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DAO NG V SNG
H thng kin thc- Mn Vt l
DAO NG V SNG
1.Dao ng iu ha:
Dao ng iu ho l mt dao ng tun hon trong i lng x m ta kho st bin thin theo thi gian theo quy lut hm cos hoc hm cocosx = acos ( t +()
+ a gi l bin cc i ca dao ng+ x l bin ( li ) thi im t+ l tn s gc, ( gi l gc lch pha ban u+ ( t - () gi l pha ca dao ng.
Chu k T: l khong thi gian nh nht m sau hin tng lp li nh c ( hoc l thi gian thc hin mt dao ng ton phn)
Tn s f : s chu k trong mt n v thi gian (1 giy)
f c tnh bng hc (Hz).
= 2 f .1. Con lc l xo
Xt mt cht im khi lng m gn u mt l xo, u kia ca l xo c c nh ti mt im trn tng.
trng thi t do ca l xo gi s cht im v tr cn bng O ca trc Ox.
Khi ta dng mt lc nn nn nn l xo v gi s khi cht im dch chuyn c mt on .
Khi l xo b nn th bn trong l xo xut hin mt lc n hi h c xu hng chng li bin dng v a cht im tr li v tr cn bng.
Phng trnh chuyn ng ca con lc
h = -k = m (( -kx =ma (( a = x = vi
Nghim phng trnh x= a.cos((t+()
Vn tc con lc v = x = -a( sin((t+()
Gia tc con lc a=v=x =-acos ((t+()
Chu k con lc T= 2(
m : khi lng con lc ( kg) ; k: h s n hi (N/m)
2. Con lc ton hc (con lc n)
Con lc ton hc :l mt cht im c khi lng m buc u mt si dy c chiu di l, u kia ca si dy c c nh ti mt im O.
thi im ban u t=0, cht im c ko lch khi v tr cn bng O ca n v v tr ca cht im c xc nh bi ta gc o.
Phng trnh dao ng ca con lc
F = ma=-Pcos( =-mg. => a =s =-g
Nghim phng trnh : s =S0cos((t +() vi
Chu k dao ng T=2( m : khi lng con lc ( kg) ; k: h s n hi (N/m)
ng dng : con lc n dng o gia tc trng trng nhng cao khc nhau
Dao ng con lc ton hc c xem l dao ng iu ha nu gc lch dy treo so vi phng thng ng l nh sao cho ( cos( (()
Ch :a) Gia tc trng trng ph thuc cao h gh =g0 (1-) - g0 gia tc trng trng ti mt t - R =6370km l bn knh tri t
b) Gia tc trng trng ph thuc nhit v lt =l0(1+(t)
- l0 chiu di con lc 00C - lt chiu di con lc t0C - ( h s n nhit
3. C nng trong dao ng iu ho1- C nng con lc l xo
Phng trnh dao ng x= a.cos((t+()
Vn tc con lc v = a( cos ((t+()Tn s gc (=
C nng ton phn E =Ed +Et
ng nng :Ed = mv2 = ma2(2 sin 2((t +() = ka2 cos 2((t +()
Th nng Et = kx2 = ka2 cos2 ((t +()
C nng E =Ed +Et = ka2Vy c nng con lc c bo ton trong sut qu trnh dao ng
2- C nng con lc n
v tr cn bng O ca cht im c vn tc cc i nn c nng ca n ch l ng nng :
E = mvo2- thi im t bt k, v tr ca con lc c xc nh bi ta gc , cao ca con lc khi l :
h = l(1-cos )
- v tr bin c nng con lc chnh l th nng cc i
E = mgh0 = mg l(1-coso)
- Vn tc cc i v tr cn bng v0 =
- Vn tc con lc v tr gc lch dy treo l ( v=
Lc cng dy:
+ Trng hp ko vt khi v tr cn bng mt gc (.
T = mg = mg(3cos( - 2cos(o)
+ VTCB Tmax = mg(3 2cos(o.
+ v tr bin: ( = (o ( Tmin = mgcos(o2. S tng hp dao ng :
Mt vt ng thi c th tham gia vo nhiu dao ng khc nhau, v d mt vt nng c treo vo ba im khc nhau bng ba l xo ng s dao ng theo mt dao ng l tng hp ca ba dao ng ring l. Vn l cn phi bit cch xc nh dao ng tng hp t cc dao ng thnh phn.
Gi s hai dao ng thnh phn l :
x1 = a1cos ( t + 1)
x2 = a2cos( t + 2)
Dao ng tng hp s l :
x = x1+ x2 = a1cos( t + 1) + a2cos( t + 2)
= acos( t + )
Dao ng tng hp cng l mt dao ng iu ha c cng tn s nh cc dao ng thnh phn
Bin ca dao ng tng hp c th tnh c theo cng thc
a =
trong bin cc i a v gc lch pha ban u c xc nh theo cc biu thc sau :
Khi 2 - 1 = 2n , (n = 0,1,2,) ta ni rng hai dao ng thnh phn cng pha. Khi a = a1 + a2
Khi 2 - 1 = 2(n + 1) , ta ni rng hai dao ng thnh phn ngc pha nhau. Khi a = Gc lch pha ban u ca dao ng tng hp tnh bng biu thctg = 3. Dao ng tt dn :
+ nh ngha : l dao ng c hc c bin gim dn theo thi gian ( do tc dng ca lc ma st.
+ Mi qu trnh dao ng c k n tc dng lc ma st u l dao ng tt dn.
+ Ma st cng ln, dao ng tt dn cng nhanh, vy dao ng tt dn ph thuc h s ma st ca mi trng.
+ng dng dng trong cc b phn gim xc cua xe, hm dao ng cc kim o trong cc my o in xoay chiu.
4. Dao ng cng bc v hin tng cng hng :
c th duy tr dao ng khng tt dn, ta phi tc ng vo h mt ngoi lc bin i theo thi gian theo qui lut iu ha F = F0cos t trong F0, l bin v tn s ca ngoi lc.
Ban u h thc hin dao ng tng hp : dao ng ring v dao ng cng bc, nhng sau mt thi gian dao ng ring tt dn, h ch cn thc hin dao ng cng bc .Bin ca dao ng cng bc ph thuc vo tn s ca ngoi lc.
Hin tng cng hng, xy ra khi bin dao ng cng bc bng vi tn s ring ca h .
ng dng :hin tng cng hng c th ph v tnh n hi ca vt cht nn trong thit k chi tit my, xy dng b t my mc ngi ta phi thit k sao cho tn s dao ng ca cc b phn ny khc xa vi tn s dao ng ring ca chng
Ch : iu kin xy ra cng hng c
Vt phi thc hin dao ng cng bc
Tn s ngoi lc phi bng tn s ring ca h
(Ch : Dao ng t do l dao ng khng c tc dng ca ngoi lc)
5. Qu trnh truyn sng :
.1. i cng v sng :1. Sng c hc: Sng c hc l nhng dao ng n hi lan truyn theo thi gian trong mt mi trng vt cht.
- Sng truyn c trong mi cht rn, lng v kh.
- Truyn sng tc l truyn pha dao ng, trong khi cc phn t vt cht khng truyn i m ch dao ng quanh v tr cn bng.
.2. Phn loi sng
+ Sng ngang : c phng dao ng vung gc vi phng truyn sng
+ Sng dc: c phng dao ng trng vi phng truyn sng
.3. Cc i lng c trng 3. Bc sng: Khong cch gia hai im trn phng truyn sng gn nhau nht v dao ng cng pha vi nhau gi l bc sng. K hiu:
+ Nhng im cch nhau mt s nguyn bc sng trn' phng truyn th dao ng cng pha vi nhau.
+ Nhng im cch nhau mt s l na bc sng trn phng truyn th dao ng ngc pha nhau.
4. Chu k, tn s v vn tc ca sng :
* Chu k ca sng dao ng bng chu k chung ca cc phn t vt cht c sng truyn qua v f = tn s dao ng ca sng
* Bc sng cn c nh ngha l qung ng sng truyn i c trong mt chu k dao ng ca chng.
* Vn tc sng l vn tc truyn pha dao ng.
5. Bin v nng lng ca sng
a) Bin sng: l bin dao ng ca cc phn t vt cht ti im sng truyn qua.
b) Nng lng:
- Qu trnh truyn sng l qu trnh truyn nng lng
- Nng lng sng ti tng im t l vi bnh phng bin sng ti im . Nng lng sng gim t l vi qung ng truyn sng.
.4. Phng trnh sng :Phng trnh sng l mt biu thc m t mi lin h gia dch chuyn u ca cc phn t ca mt mi trng l hm ca to x v thi gian t :
Sng ti ngun : u = acos t
Sng ti im cch ngun khong x lan truyn vi vn tc v xt ti thi im t l:
u = acos (t ) = acos (2(ft -)
6. S giao thoa ca sng :
+ Hin tng giao thoa l hin tng trong dao ng tng hp ti mt s im c bin cc i trong khi ti mt s im khc dao ng tng hp c bin cc tiu.
+ Hai sng kt hp phi c cng tn s v hiu s pha khng thay i theo thi gian
+ iu kin xy ra hin tng giao thoa : cc sng gp nhau phi l sng kt hp
+ Chi tit ca hin tng giao thoa
Ta xt s giao thoa ca hai sng, do hai ngun S1 v S2 l hai ngun kt hp pht ra, ti mt im M rt xa so vi khong cch gia hai ngun. Khi ta c th coi hai sng do S1 v S2 truyn ti M c cng phng. Gi r1, r2 tng ng l khong cch t M n cc ngun S1 v S2
Dao ng ti im M do hai sng to ra l:
u1= a1cos( t-2(r1/ ()
u2= a2cos( t-2(r2/ ()
lch pha
Dao ng tng hp ti M nh bit (xem V.13a) c cng tn s v chu k nh cc dao ng thnh phn v c bin cc i a c xc nh bi:
A2= a12+a22 - 2a1a2cos[(-( 2- 1)] = a12+a22 +2a1a2cos( 2- 1)
Trng hp
( hai dao ng ng pha)
Khi ( r2-r1) =
hay (r2-r1) = n = 2n= n(
Vy bin dao ng tng hp t gi tr cc i ( a= a1+a2) khi hiu ng i bng s nguyn ln bc sng
Trng hp
(2n+1)( ( hai dao ng nghch pha)
Khi (r2-r1)= (2n+1) (V.38)
Vy bin dao ng tng hp t gi tr cc tiu a= a1-a2 khi hiu ng i bng s l ln na bc sng.
7. Sng dng : Sng dng l s giao thoa ca hai sng phng c cng bin v truyn ngc chiu nhau to ra nhng bng sng v nt sng c nh trong khng gian. Sng ti v sng phn x s giao thoa vi nhau to thnh sng ng.
Gi s hai u dy c nh :
Hm sng ca sng ti v sng phn x tng ng l :
x1= acos( t kx)
x2 =- acos( t + kx)
v i k =,
Sng tng hp :
x = x1 +x2 = a[cos( t kx) - cos( t + kx)]
= 2 a.coskx.cos t
Thay , ta c :
x = (2acos)cos t
Biu thc ny cho thy :
Bin cc i ca sng khng ph thuc thi gian m ch ph thuc vo ta x ca im kho st :
A = 2acos
Ta xt hai trng hp ring ca A :
cc im tha mn iu kin , dao ng c bin cc tiu. Cc im ny gi l cc im nt ca sng
cc im tha mn iu kin, dao ng c bin cc i. Cc im gi l cc im bng
+ Cc im nt v cc im bng nm xen k nhau v cch u nhau mt on bng /4.
+ Hai nt hoc hai bng cch nhau /2
iu kin c sng dng trn dy
Hai u dy l hai nt : chiu di dy l = k /2
Mt u dy l nt, mt u l bng l = k /2 + /4.
Sng ng to ra khi sng ti b phn x ti mt mt rn P phng
.8 Sng m
nh ngha : - Sng m l sng c hc, l sng dc - C tn s t 16Hz (20.000Hz, gy cm gic m i vi tai ngi - Sng siu m : c tn s > 20000Hz, sng h m c tn s ZC: i chm pha hn u
on mch gi l c tnh dung khng khi ZL < ZC: i nhanh pha hn u
4. Hin tng cng hng in:
a) Xy ra hin tng cng hng trong mch RLC khi cng dng in trong mch t gi tr cc i
b) iu kin c hin tng cng hng in
ZL = ZChoc
i cng pha vi u
hoc
Khi Z = Zmin = R
I = Imax =
DAO NG IN T - SNG IN T
I. MCH DAO NG V DAO NG IN T1. S bin thin ca in tch trong mch dao ng:Mch dao ng l mt mch kn gm c mt cun cm L mc ni tip vi mt t in c np in.
Th nghim nh hnh v:
Khi K A: ngun in tch in cho t in.
Khi K B: t phng in qua L, trong mch c dng in i = = q(t).
Trong mch xut hin sut in ng cm ng:
e = - Li = -Lq.
nh lut Ohm cho hai u cun dy : u = e = - Lq.
Hiu in th :
u = uC =uL= ( - Lq = ( q = - q
t = (2
Ta c
q + (2q = 0
Nghim ca phng trnh: q = Qosin((t + ()
in tch trong mch dao ng dao ng iu ha vi tn s ( = gi l tn s ring ca mch.2. Nng lng trong mch dao ng:
Ta c: q = Qosin((t + ()
u = = sin(t
Nng lng in trng tc thi: W = sin2((t + ()
Nng lng t trng tc thi: Wt = Li2 = cos2((t + ()
Nng lng ca mch dao ng: W = W + Wt W = sin2((t + () + cos2((t + () = = L = const.
Kt lun:
Nng lng ca mch dao ng gm c nng lng in trng tp trung t in v nng lng t trng tp trung cun cm.
Nng lng in trng v nng lng t trng cng bin thin tun hon theo cng mt tn s chung.
Nng lng ton phn ca mch dao ng c bo ton.
Tn s dao ng ( = ch ph thuc vo nhng c tnh ring ca mch nn gi l tn s dao ng ring ca mch.
3. V sao dao ng in t tt dn:
Dao ng in t l dao ng in t tt dn v : trong thc t, cun dy bao gi cng c in tr nn nng lng in t b tiu hao dn. Mt khc, mch LC khng hon ton kn nn c mt phn nng lng bc x ra ngoi, nn nng lng dao ng cng gim.
So snh dao in t v dao ng c hc:
Con lc l xoMch dao ng in t
i lng bin thinx + (2q = 0
x = Asin((t + ()q + (2q = 0
q = Qosin((t + ()
Tn s dao ng ring
Nguyn nhn tt dnLc cn mi trng fmsin tr R
Tc nhn cng bcNgoi lc tun hon
Fn = Hosin((t + ()Hiu in th xoay chiu
u = Uosin((t + ()
iu kin cng hng( = (o( = (o
Cng hng nhnFms bR b
Nng lng
dao ng ringE = m(2A2
= kA2W = = L
II. IN T TRNG
1. S lin quan gia t trng bin thin v in t trng
Pht biu hai gi thuyt ca Maxwell:
Gi thuyt 1:
Mi t trng bin thin theo thi gian u lm xut hin in trng xoy, ng sc ca in trng xoy l nhng ng sc khp kn.
Gi thuyt 2:
Mi in trng bin thin u lm xut hin t trng bin thin.
Khi nim v dng in dch v dng in dn:
Dng in dn l dng chuyn di c hng ca cc ht mang in trong dy dn.
Dng in dch l khi nim ch s bin thin ca in trng (gia hai bn t in) tng ng vi mt dng in.
Dng in ton phn trong mch bao gi cng khp kn gm dng in dn chy trong dy dn v dng in dch trong t in.
2. in t trng
T cc gi thuyt ca Maxwell, i n kt lun:
in trng bin thin no cng sinh ra t trng v t trng bin thin no cng sinh ra in trng bin thin.
in trng v t trng c th chuyn ha ln nhau ng thi tn ti thnh mt trng thng nht gi l in t trng.
in t trng l mi trng tng tc in t, lan truyn trong khng gian vi vn tc hu hn c = 3.108 m/s.
in trng v t trng l nhng trng hp ring ca in t trng.
III. SNG IN T
1. Cch to ra sng in t Gi s ti O c mt in tch dao ng iu ha vi tn s f, n to ra ti O mt in trng bin thin vi tn s f, bin thin to ra t trng bin thin, bin thin li to ra bin thin vng khng gian ln cn bin thin v bin thin to thnh mt trng thng nht l in t trng.
in t trng lan truyn trong khng gian di dng sng gi l sng in t.
2. Cc tnh cht ca sng in t
Sng in t truyn c trong mi trng vt cht v chn khng.
Vn tc lan truyn trong chn khng c = 3.108 m/s.
Ti mi im v lun vung gc vi nhau v vung gc vi phng truyn sng.
Sng in t l sng ngang.
Tun theo cc nh lut phn x v khc x.
IV. S PHT V THU SNG IN T1. My pht dao ng iu ha dung Tranzito
Khi mch dao ng c tn s ring (o hot ng, nng lng d tr lc ban u trong t in tiu hao dn v dao ng s tt dn.
duy tr dao ng, ngi ta dng mt c cu gi l my pht dao ng iu ha dng Tranzito (T). c im ca Tranzito l khi in th VB < VE th c dng qua T.
T C c ni vi ngun P qua T.
Cun L t gn L. L ni hai u E v B ca Tranzito T.
C ngn dng mt chiu t P vo B.
Mch LC v T c b tr nh trn th c sau mi chu k dao ng ca mch, c mt dng in dao ng t ngun P li i mt ln qua T vo t in v b sung cho n mt lng nng lng ng bng lng m n b mt i.
My pht dao ng iu ha dng Tranzito to ra mt dao ng in t khng b tt dn, c tn s bng tn s dao ng ring ca mch dao ng LC.2. Mch dao ng kn mch dao ng h - ng ten:
Mch dao ng kn: l mch LC trong hai bn ca t in gn nhau, nng lng in t trng kh bc x ra ngoi xa c.
Mch dao ng h: Di xa hai bn cc ca t v ko dn cc vng dy, mt phn nng lng in t trng thot ra ngoi khng gian nhiu hn i xa hn: ta c mch dao ng h.
ng ten: Khi mch dao ng h n gii hn mi bn t lch 180o ta c ng ten.
3. Nguyn tc pht v thu sng in t:
pht sng in t ngi ta mc phi hp mt my pht dao ng iu ha vi mt ng ten LA. Khi trong my pht c dao ng in t duy tr vi tn s f th s cm ng qua LA nh LA m pht sng in t c tn s f vo khng gian.
thu sng in t ngi ta mc phi hp mt ng ten thu vi mt mch dao ng. ng ten nhn c rt nhiu sng c tn s khc nhau. Mun thu sng c tn s khc nhau, iu chnh in dung ca t trong mch dao ng fo = fst DAO NG V SNG C HC
A. CON LC L XO
1. Con lc l xo c cng k, khi lng m treo thng ng. Khi khi m v tr cn bng th :
a/ Hp lc tc dng ln m bng khng.
b/ Lc hi phc F = mg.
c/ gin ca l xo : V =
d/ Lc n hi Fh = 0
e/ Cu a v c ng.
2. Con lc l xo treo thng ng, dao ng vi bin A. Lc n hi ca l xo s :
a/ Cc i bin im dng.
b/ Cc i bin im m.
c/ Nh nht v tr thp nht.
d/ Ln nht v tr thp nht.
e/ Cu a v b ng.
3. Con lc l xo dao ng ngang. v tr cn bng th :
a/ Th nng cc i.
b/ ng nng cc tiu.
c/ gin ca l xo l
d/ Lc n hi ca l xo nh nht.
e/ Gia tc cc i
4. Chu k dao ng ca con lc l xo ph thuc vo :
a/ S kch thch dao ng.
b/ Chiu di t nhin ca l xo.
c/ cng ca l xo v khi lng ca vt.
d/ Khi lng v cao ca con lc.
e/ Chiu di t nhin v cng ca l xo.
5. Nu cng tng gp 2, khi lng tng gp 4, th chu k ca con lc l xo s :
a/ Tng gp 2.
b/ Gim gp 2
c/ Khng thay i.
d/ Tng gp 8.
e/ p s khc.
6. Khi treo 1 trng vt P = 1,5N vo l xo c cng 100N/m th l xo c 1 th nng n hi l :
a/ 0,01125 J
b/ 0,225 J
c/ 0,0075 J
d/ 0,2 J
e/ 0,3186 J
7. Mt con lc l xo khi lng m = 125g, cng k = 50 N/m ( ly ( =3,14) chu k ca con lc l :
a/ 31,4 s
b/ 3,14 s
c/ 0,314 s
d/ 2 s
e/ 0,333 s
8. Con lc l xo lm 15 dao ng mt 7,5 s. Chu k dao ng l :
a/ 0,5 s
b/ 0,2 s
c/ 1 s
d/ 1,25 s
e/ 0,75 s
9. Con lc l xo c tn s l 2Hz, khi lng 100g (ly (2 = 10). cng ca l xo l :
a/ 16 N/m
b/ 100 N/m
c/ 160 N/m
d/ 200 N/m
e/ 250 N/m
10. Khi treo vt m vo 1 u l xo, l xo gin ra thm 10cm (ly g = 10 m/s2). Chu k dao ng ca vt l :
a/ 0,314 s
b/ 0,15 s
c/ 0,628 s
d/ 0,52 s
e/ 0,235 s
11. Mt con lc l xo cng k. Nu mang khi lng m1 th chu k l 3s. Nu mang khi lng m2 th chu k l 4s. Nu mang ng thi 2 khi lng m1 v m2 th c chu k l :
a/ 25 s
b/ 3,5 s
c/ 1 s
d/ 7 s
e/ 5 s
12. Con lc l xo c cng k = 10 N/m, khi lng 100g c treo thng ng, ko con lc lch khi v tr cn bng 1 on 4 cm ri bung nh. Gia tc cc i ca vt nng :
a/ 4 m/s2b/ 6 m/s2c/ 2 m/s2d/ 5 m/s2e/ 1 m/s213. Con lc l xo khi lng m = 500g dao ng vi phng trnh x = 4cos10t(cm,s). Vo thi im t = Lc tc dng ln vt c cng :
a/ 2 N
b/ 1 N
c/ 4 N
d/ 5 N
e/ 3 N
14. Con lc l xo c cng 25 N/m, dao ng vi qu o 20 cm. Nng lng ton phn l :
a/ 1,1 J
b/ 0,25 J
c/ 0,31 J
d/ 0,125 J
e/ 0,175 J
15. Con lc l xo c cng 100 N/m, dao ng vi bin 4 cm. li x = 2cm ng nng ca n l :
a/ 0,65 J
b/ 0,05 J
c/ 0,001 J
d/ 0,006 J
e/ 0,002 J
16. Mt con lc l xo dao ng vi qu o 10cm. Khi ng nng bng 3 ln th nng, con lc c li :
a/ ( 2 cm
b/ ( 2,5 cm
c/ ( 3 cm
d/ ( 4 cm
e/ ( 1,5 cm
17. Con lc l xo c cng k = 80 N/m. Khi cch v tr cn bng 2,5cm, con lc c th nng :
a/ 5.10-3 J
b/ 25.10-3 J
c/ 2.10-3 J
d/ 4.10-3 J
e/ 3.10-3 J
18. Con lc l xo treo thng ng c phng trnh dao ng : x = Acos ((t + () con lc khi hnh v tr :
a/ Cao nht
b/ Thp nht
c/ Cn bng tho chiu dng
d/ Cn bng theo chiu m
e/ Cu c v d ng
19. Khi i qua v tr cn bng, hn bi ca 1 con lc l xo c vn tc 10 cm/s. Lc t = 0, hn bi bin im B(xB = -A) v c gia tc 25 cm/s2. Bin v pha ban u ca con lc l :
a/ 5 cm ; -(/2 rad
b/ 4 cm ; 0 rad
c/ 6 cm ; +(/2 rad
d/ 2 cm ; ( rad
e/ 4 cm ; -(/2 rad
20. Mt l xo c cng k, c ct ra lm hai on bng nhau. Mi l xo c mt cng l
a)
b) k
c) 2k
d) 4k
21. Mt lc 0,2 N nn mt l xo vo khong 2 cm. Th nng khi b nn l
a) 2.10-3J
b) 2.10-5J
c) 4.10-5 J
d) 8.10-5 J
22. Mt khi lng 1 kg treo vo u mt l xo ko dn l xo c 50 mm. cng ca l xo l
a) 0,20 N/m
b) 1,96 N/m
c) 49 N/m
d) 196 N/m
23. Dao ng iu ha c chu k l T = 0,50s v bin l 20 mm vn tc cc i ca dao ng l
a) cm/s
b) 2( cm/s
c) 4( cm/s
d) 8( cm/s
24. Mt khi lng 20 g treo vo u mt l xo, to ra mt dao ng iu ha c tn s 10 Hz. cng ca l xo l
a) 2,5 N/m
b) 8,9 N/m
c) 12,6 N/m
d) B. CON LC N
1. Chu k con lc n ph thuc vo :
a. Chiu di dy treo.
b. Bin dao ng v khi lng con lc.
c. Gia tc trng trng ti ni dao ng.
d. Khi lng con lc v chiu di dy treo.
e. Cu a v c.
2. Khi chiu di con lc n tng gp 4 th tn s ca n s :
a. Gim 2 ln.
b. Tng 2 ln.
c. Tng 4 ln.
d. Gim 4 ln.
e. Khng thay i.
3. Mt con lc n c chu k 1s khi dao ng ni c g = (2 m/s2. Chiu di con lc l :
a. 50 cm
b. 25 cm
c. 100 cm
d. 60 cm
e. 20 cm
4. Con lc n chiu di 1m, thc hin 10 dao ng mt 20s (ly ( = 3,14). Gia tc trng trng ti ni th nghim :
a. 10 m/s2b. 9,86 m/s2c. 9,80 m/s2d. 9,78 m/s2e. 9,10 m/s25. Con lc c chiu di 64cm, dao ng ni c g = (2 m/s2. Chu k v tn s ca n l :
a. 2 s ; 0,5Hz
b. 1,6 s ; 1Hz
c. 1,5 s ; 0,625Hz
d. 1,6 s ; 0,625Hz
e. 1 s ; 1 Hz
6. Mt con lc n c chu k 2s. Nu tng chiu di ca n ln thm 21cm th chu k dao ng l 2,2s. Chiu di ban u ca con lc :
a. 2 m
b. 1,5 m
c. 1 m
d. 2,5 m
e. 1,8 m
7. Hai con lc n c chiu di l1 v l2 c chu k tng ng l T1 =0,6 s, T2 =0,8 s. Con lc n chiu di l = l1 + l2 s c chu k ti ni :
a. 2 s
b. 1,5 s
c. 0,75 s
d. 1,25 s
e. 1 s
8. Hiu chiu di dy treo ca 2 con lc l 28cm. Trong cng thi gian, con lc nht lm c 6 dao ng, con lc th 2 lm c 8 dao ng. Chiu di dy treo ca chng l :
a. 36 cm; 64 cm
b. 48 cm ; 76 cm
c. 20 cm ; 48 cm
d. 50 cm ; 78 cm
e. 30 cm ; 58 cm
9. Phng trnh dao ng ca 1 con lc n, khi lng 500g : s = 10cos4t (cm,s). Lc t = , ng nng ca con lc :
a. 0,1 J
b. 0,02 J
c. 0,01 J
d. 0,05 J
e. 0,15 J
10. Con lc n dao ng ti ni c g = 10 m/s2 vi bin gc 0,1rad. Khi qua v tr cn bng, c vn tc 50 cm/s. Chiu di dy treo :
a. 2 m
b. 2,5 m
c. 1,5 m
d. 1 m
e. 0,5 m
11. Con lc n c chiu di 1m, khi lng 200g, dao ng vi bin gc 0,15rad ti ni c g = 10 m/s2 . li gc bng bin , con lc c ng nng:
a. 352.10-4 J
b. 625.10-4 J
c. 255.10-4 J
d. 125.10-4 J
e. 10-2
12. Con lc n g dy trong thang my ng yn. Cho thang my i ln chm dn u th chu k dao ng s :
a. Khng i v gia tc trng trng khng i.
b. Ln hn 2s v gia tc hiu dng gim.
c. Khng i v chu k khng ph thuc vo cao.
d. Nh hn 2s v gia tc hiu dng tng.
e. Cu a v c u ng.
13. Con lc n gm 1 vt c trng lng 4N, chiu di dy treo 1,2m dao ng vi bin nh. Ti li ( = 0,05 rad, con lc c th nng :
a. 10-3 J
b. 4. 10-3 J
c. 12. 10-3 J
d. 3. 10-3 J
e. 6. 10-3 J
14. Con lc n c khi lng m = 200g, khi thc hin dao ng nh vi bin s0 = 4cm th chu k (s. C nng ca con lc :
a. 94.10-5J
b. 10-3 J
c. 35.10-5J
d. 26.10-5J
e. 22.10-5J
15. Con lc n dao ng iu ho vi bin gc (0 = 0,15 rad. Khi ng nng bng 3 ln th nng, con lc c li :
a. ( 0,01 rad
b. ( 0,05 rad
c. ( 0,075 rad
d. ( 0,035 rad
e. ( 0,025 rad
16. Con lc n c dao ng iu ho, c chiu di 1m, khi lng 100g, khi qua v tr cn bng c ng nng l 2.10-4J (ly g = 10 m/s2). Bin gc ca dao ng l :
a. 0,01 rad
b. 0,02 rad
c. 0,1 rad
d. 0,15 rad
e. 0,05 rad
17. Con lc n c chiu di l = 2,45m dao ng ni c 9,8m/s2 . Ko lch con lc 1 cung di 4cm ri bung nh. Chn gc thi gian l lc bung tay. Phng trnh dao ng l :
a. s = 4cos(t + ) (cm, s)
b. s = 4cos( + ( ) (cm, s)
c. s = 4cos( - ) (cm, s)
d. s = 4cos2t (cm, s)
e. s = 4cos( - () (cm, s)
18. Con lc n c phng trnh dao ng ( = 0,15cos(t (rad, s). Thi gian ngn nht con lc i t im M c li ( = 0,075 rad n v tr cao nht :
a. s
b. s
c. s
d. s
e.
EMBED Equation.3 s
C. SNG C HC
1. Sng c hc l:
a. S lan truyn vt cht trong khng gian
b. S lan truyn vt cht trong mi trng n hi
c. L nhng dao ng n hi lan truyn trong mi trng vt cht theo thi gian.
d. Tt c cc cu trn u ng
2. Sng ngang truyn c trong cc mi trng:
a. Rn
b. Lng
c. Mt thong cht lng
d. Kh
e. Cu a, b ng
3. Sng dc truyn c trong cc mi trng
a. Rn
b. Lng
c. Kh
d. Cu a, b ng
e. C ba cu a, b, c u ng
4. Tm cu sai trong cc nh ngha sau:
a. Sng ngang l sng c phng dao ng trng vi phng truyn sng
b. Sng dc l sng c phng dao ng trng vi phng truyn sng
c. Sng m l sng dc
d. Sng truyn trn mt nc l sng ngang
e. Trong cc cu trn c mt cu sai
5. Tm cu ng trong cc nh ngha sau:
a. Bc sng l khong cch gia hai im gn nhau nht trn cng mt phng truyn v dao ng cng pha vi nhau.
b. Bc sng l qung ng sng truyn i c trong mt chu k
c. Nhng im dao ng ngc pha nhau trn cng mt phng truyn sng cch nhau na bc sng
d. Cu a, b ng
e. C ba cu a, b, c u ng
6. Qu trnh truyn sng l qu trnh truyn nng lng v:
a. Nng lng sng t l vi bin dao ng
b. Cng xa ngun bin sng cng gim
c. Khi sng truyn n mt im, phn t vt cht ni ny ang ng yn s dao ng, ngha l n nhn c nng lng
d. Cu a, c ng
e. C ba cu a, b, c u ng
7. Vn tc sng ph thuc:
a. Bn cht mi trng truyn sng
b. Nng lng sng
c. Tn s sng
d. Hnh dng sng
e. Tt c cc yu t trn
8. Vn tc sng l:
a. Vn tc truyn pha dao ng
b. Qung ng sng truyn i c trong mt n v thi gian
c. Qung ng sng truyn i c trong mt chu k
d. Cu a, b ng
e. Cu b, c ng
9. Cc i lng c trng cho sng:
a. Bc sng
b. Tn s
c. Vn tc
d. Nng lng
e. Tt c cc i lng trn
10. Sng m l sng c:
a. Tn s t 16 kHz n 20 kHz
b. Tn s t 20 Hz n 19 kHz
c. Tn s ln hn 20.000 Hz
d. Phng dao ng trng vi phng truyn sng
e. Ch truyn c trong khng kh
11. Trong s truyn m v vn tc m, tm cu sai:
a. Sng m truyn c trong cc mi trng rn, lng v kh
b. Vn tc m ph thuc tnh n hi v mt ca mi trng
c. Vn tc m thay i theo nhit
d. Sng m khng truyn c trong chn khng
e. Trong cc cu trn c mt cu sai
12. Trong cc kt lun sau, tm kt lun sai:
a. m sc l mt c tnh sinh l ca m ph thuc vo cc c tnh vt l l tn s v bin
b. cao l mt c tnh sinh l ca m ph thuc vo cc c tnh vt l l tn s v nng lng m
c. to ca m l mt c tnh sinh l ca m ph thuc vo cng v tn s m
d. Nhc m l nhng m c tn s xc nh. Tp m l nhng m khng c tn s xc nh
e. V c tnh vt l, sng m, sng siu m, sng h m khng khc g cc sng c hc khc
13. Trong cc nh ngha sau, nh ngha no sai:
a. Chu k sng l chu k dao ng chung ca cc phn t vt cht c sng truyn qua v bng chu k dao ng ca ngun sng
b. Bin sng ti mt im l bin chung ca cc phn t vt cht c sng truyn qua v bng bin ca ngun sng
c. Sng kt hp l sng to ra bi cc ngun kt hp. Ngun kt hp l cc ngun c cng tn s, cng pha hoc c lch pha khng i
d. Bc sng l qung ng sng lan truyn trong mt chu k
e. Sng dng l s giao thoa ca hai sng ti v sng phn x, kt qu l trn phng truyn sng c nhng nt v bng sng c nh
14. Sng ti ngun A c dng u = acos(t th phng trnh dao ng ti M trn phng truyn sng cch A on d c dng:
a.
b.
c.
d.
e.
15. Sng ti A, B c dng u = acos(t
Xt im M cch A on d1, cch B on d2. lch pha ca hai dao ng t A v t B n M ti M l
a.
b.
c.
d.
e. vi d = |d2 d1|
16. Hai sng cng pha khi:
a. (k=0, 1, 2,)
b.
(k=0, 1, 2,)
c.
(k=0, 1, 2,)
d.
(k=0, 1, 2,)
e.
(k=0, 1, 2,)
17. Cc im ng yn trong vng giao thoa tha iu kin:
a.
(k=0, 1, 2,)
b.
(k=0, 1, 2,)
c. .(k=0, 1, 2,)
d.
(k=0, 1, 2,)
e.
(k=0, 1, 2,)
18. Bin dao ng ti mt im trong vng giao thoa cc hai ngun khong d1, d2 l:
a.
b.
c.
d.
e.
19. Trong cng mt mi trng, sng m tn s 300Hz c bc sng (, sng m tn s 600Hz c bc sng (' th
a. (' = 1/2(
b. (' = (
c. (' = 2(
d. (' = 4( 20.Vn tc ca sng radio l 3x108 m/s. Cho bit tn s ca sng radio c bc sng 10m ?
a. 3,3.10-8Hz
b. 3,0.107Hz
c. 3,3.107Hz
d. 3.109Hz.
DNG IN XOAY CHIU
A. BI TP
Cu 1:
Cho mt mch in xoay chiu c tn s f = 50Hz qua mt mch in ni tip gm
R = 50(, C = 63,5(F v L = 0,31H
1. dng in v hiu in th cng pha, phi thay t in bng mt t in khc c in dung bao nhiu?
2. cng dng in v hiu in th cng pha nu KHNG thay t in, th phi mc thm mt t in khc c in dung bao nhiu v mc th no?
Cu 2:
Cho mt mch in gm in tr R = 100(, t in C = 31,4 (F, mt cun dy L mc ni tip. t vo hai u mch in mt hiu in th u = 200 cos 100(t (V)
1. Tnh L cng dng in qua mch l 1A.
2. Tnh L cng dng in qua mch t gi tr cc i.
Cu 3:
Cho mt on mch AB gm mt in tr R = 12( v mt cun cm L. Hiu in th hiu dng hai u ca R l U1= 4V, hai u ca L l U2 = 3V v hai u AB l UAB = 5V
1. Tnh in tr R0 v h s t cm L ca cun dy.
2. Tnh cng sut in tiu th trng mch.
Cu 4:
Cun dy s cp ca mt my bin th c 1023 vng, cun th cp c 75 vng. t vo hai u ca cun s cp mt hiu in th xoay chiu c gi tr hiu dng 3000V.
1. Ni hai u cun dy th cp bng mt in tr thun R = 10(. Tnh cng dng in trong cun th cp.
2. Thay v in tr thun nh trong cu trn, ngi ta ni hai u cun th cp vo mt ng c in c cng sut 2,5kW v h s cng sut cos( = 0,8 th cng hiu dng trong mch th cp bng bao nhiu?
Cu 5:
Cho mch in nh hnh v:
R0 = 20(; f = 50Hz; uAB = U0cos(t. Khi m kho K th i cng pha u. Khi ng kho K th u sm pha (/4 so vi i. Hi in dung C c gi tr bng bao nhiu?
Cu 6:
Cho mch in xoay chiu nh hnh v:
uAB = U0cos100(t (V)
L = ; C = 63,6 (F
Bit cng dng in qua mch tr pha (/4 so vi hiu in th hai u mch. in tr R c gi tr bng bao nhiu?
Cu 7:
Cho mch in nh hnh v:
R0 = 90( ; L =
u = U0cos100(t (V)
Thay i C n C0 ngi ta thy ch s Vn k cc i bng 300V. Cho bit U0 c gi tr bng bao nhiu?
Cu 8:
on mch nh hnh v
C = 31,8 (F ; L =
uAB = 200 cos(100(t) (V)
Hiu in th uAM tr pha (/6 so vi dng in qua mch v dng in qua mch tr pha (/3 so vi uBM in tr R v in tr R0 ca cun dy c gi tr bng bao nhiu?
Cu 9:
Cho on mch nh hnh v:
uAB = 100 cos(100(t) (V)
Khi K ng, Ampe k (A1) ch 2A. Khi K m dng in qua mch lch pha (/4 so vi hiu in th hai u mch. Cc Ampe k c in tr rt nh. Cho bit ch s Ampe k (A2) khi K m l bao nhiu?
Cu 10:
Cho on mch gm in tr R, cun cm L =0,5H v t C mc ni tip hai u mch c hiu in th u = U0cos100(t (V). mch c cng hng t C c gi tr bng bao nhiu?
Cu 11:
Mch in LRC c R = 40( mc vo ngun in xoay chiu c u = 80cos100(t. Hi cng sut cc i ca mch bng bao nhiu?
Cu 12:
Cho mch in nh hnh v:
R0 = 20(; L = ; C = 63,6 (F; u = U0cos100(t (V) thay i R n gi tr R1 th uAM vung pha vi uMB. Tnh R1?
A. B. L THUYT
Cu 1:
iu no sau y ng khi ni v cng dng in v hiu in th tc thi:
a) Gi tr trung bnh ca dng in xoay chiu trong mt chu k bng 0.
b) C th s dng Ampe k, Vn k, khung quay o cng hay hiu in th tc thi.
c) Tc dng ca dng in xoay chiu u ging vi dng in khng i.
d) Mi im trn on mch khng phn nhnh c cng dng in nh nhau v ht mang in chuyn ng vi vn tc nh sng (c 3(108 m/s)
e) Do i v u bin thin cng tn s nn khi dng in t gi tr cc i th hiu in th cng cc i
Cu 2:
Dng in xoay chiu c i = 2cos(314t + )(A; s). Tm pht biu sai:
a) Cng cc i l 2A
b) Tn s dng in l 50Hz
c) Cng hiu dng l 2A
d) Chu k dng in l 0,02s
e) Ti t = 0 th i = A
Cu 3:
on mch xoay chiu ch c R, pht biu no sau y sai:a) U = RI
b) P = RI2c) u cng pha i
d) I v U tun theo nh lut Ohm
e) Mch c cng hng in
Cu 4:
Cho on mch xoay chiu ch c t in C, pht biu no sau y ng:a) u sm pha so vi i
b) Dung khng ca t t l vi tn s dng in.
c) U = C(I
d) T in cho dng in xoay chiu i qua v c s np v phng in lin tc ca t in.
e) Tt c cc pht biu trn u sai.
Cu 5:
on mch xoay chiu ch c cun dy thun cm, pht biu no sau y l ng:a) i tr pha so vi u
b) U = L(I
c) Do hin tng t cm nn trong cun dy c in tr ph gi l cm khng.
d) Khi tn s dng in ln th dng in b cn tr nhiu.
e) Tt c cc pht biu trn u ng.
Cu 6:
on mch xoay chiu c in tr v t C mc ni tip. iu no sau y sai:a) i tr pha so vi u hai u mch
b) Tng tr mch Z =
c) Cng sut mch P = RI2d) H s cng sut mch c gi tr nh hn 1
e) U2 = U2R + U2CCu 7:
Tn s cng hng ca mch in c cha cun cm 50mH v mt t in 0,20 (F th bng
a) 16kHz
b) 10kHz
c) 63kHz
d) 16MHz.
Cu 8:
on mch xoay chiu khng phn nhnh gm cun dy thun cm v t in. Chn pht biu ng:a) on mch c cng hng in khi ZL = ZC
b) U = UL + UC
c) U2 = U2L + U2Cd) Cng sut tiu th trn on mch P = UI
e) U hai u mch lch pha ( so vi i, tu theo gi tr ZL v ZCCu 9:
Xt mch in xoay chiu R, L, C mc ni tip. Pht biu no sau y sai:a) Tng tr mch ch ph thuc vo R, L v C
b) Mch c tnh cm khng nu L( >
c) Mch c tnh dung khng nu L( <
d) u = uR + uL + uC
e) lch pha ca i so vi u c tnh bng cng thc tg( =
Cu 10:
Mch in xoay chiu R, L, C mc ni tip. Cng hng in xy ra th:
a) LC(2 = 1
b) Tng tr mch Z = R
c) u cng pha vi i
d) UL = UC
e) Tt c cc cu trn u ng
Cu 11:
Mch in gm R, L, C mc ni tip t vo u = U0cos2(ft. iu no sau y ng:a) Dng in xoay chiu qua t C v t in cho in tch chy qua khong gia hai bn t.
b) Dng in xoay chiu trong mch l dao ng in cng bc do hiu in th xoay chiu t vo hai u mch gy ra.
c) Khi ZL = ZC th u vung pha vi i
d) Khi C ( ( th mch c tnh dung khng
e) Khi C tng, R v L gi khng thay i th U hai u t C tng theo.
Cu 12:
Khi ni v nh hng ca in tr thun R trong mch in khng i v trong mch in xoay chiu. Pht biu no sau y sai:a) Vi ngun khng i U th I =
b) Tc dng ca in tr thun ging nhau i vi mch in khng i v mch in xoay chiu.
c) Vi mch in xoay chiu th in tr thun R c gi tr tng khi tn s dng in rt ln.
d) Vi ngun in xoay chiu u = U0cos(t th i cng pha vi u
e) Trong cc pht biu trn c mt pht biu sai
Cu 13:
nh hng ca cun cm trong mch in khng i v trong mch in xoay chiu. Pht biu no sau y sai:a) Vi mch xoay chiu cun cm cn tr dng in vi tng tr Z =
b) Vi mch xoay chiu khi t cm L tng th I gim.
c) Vi mch xoay chiu khi L( >>R th i sm pha hn u gc
d) Vi ngun in khng i cun cm ch c tc dng nh in tr thun R.
e) Trong cc pht biu trn pht biu no sai
Cu 14:
nh hng ca t in C trong mch in khng i v trong mch in xoay chiu. Pht biu no sau y sai:a) Dng in khng i khng i qua c on mch c cha t in.
b) Vi mch xoay chiu th in th hai u t tr pha so vi i
c) Khi C c gi tr rt ln dng in xoay chiu qua t d dng.
d) in tr ca t c gi hu hn i vi dng in xoay chiu v v cng ln i vi dng in thay i.
e) Trong cc pht biu trn c hai pht biu sai
Cu 15:
Trong mch in xoay chiu in nng tiu th trung bnh trong mt chu k ph thuc vo:
a) Tn s f
b) H s cng sut cos(c) Hiu in th hiu dng U
d) Cng hiu dng U
e) Tt c cc yu t trn
Cu 16:
Mch in xoay chiu no sau y khng tiu th cng sut:
a) Mch ch c L v C mc ni tip.
b) Mch ch c R v L mc ni tip.
c) Mch ch c R v C mc ni tip.
d) Mch ch c R
e) Mch gm R, L, C mc ni tip v ZL=ZCCu 17:
Mch gm R, L, C mc ni tip vo u = U0cos2(ft vi R thay i. Cng sut mch cc i khi:
a) R = ZL + ZC
b) R = 0
c) R =
d) R =
e) Tt c cc cu trn u sai.
Cu 18:
L do tng h s cng thc cos( l:
a) mch tiu th cng sut hu ch ln hn
b) cng sut hao ph trn mch gim
c) cng I qua mch gim
d) Cu b, c ng
e) C 3 cu a, b, c u ng.
Cu 19:
Trong cc pht biu sau, pht biu no sai:
a) H s cng sut cos( = 1 khi u cng pha vi i
b) Cun dy thun cm bin i nng lng in thnh nng lng in trng v ngc li, nn cng sut tiu th ca cun dy bng Zr
c) in tr R tiu th nng lng in di dng nhit
d) nng cao h s cos( ca mch c ng c in ngi ta mc ni tip hoc song song vi ng c mt t in kh bt tnh cm khng ca mch
e) Mch R, L, C tiu th cng sut ln nht khi trong mch c cng hng in
Cu 20:
Trong cu to my pht in xoay chiu mt pha, iu no sau y sai:
a) Phn cm: to t trng (nam chm)
b) Phn ng: ni xut hin in ng cm ng (khung dy)
c) Li st ca hai phn cm v phn ng lm bng cc l st mng ghp cch in vi nhau
d) gim vn tc quay ca Rto ngi ta dng Stato c p cp cc.
e) Vi my pht in ln Stato phi l phn ng d ly in ra ngoi hn
Cu 21:
Trong cc pht biu sau, pht biu no sai trong l do s dng rng ri dng in xoay chiu trong thc t:
a) Dng in xoay chiu c th cung cp mt cng sut ln
b) Dng in xoay chiu c y tc dng nh dng in mt chiu
c) Dng in xoay chiu c th chnh lu thnh dng in mt chiu d dng
d) Dng in xoay chiu c th ti i xa vi hao ph t.
e) My pht in xoay chiu cu to n gin hn my pht in mt chiu.
Cu 22:
ng c khng ng b c u im l:
a) C th thay i chiu quay d dng
b) Cu to n gin d ch to
c) C momen khi ng ln hn ng c mt chiu
d) Cu a, b ng
e) Cu a, b, c u ng.
Cu 23:
Nam chm in c tnh cht no sau y:
a) T tnh ca li st ch thc t tn ti khi c dng in qua ng dy; dng in tt th t tnh mt
b) T tnh ca li st vn cn mt thi gian di sau khi dng in qua ng dy tt
c) Cc cc N, S ca li st thay i khi chiu dng in thay i.
d) Cu a, b ng
e) Cu a, c ng
Cu 24:
Trong my pht in xoay chiu 3 pha, iu no sau y sai:
a) Phn cm l nam chm in (Rto)
b) Phn ng gm 3 cun dy ging nhau t lch nhau 1200 trn mt gi trn (Stato)
c) Khi cc Bc i din vi cun no th sut in ng xut hin trong cun dy t gi tr cc i
d) Do t thng xuyn qua cun dy lch pha nhau 1200 nn sut in ng trong 3 cun dy cng lch pha nhau 1200e) My pht in xoay chiu 3 pha to ra 3 dng in 1 pha.
Cu 25:
Trong cch mc in 3 pha, iu no sau y sai:
a) Trong cch mc no ta cng c Ud =
b) Trong cch mc hnh sao cc ti khng cn i xng
c) Trong cch mc tam gic cc ti cn i xng
d) Nh c cch mc dng in 3 pha nn ngi ta tit kim c dy dn khi truyn ti
e) Trong cc cu trn c 1 cu sai
Cu 26:
ng c khng ng b 3 pha hot ng c l nh:
a) Hin tng cm ng in t
b) T trng quay ca dng in xoay chiu 3 pha
c) Hin tng cm ng in t v vic s dng t trng quay
d) Rto ca ng c l rto on mch
e) Tt c cc cu trn u ng
Cu 27:
So snh cu to ca ng c 3 pha v my pht 3 pha, iu no sau y sai:
a) Rto ca ng c l hnh tr c tc dng ging nh cun dy qun trn li thp khc vi rto ca my pht in l nam chm in
b) Rto ca ng c ging rto ca my pht in v cng l cun dy qun trn li thp.
c) Stato ca ng c ging rto ca my pht in v cng l 3 cun dy ging nhau t lch nhau 1200 trn mt gi trn
d) C th bin ng c thnh my pht in bng cch thay rto tr st bng nam chm c cng trc quay.
e) Trong cc cu trn c mt cu sai.
Cu 28:
Dng in xoay chiu 3 pha l h thng gm 3 dng in xoay chiu mt pha cnhng chn cu ng vi nh ngha trn:
a) C cng tn s nhng lch pha nhau
b) C cng bin nhng khc pha
c) C cng bin nhng khc tn s
d) C cng bin nhng lch pha nhau v thi gian l 1/3 chu k
e) Tt c u sai
Cu 29:
Khi my bin th c mch th cp h, pht biu no sau y sai:
a) Hiu in th t l vi s vng dy mi cun
b) Cng dng in cun th cp I2 = 0
c) Cng sut tiu th trong cun s cp gn bng 0
d) Cng sut tiu th trong cun th cp gn bng 0
e) Sut in ng cun th cp t l vi s vng dy cun th cp
Cu 30:
Ngi ta mc cun s cp my bin th vo mng in xoay chiu v:
a) in tr thun ca cun s cp rt nh nn hao ph nhit khng ng k
b) Dng in s cp rt nh v cun ny c cm khng ln khi my bin th chy khng ti.
c) Cng sut v h s cng sut ni cun th cp bng cng sut v h s ni cun s cp.
d) Cu a, b ng
e) Cu b, c ng
Cu 31:
Thc t trong truyn ti in nng ngi ta thc hin:
a) To ra hiu in th thch hp vi yu cu
b) Gim bt in tr dy dn bng cch tng tit din
c) Tng cng sut cn truyn ti ln nhiu ln
d) in nng c to ra nh my c tng th ri a ra dy dn ti i. Trn ng truyn ti, in th c h th dn tng bc thch hp vi yu cu
e) Cu a, b, c ng
Cu 32:
Dng in mt chiu cn thit v:
a) Cc thit b in t hot ng c nh in p mt chiu
b) Dng cung cp cho ng c in mt chiu
c) Dng m in c in
d) Cu b, c ng
e) C a, b, c ng
Cu 33:H s cng sut trong mt mch in l
a) R.Z
b)
c)
d) Cu 34:. Khi c cng hng iu no sau y ng
a) R = Z
b)
c) P = I .U
d) Cu 35: Mt vn k trong mt mch in ch 50V v mt ampe k mc ni tip ch 5A. Cng sut tiu th trong mch th
a) Nh hn hay bng 250W
b) Bng 250W
c) Ln hn hay bng 250W
d) C th nh hn, bng hay ln hn 250W.
Cu 36:Mt in tr, mt t in, v mt cun cm c mc ni tip trong mt ngun in xoay chiu tn s f. in th hiu dng ca mch gm cc thnh phn UR = 10V, UC=20V, UL=14V. in th hiu dng ca ngun in bng
a) 8V
b) 12V.
c) 16V
d) 15V.
Cu 37:Gc lch pha ca mt mch in trong cu 36 l
a) -310
b) +310
c) -530
d) +530Cu 38: Tn s cng hng ca mch in trong cu 36 th
a) Nh hn f
b) Bng f
c) Ln hn f
d) Khng iu g trn ng, m ty thuc vo in th trong mi thnh phn.
DAO NG IN T
1. Mt s bin i in trng sinh ra
A. Mt t trng
B. Sng in t
C. Sng m
D. Khng phi iu no trn
2. Trong mt sng in t, in trng
A. Song song vi t trng v phng truyn ca sng
B. Vung gc vi t trng v phng truyn ca sng
C. Song song vi t trng v vung gc vi phng truyn ca sng
3. Trong chn khng vn tc truyn ca sng
A. Ty thuc vo tn s ca song
B. Ty thuc vo bc sng
C. Ty thuc vo mt hng s ph cp
D. Tt c u sai
4. Tn s cc i c tm trong
A. Tia X
B. Tia cc tm
C. Sng truyn thanh
D. Sng rada
5. Tnh nng no ca m tng ng vi mu ca nh sng
A. Bin
B. Cng hng
C. Dng sng
D. cao ca m
6. C hai tia ( v sng truyn thanh u l sng in t v
A. Tia ( c bc sng ngn hn sng truyn thanh
B. Tia ( c bc sng di hn sng truyn thanh
C. Tia ( v sng truyn thanh c bc sng bng nhau
D. Khng phi iu no trn
7 Mt mch dao ng gm cun cm L v t in C1 th dao ng vi tn s 3 kHZ. Nu thay t C1 bng t C2 th tn s dao ng ca mch l 4 kHz.Nu ghp C2 ni tip vi t C1 th tn s dao ng ca mch l bao nhiu ?
A. 5 kHz.B. 7 kHz.
C. 1 kHz.
D. 12 kHz.
8. Pht biu no sau y l sai khi ni v nng lng ca mch dao ng LC l tng ?
A. Nng lng in t ca mch dao ng bin i tun hon theo thi gian.
B. Nng lng in t ca mch dao ng bng nng lng t trng cc i cun cm.
C. Nng lng in trng v nng lng t trng cng bin thin tun hon cng tn s.
D. Nng lng in trng cc i bng nng lng t trng cc i.
9. Mt mch dao ng in t c tn s f = 0,5.106 Hz , vn tc nh sng trong chn khng l 3.108 m/s . Sng in t do mch pht ra c bc sng l
A. 600m B. 0,6m; C. 60m;
D. 6 m.
10.Mch dao ng ca mt my thu v tuyn in c cun cm L = 25F. thu c sng v tuyn c bc sng 100m th in dung ca t in c gi tr l
A. 112,6pF. B. 1,126nF.
C. 1126.10-10F.
D. 1,126pF.Tn s ca dao ng in t trong khung dao ng tho mn h thc no sau y?
C. f = .E. f = .
B. f = .F. f = .
11. Tnh cht no sau y khng phi l tnh cht ca sng in t?
A. Sng in t l sng ngang.
B. Nng lng sng in t t l lu tha bc 4 ca tn s.
C. Sng in t khng truyn c trong chn khng.
D Sng in t c th giao thoa, khc x, phn x.
12. Mt mch dao ng LC gm cun dy thun cm c L = 2/mH v mt t in
C = 0,8/(F). Tn s ring ca dao ng trong mch l
A. 50kHz. B. 25 kHz. C. 12,5 kHz.
D. 2,5 kHz.
13. Cng dng tc thi trong mch dao ng LC l i = 0,05sin2000t (A), in dung ca t bng 5(F. t cm ca cun dy l A. 2,5.10-4 H
B. 5.10-8 H C. 5 H D. 0,05 H 14. Cng dng tc thi trong mch dao ng LC l i = sin200t (A), in dung ca t bng 10(F. in tch cc i trn t l
A.10-3 C
B. 10-6 C
C. 5.10-6 C D. 5.10-3 C
15. Mch dao ng in t gm mt t in c C = 0,125 (F v mt cun cm c L = 50(H. in tr thun ca mch khng ng k. Hiu in th cc i gia 2 bn t l 3V. Cng dng in cc i trong mch l
A. 7,5mA
B. 15mA
C. 7,5A
D. 0,15A
16. Mt t in c in dung 10(F c tch in n mt hiu in th xc nh. Sau ni hai bn t in vo 2 u mt cun dy thun cm c t cm 1H. B qua in tr ca cc dy ni, ly 2 = 10. Sau khong thi gian ngn nht l bao nhiu (k t lc ni) th in tch trn t c gi tr bng mt na gi tr ban u?
A. 3/400s
B. 1/300s C. 1/1200s
D. 1/600s
17.in tch hiu dng trn t v dng in hiu dng qua cun cm ca mt mch dao ng ln lt l Q = 0,16.10-11 C v I = 1mA. Mch in t dao ng vi tn s gc l
A. 0,4.105 rad/s
B. 625.106 rad/s
C. 16.108 rad/s
D. 16.106 rad/s
18.Cng thc tnh nng lng ca mch dao ng in t LC l
A. B. C.
D.
19. Mch dao ng c t C=1000pF v L=2,5(H. Nu hiu in th cc i 2 u t l 2,828V th cng dng in hiu dng qua mch l
A. 40mA
B. 0,4A
C. 0,2A
D. 20mA
20. Mt mch thu sng c L=10(H, C=1000/2 pF thu c sng c bc sng l
A. 0,6m B. 6m C. 60m
D. 600m
B
A
C
2
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PAGE 26Dao ng v sng c hc
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