Upload
dinh-tong
View
45
Download
1
Embed Size (px)
Citation preview
L L & KHO KH Phin bn 1.0
Tuyn tp cc cu hi vt l kh nhn t cc thi th i hc trn ton quc km li gii chi tit v bnh lun.
GSTT GROUP 11/12/2013
GSTT GROUP | 1
ng bao gi b cuc cc em nh
Anh ch yu cc em nhiu lm!
L l & kho kh 1.0 | 2
Phn 1: bi Cu 1: Hai dao ng iu ha cng phng, cng tn s, bit phng trnh x1 = A1cos(t /6) cm v x2 = A2cos(t ) cm c phng trnh dao ng tng hp l x = 9cos(t + ). bin A2 c gi tr cc i th A1 c gi tr:
A. 18 3 cm B. 7cm C. 15 3 cm D. 9 3 Christmas
Cu 2: Mt vt dao ng iu ha vi phng trnh x = A.cos(t). T s gia tc trung bnh v vn tc trung bnh khi vt i c sau thi gian 3T/4 u tin k t lc bt u dao ng l A. 1/3 B. 3 C. 2 D. 1/2
Cu 3: Mt con lc l xo nm ngang gm vt nng tch in q = 20C v l xo c cng k = 10N/m. Khi vt ang nm cn bng, cch in, trn mt bn nhn th xut hin tc thi mt in trng u trong khng gian bao quanh c hng dc theo trc l xo. Sau con lc dao ng trn mt on thng di 4cm. ln cng in trng E l: A. 2.10
4 V/m. B. 2,5.10
4 V/m. C. 1,5.10
4 V/m. D.10
4 V/m.
Cu 4: Mt con lc n c chiu di l = 64cm v khi lng m = 100g. Ko con lc lch khi v tr cn bng mt gc 60 ri th nh cho dao ng. Sau 20 chu k th bin gc ch cn l 30. Ly g = 2 = 10m/s2. con lc dao ng duy tr vi bin gc 60 th phi dng b my ng h b sung nng lng c cng sut trung bnh l: A. 0,77mW. B. 0,082mW. C. 17mW. D. 0,077mW.
Cu 5: Mt con lc l xo dao ng iu ha theo phng ngang vi nng lng dao ng 1J v lc n hi cc i l 10 N. Gi Q l u c nh ca l xo, khong thi gian ngn nht gia 2 ln lin
tip Q chu tc dng ca lc ko 5 3 N l 0,1s. Qung ng ln nht m vt i c trong 0,4s l
A. 60cm. B. 50cm. C. 55cm. D. 50 3 cm.
Cu 6: Mt con lc l xo gm l xo c cng k = 2N/m, vt nh khi lng m = 80g, dao ng trn mt phng nm ngang, h s ma st trt gia vt v mt phng ngang l = 0,1. Ban u ko vt ra khi v tr cn bng mt on 10cm ri th nh. Cho gia tc trng trng g = 10m/s2. Tc ln nht m vt t c bng A. 0,36m/s B. 0,25m/s C. 0,50m/s D. 0,30m/s
Cu 7: Mt si dy n hi c treo thng ng vo mt im c nh, u di ca dy t do.
Ngi ta to sng dng trn dy vi tn s b nht l f1. c sng dng trn dy phi tng tn s
ti thiu n gi tr f2. T s f2/f1 l:
A. 1,5. B. 2. C. 2,5. D. 3.
Cu 8: Cho mch in xoay chiu nh hnh v. in dung C c gi tr thay i c v cun dy thun cm. iu chnh gi tr ca C v ghi li s ch ln nht trn tng vn k th thy UCmax = 3ULmax. Khi UCmax gp bao nhiu ln URmax?
A. 3
8 B.
8
3 C.
4 2
3 D.
3
4 2
Cu 9: Cho mch in xoay chiu nh hnh v. in dung C c gi tr thay i c v cun dy thun cm. iu chnh gi tr ca C th thy: cng thi im s, ch ca V1 cc i th s ch ca V1 gp i s ch ca V2. Hi khi s ch ca V2 cc i th s ch ca V2
gp bao nhiu ln s ch V1?
A. 2 ln. B. 1,5 ln. C. 2,5 ln. D. 2 2 ln Cu 10: Gi s ban u c mt mu phng x X nguyn cht, c chu k bn r T v bin thnh ht nhn bn Y. Ti thi im t1 t l gia ht nhn Y v ht nhn X l k. Ti thi im t2 = (t1 + 2T) th t l l
Cu 11: Mt con lc l xo c cng k = 40N/m u trn c gi c nh cn pha di gn vt m. Nng m ln n v tr l xo khng bin dng ri th nh vt dao ng iu ha theo phng thng ng vi bin 2,5cm. Ly g = 10m/s2. Trong qu trnh dao ng, trng lc ca m c cng sut tc thi cc i bng
A. 0,41W B. 0,64W C. 0,5W D. 0,32W
Cu 12: Mt con lc l xo t trn mt phng nm ngang gm l xo nh c mt u c nh, u kia gn vi vt nh c khi lng m. Ban u vt m c gi v tr l xo b nn 9cm. Vt M c khi
GSTT GROUP | 3
lng bng mt na khi lng vt m nm st m. Th nh m hai vt chuyn ng theo phng ca trc l xo. B qua mi ma st. thi im l xo c chiu di cc i ln u tin, khong cch gia hai vt m v M l: A. 9 cm. B. 4,5 cm. C. 4,19 cm. D. 18 cm.
Cu 13: Mt CLLX nm ngang gm l xo c cng k = 20N/m va vt nng m = 100g .T VTCB
ko vt ra 1 on 6cm ri truyn cho vt vn tc 20 14 cm/s hng v VTCB .Bit rng h s ma st gia vt v mt phng ngang l 0.4 ,ly g = 10m/s2. Tc cc i ca vt sau khi truyn vn tc bng :
A. 20 22 cm/s B. 80 2 cm/s C. 20 10 cm/s D. 40 6 cm/s
Cu 14: Mt vt dao ng iu ha vi phng trnh li : x = 4cos(8t 2/3) cm. Thi gian vt
i c qung ng S = (2 + 2 2 ) cm k t lc bt u dao ng l: A. 1/12 B. 5/66 C. 1/45 D. 5/96
Cu 15: Mt con lc l xo gm vt m1 (mng, phng) c khi lng 2kg v l xo c cng k = 100N/m ang dao ng iu ha trn mt phng nm ngang khng ma st vi bin A = 5cm. Khi vt m1 n v tr bin th ngi ta t nh ln n mt vt c khi lng m2. Cho h s ma st gia m2 v m1 l = 0,2 v g = 10m/s
2. Gi tr ca m2 n khng b trt trn m1 l A. m2 0,5kg B. m2 0,4kg C. m2 0,5kg D. m2 0,4kg Cu 16: Mt con lc l xo treo thng ng, khi vt v tr cn bng l xo gin 4cm. Kch thch cho vt dao ng iu ha th thy thi gian l xo b nn trong mt chu k l T/3 (T l chu k dao ng ca vt). gin v nn ln nht ca l xo trong qu trnh vt dao ng l: A. 12 cm v 4 cm. B. 15 cm v 5 cm. C. 18 cm v 6 cm. D. 8 cm v 4 cm.
Cu 17. Mt si dy n hi cng ngang, ang c dng dng n nh. Trn dy A l mt nt, B l im bng gn A nht, AB = 14cm. C l mt im trn dy trong khong AB c bin bng mt na bin ca B. Khong cch AC l A. 14/3 cm B. 7 cm C. 3,5 cm D. 1,75 cm
Cu 18. Hai im A, B nm trn cng mt ng thng i qua mt ngun m v hai pha so vi ngun m. Bit mc cng m ti A v ti trung im ca AB ln lt l 50 dB v 44 dB. Mc cng m ti B l A. 28 dB B. 36 dB C. 38 dB D. 47 dB
Cu 19: Ti O c 1 ngun pht m thanh ng hng vi cng sut khng i. Mt ngi i b t A n C theo mt ng thng v lng nghe m thanh t ngun O th nghe thy cng m tng t I n 4I ri li gim xung I. Khong cch AO bng:
A. AC 2
2 B.
AC 3
3 C.
AC
3 D.
AC
2
Cu 20: Cho hai mch dao ng l tng L1C1 v L2C2 vi C1 = C2 = 0,1F, L1 = L2 = 1 H. Ban du tch in cho t C1 n hiu in th 6V v t C2 n hiu in th 12V ri cho mch dao ng. Thi gian ngn nht k t khi mch dao ng bt u dao ng th hiu in th trn 2 t C1 v C2 chnh lch nhau 3V?
A. 610
6
s B. 610
3
s C. 610
2
s D. 610
12
s
Cu 21: Mc vo hai u on mch RLC ni tip gm mt ngun in xoay chiu c tn s f thay i C. Khi tn s f1 = 60Hz, h s cng sut t cc i cos1 = 1. Khi tn s f1 = 120Hz, h s
cng sut nhn gi tr cos2 = 2
2. Khi tn s f3 = 90Hz th h s cng sut ca mch bng
A. 0,874 B. 0,486 C. 0,625 D. 0,781
Cu 22: t in p u = U 2 cos(t + ) (V) vo hai u mch RLC ni tip, cun dy thun cm, in dung C thay i C. Khi in dung c C = C1, o in p hai u cun dy, t in v in
tr ln lt UL = 310V v UC = UR = 155V. Khi thay i C = C2 UC2 = 155 2 V th in p hai u cun dy khi bng A. 175,3V. B. 350,6V. C. 120,5V. D. 354,6V
Cu 23: Cho on mch RLC ni tip, cun dy thun cm v in tr R thay i C. t vo hai u on mch in p xoay chiu c gi tr hiu dng U = 200V. Khi R = R1 v R = R2 th
mch c cng cng sut. Bit R1 + R2 = 100. Cng sut ca on mch khi R = R1 bng A. 400W. B. 220W. C. 440W D. 880W
L l & kho kh 1.0 | 4
Cu 24: Mt on mch xoay chiu gm 3 phn t mc ni tip: in tr thun R, cun dy c (L; r) v t in c in dung C. t vo hai u on mch mt in p xoay chiu, khi in p tc
thi hai u cun dy v hai u t in ln lt l: ud = 80 6 cos(t +
6) V, uC = 40 2 cos(t
2
3)V, in p hiu dng hai u in tr l UR = 60 3 V. H s cng sut ca on mch trn
l
A. 0,862. B. 0,908. C. 0,753. D. 0,664.
Cu 25: Cho mt mch in xoay chiu AB gm in tr thun R = 100, cun dy thun cm L, t in c in dung C. t vo hai u on mch mt hiu in th xoay chiu u =
220 2 cos100t (V), bit ZL = 2ZC. thi im t hiu in th hai u in tr R l 60(V), hai u t in l 40(V). Hi hiu in th hai u on mch AB khi l:
A. 220 2 (V) B. 20 (V) C. 72,11 (V) D. 100 (V)
Cu 26: t in p u = U 2 cos(2ft) vo hai u on mch gm in tr thun R, cun cm thun L v t in C mc ni tip. Bit U, R, L, C khng i, f thay i C. Khi tn s l 50Hz th dung khng gp 1,44 ln cm khng. cng sut tiu th trn mch cc i th phi iu chnh tn s n gi tr bao nhiu?
A. 72Hz B. 34,72Hz C. 60Hz D. 50 2 Hz
Cu 27: t in p xoay chiu c u = 100 2 cos(t) V vo hai u mch gm in tr R ni tip vi t C c ZC = R. Ti thi im in p tc thi trn in tr l 50V v ang tng th in p tc thi trn t l
A. 50V. B. 50 3 V. C. 50V. D. 50 3 V.
Cu 28. Cho mch in xoay chiu RLC c CR2 < 2L. t vo hai u on mch mt in p
xoay chiu c biu thc u = U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr
ca in p hiu dng gia hai bn t t cc i. Khi UL = 0,1UR. Tnh h s cng sut ca mch khi .
A. 1
17 B.
1
26 C.
2
13 D.
3
7
Cu 29. Cho mch in AB gm in tr thun R, cun thun cm L v t C ni tip vi nhau theo th t trn., v c CR2 < 2L. t vo hai u on mch mt in p xoay chiu c biu thc u =
U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr ca in p hiu dng
gia hai bn t t cc i. Khi Cmax
5UU
4 . Gi M l im ni gia L v C. H s cng sut
ca on mch AM l:
A. 2
7 B.
1
3 C.
5
6 D.
1
3
Cu 30. Cho mch in xoay chiu RLC c CR2 < 2L. t vo hai u on mch mt in p
xoay chiu c biu thc u = U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr
ca in p hiu dng gia hai u ca cun cm t cc i. Khi Lmax
41UU
40 . Tnh h s
cng sut ca mch khi .
A. 0,6 B. 0,8 C. 0,49 D. 3
11
Cu 31. Cho mch in AB gm in tr thun R, cun thun cm L v t C ni tip vi nhau theo th t trn., v c CR2 < 2L. t vo hai u on mch mt in p xoay chiu c biu thc u =
U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr ca in p hiu dng gia hai bn t t cc i. Gi M l im ni gia cun cm v t. Ngi ta dng vn k V1 theo di gi tr ca UAM, vn k V2 theo di gi tr ca UMN gi tr ln nht m V2 ch l 90V.
Khi V2 ch gi tr ln nht th V1 ch gi tr 30 5 V. Tnh U.
GSTT GROUP | 5
A. 70,1V. B. 60 3 V C. 60 5 D. 60 2 V
Cu 32. Cho mch in RLC mc ni tip, trong RC2 < 2L. t vo hai u on mch in p
xoay chiu u = U 2 cos 2ft, trong ng U c gi tr khng i, f c th thay i c. Khi f = f1
th in p hiu dng trn t c gi tr cc i, mch tiu th cng sut bng 3
4 cng sut cc i.
Khi tn s ca dng in l f2 = f1 + 100Hz th in p hiu dng trn cun cm c gi tr cc i.
a. Tnh tn s ca dng in khi in p hiu dng ca t cc i.
A. 125Hz B. 75 5 Hz C. 50 15 Hz D. 75 2 Hz.
b. Tnh h s cng sut ca mch khi in p hiu dng gia hai u cun cm cc i.
A. 3
2 B.
1
3 C.
5
7 D.
2
5
Cu 33. Dng d kin sau tr li cc cu hi:
Cho mch in nh hnh v. C ba linh kin : in tr, t, cun thun cm c ng trong ba hp kn, mi hp cha mt linh kin, v mc ni tip vi nhau. Trong : RC2 < 2L.
t vo hai u on mch mt in p xoay chiu c biu thc u = U. 2 .cos t, trong U
khng i, c th thay i c. Tng dn gi tr ca t 0 n v theo di s ch ca cc vn k v am pe k, ri ghi li gi tr cc i ca cc dng c o th thy gi tr cc i ca V1 l 170V, ca V2 l 150V, ca V3 l 170V, ca A l 1A. Theo trnh t thi gian th thy V3 c s ch cc i u tin.
a Theo th t t tri sang phi l cc linh kin: A. R, L, C B. L, R, C C. R, C, L D. C, R, L
b. Theo trnh t thi gian, cc dng c o c s ch cc i ln lt l: A. V3, V2, A, V1
B. V3, sau V2 v A ng thi, cui cng l V1 C. V3 sau l V1, cui cng l V2 v A ng thi. D. V3 v V1 ng thi, sau l V2 v A ng thi.
c. Tnh cng sut tiu th trong mch khi V1 c s ch ln nht. A. 150W B. 170W C. 126W D. 96W
Cu 34. Cho mch in RLC mc ni tip, trong RC2 < 2L. t vo hai u on mch in p
xoay chiu u = U 2 cos 2ft, trong ng U c gi tr khng i, f c th thay i c. Khi f = f1
th in p hiu dng trn t c gi tr bng U, mch tiu th cng sut bng 3
4 cng sut cc i.
Khi tn s ca dng in l f2 = f1 + 100Hz th in p hiu dng trn cun cm c gi tr bng U.
a. Tnh tn s ca dng in khi in p hiu dng ca t cc i.
A. 50Hz B. 75Hz C. 50 2 Hz D. 75 2 Hz.
b. Tnh h s cng sut ca mch khi in p hiu dng gia hai u cun cm cc i.
A. 6
7 B.
1
3 C.
5
7 D.
2
5
Cu 35. Cho mch in nh hnh v:
t vo hai u on mch in p xoay chiu c biu thc u = U0cos t (V) trong , U0 c gi tr khng i, c th thay i c. iu chnh in p hiu dng trn t c gi tr cc i, khi uAN lch pha gc 71,57
0 (tan 71,57
0 =3) so vi uAB, cng sut tiu th ca mch khi
l 200W. Hi khi iu chnh cng sut tiu th ca mch t cc i th gi tr cc i bng bao nhiu? Bit rng h s cng sut ca on mch AN ln hn h s cng sut ca on mch AB.
L l & kho kh 1.0 | 6
Cu 36. Cho mch in nh hnh v:
t vo hai u on mch in p xoay chiu c biu thc u = U0cos t (V) trong , U0 c gi tr khng i, c th thay i c. iu chnh in p hiu dng trn t c gi tr cc i, khi uAN lch pha gc so vi uAB. Tm gi tr nh nht ca .
Cu 37. Cho mch in xoay chiu nh hnh v, trong cun dy c in tr thun r. t vo hai u on mch in p xoay chiu c biu thc u = U0cos t (V), trong U0 khng thay i, c th thay i c. iu chnh gi tr ca in p hiu dng ca on MB t cc i th gi tr cc i ng bng U0, cng sut tiu th ca on mch khi l 182W, in p hiu dng ca on AM khi l 135,2V.
a. Tnh r.
b. Tnh U0.
Cu 38. Cho mch in xoay chiu RLC ni tip, trong L l cun thun cm, RC2 > 2L. t vo hai u on mch in p xoay chiu
c biu thc 0
u U .cos t V trong
U0 khng i, cn c th thay i c. Ban u tn s gc ca dng in l , h s cng sut ca on mch MB bng 0,6. Khi tng tn s ca dng in ln gp i th in p gia hai u cun cm t cc i. Hi t gi tr , phi thay i tn s ca dng in th no :
a. Cng sut tiu th trn on mch t cc i.
b. in p hiu dng trn t t cc i.
Cu 39: Trong th nghim Y-ng v giao thoa nh sng, Ngun pht ng thi 2 bc x n sc
1 =0,64m () v 2 =0,48m (lam). Trn mn hng vn giao thoa, trong on gia 3 vn
sng lin tip cng mu vi vn trung tm c s vn sng v vn lam l:
A. 4 vn , 6 vn lam. B. 6 vn , 4 vn lam. C. 7 vn , 9 vn lam.
D. 9 vn , 7 vn lam
Cu 40: Mt cht im dao ng iu ha trn trc Oy. chnh gia khong thi gian ngn
nht khi vt i t v tr bin n v tr cn bng th tc l 40m/s. Khi vt c li 10cm th
tc ca vt l 30m/s. Chu k dao ng l:
A. B. C. D.
Cu 41: Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch ngoi
RLC ni tip. B qua in tr dy ni, coi t thng cc i gi qua cc cun dy ca my pht
khng i. Khi Rto ca my pht quay vi tc n0 (vng/pht) th cng sut tiu th
mch ngoi t cc i. Khi Rto ca my pht quay vi tc n1 (vng/pht) v n2
(vng/pht) th cng sut tiu th mch ngoi c cng mt gi tr. H thc quan h gia n0,
n1, n2 l:
A. B. C. 2 2
2 1 20 2 2
1 2
n .nn =
n +n D.
2 22 1 20 2 2
1 2
n .nn = 2
n +n
Cu 42: Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch AB gm
in tr thun mc ni tip vi mt cun dy thun cm. B qua in tr ca my pht. Khi
roto quay u vi tc n vng/pht th cng dng in hiu dng trong on mch l
1(A). Khi roto quay vi tc 3n vng/pht th cng dng in hiu dng trong on
mch l 3(A) . Nu roto quay u vi tc 2n vng/pht th cm khng ca on mch l:
A. R / 3 B. 2R 3 C. R 3 D. 2R / 3
Cu 43: Mt con lc n gm mt qu cu khi lng m = 250g mang in tch q = 10-7C c treo bng mt si dy khng dn, cch in, khi lng khng ng k chiu di 90cm trong
L R C
M N A B
C L,r
A B M
GSTT GROUP | 7
in trng u c E = 2.106 V/m ( c phng nm ngang). Ban u qu ng yn v tr cn bng. Ngi ta t ngt i chiu ng sc in trng nhng vn gi nguyn ln ca E, ly g = 10m/s2. Chu k v bin dao ng ca qu cu l:
A. 1,878s;14,4cm B. 1,887s; 7,2cm C. 1,883s; 7,2cm D. 1,881s;14,4cm Cu 44: Trong giao thoa Y-ng c a = 0,8mm, D = 1,2m. Chiu ng thi hai bc x n sc
= 0,75m v = 0,45m vo hai khe. V tr trng nhau ca cc vn ti ca hai bc x trn
mn l:
A. 0,225(k+1/2)mm (k = 0; 1; 2; 3....) B. 0,375(k+1/2)mm (k = 0; 1; 2;
3....)
C. 2(2k+1)mm (k = 0; 1; 2; 3....) D. 1,6875(2k+1)mm (k = 0; 1; 2;
3....)
Cu 45: on mch AB theo th t gm cc on mch AM, MN v NB mc ni tip. on
mch AM cha in tr thun R, on mch MN cha t in C, on mch NB cha cun dy
khng thun cm r, L. t vo A, B in p xoay chiu Bit in p
hiu dng , , in p gia 2 im M, B lch pha 90o so vi in p
gia 2 im A, N. H s cng sut ca on mch AB l:
A. 0,642 B. 0,5 C. 0,923 D. 1
Cu 46: Cho on mch RLC ghp ni tip, cun cm thun c t cm L thay i: R = 120, 410
C F0,9
, in p hai u on mch u = Uocos100t(V). iu chnh L = L1 th ULmax = 250V.
Tm gi tr ca L LU 175 2 (V)?
A. 3,09
L H
B. 0,21
L H
C. 3,1
L H
D. 2,5
L H
Cu 47: Khi thc hin giao thoa vi hai ngun kt hp O1O2 cch nhau 12 cm v c phng trnh
1
u =3cos(40t + )cm
6;
2
5u =3cos(40t - )cm
6. Vn tc truyn sng 60cm/s. Tm s im dao
ng vi bin 3 cm trn on O1O2?
A. 16 B. 8 C. 9 D. 18
Cu 48: on mch xoay chiu AB gm mt cun dy mc ni tip vi mt in tr R, UAB =
150 2 V. in p hiu dng gia hai u in tr v hai u cun dy ln lt l 70V; 170V.
Cng sut tiu th l 75W, gi tr ca R l:
A. 65,3 B. 140 C. 160 D. 115,7
Cu 49: Mt on mch ni tip gm cun dy c in tr thun r = 100 3 v t cm L =
0,191 H, t in c in dung C = 1/4(mF), in tr R c gi tr thay i c. in p t vo
hai u on mch u = 200 2 cos(100t) V. Thay i gi tr ca R cng sut tiu th trong
mch t cc i. Xc nh gi tr cc i ca cng sut trong mch.
A. 200 W B. 228W C. 100W D. 50W
Cu 50: Trong mch dao ng c T=0,12s. Ti thi im gi tr in tch v cng dng
in l 01Q 3
q =2
, . Ti thi im (trong gi tr mi ca
chng l 02
Qq =
2, Gi tr ln nht ca l:
A. 240,12s B. 240,24s C. 241,33s D. 241,45s
Cu 51: Cho mch in xoay chiu AB nh hnh v, trong in tr R =
20, cun dy c in tr thun r =10, t cm L = 1/ H, t in c
in dung C thay i c. in p gia hai u on mch c biu thc
uAB = 120 cos100t (V). Ngi ta thy rng khi C = Cm th in p
hiu dng gia hai im M v B t cc tiu l U1min. Gi tr U1min khi l:
L l & kho kh 1.0 | 8
A. 40 V B. 40 V C. 60 V D. 60 V
GSTT GROUP | 9
p n Cu 1:
V gin vect nh hnh v v theo nh l hm s sin:
22
A A Asin= A =
sinsin sin
6 6
, A2 c gi tr cc i khi sin c
gi tr cc i bng 1 = /2
A2max = 2A = 18cm A1 = 2 2 2 2
2A A = 18 9 = 9 3
Cu 2:
Vn tc trung bnh: 2 1tb2 1
x xv =
t t
, 2 1x = x x l di. Vn tc trung bnh trong mt chu k lun
bng khng
Tc trung bnh lun khc 0: tb
2 1
Sv =
t t trong S l qung ng vt i c t t1 n t2.
Tc trung bnh: tocdoS 3A 4A
v = = =3Tt T
4
(1); 3T
4 chu k u vt i t x1 = + A (t1 = 0) n x2 = 0
(t2 = 3T
4 ) (VTCB theo chiu dng)
Vn tc trung bnh: 2 1van toc tb2 1
x x 0 A 4Av = = =
3Tt t 3T0
4
(2). T (1) v (2) suy ra kt qu bng 3.
Cu 3:
V chiu di on thng dao ng l 4cm nn suy ra bin A = 2cm. Khi vt m dao ng hp ca lc in trng v lc n hi gy gia tc a cho vt. Ti v tr bin, vt c gia tc cc i. Khi ta c: F Fh = m.amax
qE kA = m.2.A = m.k
m.A qE = 2kA E = 2.104 V/m
Cu 4:
0 = 60 = 0,1047rad v T = 2
g
l = 2
2
0,64
= 1,6 (s)
C nng ban u W0 = mgl(1 cos0) = 2mglsin2
2
0 mgl
2
2
0
C nng sau t = 20T: W = mgl(1 cos) = 2mglsin2 2
mgl2
2=mgl
8
2
0
gim c nng sau 20 chu k: W = mgl(2
2
0 8
2
0 ) = mgl8
3 20 = 2,63.103J
Cng sut trung bnh cn cung cp con lc dao ng duy tr vi bin gc l 60
Ptb = 3
3W 2,63.10= = 0,082.1020T 32
W = 0,082mW.
Cu 5:
21 k = 50 N / mkA = 12
A = 20 cmkA = 10
v kx = 5 3 x =10 3cm
max
Tt = 0,1= T = 0,6s S = 2A + A = 60cm
6
Cu 6:
Vt c tc cc i khi gia tc bng 0; tc l
lc hl dh msF = F + F = 0 ln u tin ti N
L l & kho kh 1.0 | 10
ON = x kx = mg x = mg/k = 0,04m = 4cm Khi vt i c qung ng S = MN = 10 4 = 6cm = 0,06m
Theo L bo ton nng lng ta c: 2 2 2
maxmv kx kA+ = mgS2 2 2
(Cng ca lc ma st Fms =
mgS)
2 2 2
maxmv kA kx= mgS2 2 2
06,0.10.08,0.1,02
04,0.2
2
1,0.2
2
08,0 222max v
= 0,0036 2maxv = 0,09 vmax = 0,3(m/s) =
30cm/s.
Cch 2:
gim bin sau na chu k 1 22mg 2.0.1.0,08.10
A A = = = 0,08m = 8cmk 2
Sau na chu k u tin bin cn li A2 = 2cm Tc ln nht t c ti v tr cn bng mi
1 2 1 2max
A + A A + Ak 2 10 + 2v = = = = 30
2 m 2 0,08 2cm/s
Cu 7:
Si dy 1 u c nh, 1 u t do nn v
(2k 1) f (2k 1).4 4
ll
1
vk 1 f
4l v 22 1
1
fvk 2 f 3. 3f 3
4 fl Ch :
Tn s ti thiu bng k 1 kf f
2
Cu 8:
V C bin thin nn: 2 2Cmax LU
U R ZR
(1)
Lmax max L L L
min
U UU I .Z .Z .Z
Z R (2) (cng hng in)
v RmaxU U (3) (cng hng in)
2 2
LCmaxL
Lmax L
R + ZU(1)= 3 = R = Z 8
(2) U Z (4)
2 2
LCmax
Rmax
R + ZU(1)=
(3) U R (5)
T (4) v (5) 8
3
U
U
maxR
maxC
Cu 9:
Khi V1 cc i th mch cng hng: UR = U = 2UC = 2UL hay R = 2ZL (1)
Khi V2 cc i ta c:R
ZRUU
2L
2
maxC
theo (1)
2 2
L L
Cmax
L
U 4Z + Z U 5U =
2Z 2 (2)
Khi li c: L
2L
2
CZ
ZRZ
theo (1) ta c: ZC = 5ZL = 2,5R Z = R 5 (3)
Ch s ca V1 lc ny l RUR U
U = IR = =Z 5
(4)
GSTT GROUP | 11
T (3) v (4) ta c: Cmax
R
U 5= = 2,5
U 2
Cu 10:
p dng cng thc nh lut phng x ta c: 1
1 1
1
1
tY t01
t
1X 1 0
N N (1 e )N 1= = = k e =
N N N e k +1
(1)
2 1
2
2 1 1
2
t (t +2T)Y 02
2 t (t +2T) t 2T
1X 2 0
N N (1 e )N (1 e ) 1k = = = = = 1
N N N e e e e
(2)
Ta c ln2
2 T2T 2ln2T
1e = e = e =
4
(3). Thay (1), (3) vo (2) ta c t l cn tm:
2
1k = 1 = 4k + 3
1 1.
1+ k 4
.
Cng sut tc thi ca trng lc Pcs = F.v = mg.v vi v l vn tc ca vt m
Pmax = mg.vmax = mg.2kA
m = gA mk = gA
kAk
g; (v A = l0)
Pmax = kA Ag = 40.2,5.102
10.10.5,2 2 = 0,5W.
Cu 12:
Khi qua v tr cn bng, vn tc 2 vt l v p dng nh lut bo ton c nng cho qu trnh hai vt chuyn ng t v tr l xo b nn
l n khi hai vt qua v tr cn bng:
2 21 1 kk( ) = (m + M)v v = 2 2 m + M
l l (1)
n v tr cn bng, vt m chuyn ng chm dn, M chuyn ng thng u, hai vt tch ra, h con lc l xo ch cn m gn vi l xo.
Khi l xo c di cc i th m ang v tr bin, thi gian chuyn ng t v tr cn bng n v tr bin l T/4
Khong cch ca hai vt lc ny: 2 1
Tx = x x = v. A
4 (2), vi
mT = 2
k;
mA = v
k,
T (1) v (2) ta c: k 2 m m k 1 1
x = . . . . = . = 4,19cm1,5m 4 k k 1,5m 2 1,5 1,5
l l l l
Cch 2
Khi h vt chuyn ng t VT bin ban u n VTCB: CLLX (m + M = 1,5m): vmax =
kA = A
1,5m
Khi n VTCB, hai vt tch khi nhau do m bt u chuyn ng chm dn, lc ny M chuyn ng thng u vi vn tc vmax trn. Xt CLLX c vt m (vn tc cc i khng thay i):
vmax = k
A'' = A'm
= k A 9
A A' = = cm1,5m 1,5 1,5
L l & kho kh 1.0 | 12
T khi tch nhau (qua VTCB) n khi l xo c chiu di cc i th m n v tr bin A, thi gian
dao ng l T' 2
t = = =4 4' 2'
; vi k
' = = 1,5 t =m .2 1,5
. Trong thi gian ny, M
i c qung ng:
s = vmax.t = 4,5
A. = cm.2 1,5 1,5
khong cch hai vt: d = s A 4,19 cm
Cch 3
Sau khi th h con lc l xo dao ng iu ha, sau khi hai vt t vn tc cc ai th M tch ra chuyn ng thng u, cn m dao ng iu ha vi bin A
22
max(m + M)vk( ) =2 2
l maxv = l
k
m + M = l
k
1,5m
22
maxmvkA =2 2
A = maxvm
k = l
k
1,5m
m
k =
1,5
l = 7,348 cm
Sau khi tch nhau vt m dng li v tr bin sau thi gian t = T
4 =
2
4
m
k khi M i c
qung ng S2 = maxv t = lk
1,5m .
2
4
m
k =
.
2 1,5
l = 11,537 cm
Khong cch gia hai vt khi l S = S2 A = 11,537 7,348 = 4,189 = 4,19 cm
Cu 13: Vt c tc cc i khi gia tc bng 0;
tc l lc hl dh msF = F + F = 0 ln u tin ti N
ON = x kx = mg x = mg/k = 0,02m = 2cm Khi vt i c qung ng S = MN = 6 2 = 4cm = 0,04m
Ti t = 0 x0 = 6cm = 0,06m, v0= 20 14 cm/s = 0,2 14 m/s Theo L bo ton nng lng ta c:
2 2 22
max 0 0mv mv kxkx+ = + mgS2 2 2 2
(Cng ca Fms = mgS)
2 2 2 2
max 0 0mv mv kx kx= + mgS2 2 2 2
2 2 2 2
max0,1v 0,1(0,2 14) 20.0,06 20.0,02= + 0,4.0,1.10.0,042 2 2 2
= 0,044 2maxv = 0,88
vmax = 2204,088,0 = 0,2. 22 (m/s) = 20 22 cm/s.
Cu 14:
Vt xut pht t M n N th i c qung ng S = 2 + 2 2 . Thi gian:
T T 5t = + = (s)
12 8 96
Cu 15: vt m2 khng trt trn m1 th lc qun tnh cc i tc dng ln m2 c ln khng vt qu lc ma st ngh gia m1 v m2 tc l
msn qtmaxF F 2 2 maxm g m a 2
2
1 2
kg A g A m 0,5(kg)
m + m
Cch 2
Sau khi t m2 ln m1 h dao ng vi tn s gc = 1 2
k
m + m 2 =
1 2
k
m + m
GSTT GROUP | 13
m2 khng trt trn m1 th gia tc chuyn ng ca m2 c ln ln hn hoc bng ln gia
tc ca h (m1 + m2); vi a = 2x. Lc ma st gia m2 v m1 gy ra gia tc ca m2 c ln: a2 =
g = 2m/s2 iu kin m2 khng b trt trong qu trnh dao ng l
amax = 2A a2; suy ra
1 2
kAg
m + m g(m1 + m2) kA 2(2 + m2) 5 m2 0,5kg.
Cu 16: Thi gian l xo nn l T/3. Thi gian khi l xo bt u b nn n lc nn ti a l T/6. nn ca l xo l A/2, bng gin ca l xo khi vt v tr cn bng. Suy ra A = 8cm. Do gin ln nht ca l xo A/2 + A = 4cm + 8cm = 12cm, cn nn ln nht A/2 = 4cm
Cu 17.
= 4.AB = 46 cm Dng lin h gia H v chuyn ng trn
u: AC = 30
.360
= 14/3 cm
Cu 18. T cng thc I = P/4d2
Ta c: 2A M
M A
I d= ( )
I d v LA LM = 10.lg(IA/IM) dM =
0,6
A10 .d
Mt khc M l trung im cu AB, nn ta c: AM = (dA + dB)/2 = dA + dM; (dB > dA) Suy ra dB = dA + 2dM
Tng t nh trn, ta c: 2 0,6 2A B
B A
I d= ( ) = (1+ 2 10 )
I d v LA LB = 10.lg(IA/IB)
Suy ra LB = LA 10.lg0,6 2(1 2 10 ) = 36dB
Cch 2
Cng m ti im cch ngun m khong R; I = 2
P
4R = 10
L.I0; vi P l cng sut ca ngun;
I0 cng m chun, L mc cng m R = 0
P
4.I L1
10
M l trung im ca AB, nm hai pha ca gc O nn: RM = OM = B AR R
2
(1)
Ta c RA = OA v LA = 5 (B) RA = 0
P
4.I LA1
10=
0
P
4.I 51
10 (2)
Ta c RB = OB v LB = L RB = 0
P
4.I LB1
10=
0
P
4.I L1
10 (3)
Ta c RM = OM v LM = 4,4 (B) RM = 0
P
4.I LM1
10=
0
P
4.I 4,41
10 (4)
T ta suy ra 2RM = RB RA 2 4,410
1=
L10
1
510
1
L10
1 =
510
1 + 2
4,410
1
L10 = 9,4
4,4 5
10
10 + 2 10
L
210 = 5,22,2
7,4
10.210
10
= 63,37 8018,1
2
L L = 3,6038 (B) = 36
(dB)
L l & kho kh 1.0 | 14
Cu 19: Do ngun pht m thanh ng hng. Cng m ti im cch
ngun m R l 2
PI =
4R. Gi s ngi i b t A qua M ti C IA
= IC = I OA = OC Gi thuyt: IM = 4I OA = 2.OM. Trn ng thng qua AC IM t gi tr ln nht, nn M gn O nht OM vung gc vi AC v l trung im ca AC
AO2 = OM
2 + AM
2 =
2 2AO AC+
4 4 3AO2 = AC2 AO =
AC 3
3
Cu 20:
Hai mch dao ng c1 2 1 2
C = C ; L = L nn 1 2
1
1 = = =
L C1
Khi cho hai mch bt u dao ng cng mt lc th hiu in th gia hai bn t ca mi mch dao ng bin thin cng tn s gC.
Ta biu din bng hai ng trn nh hnh v Ti thi im t k t lc bt u dao ng, hiu in th trn mi t l u1, u2
Theo bi ton: u2 u1 = 3V (1) T hnh v, ta c: 02 2
01 1
U u= = 2
U u (2)
T (1) v (2), ta c: 6
01
1
U 10u = 3V = = t = = = (s)
2 3 3 3
.
Cch 2: Phng trnh hiu in th: 1 2
u = 6cos(t); u =12cos(t)
V hiu in th bin thin cng tn s, c ngha l khi u1 gim v 0 th u2 cng gim v 0.
Do , ta c: 2 1
u u = 3 12cost 6cost = 3 1
cost = t = + k22 3
V hiu in th trn mi t ang gim nn ta chn h nghim
t = + k23
Thi gian ngn nht nn ta chn k = 0. Vy 6 10
t = t = = (s)3 3 3
Cu 21:
Khi cos1 = 1 ZL1 = ZC1120L = 1
120.C LC =
2
1
(120) (1)
Khi cos2 = 2
2 2 = 45
0 tan2 =
L2 C2Z Z
R
= 1 R = ZL2 ZC2
tan3 = 2
L3 C3 L3 C3
2
L2 C2
1180L
Z Z Z Z 4 (180) LC 1180C= = = .1R Z Z 3 (240) LC 1
240L240C
tan3 =
2
2
2
2
(180)1
4 4 5 5(120)= =
(240)3 3 4.3 91
(120)
(tan3)2
= 25/91 2
3
1 25 1061
cos 81 81
cos3 =
0,874.
Cch 2
GSTT GROUP | 15
T/h 1: ZL1 = ZC1
T/h 2: f2 = 2f1 ZL2 = 4ZC2 v cos2 = 2
2 2 = 45
0 R = ZL2 ZC2 ZC2 = R/3
2
3C =
2f R
T/h 3: f3 = 1,5.f1 ZL3 = 2,25.ZC3
32 2 2 2 2
2C 2
2
3
R Rcos 0,874
R (1,25) Z (2 f ) RR 1,5625
(2 f )
Cu 22:
2
2 2L LL
Z = 2R U155 2 = + U 155 2
2U =155 2
LU = 350,6V
Cu 23:
P1 = P2 1
2 2
1 L C
R
R + (Z Z )= 2
2 2
2 L C
R
R + (Z Z ) (ZL ZC)
2 = R1.R2
P1 = 2
1
2 2
1 L C
U R
R + (Z Z )=
2
1
2
1 1 2
U R
R + R R=
2
1 2
U
R + R = 400W.
Cu 24:
d C
2 5 = + =
6 3 6 uC chm so vi i mt gc /2 vy ud nhanh pha so vi i mt gc /2
tand = tan
3= L
r
U
U nn L rU = 3U m
2 2 2 2
d r L rU = U + U = 4U
r LU = 40 3 (V) v U =120 (V)R rU + Ucos = = 0,908U
Cu 25: Ta c hiu in th hai u on mch thi im t l: uAB = uR + uC + uL = 20(V); (v uCv uL ngc pha nhau) Cu 26:
Khi f = f1 = 50 (Hz): ZC1 = 1,44.ZL1
1
1
2f .C= 1,44.2f1L LC = 2 2
1
1
1,44.4 f (1)
Gi f2 l tn s cn iu chnh cng sut tiu th trn mch cc i. Khi f = f2 th trong mch xy
ra cng hng: ZC2 = ZL2 2
1
2f .C = 2f2.L LC = 2 2
2
1
4 f (2)
So snh (1) v (2), ta c: 2 2
2
1
4 f =
2 2
1
1
1,44.4 f f2 = 1,2.f1 = 1,2.50 = 60 (Hz)
Cu 27:
T ZC = R U0C = U0R = 100V m Ru 50i = =
R R cn 0R0
UI =
R
p dng h thc c lp trong on ch c t C:
2R2 22
C C
2 2 220R0C 0
u( )
u ui R+ = 1 = 1UU I 100
( )R
2
C Cu = 7500 u = 50 3V ; v ang tng nn chn Cu = 50 3V
Cch 2 R = ZCUR = UC.
Ta c: U2 = UR
2 + Uc
2 = 2UR
2 UR = 50 2 V = UC. Mt khc:
CZtan =R
= 1
=
4
T ta suy ra pha ca i l (
t +4
).
R L, r = 0 C A B
M N
L l & kho kh 1.0 | 16
Xt on cha R: uR = U0Rcos(
t +4
) = 50cos(
t +4
) = 2
1
V uR ang tng nn u'R > 0 suy ra sin(
t +4
) < 0 vy ta ly sin(
t +4
) = 2
3(1)
v uC = U0C.cos(
t +4
2) = U0C.sin(
t +
4) (2) Th U0C = 100V v th (1) vo (2) ta c uC =
50 3 V
Cu 28:
Ta c: L1 2
R 1
U 0,5tan 0,1 tan 5
U tan
H s cng sut ca mch l : 2 2
2
1 1cos
1 tan 26
Cu 29:
Ta c: Cmax C
5U 5ZU Z
4 4 .
Khng lm nh hng n kt qu bi ton, c th gi s ZC = 5, Z =
4. Khi : 2 2LZ 5 4 3
L C LR 2.Z . Z Z 2.3. 5 3 2 3 . Suy ra: ZAM =
2 2
LR Z 12 9 21
H s cng sut ca on mch AM
1AM
R 2 3 2cos
Z 21 7
Cu 30:
Tng t trn, c th gi s: Z = 40, ZL = 41.
Khi : 2 2CZ 41 40 9
C L CR 2.Z . Z Z 2.9. 41 9 24
H s cng sut ca mch khi : R 24
cos 0,6Z 40
Cu 31:
GSTT GROUP | 17
Bn gin vc t, ta c:
22y 90 30 5 60V
x = 90 y = 30V
2 2 2 2U 90 x 90 30 60 2V
Lu : Nu cn tnh UR khi th ta c:
RU v 2.x.y 2.60.30 60V
H s cng sut ca mch khi l: RU 1
U 2
Cu 32:
a. Hai tn s f1 v f2 tho mn cng thc: 2 2 2
1 2 Rf .f f . Vy tn s ca dng in in p hiu
dng trn in tr t cc i l: R 1 2f f .f (*)
Khi iu chnh f cng sut tiu th trn mch cc i th trong mch xy ra cng hng. H s
cng sut khi bng 1. V cng sut tiu th ca mch c tnh bng biu thc: 2
max
UP
R
Trong cc trng hp khc th cng sut ca mch c tnh bng biu thc: 2 2 2 2
2 2 2
max2 2
U U R UP I .R .R . .cos P .cos
R RZ Z
ng vi tn s f1, cng sut tiu th trn mch bng 3
4 Pmax. Vy ta
suy ra h s cng sut khi Ucmax l 3 3
4 2 ( trn hnh v, h s
cng sut ca mch khi ny c gi tr bng 1
cos .
Khng lm nh hng n kt qu, c th gi s v = 3 , z = 2. Khi ta suy ra y = 1.
Theo cng thc ca phn l thuyt trn th ta c: 2v 3
x 1,52.y 2
Theo t l trn hnh v th khi tn s dng in l f1 th t s gia dung khng v cm
khng ca mch l : C1
L1
Z x y 2,5 5
Z x 1,5 3
V khi tn s ca dng in tng t f1 n f2 th in p ca t v ca cun cm i gi tr cho nhau, nn cm khng v dung khng trong mch cng i gi tr cho nhau. Nn tn s
f2 th ta c: L2
C2
Z 5
Z 3 . Hay L2 2
L1 1
Z f 5
Z f 3
L l & kho kh 1.0 | 18
Mt khc: f2 = f1 + 100 (Hz)
Gii h phng trnh ta suy ra: f1 = 150Hz, f2 = 250Hz
Thay hai gi tr f1 v f2 trn vo(*) ta c: Rf 150.250 50. 15 Hz
b. H s cng sut ca mch khi in p gia hai u cun cm t cc i cng bng h s
cng sut ca mch khi in p gia hai u t in t cc i v bng 3
2
Cu 33:
a. Khi tng dn t 0 n th UC t cc i u tin. Theo , V3 c s ch cc i u tin. Vy Z l hp cha t.
Do Lmax CmaxU U . M s ch cc i ca V1 v V3 bng nhau. Nn ta suy ra X l hp cha
cun cm.
Cui cng, Y l hp cha in tr thun.
Vy theo th t t tri sang phi l cc linh kin: L, R, C. Chn p n B.
b. Khi I t cc i th UR cng t cc i nn A v V2 ng thi c s ch cc i. Theo trnh t thi gian, cc dng c o c s ch cc i ln lt l: V3 , sau V2 v A
ng thi, cui cng l V1. Chn B.
c. V2 c s ch cc i Rmax ABU U . Vy ta c UAB = 150V.
Khi V2 (v ng thi A) c s ch cc i th cng sut tiu th
trn mch ln nht v bng: max maxP U.I 150.1 150W
Khi V1 c s ch cc i th ta c gin vc t nh hnh bn:
Ta c: 2 2CU 170 150 80V
RU 2.80. 170 80 120V
H s cng sut ca mch l 2
120cos cos 0,8
150
Cng sut tiu th ca mch khi l: 2
2 2 2
max
UP .cos P .cos 150.0,8 96W
R
Cu 34:
a. Cng sut tiu th ca on mch c tnh bng cng
thc: 2max
P P .cos
Theo , khi f = f1 th UC = U v c
2 3 3cos cos4 2
. Gin vc t ca mch khi
c dng nh hnh v:
trn hnh v: ta c = 300, = 600, OB = MB. Suy ra tam gic OMB l tam gic u. Vy UC = 2UL.
Suy ra: 1
1
12 f L
2 f C
ng vi hai tn s f1 v f2 th UL v UC i gi tr cho nhau nn ZL v ZC cng i gi tr cho nhau, ta c:
UC
UAB
UL
UR O
B
M
GSTT GROUP | 19
ZL2 = ZC1 = 2ZL1. Suy ra f2 = 2f1.
Mt khc, f2 = f1 + 100 Hz
Suy ra: f1 = 100Hz, f2 = 200Hz.
Tn s ca dng in khi UC = U gp 2 ln tn s ca dng in khi Ucmax. Vy khi Ucmax th tn s ca dng in l:
1C
f 100f 50 2 Hz
2 2
b. ng vi tn s f2, UL = U, gin vc t ca mch nh hnh v:
Khng lm nh hng n kt qu, c th gi s: ZL = ZAB
= 2 . Khi , ZC = 1 , R = 3 .
ng vi tn s fL = f2. 2 th in p trn t t gi tr cc
i. Lc , cm khng ca mch tng ln 2 ln, dung khng
ca mch gim i 2 ln. Gin vc t nh hnh v c.
Trn gin ny, ta c: OH = 3 , HM = 1 3
2 22 2
Suy ra: MO = 9 15
32 2
H s cng sut ca mch khi l:
OH 3 6 2cos
MO 15 515
2
Cu 35:
Khi UC t cc i th gin vc t ca mch nh hnh v.
Ta c:
01 21 2
1 2
tan tantan tan71,57 3
1 tan . tan
(1)
Mt khc, ta c: 1 2
tan .tan 0,5 (2)
V v h s cng sut ca on mch AN ln hn h s
cng sut ca on mch AB nn ta c: 1 2
(3)
T (1),(2),(3) ta suy ra: 1 2
1tan , tan 1
2
H s cng sut ca on mch AB l
2
2cos cos cos
4 2
Cng sut tiu th ca on mch c tnh bi cng thc:
2max max
1P P .cos P .
2
Theo th P = 200W. Suy ra Pmax = 400W.
ZC
L l & kho kh 1.0 | 20
Cu 36:
Gii:
Khi UC t cc i th gin vc t ca mch nh hnh v.
Ta c: 1 2
tan .tan 0,5
1 2 1 21 2 1 2
1 2
tan tan tan tantan tan 2. tan tan
1 tan .tan 1 0,5
V 1, 2 l nhng gc nhn, nn tan ca chng l nhng s dng.
Theo bt ng thc Cosi ta c:
1 2 1 2
1tan tan 2. tan . tan 2. 2
2
Vy thay vo biu thc trn ta c:
0tan 2 2 70,53
Vy khi UC t gi tr cc i th uRL sm pha hn uAB mt gc ti thiu bng 70,530.
Cu 37:
a. Khi tn s gc l , h s cng sut ca on MB l 0,6. Khng lm nh hng n kt qu c th gi s khi : R = 6, ZMB = 10. Suy ra ZC = 8.
Khi tng tn s ca dng in ln gp i (n = 2) th dung khng ca mch l 'CZ 4 ,
in p hiu dng trn cun cm t cc i. Lc gin vc t ca mch nh hnh v.
Ta c: 2 2
C
R 6x 4,52.Z 2.4
Cm khng ca mch khi ny l : 'LZ 4 4,5 8,5
T l gia cm khng v dung khng ca mch l:
'2L
'C
Z 8,5 172 L.2 C 4 .LC
4 8Z (1)
Khi iu chnh cng sut tiu th ca mch t gi tr cc i th trong mch xy ra cng hng. Lc t s gia cm khng v dung khng ca mch l:
''" " "2L
"C
ZL. C .LC 1
Z (2)
Chia hai v ca (1) cho (2) ta c: ""
2 17 32.
8 17 .
Vy t tn s gc , mun cho cng sut ca mch t cc i th phi tng tn s gc ln 32
17
ln.
b. Gi l tn s gc khi in p trn t t cc i. Ta c:
x
y
v
ZRL
2
1
O
Z
x
6 O H
Q
4
Z
ZRC
GSTT GROUP | 21
2"2
"'
'
32. 1617 .2. 17
Vy t gi tr tn s gc , mun cho in p hiu dng trn t t cc i th phi gim tn s
gc xung n gi tr 16.17
( tc l gim bt i mt lng 17
)
Cu 38:
a. iu chnh Ucmax th gin vc t ca mch nh hnh v:
Ta c: 2 2 2 20
x U U 2U U U
0y U x U 2 1
v 2xy 2U.U 2 1 U. 2 2 2 (*)
in p hiu dng ca on AM l:
2 2 2 2rLU x v U U 2 2 2 U 2 2 1 =135,2 (V)
Suy ra: U = 100(V). Thay vo (*) suy ra v = 91(V)
Ta c: 2 2v 91
P 182 r 45,5r r
b. Gi tr ca U0
0U U. 2 100 2 V
Cu 51:
Cu 50:
Khong cc ca 2 vt:
Khi 2 vt gp nhau:
Ban u vt (v tr gp nhau). Gc quay:
Vy vt qua 6 ln (k c ban u)
Cu 49:
Ta c
x
y
v
UrL
2
1
O
U
U0
L l & kho kh 1.0 | 22
Cu 48:
Cu 47:
Gi s bin dao ng ca phn t M l 3 cm, ta s c phng trnh sng ti M l:
Cu 46:
Ta c R=120 v ZC=90
GSTT GROUP | 23
Cu 45:
Ta c gin vecto:
C MB AB NBU 50 2(V);U U 130(V) OE 50 2;OP OQ 130
t R C ANU x U x U x 2 EP EQ x;PQ x 2
Gi F l trung im ca PQ ta c x 2
OF PQ;EF PF QF2
C
2 2
2 2 2 2x 2 x 2PF OF OP 50 2 130 x 70(V)2 2
Vy 2 2 2OP PE OE
cos 0,9232OP.PE
Cu 44:
Khi vn ti ca 2 bc x trng nhau th
Ta c th vit:
Ch : Vi cc bi ton c cc i lng thay i v mi lin h gia chng, ta c th th loi tr cc p n sai, nh vy c th rt ngn thi gian lm bi.
Cu 43:
uR
uC
ur
uL
uAB
uAN
uMB
O
P
Q
E
F
L l & kho kh 1.0 | 24
Khi i chiu th v tr cn bng mi i sang pha bn kia (hnh v).
Bin chnh bng:
Cu 42:
Khi roto quay vi tc n (vng/pht) th 2 2
L
UI 1(A)
R Z
Khi roto quay vi tc 3n (vng/pht) th 2 2
L
3UI 3(A)
R 9Z
2 2 2 2L L
L2 22 2LL
R 9Z R 9Z13 R 3Z
R Z33 R Z
Khi roto quay vi tc 2n (vng/pht) th L1 L
2RZ 2Z
3
Ch : Khi thay i tc quay ca roto th tn s ca dng in thay i hiu in th gia 2 u on mch cng thay i.
Cu 41:
Gii theo phong cch t lun nh:
Cc em cn nm c cc tr tam thc bc 2 mi hiu r li gii bi ton ny
Cu 40:
Ti A: A AA 2
x = ;v = 40 (m / s)2
Ti B: B Bx =0,1 (m);v =30 (m/s)
GSTT GROUP | 25
2 2
2
A 40 40 2(1) = A =
2 thay vo (2) ta c:
22
2300 2300 2 0,2= 0,1 = = T = (s)
0,1 T 2300
Cu 39:
Cc im c mu ging vn trung tm (hay c cc vn sng trng nhau) th tha mn
1 21 1 2 2
2 1
k 3x k k
k 4
V 1 2 1 2k ,k Z k 3;k 4
Trong khong gia 3 vn sng lin tip cng mu vi vn trung tm, ly k1 l 3, 6, 9 th
k2 l 4, 8, 12.
Cc vn sng n sc nm trong khong gia 3 vn sng lin tip ng vi k1 l : 4,
5, 7, 8
Cc vn sng n sc lam nm trong khong gia 3 vn sng lin tip ng vi k2 l :
5, 6, 7, 9, 10, 11
Vy c 4 vn v 6 vn lam
Phin bn 1.0 tm thi 51 cu , cc em ch i phin bn 1.1 sau y mt tun na nh (d kin l 20/11 ra mt)
Mt ln na, anh ch mong cc em lun lun n lc phn u, ng bao gi nn ch!
Ngoi ra, xem thm cc hot ng, bi ging ca anh ch GSTT GROUP, cc em vo nhng knh sau: Website : gstt.vn
Knh Youtube : youtube.com/luongthuyftu
Facebook tin tc : https://www.facebook.com/SHARINGTHEVALUE. Facebook h tr hc tp: https://www.facebook.com/onthidaihoccungthukhoa