ngang mc nc bin v nhit ti = 10c. Cho bit h s ndi X = 2.10"5 . a ng h ln cao h so vi mc nc bin vinhit t2 = 6c th ng h chy ng. Xc nh cao ca h. so vi mc nc bin. .



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9/291

Li g i i

+ ng h chy ng thi phi chy chm li, hay AT > 0.

+ Theo cng thc (12): = -M t2 - t) + 2 R

AT '86400 (giy)

v-5

1 , , ^ h ^(t2 ti ) + -------2 R 86400

- .2 .1(T5 (6 c - 10 C) + ----- -------=2 6400 (km) 86400

h = 0,736 (km) = 736 (m).

4. Chu k dao ng ca con lc n hay i rong h quy chiu phi qun tnh

a) Con lc on treo vo trn t, tu ha. t, tu ha chuyn ng theophng ngang vi gia tca. -*

+ Ti v tr cn bng mi:

+ P + Fqt = 0

t p + Fqt = P '

P' = mg' gi l trng lc hiu dng.

V p* = p2 + F2

g'= V2 + a2

+ Nu bi ton cho gc lch a:

g_ p _ mg _ cosa = = => g =P' mg' cos a

(13)

V d 16. Mt con c n dao ng iu ha vi chu k T = 2 (giy),ti ni g = 9,8 (m/s2). Treo con c vo trn mt toa xe v xe chuynng nhanh dn u vi gia tc a = 2 (m/s2). Tnh chu ki dao ng mi ca con lc. $

Li g i i

-- T = 2tc


10/291

V d 17. Mt con lc n dao ng iu ha vi chuk T = 2 (s) tini g = 10 (m/s2). Treo con lc vo trn xe t-v choxe chuyn ngnhanh dn u th thy dy treo con ic hp vi phng, thng ng mt gc a = 9. Tnh gia tc ca xe v chu k dao ng mi T.

Li g i i

ma atana - - =

p mg g=> . = gtana = 10 Xtan9 = 10 X0,158 = 1,58 m/s2.

g' = _JL_ = _ i _ = 10,125 na/s2cosa cos9

_ 10=> T - * 2 (giy) = 1,987 (giy).

11ZD

b) Con lie n treo vo trn thang my, thang my chuyn ng theophng thng ng vi gia tc a.

Cng thc tnh chu k mi ca con lc T.

T = 2nJ- (14) a

Lu Trng hp thang my chuyn ng ln:

- Nu chuyn ng nhanh dn u: V cng chiu vi a, lcqun tnh

Fqt hng xuig, dy cng hn: g =g+ a.

- Nu chuyn ng chm dn u: V ngc chiu .a,. lc qntnh Fqt

hng ln -> dy chng: g = g - a. V d 18. Mt con lc n dao ng iu ha vi chu k dao ng T = 2 (giy).

Treo c on lc vo trn thang my v cho thang my chuyn ng nhanh

dn u vi a = theo hng i ln. Xc nh chu k dao ng mi

ca con lc.

Li g

g' =-g + a = g + =a = g + 8. = M10 10

h o

T :: 27tJ~ = 271 k r = Tj^f- = 2 (giy) X J = 1,9 (giy).

V d 19. Mt con lc n dao dng iu ha vi T = 2 (giy) ti nig = 9,8 (m/s). Treo con lc vo trn thang my v cho thang my i xung chm dn u vi gia tc 2 m/s2. Xc nh chu Td dao ng mi ca coa lc.

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11/291

Li g i i

Thang my i xung chm dn u -> V v a ngc chiu, lc qun

tnh Fqt hng xung dy cng.

g = g + a = 9,8 + 2 = 11,8 (ra/s2)

T' = 2tt

T' = T = 1,82 (giy)

Vy T = 1,82 (giy).

5. Chu k dao ng ca con lc n thay i trong in trng

a) in trng E hng theo phng nm ngang

+ Ti v tr cn bng mi ___E

^ + P + Fqt = 0 F = p + Fqt

g = V2 + a2

m Vq2E2

m

g' = g j l +q E

g2m2(15)

+ La :Nu bi ton cho gc l ch a th : COSa = =P' mg'

cos a(15')

V d 20. Mt con lc c chiu di = 10 (cm). Treo qu cu c khilng m = 10 (g), mang din tch q = 100 (iC). Con lc c treo giahai bn t c tch in t thng ng c in trng u theophng ngang vi E.= 400 (V/m). Cho g = 10 m/s. Hy xc ih chu kdao ng mi ca con lc.

Li g i i

TT ' = 2 M ?

. H q2E2 _ L O^IO-6)2 X4002Trong : g = g 1 + \ T = 1 0 +102 Xc o -2)2

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12/291

g = 10,77 => T' = 2c

10,77= 0,6 (giy).

b) in trng E hng theo phng thng ng t trn xung

+ Ti v tr cn bng: "+ p + F = 0

t p + F =P'

g + - g

Trong : a = s= m m

a l i lng i s, ph thuc du cain tch q, nn c th vit

g' = g + a

T = 2x 1 = 2n

= 2til

i g i + qEni H i m *

= 271 -

1 +qE

msr

T =T

1 +qE

mg

(16)

V d 21. Mt con lc n c chu k dao ngT. t con lc vo i

trng E hngthng ngxungdi.- Khi truyn cho qu cu in tch qi th chu k dao ng ca con lc ' T1 = 5T.

- Khi truyn cho qu cu in tch q2 th diu k dao ng ca con lc

, CoXc nh t s

%

Theo cng thc (16):

+ ng vi qj: T =

Li g i i

=5T

1 + qiEmg

r A E = 1 x q jE = J _ ( E = _ 24

mg 5 mg 25 mg 25

12



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13/291

+ n g vi q2-' T2 =

Y m

qE 24

1 2E = 49 q^E = 24 jng_ =


14/291

+ Chu k dao ng ca con lcvng inh l:

T = Tl + (17)2

+ Nng lng con lc cI bng nng lngcacon lc c2'.

W !=W 2 => = ' (18)l 2 . 01

V d 23. Con lc n c chiu di dy treo 1 ~ 1 (m), dao ng iuha vi chu k Ti = 2 (giy). Trn ng thng ng cch im treomt on 36 (cm) ngi ta ng mt ci inh. Khi dao ng dy treovng vo inh. Chu k dao ng ca con lc vng inh l bao nhiu.

i chu k Ti = 2 (giy). Trn nglon 36 (cm) ngi ta ng mt ci (vo inh. Chu k dao ng ca con

Li g i i

% = = 2 (giy)

~ = => T2= 0,8Ti = 0,8 X 2 =T2 0,8

= => T2 = 0,8Ti = 0,8 X2 = 1,6 (giy)0,8

=> T = Tl +-S - = = 1,8 (giy)2 2

Vy T = 1,8 (giy).

7. Con lc in tng qut: 0 < a 0 < 90+ Vn tc ca vt khi i qua v tr li gc a:

V2 = 2gZ(cosa - coscco) (19)

+ Lc cing ca si dy khi vt i qua v tr c li gc ct:

= mg(3cosa - 2cosa0) (20)

V d24. Mt con lc n c dy treo di / = 1 (m)dao ng ni cgia tcc g = 9,8 (m/s2). Bit li gc cc i ao = 30. Hy tnh tc ca vt kh qua v tr cn bng.

Li g i i

Theo cng thc (19):

V2 - g/(cosa - cosccq) = 2x 9,8 X l[cosO - cos30] = 2 X 9,8 X 1

V = 1,62 (m/s).

i - 2

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15/291

V d 25. Mt con lc n c qu cu khi lng m = 50 (g). a qucu ra v tr c li gc o = 60 v bung khng vn tc ban u. Bit gia tc g = 9,8 (m/s). Xc nh. lc ciig ca dy treo khi vt qua v trc li gc a = 30.

Li g i i

Theo cng thc (20):

'= mg(3coscc - 2coscc0) = 5.lo'*2 X 9,8[3cos30 - 2cos603

= 5.10 X 9,8 V 13, - 2 . -

2 2= 0,78 (N).

8. Cng thc v rong cng mt khong thi gian t, con lc c chu k dao

ng T thc hin rti (dao ng), cn con lc c chu k dao ng 2 thchin n2 {dao ng)

t niTi = 112T2 (21)

V d 26. L xo c cng k = 80 (N/m). Ln lt gn hai qu cu ckhi lng ni! v m2 vo l xo, ri kch thch cho vt dao ng. Trong cng mt khong thi gian con lc c khi lng .m.1 thc hin. 10 (daong), cn con lc c khi lng m2 thc hin 5 (dao ng). Khi gn c

hai vt 1! v m2 vo l xo tl h dao iag vi chu k T = (giy).

Hy xc nh nil v m2.

Li g i i

+ Theo cng thc (21): 111T1 = n2To => 10TX= 5T2

2T] = T2 => 2-Jrn = ^m2 => 4mi =:

_ _ k _ 80 - c ,=> ffij+n u = - = = 5 (kg)16 16

( VkV +m2 = --

4m1 = m2

mi + m2 = 5 (kg)

m2 = 1 (kg)

m2 = 4 (kg).

V d 27. Hai con lc dn c chiu di chnh lch nhau 22 (cm). Trongcng mt khong thi gian, C011 lc c chiu di dy treo l thc hinc 20 (ao ng), cn C011 lc c chiu di y treo l2thc hin c24 (dao ng). Tm chiu i ! v l2ca cc con lc.

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16/291

Li g i i

Theo cng thc (21): niTi = iijjTjj => 20Tj = 24T2

20 X2 t - 24X

5-J =6 ^ => 25 = 36 2

Suy ra Zi >2hay /i - li = 22 (cm).

l \ - 2 - 22 (cm) Zj - 72 (cm)

25L - 36 2 z2 = 50 (cm).

9. Lc n hi

+ Fwx = k(A/ + -A) ' (22)

+ Fmin - 0 khi A > AZ (23)

+ Fmin= k(Al- ) khi AZ > A (24)

Lu :lc hi phc f = -kx hay If I = k [ XI.

V d 28. Mt vt c khi lng ra = 0,25 (kg) c treo vo l xo c cng k ss 10 (N/m). Bit con lc l xo dao ng vi bin A = 10 (cm)

v gia tc g = 10 (m/s2). Tnh ln cc i ca lc n hi.

Li gi i

Fmax = k(Al+ A)

7 mg 0,25 X10 __ . .AZ = = = 0,25 (m)k 10

Vy F** = 10 X(0,25 + 0,1) = 3,5 (N).

V d 28. Mt con lc l xo c khi lng m = 400 (g) dao ng iuha vi chu k T = 0,6 (giy) v biu A = 8 (cm). Xc nh ln

ca lc hi phc ng vi pha dao ng .6

Li g i i

f = kxlX = Acos(wt + ) - 8CDS = 4 V3 (cm)

6

2 k _ 1 _ 20) = => k= mu>m

f i =mw2lx| = 0,4 X x4"v/3.1Cf2 = 0,4 X X 4V3 .IO2I t ; 0,6

Vy f[ = 1,84 (N).

16



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17/291

V d 30. Mt con lc l xo treo thng ng c khi lngm = 0,1 (kg)v cng k = 40 (N/m). Bit bin A = 3 (cm) v gia tc g =10(m/s2).Xc inh ln cc i v cc tieu ca lc n hi.

L g i i

. mg 0,1x10+ A/ = =

---- - = 2,5 (cm)k 40

VA>AZ => F |min = 0(N )

+ IFI aw, = k(Ai + A) = 40(2,5 + 3).10-2 = 2,2 (N)

Vy !Fl min= 0 (N) v IF max = 2,2 (N).

V d 3.1. Mt con lc l xo treo thng ag, dao ng ti ni g = 10 (m/s2)vi bin A = 10 (cm). Trong qu trnh dao ng t s ln cc i

7v cc.tiu ca lc n hi bng . Hy xc inh chu k dao 'ing T.

3

Li g i i

1 F i ma* = k(AZ + A)

|F |min = k(AZ-A)

lFLX = A/ + A 7A - A 3F| lr

= -- hayA/*1

A/-10

Suy ra Al= 25 (cm).

- - I -

10

Y0,25

= 2 (rad/s)

_ 2n 2nT 1 (giy)

C 2ji

Vy T = 1 (giy).

1. Nng lng ca con lc l xo v con lc n

a) Con lc l xo

TTT TT7 my2 kx2 1. -2w = w + w t= -+ --= -kA

2 2 2Trong : W = Wsin2(wt +


18/291

mv2 m 2 a2 .- 2 / X \+ w rt = - = (0 A sin (o t + (p)2 2

=> w = (02A 2 cos2 (cot + (p) + ^ a)2A2sin 2(cot + (p)2 2

= V A* = 2 - a I? = .< x (26)2 2 2 0

c) Chu k T, vn tc gc (' v tn s" ao ng f ' ca ng nng v th nng trong dao ng iu ha:

C = 2o)

f = 2f

TT = (27)

2

d) Sau nhng khong thi gian t thi ng nng w bng th nng w t

w .. whay w X - .2 2 2

V d 34. Mt con lc l xo dao ng iu ha vi cnng w = 0,12 (J).Bit o cng ca l xo k = 150 (N/m) v khi con lc c li 2 (cm) th

tc ca vt l 10jV3 (cm/s). Tnh bin A v chu k T.

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19/291

T I7 _ *2 _ *2 _ 2W _ 2x0,12+ w = -k A => A = = - r => A = 4 (cm)

2 k 150

+ V2 = 0)2(A2 - X2) => 0) = v = = - - Tt

+ T = = = 0,4 (giy)co 5 h

Vy A = 4 (cm) v T = 0,4 (giy).

V d 35. Mt con lc n gm qu cu c khi lng m = 100 (g) treovo si dy di 1 = 1 (m). Cho gia tc g = 10 m/s2. a con lc lch mt gc (Xo = 0,2 (rad) so vi phng thng dng ri bung nh. C nngca con lc khi dao ng bng bao nhiu ?

Li g

w X ^ a = 0j* xl0>


20/291

11. Cng thc v tng hp dao ng

a) Hi dao dng ngpha

A A i +Ao

= O; =


21/291

V d 40. Hai phng trnh dao ng iu ha:

Xj = 7cos nt+ (cm); Xo = 3cosf Tt + i (cm).I 4 j l 4 )

Lp phng trnh ao ng tng hp.

Li g i i

A V _ nAi > A2 tan = ---------------- ----- -

Aj COS + A2 cos


22/291

V d 42. Hai dao ng iu ha:

Xj = 3cos 2rct + j (cm); x2 = 5cos^2jit + j (cm).

Xc nh bin dao ng tng hp A.

Li g i i

A2 = + A | +- 2A1A2cos(q>! - (p2) = 32 + 52 + 2 X 3 X 5 c o s -3

= 9 + 25 + 15 = 49

=> A = 7 (cm).

12. ao ny tt dn

+ Gi AA l gim bin sau mi chu k

AA = (33)

k+ Gi A' l bin sau mt chu k

A = A - AA (34)

V d 43. Mt con lc l xo dao ng tt dn ti ni g = 10 (m/s2).Bit h s' ma st = 0,1 v khi lng ca qu nng m - 10 (g) vk = 10 N/m. Hy xc nh d gim bin trong mt chu k.

Li g i i

k.AA = 4)img = 4 X 0,1 X 1CT2 X 10 = 0,04 (m)

=> AA = _ 0,004 (m) = 0,4 (cm).10 N/m

V d 44. Mt con lc n dao ng tt dn chm, c sau mi chu kth bin gim i 2%. Nng lng b mt mt trong mt dao ng l bao nhiu phn trm ?

Li g i i

A- . f 2 ^ \ A . i ! x A -, 100 J 100

W = k A '2 ; w = -k A 22 2

1 f 98 982 1,V = k ---- .W = - k .A = - k A z =0,96W

2 U ) 1002 2

AW = w = 0,96W = 0,04W hay AW = 4%.

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23/291

13. Dao ng cng bc. Cng hng c

+ Gi T l chu k ca ngoi lc tc dng ln con lc dao ng tt dn.

+ Gi khong cch gia 2 ln lin tip m ngoi lc tc dng ln con lc.

+ .Gi T0 l chu k dao ng ring ca con lc:

To = T

Vy vn tc ca xe m con lc treo trn l:

V= A (35)

V d 45. Mt con lc em c chiu di dy treo 39,2 (cm) c treoln trn mt toa xe la. Con lc b tc ng mt ngoi lc kh gp ch ni ca hai thanh ray. Bit khong cch gia hai ch ni lin tip l 12,5 (m). Cho g = 9,8 (m/s2). Khi bin ca con lc c ln cc ith vn tc ca xe la l bao nhiu ?

Li g i i

\ 0,392+ T = 2^ g = = ^ 2S (giy)-

+ Khi bin ln nht: T T0 =: 1,25 (giy)

_ _ 12,5 (m) 1/w_ , ,=> V= - = 10 (m/s)

T 1,25 (giy)

Vy V = 10 (m/s).

V d 46. Mt con lc l xo c qu nng khi lng m = 100 (g) daong tt dn chm. Tc ng vo con lc ngoi lc bin thin tun hon vi vn tc gc o = 2 (rad/s) th bin dao dng ca con lc c ln cc i. Xc nh cng k ca con lc.

L i g i

Khi bin cc i th vn tc gc ca dao ng co= Q= 20 (rad/s).

M = f Vy = 20

k = m X202 = 0,1 X4Q0 = 40 (N/m)

Vy k = 40 (N/m).

4. Cng thc v lch pha gia hai im trn phng truyn sng

2nx 2tdA = hay A


24/291

V d 47. Mt sng c c tn s f = 20 (Hz) lan truyn vi vn t100 (cm/s). Xc inh lch pha gia hai im cch nhau 25 (cm).

Li g i i

2i 27t 27f 271X25 (cm)X20 (Hz) ,Acp - = = = ---------------------- - 17T(rad/s).

X V V 100 (cm/s)

f

V d 48. Sng c,c tn s f = 80 (Hz) lan truyn trong mi trng cvn tc 4 (m/s). Dao ng ca cc phn t vt cht ti hai im trn mphng truyn sng cch ngun ln lt l di = 20 (cm) v d2 = 21,25 (cmlch pha nhau mt gc l bao nhiu.

Li g i i

2nd 27r(d2 - d1) 27i(d2 - d 1)f 2tu(21,25 - 20).80 71,Acp= = ----- ----- = ----- ----- = --------------------------------

k X V 400 2

V d 49. Mt sng .c c tn s f = 136 (Hz) truyn trong khng khtheo mt phng c'tc . 340 (m/s). Xc nh khong cch gn nhgia hai im c dao ng cng pha.

Li g ii

* 27td+ Ao =

+ ng pl a: Acp = + 2k7T

Gn nht: k = 0 o A(p = = Jt

J _ ^ _ v 340 _ * nr / \d = = - - = 1,25 (m).2 2f 2 X136

V d 50. Phng trnh sng ti ngun 0 trn mt nc Uo = Acos27Tt (cmim M trn phng truyn sng cchngun0,3 (m.) ao ng lch ph

vi ngun mt gc v gn ngun nht.Xc nh, tc truyn sng.4

Li g i i

Acp = M cp = + 2kxX 4

Gn nht: k = 0' '=>' A = 4

2nd~T~

Vy V = 2,4 m/s.

Suy ra = =>d = = => V = 8df = 8 X0,3 X 1 = 2,4 m/sX 4 8 8f

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25/291

* Y d 51. Mt dy dn lai lt i dao dng vi tn s trong khong 22 (Hz) dn 26 (Hz) theo phng vung gc vi si dy. Bit tc truynsng V = 3 (ra/s). Mt im M cch ngun sng mt khong 30 cm) dao

71ng lch pha so vi ngun Aq> = (2k + 1). Xc dinh tn s (lao ng

2ca dy.

2/U+ A

=

2 V

f = (2k + l)v _ (2k + l) 300 _ 2 5(2k + 1)4d 4 30

Theo gi thit: 22 < f f = 2,5(2k + 1) = 2,5[(%x 4) + 1] = 22,5 (Hz).

15. Cng hc v' s' cc i v cc tiu trong giao hoa ca sng c

a) Cc di: - < (2k + 1) - < z2 2

Hay -21< (2k + 1) < 2 (38)

Lu : Trng hp hai sng ngc pha (37) dng cho cc tiu, cn (38)dng cho cc i.

c) Mt phng php kh c

Gi N l skhong vn i v mt pha ng trung trc ca SiS2:

N = I - ~

2i 2 X 2

Trong : m l s nguyn, n l s thp phn.

_ N = = m, nA.

(39)

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26/291

+ n =: 0 th c 2(m - 1) + 1 cc i

+ 1 < 11< 9 th c 2m + 1 cc i

+ n < 5 th c 2m cc tiu

+ n > 5 th c 2(m + 1) cc tiu.

* V d 52. Hai ngun kt hp Si v S2 trnmt nccch nhau 13 (cm)dao ong vi phng trnh Uj = u2= 2cos407t (cm).Tc truyn sng

V = 80 (cm/s). S im dao ng vi bin cc i trn on SjS2 lbao nhiu ?

Li g i i

* Cch 1:Theo cng thc (37):

V- l k = 3; 2; 1; 0. vy c 7 v tr cc i.

* Cch ::Theo cng thc (39): N = = = ^ x w3,2X V 80 '

Vy s cc i: (3 X2) + 1 = 7.

* V d 53. Hai ngun s.ng kt hp Si, s 2 cch nhau 21 (cm), cng pha ;ao ng vi chu ki T = 0,5 (giy), v tc truyn sng 8(cm/s).Xc 'nh s im khng dao ng gia Si v s 2.

Li g i i* Cch J:Theo cng thc (39):

-21


27/291

* Cch 1: -21< (2k + 1).< 21

-42 (cm) < (2k + 1)8 (cm) < 42 (cm)

-3,1 < k < 2,1 => k =-3; 2; 1 v 0

Vy c 6 v tr khng dao ng.

* Cch 2 :Theo cng thc (39): N = = = 2,6X 8

S cc tiu: 2(m + 1) = 2(2 + 1) = 6 v tr.

6. Cng hc v sng dng c hai u c' nh

iu kin c sng dng: = d = (40)

Vi k = s' bng = s nt - 1.

* V d 55. Mt dy n hi AB di 2,1 (m), u B c nh, cn u Agn vi mt nhnh m thoa dao ng vi tn s' f = 10 (Hz), to sng dng v tc 7 (m/s). Hy xc nh s ut.

Lr g i i

S' nt bng (k + 1) = 6 + 1 = 7-

Vy c 7 (nt).

V d 56. Mt dy n hi AB di ,u B cnh, cn u Agn vimt nhnh m .thoa dao ng vi tn s f, to sng dng. Trn dy cmt bng sng. dy c ba bng sng th tn s dao ng l bao nhiu ?

V d 57. Mt dy n hi AB di 1 - 2 (m), u B c nh, cn du Agn vi mt thanh m thoa dao ng vi tn s f = 20 (Hz), to sngdng. Bit tc d truyn sng bng 10 (m/s). S bng trn dy l baonhiu ?

kX _ kv

2 - 2f=> k =

V

Li g i i

2l _ 2 X 2,1 (m) X 10 (Hz)

V 7 (m/s)

Li g i i

Li g i i

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28/291

2Zf 2 X2 (m) X 20 (Hz) k = = ---- ---------- = 8

V 0 (m/s)

Vy c 8 (bng).

V d 58, Mt si dy n hi AB di l - 1,2 (ra), u B cnh, cu A gnvi mt nhnh m thoa, to sng' dng. Ngi ta thy c

v tr khng dao ng (k c v B). Xc nh bc sng X.Li g i i

_ k ^ 1 _ 21 - 2 x.l,2 '(m)- ^l = ~~ => K = = ----- -------- = 0,8 (m)2 k 3

Vy ?L= 0,8 (m).

Lu : i vi' sng dng thi gian gia hai ln dy dui thng ha

i ' ( Tgia hai ln to bng sng lin tip l mt na chu k It =

V d 59. Trong th nghim to. sng dng vi si dy n hi d - 1,2 (m) vi hai u c' nh, th thy c 4 nt sng. Bit khong thigian gia hai ln si dy dui thng lin tip l 0,05 (s). Xc nh t truyn sng.

Li g i i

= v x T => V=T

rp A ' _ 21 _ 2 x !>2 _ n Q \

Trong : X = = ------ -------- = 0,8 (m)k 3

T = 2t = 2 X0,05 = 0,1 (giy)

= > v = - = = 8 (m/s)T 0,1

Vy V= 8 (m/s).

17. Cng thc v sng dng mt u t do, m u C nh v sng dngtrong ct khng kh

= d = (2k + 1)- (41)4

Vi k = snt - 1

k = s' bng - 1.

V d 60. Mt dy n hi di 1 - 1 (m). u B t do, u A gnvi mt nhnh m thoa dao ng vi tu s f = 20 (Hz), quan st trndy thy c 3 bng sng. Xc nh tc truyn sng.

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29/291

L i g ii

X

4

4Zf 4x 1 (m) X 20 (Hz)

= (2k + 1 )- = (2k + 1)4 4f

= 16 (m/s)(2k + 1) (2 X2) + 1

Vy V = 16 (m/s).

V d 61. Mt dy n hi AB di , u B t do, cn u A gn vimt nhnh m thoa, dao ng vi tn s f = 5 (Hz). Quan si trn dy c 6 nt. Bit tc truyn sng V = 1 (m/s). Xc nh chiu di cady.

L i g i i

= (2k +1 ) - =(2k + 1) = [(2 X5 ) + 1] X - L - = 1 1 ^ 0 ,5 5 (m)

4 4f 4 x 5 20V =0,55 (m).

V d 62. Mt m tho t gn ming ng thy tinh d h c chanc, ao ng vi tn s f = 680 (Hz). Bit tc truyn sng trongnc vV= 340 (m/s). Xc nh chiu di ngn nht ca ct khng khtrong ng thy tinh ming ng m nghe c ln nht.

Li g i i

= (2k + 1). Ngn nht k = 0

4 = (2k +i ) X = f = L - = (m)

4f 4 f 4 X680 8

Vy = - (m).8

18. Cng hc v sng m

+ Gi p l cng sut ca ngun m.

+ I l cng m ti v tr cch ngun mt khong R:I = _ L - (42)

4nR

V d 63. Loa ca mt my thu thanh c cng sut p = 1 (W). Cng m ti mt im cch loa R = 10 (m) l bao nhiu ?

Li g i i

I = L = 8.1CT4 (W/m2).4nR2 471X10

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30/291

V d 64. Nu mc cng m ti mt im cch ngun m R = 112 (cm)l 73 (dB) th cng sut ca ngun m l bao nhiu ?

Li g i i

L (dB) = lOlg = 73 (dB) => I = I0.107'3 = lo -57lo

p = 4 tiR2.I = 4 X 3 , 1 4 .1122 X lo '57 = 3,15 (W)

Vy p = 3,15 (W).

19. Cng thirD v lch pha ca u so vi i

u = u 0cos(cot + (pu)

i = IoCOSCOt + (pi)

Gi (p) lch pha ca u so vi i

= (pu - (>i (43)

V d 65. Cho on mch ch c cun thun cm L vi L = (H). t71

in p xoay chiu vo hai u on mch u = 200-\/2cosl007rt (V). Lpbiu thc cng dng in i.

Li g i i

i = Iocos(100nt + cpi)

z = L-= -.1007C = 100 (Q)%

Xo = = 2 V2(A)z 100

cp = (pu - cpi = -cp =

Vy i = 2-V2cosl00it - j (A).

V d 66. Cho on mch ch c cun thun cm L. t in p xoaychiu u = UoCoslOOit (V) vo hai u on mch.

Ti thi im t: u = 60V2 (V), i = V (A). Bit L = (H). Lp i.

Li g i i

i= I0cos^lOOjct-j hay i = IoSnl007Tt

cos2100rt + sin2100rrt = =>u2. I*z2 ' 2

2 ;2 2 24---= 1

T

0 10

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31/291

2.60

I? X 60 2 11

Vy i = 2-V2cos| 1007it - -j(A ).

V d 67. Cho on mch R, L, c mc ni tip vi R = 100{Q),L =:(H),71

q-4c = (F). t in p u = 200V2cosl00rct (V)gia hai u on

mch. Lp i.

Li g i i

i = I0cos(1007lt + (pi)

+ z = tr 2 + (Zl - z c )2 *=Vioo2 + (100- 200)2 = 100V2(Q)

n 200V2! = = 2 (A) 100V2 ZL - zc 100200 7t

+ tan

- cp' tpi = -


32/291

+ (Fa) = (Fa)i - - = lOOttt + - - - = { lOOnt- - ic 2 4 2 V 4)

Vy uc = 400cos^L007it - j(V)..

V d 69i Cho on mch R?L, c ni tip vi R = 100 (Q), L = (H)410-4 , .

c = r (F). t in p xoay chiu vo gia hai u oan mach th2t '

Ur = 200cos|^1007t + 1(V). Lp biu thc u.

Li g i i

+ i = I0cos(1007t + i)

Ur = 200 =R 100

(Fa)i = (Fa)UR = (OOxt + j) = l00:rt +

i = 2cosl007rt + j .

+ 0 = I0Z = Iq^R2 +(Zl -Z c )2 = 2V1002 + (100 - 200)2 = 200V2(V)

ZL- Z C 100- 200 , 7t.+ tan = = --------- = -1 > = R 100 4

M = u - i (pu = cp +


33/291

' _ ZL OV rr 71tantp = = V3 => =

R 50 3

n/r:, - n It 7M (p =


34/291

V d 73. Cho on mch R, L, c mc ni tip vi R = 100 (Q), L = (H),71

10" c = - (F). t in p u = 200V2cosa>t(V). Thay di vn tc gc C

2r.th thy cng sut tiu th trong mch t cc i. Xc nh (0v Pmax.

Li g i i

+ Zl = Z(

L(0 = => >= == = = = IOOJ1V2 (rad/s)Cu> Vl | 10"*

i n ' 2 t i

+ Pmax = = = 400 (W)R 100

Vy I) = IOO71V2(rad/s) v Pmax = 400 (W).

b) Cng d ng in trong m ch cc i

= M5)

L, r C V d 74. Cho on mch: 0 nrs'flX'7tyf-a

Vi r =: 50 (Q), L = (H). t din p u = 10oV2cosl007rt(V) vo gia71 ",

hai u on mch th cng dng in trong mch t gi tr cci. Xc nh c v Imax-

Li g i i

+ z c = Z|. = L) = X1007T = 100 ()T

c = - - = 51 (F)ICOXIOOt 71

+ U m = v . m r n . 2 Wr 50 (Q)

Vy c = (F) v 1 ^ = 2(A).

R C V d 75. Cho on mch xoay chiu: &d I I 4

R = 100 (Q), L = - (H), r = 25 (), c = (F).71 2n

t in p u = 200V2cosa)t (V) vo gia hai u on mch. Thay i

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35/291

co th thy cng dng din trong mch t cc i. Xc nh tn s f v Iroax.

Li g i i

+ = = > f= = ---- => f =50V2(Hz)VLC 271-y/LC _ 1 10-4

2ttJ - 1 71 2t

u 200 200 , A \+ Imax = ------ = ------- = = 1,6 (A)

R + r 100 + 25 125

Vy f = 50 J2(Hz) v Inax =1,6 (A).

V d 76. Cho on mch R, L, c mc ni tip, vi R = 60 (Q). t

u = 12oV2coscor (V) vo gia hai u on mch. Khi thay i L th cng dng din trong mch t cc i. Xc nh cng dng in cc i Imaj(.

Li g i iIm s i= H = i ? 2 i X > = 2 ( A ) .

R 60 (Q)

c) i n p gia h ai u in tr t gi tr cc

Uniax = u (46)

V d 77. Cho mch in R, L, c mc ni tip, vi R = 100 (Q), L = (H)71

v c thay i c. t in p u = 20.(W2cosl0(kt(V) vo hai uon mch th in p gia hai u in tr t gi tr cc i. Xcnh c v Uftnax.

Li g i i

+ z c = ZL = Leo = - X1007 = 200 (Q)n

1 1CT4 _c ---------------=


36/291

+ Zl = Zc

T _ 1 _ , , _ 1Lo = --- ( = ----CD. VLC

Hay f = * = -, - L .......- = 50>2 (Hz)2rcV x 2 ( T 1 0 J

V 2tt tu

Li g ii

+ Rraax = = 200 (V)

Vy f = 50V2 (Hz) v Rmax = 200 (V).

V d 79. Cho on mch (nh hnh. bn). R L r

_ 104 *

Bit R = 50 (), T = 50 (fi), c = (F)..7 ,

t in p u = 200V2cosl007tt (V) vo gia hai uon mch thUR= 200 (V). Xc nh L v cng sut tiu th p.

Li g i i

+ R.= u = 200 (V) o ZL=Zc, L = = = (H )C lQOn 7C

2 200- = = 400 (W)(R + r) (50 + 50)

Vy L = (H) v pmax = 400 (W).71

d) Khi c thay i th in p gia hai u cun thun cm L gi tr cc di

U Lmax = ^ .Z L (47)R

V d 80. Cho on mch R, L, c mc n tip, vi R = 50 (), L = (HTtv c thay i c. t in p u = 200V2cosl00ttt (V) vo gia hau on mch th in p gia hai u cun thun cm L t gi trcc i. Hy xc nh, c v Lmax.

Li g i

+ Zc = ZL x> C = = --------------- = (F)Zr .ca 100 X100ji 71



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37/291

+ I W = ^ = 255x100 = 400 (V)R 50

Vy c = (F) v ULmax = 400 (V).71

e) Kh L thay i th in p gia hai t in c t gi tr cc

cma* = 2c (48)K

_ 104 V du 81. Cho oan mach R, L, c mc ni tip vi R = 100 (Q), c = - (F).

2n

t in p u = 200>/2cosl007t (V) vo gia hai u on mch th c

t gi tr cc i. Xc dnh L v cmax-

Li g i i

+ ZL= Zc= i ------ = 200(Q)Cto KT4 X 100712n

L = 2= 2 = (H)0) lOOn %

+ t w = .z c = X 200 = 400 (V)R OO

Vy L = (H) v Ucmax = 400 (V).n

f) H s cn g su t cos

/2(Hz).

37



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38/291

L, r c V d 83. Cho on mch: gHyfflf'- - >B 0

Bit L = (H), r = 100 (Q), c thay i c.2k

t in p u = 100V2cosl00rct(V) vo gia hai u on mch. Thayi c t-i thy cng dng in trong mch t gi tr cc i. Xcn h h s cng sut C0S(J> v c.

Li g i i

+ => R = z

cos

L = (H)71

+ (cOS


39/291

V d 85. Cho mch, in R, L, c mc ni tip, vi R = 100 (2), L = (H)71

v c thay i c. t in p u = 200V2cosl00tt (V). Kh c thay dith in p gia hai u t in t gi tr cc i. Xc nh c vUcmax-

Li g i i

_ R 2 + Z l+ Z c =

ZL = L(0 = . 10071 = 1 00 (2)71

z c = 100 +10 2 = 200 (Q)100

c = = ( F )1 c n

z c( 200 XlOOt 2n

+ Uo = IZc = .Zc = - u ,=.zcz Vr 2 + l - Z c >2

200 X 200 r-= 200V2 (V)V1002 + (100 - 200)2

Vy c = H i (F) v Gmax = 200V2(V) ' Gmax

b) Mch in R, L, c ni tipL thay i th ULt gi tr cc i: Lmax-

_ R 2 + Z gZl = -- (51)

Lc

Vd 86. Cho on mch xoay chiu R, L, cmc ni tip vi R = 50 (Q);1 0 - 3

c = (F) v L thay i. t in p u = 100-\/2cosl007rt (V) vo gia5n

hai u cun thun. cm L t gi tr cc i ULmax. Xc nh L v Lroax-

Li g i i

R 2 + z + ZT =

zc = 77- = T 1 = 50 (n)Cco xo-3

X IOOtt 5i

39



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40/291

502 + 502

^ k T2- i0

L - . S t . m . l a oC 107I 71

+ UL = IZL= H.Zl = u = 3 ^- rj L ----1- ^ >Lz VR + (ZL - Z C)2

- = 100 = =.100 = 100V2(V)V502 + (100 - 50)2

Vy L = (H) v ULmax =100^2 (V).71

c) Mch i n R, L, c mc ni tipKhi (0thay i th UCmax:

C2 = - -5 (52)LC 2L '

V d 87. Cho on mch R, L, c mc ni tip, vi R - 100 (Q); L = - (

v c = (F). t in p u - UVsicoscot (V) vo gia hai u o2 T

mch. Khi 0) thay i th din p Uc t gi tr cc i. Xc nh co.

Li g i i

1 T>2 1 1/1 2 iM_2C02 = = - 4 _ - ^ = 2 . 10v - ^ = .1 0 ^LC 2L 1 O-4 9 _L 2 2_ . _ l.

it 2% l

= lOOnJ=

Vy ) =OTC -(rad/s).

d) Mch i n R, L, c mc ni tip

Khi (0thay i th ULdt gi tr cc i.

(53)C2 2

_ 2 V d 88. Cho on mch R, L, c mc n tip, vi R = 100 (ft); L = (H

71

v c = (F). t in p u = Vcosot (V) vo gia hai u o2iz

mch th in p U l t gi t cc i. Xc nh C.

40



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41/291

hi gii

_ _ 100 1. T _ R2C2 2 10-4

1j(J 2n 10-4 10-4 7.10'

C2 2 n 2rt 2 712 8t i2 8rc2

S%2 2a/2 .ttX 10 0- => ( = ---= -7.10 v7

/2Vy co = 200kJ(rad/s).

22. Cng thc v bi ton cc tr khi R thay i

a) on mch R, Ly c mc n' tip

Khi R thay di th cng sut ton mch t gi tr cc i Pmax:R = Z l - Z c (54)

2 R jr L C(55)Pmax 2R

7T V2 R v coscp.



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42/291

+ p -max 2R 2

L i gi i

200Trong R = Zc = 100 (Q) => = 200 (W)

2x10 0

c 100 . n _V2+ tancp - = -1 =>

COSP = - R 100 4 2

V2Vy pmax = 200 (W); R - 100 (2) v cos = .

2

V du 91. Cho oan mach xoay chiu: R L, r ci K Z M W 11 *

_ 14 10 _Bit L = (H), c = (F) v r = 10 (f).

71 71

t iin p u = 100V2coslOOrct (V) vo gia hai u on mch th cng sut trong mch t gi tr cc i. Xc nh R v Pmax-

Li g i i

+ R + r - IZ L - z c .l = 1140 - 100 = 40 ( )=> R = 40 - r = 40 - 10 = 30 (Q)

2 1002+ Pmax - ------ = = 125 (W)max 2(R + r) 2(30+10)

Vy }l = 30 (O) v pmax = 125 (W).

b) o n mch R, L, r, c mc ni tip

Khi R thay i th cng sut ca in L r ctr R (it gi tr cc i Pftmax: 1 / 1 'TTT' I---- *

R = 2+ (ZL - z c (57)

2Rmax 2(R + r)

(58)

R / U r V d 92. Cho on mch xoay chiu: 0 *

t in p u = 100V2cosl007it(V) vo hai u on mch.

2 _Bit r = 30 (Q); L = (H). Thay i R th cng sut tiu thu trn in

5irtr R (lt gi tr cc i. Xc nh R v PRmax-

Li g i i

+ R = / r 2 + z

42



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43/291

z = Leo = .IOOt = 40 (Q)5 7t

R = V302 +402 = 50 (Q)

u 2 1002+ PRmax = -----= ----- = 62,5 (W)

Rmax 2(R + r) 2(50 + 30)

Vy R = 50 (Q); PRxnax = 62,5 (W).Lu :Cng thc (57), khi khng c t c h R = .- r2 + z | .

V d 93. Cho on mch xoay chiu: *^25''YT -

t in p xoay chiu u = lOoVcoslOOTt (V) vo hai u on mchth cng sut tiu th trn in tr R t gi tr cc i.

1 10-3Bit r = 25 (Q); L = - (H); c = (F). Hay xc inh R v PRmax-

2ti 7,5tc

Li g i i

+ R = ^r2 +(ZL- Z c )2

Zt = Lco = .lO O it = 50 (n)2K

z c ~ J _ = ---J ------ = 75(Q) => R = V252 + (50 - 75)2 = 25V2 (Q)C o 1CT3

.100*7,5*

u 2 _+ PR = r ~ = ----- ------ = M W )

u 2 _ ' 10Q2 _ 200

2(R + r) 2(25^2 + 25) ~ (V2+ 1)

Vy R = 25V2(Q) v PEmsJt = ! (W).. (V2+1)

3. on mch R, L, c mc ni t ip: W Y '-Khi R thay i th cng sut tiu th trong mch k h E L g thay i.

Khi R = Ri hay R = R2thi cng sut tiu th khng i:Ri X R2 = (ZL- Zcf (59)

Lu : + Khi on mch khng c t in c

Rj XR2 = Zl

+ Khi on mch khng c cun L

r 2 Xr 2 =Zq

43



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44/291

+ Khi cun cm c in tr hot ng r

(Ri + r)(R2+ r) = (ZL- z c)2-L

V d 94. Cho on mch xoay chiu: 0I / 1 OTOP 0

Vi L = (H). t in p xoay chiu gia hai u on md71

u = 200V2cosl07tt (V). Khi R = R] = 50 (Q) v khi R = R2 th cnsut trong mch khng i p. Xc nh R2 v cng sut p.

Li g i i

+ Rx xR2 =z

_ _ z 1002Ro = = = 200 (Q)

Ri 50

T3_ 2d _. 2 _ u 2 _ 002 X50+ p = IR, = - - R i = -.R-, = - - - = 40 (W)

7 ' B + z 1 S & + 1002

Vy R = 200 (5) v p = 40 (W).R / C

V d 95. Cho on mch xoay chiu: 0 ------ I #2Q-3

vi C = (F). t in p xoay chiu u = 100-V2cosl007t (V) gi5rt

hai u on mch. Khi thay di R vi R = Ri = 40 (>) v R = R2 thcng sut trong mch khng i. Xc nh R2v cng sut p,

Li g i i+ Ri XR2 = Zq

Zc = =---- --------- = 50 (Q)c Cco 10-3

-IOOtt 5tt

+ 40 X Ra = 50 2 => Rs = 62,5 (Q)

97j5(W)Z R +z 40 + 50

Vy R2 = 62,5 (Q); p = 97,5 (W).RJf L

V d. 96. Cho mch in xoay chiu: *z_}rtCs-*"Vi L = (H); c = (F). t in p u = V2cosl007t (V) vo gi

71 2n . hai u on mch. R thay i c. Khi R = Ri v khi R = R2 th cnsut trong mch khng i. Bit khi R = Ri cng dng in tronmch ln gp hai ln khi R = R2. Xc nh Rj v R2.

44



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45/291

+ Ri XR2 = (ZL Zc)2

ZL = Lc = -.100* = 100 (D)

zc = T- = z r------= 200 (Q)Co 10-47.10071

2t

Rl XRa = (100 - 200)2 = 1002 (1)

+ M li = 2I2: P = p 2

. i?Ri = i|R2

. l ^ = r 2 => R2 = 4R1 (2)

T (1) v (2) ta c: Rj = 50 (Q); R2= 200


46/291

R w V d 98. Cho on mch xoay chiu: I i ' H i ' 0

t u = uVcoslOOrtt (V) vo gia hai u on mch.

+ Khi L = Lj th lch pha ca u so vi i l vi costpi = 0,5 v cngsut tiu th Pi = 100 (W).

+ Khi L = L2 thi COSCP2= 0,6 v cng sut tiu th P2. Xc nh P2. .

Li g i ip j c o s 2 2

( W) = ^ => p 2 = 144 (W).p2 (0,6)

R Lyr V d 99. Cho on mch xoay chiu: 0I-----

t in p u = uVcoslOOtt (V) v R = lOO-v/3 (H).

Khi L = u th = I01>/2coslOOTit - -j(A ) v Pi = 100 (W).

Khi L = L2 th cng sut tiu th p2 = 150 (W). Hi cm khng bzbng bao nhiu ?

Li g i i

Pj COS2 ip:

P 2 COS2 (f>2

V2

100 C0S2 (P! [ 2 j 73 = - = => COS (09 =

150 cos 2 cos tp2 2

q = => tanq = ^ 2L = Rtan2 = lCK)V3tan6 R 6

L = V. = 100 (Q)2 v3

Vy L2 =- (H) . n

R y*V d 100. Cho on mch xoay chiu: * ---- J /TT *

t in p xoay chiu u = uVcoslOOrtt (V) vo gia hai u onmch. Khi c = C th cos(pi = 0,5 v cng sut tiu th Pi =: 100 (W).

Khi c = C2 th COS2= 0,5 v cng sut tiu th p2. Xc nh p2.

46



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47/291

=> P2 = 144(W ).p2 COS tp2 p 2 0,6 :

R U V d 101. Cho on mch xoay chiu: *L J /TT *

vi R = I00V3(2). t in p u = U-\/2cosl00it (V) vo gia hai u

or mch.- Khi c = Ci th cng sut tiu th trong mch P = 100 (W).

1CT4 '- Khi c - C2 - (F) th cng sut tiu th P2 - 150 (W).

K

Hi in dung Ci bng bao nhiu ?

Li g i i

Pj _ COS2 (p1 100 _ COS2 ! _ 2

P 2 c o s 2 5 1 5 0 COS2 9 2

Li g i i

* - ZC2 -10 1 ' 7:tancp2 = - ' = -7= = = => 92 =

B 100V3 V 6

71' -v/3 COS2 9 , 2costp2 = COS = => -------- -

I 6j 2 (S 3

2

2 . 2 3 1 V ' 71COS (Pi = ----------- = => coscp1= => ZCi = R = 100V3

1CT4V y (?! = p - (F ) .

W3

V d 102. Cho on mch xoay chiu: * IJ I #

t in p xoay chiu u = UVcoslOOTt (V) vo gia hai u on mch.

V- Khi L = Li th coscpi = v cng sut tiu th Pi = 100 (W).

- Kh.i L = L2 th u c t gi tr cc i. Xc nh cng sut tiu th p2-

Li g i i .

Px _ cos2^ -

P 2 COS2 (>2



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48/291

R L c V d 103. Cho on mch xoay chiu: ff 'Tflfinr I---- 0

t in p xoay chiu u = V2coscot (V) vo gia hai u on mc

Khi CD= 1 th ij = I01cosot + j (A) v Pi = 200 (W).

Khi (0= to2 th cng dng in trong mch t gi tr cc i l 2 mXc nh cng sut tiu th khi (0= 0)2.

Li g i i

Khi I2max th xy ra hin tng cng hng: coscp2= 1v p2max-

COS2v r 2

COS (p2 ^

200 (W) 1

2 max

, p2roax = 800 (W).4

25. Cng thc vung pha: tani = - (61)tan2

N A Jj* R C g Vi du 104. Cho mach din xoay chiu: I' #M N

t in p u = uV2cos)t(V) vo gia hai u on mch th th

uan lch pha so vi UMB. in tr R, cm khng L v dung khng2

lin h vi nhau bi biu thc th no ?

Li g i i

----ta nMB

Zl 1R Zc_

" R

R2 = ZLZc = Leo X = - => L, = CR2.Cu c

V d 105. Cho on mch xoay chiu:

t in p u = VcslOOrt (V) vo gia hai u on mch. Ng



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49/291

ta thy din p gia hai du cun dy lch pha ^ so vi in p giaz

hai u on mch. Bit R = 80 (Q); r = 20 (Ci); L = (H). Hy xc71

nh in dung c.Li g i i

tantp = tan(pLy

~L ~ Zc __(R + r) Zl

r

(R + r)r = Z tfc - z

. 100*20 + 10 ^ _ W_ _zc = : k. = =120

(Q) => c = -(F).ZL 100 1,271

Lu :Mch R, L, c mc ni tip:

+ Khi c thay i m c t gi tr cc i th Ur,I nhanh pha hn u mt- TC '

g


50/291

Suy ra u A M v Umb ng pha hay (pAM=


51/291

1 R2C2+ Khi (0 = (02 th ULmax: T- - LC---------

14 2

+ Khi Rmitt th xy ra cng hng: t) =LC

2 1 r 2>


52/291

X =3.108m/s X 271V2 .IO3 X70.1(r12 = 705 (m)

x2= 3.108 m/s X 2W 2 .IO"3 X 5 3 0 .10-12 = 1940 (m)

Vy mch dao ng bt c sng t 705 (m) n .1940 (m).

V d 109. Mt mch dao ng LC.

- Khi c = Ci th tn s dao ng ring bng fi = 60kHz.

- Khi c c 2 th tn s dao ng ring bng f2 = 80kHz.

c cKhi mch dao ng c in dung bng - 2 th tn s' dao n

Ci + c 2ring bng bao nhiu ?

Li g i i

Li g i i

Theo cng thc (70): f = + f | = V602 + 802 = 100 (kHz)

Vy f = 100 (kHz).

V d 110. Mt mch dao ng LC.

- Khi c = Ci th chu k dao ng Tx = 30 C|as),

- Khi c = C2th ch.u k dao ng T2 = 40 (lis). ,

Nu C mc song song vi C2thi chu k dao ng bng bao nhiu ?

Li g i i

- Theo cng thc (66): T = + TI = V3 0 2 + 402 = 50 (ns).

V d 111. Mt mch dao ng LC, bt c sngc bc sngX =250 (m bt c sng c bc sng 50 (m) th phi mc thm t C' vi t thnh b t. Xc nh Ctheo c.

Li g i

X = 3.10 X 2nVLC = 250 (m)

xh - 3.108 X 2^^LCb = 50 (m)

V250 . J l = 25

& 50 c

C.C-_ , r CxCb '2% cc > Cb: Mc ni tip C vi c => C = = 7- -

c c _ 24- ^ 25

Vy C' = ~24

52



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53/291

V d 112. Mt mch dao ng LC vi c l t in phng, dao ngvi tn s dao ng ring f = 100 (kHz). Khi gim khong (ch gia hai bn t 2 ln th tn s dao ng ring 3 bao nhiu ?

Li g i i

+ f =

P

+ f ' = = = L => - = V2 2jVLC' / 2ss0S f'

27iJL.-----V d

=> '= = ^ = 50V2 (kHz).V2 V2

29. Cng thc v nng ing trong khung dao ng

^'22 '1 Qo- = -L = - c u (72)

w = -Li2 + -Cu2 = - (73)2 2 2 c

V d 113. Mt mch dao ng LC. Trong qu trnh dao ng o ccng dng in cc i bng 10 (A) v in tch cc i trn bn

t bng 107 (C). Hi mch ao ng ny bt c sng c bc sngbng bao nhiu ?

Li g i i

X= 3.108 (m/s) X WLC

i i 4 = - 2 2 c

LC = 2 l ^ Vl c = S . = 1 2 ^ = 10-*I| 1 10(A)

=> A. = 3.108 X2rc X 10"8 = 6ji (m).

V d 114. Mt mch dao ng LC. Trong qu trnh dao ng, o c cng dng in cc i qua cun dy Lbng 18 (mA). Cng dng in qua cun dy khi nng lng in trng bngnng lngt trng l bao nhiu ?

Li g i i

W = Wdin+Wt= 2Wt

53



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54/291

L I = 2 X L i 2 2i22 0 2

1 = % = l = 9 V2 (mA).V2 V2

30. Cng thc tnh sau nhng khng thi gian ngn nht th

TI7 w ^inmax Wtmw bl = w u,= ^ - - 2 | = - * 2 2 - :

t = (74)4

V d 115. Mt mch dao ng LC, dao ng vi tn s' f = 2 (MHz).Hi sau nhng khong thi gian ngn nht bng bao nhiu thi nnglng in trng bng na gi tr cc i ca n ?

hi gi i

t = - = ^ - = = 0,125.1c)-6 (giy)4 4f 8.10

hay t = 0,125 (ns).

31. Nng lng mt mt trong dao dng tt dn

p = I2R (75)

V d 116. Mch dao ng gm cun dy L = 20 (mH) v in trthun R = 2 (Q). Bit in dung c = 2000 (pF) v u 0 = 5 (V). cn phi

cung cp cho mch cng sut l bao nhiu duy tr dao ng ?

Li g i i

1 4 = CUi2 Z

'4-42_ = A V2 V2

= 5 ,1 ^ 1 = 5.10'2 (A)

I = M ^ . 1 0 _2(A)

p = I2R = ^ - 10_2J x 2 = 2,5.10"3 (W) hay p = 2,5 (mW).

32. Cng thc rong hin tng n sc

Nu gc chit quang A nh v gc ti c nh, th gc lch D:

D = (n - 1)A (76)-

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V d 117, Mt lng knh c gc chit quang A = 6, chiu chm tiasng trng, hp vo cnh lng knh theo phng vung gc vi mtphng phn gic ca gc chit quang. Trn mn, t song song vi mt phng phn gic ca gc chit quang v cch ,n mt khong d = 1 (m),ta thy c quang ph lin tc. Bit chit sut ca lng knh vi mu l 1,61 v mu tm l 1,68. Xc nh b rng quang ph.

Li g i i

+ MN l b rng quang ph.

+ AH = d = 1 (m)

+ tanD = = D =p HM = dD = d(nd- 1)Ad

Tng t HN = d(nt - 1)A

+ MN = HN - HM = d(nt - n)

= 103 (mm) X0,07 X6 X = 7,3 (mm).180 V d 118. Mt lng knh c gc chit quang A = 5. Chiu nh sng

trng vo mt bn lng knh di gc ti i nh. Bit n


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st thy hnh nh giao thoa v b rng vng giao thoa bng 13,8 (roXc nh s" vn sng v ti trn mn.

Li g i i

D k 2.103(mm) X0,6.10_3(mm) ,1 = -------------- - --------:---- = 1,2 (mm)

a 1 (mm)

Theo cng thc (77): N = - - = 5,72i 2 X1,2

+ N = m = 5 vn sng v mt pha

Vy s vn sngl (5 X2) + 1 - 11 vn sng (k c vn sng trungtm)

+ n > 5: s vn ti N = 5 + 1 = 6 v mt bn

Vy s vn ti l: 6 x 2 = 12.

V d 12. Thc n giao thoa bi khe I-ng vi b sng n

X =0,59im. Bit khong cch gia hai khe a= 1 (mm) v khong cch thai khe n mn D = 2 (m). Mt im M trn mn, cch vn sng trutm mt khong 7,67 (Tiro). Ti M l vn sng hay vn ti th my

Li g i i

DX 2.103(znm) XO ^.H r^ m m ) ___+ 1 ---- ------------------------------------- = l, lo (nun)

a 1 (mm)

+ Theo cng thc (77): N = = = 6,5i 1,18

C 7 vn ti, hay ti M l vn ti th 7.

34. Cng thc nh khong vn trong cc mi trng

Gi imt l khong vn trong mi trng v n l chit sut tuyt i mi trng.

imt= - (78n

V d 121. Khi thc hin giao thoa I-ng trong khng kh thkhon

vn i = 1 (ram). Nu thc hin giao thoa i nc th khong vn/ 4bao nhiu ? Cho bit chit sut ca nc n = .

3

+ Trong khng kh i =

Li g i i

D D XcT

a a

+ Trong nc iH20 = h2o =a a iH20 V

= = n

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57/291

+ c l vn tc nh sng trong khng kh

+ V l vn tc nh s ng trong mi trng

=> iH0 = = 1 (nun) _ Qyg (nun). Vy i = 0,75 (mm).2 n 4

3

35. Cng thc tnh b rng quang ph7bc k

A x = k - a - t ) (79)a

V d 122. Thc hin giao thoa I-ng vi nh sng trng:

0,4 (jaii) < A. < 0,76

Bit khong cch gia hai khe Si v S2 l = 0,3 (mm) v khong ccht hai khe n mn l D = 2 (m). Hy xc nh b rng quang ph bc

hai.

Li g i i

Theo cng thc (79):

Ax = 2 X ( i p i ) - 0,4). l -3 (mm) = 4,8 (mm).0,3 (nun)

36. Cng thc v cc vn sng ca cc nh sng n sc trng nhau

k-lA.1 = k2 -2 = 3 3 = (80)

V d 123- Thc hin giao thoa vi khe I-ng. Ngun sng s pht rang th&i h.ai bc x A.! = 0,6 (|jan) v x2 = ,48 (um). Hi ti v tr vnsng bc 4 ca bc x ! l vn sng th my ca bc x 2?

Li g i i

Theo cng thc (80): .! = kik => = 1- - = t = 5x2 0.48

Vy ti v tr trng nhau l vn sng th 5 ca bcx 2.

V d 124. Thc hin giao thoa vi khe I-ng. Khong cch gia hai khel a = 2 (mm) v khong cch t hai khe n mn D = 2 (m). Ngun sng s pht ra ng thi hai bc x Xi - 0,5 (jn) v X2 = 0,6 (um).Khong cch, ngn nht gia hai vch trng bng bao nhiu ?

Li g i i

ki/tj ]s.2 -2

0,5ki = 0,6k2

5ki = 6k2

57



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K-l

0 06 512 10

Khong cch ngn nht chn k = 0 v ki = 6.D\, 2.103(mm) X _____

X = k j = 6 X ---------------------- ----------------------- =: 3 ( m m ) .2 (mm)

37. Cng thc tnh bc sng hay s' bc x ca bc x dn sc c vn sng

rng nhau i mt im M i vi nh sng trng

0,4um < 0,76 umkD

(81)

V d 125. Thc hin giao thoa vi nh sng trng: 0,4jiin < X< 0,76 n.Bit khong cch gia hai khe a 0,3 (mm) v khong cch t hai khen mn D = 2 (m). Mt im M trn mn, cch vn sng trung tmmt khong 20,26mm. Xc nh bc sng c cc bc x c vn sngtrng nhau ti M.

Li g i i

Theo cng thc (81): 0,4^m ^ ^ 0,76 111

- . 0,3.20,26mm - -0,4ura < ------ ---- < 0,76(imk.2.10

- 3,04.10"3imn - __ 0,4fim < --------------- < 0,76|im

0,4nm V = 8.4-105 m/s.

V d 135. Chiu bc x vo catot t bo quang in th cc quangelectron bt ra v trit tiu dng quang in bi U a k ^ -0,85 (V). KhiU a k = 0.85 th vn tc ca quang electron khi ti anot bng bao nhiu ?

62



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Li g i i

+ Trit tiu dng quang in A K ^ -0,85 (V) ngha l u h = -0,85 (V).

+ Theo cng thc (86): mvmax + eUj.

^ v . = |2 x 1 , 6 . 1 0 - ^ 0 3 5 . 0 , 8 1 = 5>9 lo5

V m y 9,1-10

43. Cng hc tnh cc i lung kh electron chuyn ng ong t trng u B

B = m vmax (87)e x R max

V d 136. Chiu bc x ,533jn vo catot t bo quang din th cc

quang electron bt ra. Cho A = 3.10"19 (J). Tch chm hp quang electronv cho di vo t trng u B, sao cho cc electron chuyn ng c

phng vung gc vi ng sc t, th qu o ca electron l vng trn v bn knh ln nht bng Rma= 22,75 (mm). Hy xc nh B.

Li g i i

_ f>625-1 0 -3.10* 3 10_19

l 0,533.10-6 '),------------- ------------- = 4.10 m/s.9,1-10

B - ^ f j !_ , 1(r. ( T ) ,1,6.1019 x22,75.103

44. Cng hc v nng lng nguyn t H2 trng thi th n

En= _ 13,6(eV) (88)

nr

V d 137. Nng lng ion-ha nguyn t H2 l nng lng cn thit

a electron t trng thi c bn ln v nguyn t. Hy xc nhnng lng .

Li g i i

+ Nng lng trng thi c bn: n = 1 => Ei = -13,6 (eV).

+ Nng lng v nguyn t: n = co => E = 0

=> Eion-ha , E i = 0 (13,6) 13,6 (eV).

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64/291

V d 138. Ch.0bit nng lng nguyn t H2 trng thi n:~ 13,6 (eV)

n 2n

Hy xc nh -bc sng- ca vch Hp ?

Li g i i

Vch Hp c bc sng.k2trong dy Ban-me.13,6 { 13,6'

Ji' = E4 - E 2 =Artv 4 2

eV

- L - - L | = 13,6eV X - * 1 16

X42 he X16 6,625.10 .3.10 X16

13,6 X 1,6.10~19 x3= 0,486.106 (m)

13,6 (eV) X3

Hay 42=0,486 (pm).

V d 139. Kch thch nguyn t H2 trng thi c bn bi bc x bc sng X= 0,1218m. Hy xc nh bn knh qu o nguyn t H trng thi kch thch.

Li g i ihc

-- Nng lng kch thch = hf =

+ Khi chiu vo th nguyn t H2 chuyn t trng thi c bn n = 1 ltrng thi m.

ri T-. hcEm- E ~ z = ~

- 13,6 (eV) 0,1218.10

=> Ejn = 16;4.10-19 (J) - 13,6 (eV) = 10,2eV - 13,6eV = -3 ,4 (eV)

13,6eVM Em= -

2_ -13,6_(eV) _ . _ _ _=> m = ----- = 4 => m = 2

m . -3,4 (eV)

+ rm= r2 = 2a X ro = 4 X5,3.icru (m) = 2,12.10"10 (m).

V d 140. Khi nguyn t H2 chuyn t trng thi m xung trng thn = 2 th pht ra bc x = 0,486|im. Xc nh nng lng nguyn tH2 trng, thi m. .

Li g i i

E -E 2 = ^K

0,4&6.10-6- 3,4 (eV).

64



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65/291

Em= +2,55eV 3,4eV = -0,85 (eV)

Vy Em= -0,85 (eV).

V d 141. Trong quang ph vch nguyn t H2, bc sng th 1 v th2 ln lt l = 0,1218 (um) v 2 = 0,1026 (^m). Hy xc (lnh bcsng ln nht trong dy Ban-me.

Lu : Cng thc (89) khng cn d liu ca bi ra, m trc tip xcnh bc sng trong bi nu ra, nn bi sau ch nu cch su dng.

V d 142. Trong quang ph vch nguyn t H2. Hy xc dnh bcsng di nht trong dy Lai-man.

X21=--- _-- =0,1216.10-6(m).1,1.107 X 3

46. Cng thc dng xc nh s' vch quang ph trong nguyn t H2

V d 143. Kch thch nguyn t H2 trng thi c bn bi bc x cnng lng = 12,1 (eV). Hi nguyn t H2 c th pht ra ti a mybc x.

+ max trong dy Ban-me l ?-32.

E31 E32+ E21 => E2El (E3E2) + (E2E i)

hc _ hc hc _ X,21 X\ 3l _ 0,1218 X 0,1026

31 32 -21 21 ~ -31 0,1218 0,1026

Vy -32= 0,6562 (^m).

45. Cng thc R-bc: (89)

Li g i i+ Bc sng di nht trong dy Lai-man l 21-

N(s


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2 -13,6 _ 0m ---- -= 9 => m = 3

-1,5

Theo cng thc (90): N (s' vach) = (3 - 1). = 3 (vach).2

47. Cng thifc xc nh s' nguyn t, khi lng cn i sau thi gian t v phng x ti thi m t

N(t)= N02~T = N 0e-Xt (91)

tm (t) = m 2 T = nioe"^ (92)

tH(t) = H02_T = H0e_X (93)

Trong : N(t), m(t), H(t): s nguyn t, khi lng v phng xti thi im t.

No, mo, H0: s nguyn t, kh.i lng v phng x banu (t = 0).

. _ ln2 _ 0,693 , V -r , .k= ---- = -------- l hng so phng x.

T l chu k bn r.

V d 144. Ht nhn 2 Po l cht phng x, c khi ng ban u

mo = 1 (g) v chu k bn r T = 138 (ngy). Hy xc nh s" nguyn tcn li sau t = 414 ngy.

Li g i i

Theo cng thc (91):4 1 4

x2 138

3

N(t) = N0.2~T = Na ,P1q.2 T = 6j023'1Q23 x l(g ) A 210(g)

N(t) = 0,359.1021 (ngun t).

48. Cng thc xc nh s' nguyn t hay khi lng b phn r sau hi gian t

AN(t) = N0(l - 2~T) = N0(l - e"Xt) (94)

Am(t) = m0(l - 2 T) = m od - e-^ ) (95)

V d 145. Ht nhn 286Rn l cht phng x, c khi lng ban u

Iio = 2 (g) v chu k bn r T = 3,8 (ngy). Hy xc nh s' nguyn tb phn r sau thi gian 19 ngy.

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Theo cng thc (94):

AN(t)= N0(l - 2~T) = * m (1 - 2~T) = 6>023-.1Q23 x 2 (1 - 2 ~ ^ )y A 222

AN(t) = 5,25.1021 (nguyn t).

V d 146. Ht nhn Na24 l cht phng x, c khi lng ban u

mo = 0,24 (g) v chu k bn r T = 15 (h). Khi lng Na24 cn li sau thi gian t = 45 (h.) bng bao nhiu ?

Li g i i_t_ 4 5

Theo cng thc (95): Am(t) = m0 (1 - 2 T) = 0,24 X(1 - 2 15) = 0,21 (g).

9. Cng thc v bi ton phn trm

+ S' nguyn t b phn r bao nhiu phn trm l

Li g i i

+ Khi lng b phn r bao nhiu phn trm l

. N0

Am(t)

+ phng x b phn ra bao nhiu phn, trm l

mo

H -H(t)

H0

Vy: = Hq H(t) = 1 - eu = 1 - 2 T (96)-N0 m0 Ho

. . H Ngoi ra: (97)

H(t) = XN(t) V d 147. Co55 l cht phng x, c chu k bn r T = 18 (h). Hi sau

thi gian t = 1 (h), s nguyn t b phn r bao nhiu phn trm ?

Li g i i

Snguyn t b phn r AN(t):0 ,6 9 3 x

ii i = 1- e"X = 1 - e" 18 * = 1 - 0,962 = 0,038 = 3,8%.N0

V d 148. Ht nhn 2| Po l cht phng x, c chu k bn r T = 139 (ngy).

Hi sau bao lu th mt lng cht phng x c phng x gim 90%.

Li g i i

Theo cng thc (96): = 1 - e-xt = 90% = 0,9H0

e_xt =0,1 => x.t = lnO,l = 2,3 => t - - ^ - x .1 3 8 = 458(ngy).0,693

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Lu :Bi ton phn trm lu nhng ssau:

- Sau thi gian t - T, gim 50%.

- Sau thi gian t = 2T, gim 75%.

- Sau thi gian t = 3T, gim 87,5%.

- Sau thi gian, t = 4T, gim 93,75%.

- Tng t.

V d 149. Mt khi cht phng x sau thcfi gian t = 24 (ngy) th phng x gim 87,5%. Tnh chu k bn r ca khi cht phng x .

Li g i i

Theo cng thc (96): = 87,5%Ho

_ _ _ t 24 _ Suy ra t = 3T => T = = = 8 (ngy). Vy T = 8 (ngy).

3 3

50. Cng thc v bi ton s' ln n

n = 2 - = (98)N(t) m(t) H(t)

V d 150. Ht nhn Na24 l cht phng x, pht ra T, ri tr thnhMg24. Sau thi gian t = 105 (h) th s' nguyn t ca mt lng chphng x gim 128 ln. Hy xc nh chu k bn r ca Na24.

Li g i Theo cng thc (98): = 128 = 27

N(t)

L M M N(t) = N02~T ^ S- = 2T

0 N(t)

Suy ra 7 = - => T = - = = 15(h). Vy T = 15(h).T 7 7

V d 151. Mt mu g c c phng x t hn phng x ca mu

g cng khi lng va mi cht 4 ln. Bit ng v cacbon C14 c chuk bn r T 5600 (nm). Hy tnh tui ca mu g c.

Theo cng thc (98):

Li g i i

Ho .a t H(t)

IrfcO _=^ U = I n 4 = > -^ -x t = 21n2 => t = 2T = 2 X 5600 - 11200 (nm)

T

Vy t = 11200 (nm).

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51. Cng thc v bi ton c s' xung pht ra t l vi s' nguyn t b phn r

Thng th s xung pht ra n bng s" nguyn t b phn r nn:

n (xung) = AN(t) (s nguyn t b phn r) (99)

V d 152. Ngi ta dng my m xung, m p pht ra t cht

phng x. Trong thi gian tj = 2 (h), k t t = 0, my m c rij (xung), cn trong thi gian 2 = 2ti k t t = 0 , my m c n2= 1 ,811! (xung).Xc nh, chu k bn r ca cht phng x.

Li gi i

+ 'i vi ti: ANi(ti) = N0(l - ) = Iij (xung)

+ i vi t2: AN2(t2) = N0(l - e_Xt2) = n2 (xung)

AN2(t2) 1 - e~U '

ANiftj)

t: e-Ul - X, iu kin X > 0 v X # 1.

1 - Y2---------= 1,8 => 1 + X = 1,8 => X = e" t[ - 0,8 => -A.ti = Iu0,8

1 - X

x 2 (h) = 0,223 => T = 6,21 (h).T

V d 153. Ngi ta dng my m xng, mp~phtrat chtphngx. Trong thi gian ti = 1 (h) k t t = 0, my m.cnt= 380 (xung).

Cng trong thi gian 2= 1 (h), nhng sau t = 30 (h) k t t =0, my m c 90 (xung). Hy xc nh chu k bn r T ca cht ph ng x.

Li g i i

ANiti) = N0(l - eUl) = 360 (xung)

AN2(t2) = N2(t)(l - e~u2) = 90 (xung)

ANO ) N (1 - e~Ul) N0 = 360 =

^ N2(t2) ~ N2(t) N0e'xt 90

:=> ext = 4 => xt =ln4

Xt = 21n2 => T = - = = 15(h ).T 2 2

52. Cng thc v t l nguyn , l khi lng

+ Gi Nx l s" ht nhn COD. to thnh.

+ Gi AN(t) l s' ht nhn m b phn r.

=> Nx = AN(t).



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+ Gi N(t) l s ht nhn m cn li.

Vy t ] s nguyn t b phn r v s cn li l:

Nx ' AN(t)(100)

N(t) N(t)

Tomg l : T l ki lng b phn r v cn li l:

mx = AN(t) XAx (101)

m(t) N(t) XATrong (l: Ax l s' khi ca ht nhn con, A l s khi ca ht nhnm.

V d 154. 21Po l cht phng x, pht ra a, ri tr thnh ch 2g|Pb.

Bit chu k bn r ca 284Po l T = 138 (ngy). Hi sau bao lu th t

l khi lng gia ch v Po bng 0,4063.

Li g i i

Theo cngthc (101):

mp_b = AN(t)x 206 = - e ^ x 2^6 = u _ 2^6 =

mP N(t) X210 e"xt 210 210

=> eu ~ 1 = 0,4063 X = 0,41 eu = 1,41 = V2206

=> .t = I11V2 => t = = = 69 (ngy). Vy t = 69 (ngy).

53. Cng thc v ng nng, khi lng, vn tc ca cc ht trang hin tng

phng x+ Gi V , K v m l vn tc, ng nng, khi lng ca tia phng x.

+ Gi Vx, Kx, mx l vn tc, ng nng, khi lng ca-ht nhn con.

+ V gi Q l nng lng ta ra ca phn r.

(102)vx Kx m

K + KX= Q (103)

Cng thc (102) v (103) thng i vi nhau xc nh ng nng catia phng x v ht nhn con.

V d 155. Ra226 pht ra a ri tr thnh nguyn t khc. Hy xcnh ng nng ca ht a v ht X.

Cho bit: niRa = 225,977 (u)

mx = 221,970 (u)

ma = 4,0015 Cu) v u = 931 (MeV/c2).

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+ = _ 221,970 _ 55 47Kx ma - 4,0015

+ Q - [niRa - m0 -m x ]c2

= [225,977 - 4,0015 - 221,97].931 = 5,12 (MeV)

Li g ii

M Q = K + Kx 2 - = 55,47 Ka = 5,03 (MeV)

K : + Kx = 5 , 1 2 " K^ :09(MV)-

4. Cng thc v nng lng ta ra trong phn ng h nhn v hin tng

phng x

+ Gi (AE)s l nng lng lin kt sau phn ng.

+ Gi (AE)t l nng lng lin kt trc phn ng.

=> Nng lng ta ra bng:

Q = (AE)sau - (AEXruc (104)

V d 156. Cho phn ng nhit hch: D + T ->n + X

Bit ht khi D, T, X ln lt l:

Amp = 0,0024 (); Am? = 0,0087 (u); Amx = 0,0305 (u)

v 11= 931,5 (MeV/c2).

Li g i i

Theo cng thc (104):

Q = (AE) - (AE)tntc = Amx.c2 - (AmD+ AmT).c2

- [A m x - A e l q - A t q t ] . c 2

= [0,0305 - 0,0024 - 0,0087].931,5.c2

= 18,07 (MeV).

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Chng II. BI TP P DNG THAM KHO

1. Mt vt dao ng iu ha quanh v tr cn bng o vi bin A. Khitc ca vt bng mt na tc cc i th li ca vt l

A. x = M B. X = c . X = a V5 D .x = M .2 2 3

2. Mt con lc l xo treo thng ng. Ti v tr cn bng l xo dn 2,5 (cm).

Ko vt xung pha di 2 (cm) ri truyn cho n vn tc 40V3(cm/s)

hng theo chiu dng. Coi dao ng l iu ha v cho g = 10 (xo/s). Bin dao ng ca con lc l

A. 2,5 (cm) B. 4 (cm) c. 5 (cm) D. 8 (cm).

3. Mt con lc l xo dao ng iu hoa vi phng trnh X = Acos(ot + cp).

Ti thi im pha dao ng bng th vt c li X = - 5.1/3(em).6

Bin dao ng ca con lc l

A. 10 (cm) B. 10>/3 (cm) C. 20 (cm) D. 12i/3(cm).

4. Ln lt gn hai qu cu c khi lng mi v m2 vo l xo treo thngng th dbtu ld dao ng tng ng l Ti = 0,6 (giy) v T2 = 0,8 (giy). Nu vt c kh.i lng m = mi + m 2th chu ki dao ng ca Cn lc l

A. 1,4(s) B. 1,2 (s) c . 1,0 (s) D. 0,48 (s).

5. Hai con lc n c chiu di dy treo ln lt l \ v 1%th chu k daong tng ng l Tx = 1,2 (giy) v T2 = 1,6 (giy). Nu con lc cchiu di l = l2 - l\ th chu k ao ng l

A. 0,5 (s) B. 1,15 (s) c . 1,4 (s) D. 1 (s).

6. Hai con lc em c chiu di dy treo ln lt l \ v 1%th chu k daong tng ng l Ti v T2.

- Nu con lc c chiu diI+2 th chu k dao ng T = 2,7 (giy).

- Nu con lc c chiu di - 2th chu k dao ng T =: 0,9 (giy).

Hi chu k dao ng T v T2ca hai con lc trn l bao nhiu ?

A. Ti = 3,6 (giy); T2 = 1,8 (s) B. Ti = 1,8 (giy); T2 = 2 (s)

c. Ti = 2 (giy); T2 = 1,8 (s) D. Ti = 1,2 (giy); T2 = 2,4 (s).

7. Gn vt m ln lt vo hai l xo c cng tng ng l ki v k2 thitn s dao ng l fi = 3 (Hz) v f2 = 4 (Hz). Nu mc hai l xo ni tip nhau thi tn s dao ng l

A. 5 (Hz) B. 4,8 (Hz) c. 2,5 (Hz) D. 2,4 (Hz).

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8. Hai con lc l xo c cng ln lt l ki v k2. Gn vt m vo l xoc cng ki th chu k dao ng Ti = 1,2 (giy).Gn vt m vo l xo c cng k2 th chu k daong T2 = 1,6 (giy). Nu gn hai l xo nh hnh vth chu k dao ng bng

A. 1,92 (giy) B. 0,96 (giy) c. 1,5 (giy) D. 3 (gi.y).9. Mt con lc n ni nhit ti = 0c th chu k dao ng Ti - 2 (giy).Bit h s" n di ca dy treo X = 2.10"5(l/). Ti ni , nhngnhit d t2= 20c th chu kdao ng ca con lc

A. tng thm 2.1CT4 (giy) B. gim 2.10-4 (giy)

c. tng thm 4.1CT4 (giy) D. gim 4.10-4 (giy).

10. Trn mt t, c v tr ngang mc nc bin con lc n dao ng iuha vi chu k T = 2 (giy). a con c ln cao h = 9,6 (km) so vimc nc bin. Bit R = 6400 (km). Chu k dao ng mi ca COI) lc s

A. gim 3.10-3 (s) B. gim 6.10-3 (s)

c. tng 3.1CT3 (s) D. tng 6.103 (s).

11. Mt con lc n cao hi = 1,6 (km) so vi mc nc bin th chu kdao ng T = 2,0005 (giy). a con lc ln cao h2 = 3,2 (km) sovi mc nc bin th chu. k dao ng T2 l (cho R = 6400km)

A. T2 = 1,9995 (giy) B. T2 = 2,001 (giy)

c. T2 = 2,0005 (giy) D, T2 = 1,999 (giy).

12. Mt hg h qu lc thc hin bi con lc n ti ni nhit ti = 30cth ng h chy ng. Bit h s n di ca dy treo X - 2.10~5(l/).Ti ni nhng nhit l t2 = 10c th trong mt ngy n h s

A. chy nhanh 17,28 (s) B. chy chm 17,28 (s)

c. chy chm 8,64 is) D. chy nhanh 8,64 (s).

13. Ti ni ngang mc nc bin, vi nhit ti = 15c th chu k dao dngca ng h qu lc l T] = 2 (giy). a ng h ln cao h = 4,8 (km)so vi mc nc bin th chu k dao ng T2 = 2 (giy). Bit h s ndi ca dy treo = 2.1CT4 (1/) v bn knh Tri 't R = 6400km.

Xc nh nt cao hA. t2= 17,5c B. t2 = 10c c . t2X 7,5c D. ts = 5C.

14. Con lc n dao ng vi chu k T = 2 (giy) c treo ln trn mtchic xe. Cho xe chuyn ng trn mt ng nm ngang vi gia tc

a = g-Js th chu k dao

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