July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 1 II. DAO NG C A. TM TTLTHUYT 1. Dao ng iu ha: * Dao ng c, dao ng tun hon + Dao ngc lchuynngqua lica vtquanhvtrcn bng. +Daongtunhonldaongmsaunhngkhongthigianbngnhau,gilchuk,vttrlivtr c theohngc. * Dao ng iu ha + Dao ngiuha ldao ngtrong li ca vtl mthmcsin(haysin)ca thigian.+ Phngtrnhdao ng:x = Acos(et+ ). +imPdaongiuhatrnmtonthnglunluncthccoilhnhchiucamtimM chuynngtrnu trnngtrnc ngknhl on thng. * Cc i lng c trng ca dao ng iu ho Trongphngtrnhx = Acos(et + ) th: + A l bin dao ng, l gitrcc ica li x;nvm,cm. A lunlundng. + (et + ) l pha cadao ngtithiimt; nvrad. + l pha ban uca dao ng;nvrad. + e trongphngtrnhx = Acos(et+ ) l tns gc cadao ngiuha;n vrad/s. + Chuk T ca dao ngiuha lkhongthigian thc hinmtdao ngton phn;n vgiy(s). + Tns f cadao ngiuhal s dao ngtonphnthc hinc trongmtgiy;n vhc (Hz).+ Linh giae, T vf:e = Tt 2 = 2tf. CcilngbinAvphabanuphthucvocchkch thch banu lm cho h daong, cn tns gc e (chukT, tn s f)chphthucvo cu to ca h dao ng. * Vn tc v gia tc ca vt dao ng iu ho + Vntc lo hmbc nhtca li theothigian:v = x' = - eAsin(et+ ) = eAcos(et + + 2t)Vntc ca vtdao ngiuha binthiniuha cngtns nhngsmpha hn 2t so vivili.Vtrbin(x = A), v = 0. Vtrcn bng(x= 0), |v|= vmax = eA. + Giatc l o hmbc nhtca vntc (o hmbc 2 ca li) theothigian: a = v' = x= - e2Acos(et + ) = - e2x. Giatccavtdaongiuhabinthiniuhacngtnsnhngngcphavili(smpha 2t so vivntc). Vc t giatc ca vtdao ngiuha lunhngv vtrcn bng,c lntlvi lncali.- v trbin(x = A), giatc c lncc i:amax = e2A. - v trcn bng(x = 0), giatc bng0. + Lc tc dnglnvt dao ngiuha F = ma= - kx lunhngv vtrcnbng,gillc ko v. +thdaongiuha(li,vntc,giatc)lnghnhsin,vthngitacngidaongiu ha ldao nghnhsin. +Phngtrnhdaongiuhax=Acos(et+)lnghimcaphngtrnhx+e2x=0.l phng trnhnglc hcca dao ngiuha. 2. Con lc l xo: Conlclxogmmtlxoccngk,khilngkhngngk,mtugncnh,ukiagn vivt nngkhilngm c t theo phngnganghoc treo thngng. *Phngtrnhdaong: x = Acos(et + ); vi: e =mk; A = 20 20|.|

\|+evx ; xc nh theo phng trnh cos = Ax0; (lynghim(-) nuv0 > 0; lynghim(+) nuv0 < 0). July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 2 * Chu k, tn s ca con lc l xo: T = 2tkm; f = 12t mk. * Nng lng ca con lc l xo: +ngnng:W= 21mv2= 21me2A2sin2(et+).Thnng:Wt= 21kx2= 21kA2cos2(et+).ngnng, thnngca vt dao ngiuha binthintunhonvitns gce = 2e, tn s f= 2f,chuk T = 2T. + C nng:W = Wt+ W = 21k A2 = 21 me2A2 = hngs. 3. Con lc n. Con lc vt l: Conlcngmmtvtnngtreovosidykhnggin,vtnngkchthckhngngksovi chiudi sidy, sidy khilngkhngngk so vikhilngca vtnng. * Phng trnh dao ng (khi o s 100): s = S0cos(et + )hoco = o0 cos(et + ); vio = ls; o0 = 0Sl. * Chu k, tn s, tn s gc ca con lc n: T = 2tgl;f = t 21lg; e = lg. * Lc ko v khi bin gc nh:F = -slmg. * ng dng: Xc nhgiatc rit do nho chuk v chiudica con lcn:g= 224Tl t. * Nng lng ca con lc n: + ngnng: W = 21mv2. Thnng:Wt = mgl(1- coso) = 21mglo2 (o s 100, o (rad)). + C nng:W = Wt + W = mgl(1- coso0) = 21mglo20.C nngca conlc nc bo tonnub qua ma st. * Con lc n chu tc dng thm lc khc ngoi trng lc Nungoitrnglcra,conlcncnchuthmmtlc F khngikhc(lcintrng,lcqun tnh,lcyAcsimet,...),thtrnglcbiukintcdnglnvtsl: ' P = P + F ,iatcritdobiu kinl: ' g= g+ mF. Khi chuk dao ngcacon lcnl:T = 2t' gl. * Con lc vt l: Con lc vtll mtvtrn quayc quanhmttrcnmngangc nh. +Phngtrnhdaongcaconlcvtl:o=o0cos(et+);vie = d mgI; trong m l khi lng ca vtrn, d lkhongcch t trngtmca vtrn n trcquaycn I l momenquntnhcavt rn. + Chuk, tns ca con lcvtl:T = 2tdImg, f = 12td mgI. +ngdngcaconlcvtl:Gingnhconlcn,conlcvtldngogiatctrngtrnggnit con lc. 4. Dao ng tt dn, dao ng cng bc: * Dao ng tt dn +Khikhngcmast,conlcdaongiuhavitnsring.Tnsringca con lc ch ph thuc vo cc c tnhcacon lc. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 3 +Daongcbingimdntheothigiangildaongttdn.Nguynnhnlmttdndaongl dolcmastvlccncamitrnglmtiuhaocnngcaconlc,chuynhadncnngthnh nhitnng.Vthbin ca conlc gimdn v cuicngcon lcdngli. +ngdng:ccthitbngcatng,ccbphngimxccat,xemy, l nhng ng dng ca dao ngtt dn. * Dao ng duy tr Nutacungcpthmnnglngchovtdaongcmastblistiuhaovmastmkhng lm thayi chukringcan thdao ngs ko dimiv c gil dao ngduytr. * Dao ng cng bc + Dao ngchutc dngca mtngoilc cngbc tunhongildao ngcngbc.+ Dao ngcngbc c bin khngiv c tns bngtns ca lc cngbc. +Bincadaongcngbcphthucvobincalccngbc,volc cn trong h v vo s chnhlchgiatnscngbcfvtnsringf0cah.Bincalccngbccngln,lccn cngnhv s chnhlchgiafv f0 cngtthbin ca dao ngcngbc cngln. * Cng hng +Hintngbincadaongcngbctngdnlnngitrccikhitnsfca lc cng bc tinn bngtns ringf0 ca h dao nggil hintngcnghng. + iukinf = f0 giliukincnghng. +ngcongbiudinsphthuccabinvotnscngbcgilthcnghng.Ncng nhnkhilccn ca mitrngcngnh. + Tmquantrngca hintngcnghng: Tanh,cu,bmy,khungxe,...ulnhnghdaongvctnsring.Phicnthnkhng chochngchutcdngcacclccngbcmnh,ctnsbngtnsringtrnhscnghng, gydao ngmnhlmgy,. Hpncanghita,vilon,...lnhnghpcnghngvinhiutnskhcnhaucadynlm cho tingn ngheto, r. 5. Tng hp cc dao ng iu ha: +Midaongiuhacbiudinbngmtvctquay.Vctny c gc ti gc ta ca trc Ox, cdibngbindaongA,hpvitrcOxmtgcbanuvquayuquanhOtheochiu ngcchiukimngh vitc gce. +PhngphpginFre-nendngtnghphaidaongiuhacngphng, cngtns:Lnltvhaivctquay 1A v 2A biudinhaiphngtrnhdaong thnhphn.Sauvvcttnghpcahaivcttrn.Vcttng A = 1A +2A l vc t quaybiudinphngtrnhca dao ngtnghp. +Numtvtthamgiangthihaidaongiuhocngphng,cngtnsviccphngtrnh:x1=A1cos(et+1) v x2 = A2cos(et + 2), th dao ng tng hp sl: x = x1 + x2 = Acos(et + ) vi A v c xc nhbicc cngthc:A2 = A12 + A22 + 2 A1A2 cos (2 - 1) v tan = 2 2 1 12 2 1 1cos cossin sin A AA A++. Binvphabanucadaongtnghpphthucvobinvphabanucaccdaong thnhphn. + Khix1 v x2 cngpha (2 - 1 = 2kt) th dao ngtnghpc bin cc i:A = A1 + A2. + Khix1 v x2 ngcpha (2 - 1 = (2k + 1)t) th dao ngtnghpc bin cc tiu:A = |A1 - A2| .+ Trnghptngqut:A1 + A2> A > |A1 - A2|.B. CC DNG BI TP 1. Tm cc i lng c trng trong dao ng iu ha. * Cc cng thc: + Li (phngtrnhdao ng):x = Acos(et+ ). + Vntc:v = x = - eAsin(et+ ) = eAcos(et+ + 2t). + Giatc:a = v= - e2Acos(et + ) = - e2x; amax = e2A. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 4 + Vntc v smpha 2t so vili x; giatc a ngcpha vili x (smpha 2t so vivntc v).+ Linh giatn s gc, chuk v tns ca dao ng:e = Tt 2 = 2tf. + Cngthc c lp:A2 = x2 + 22ve= 2 22 4v ae e+ . + vtrcn bng:x = 0 th|v| = vmax = eA v a = 0. + vtrbin:x = A thv = 0 v |a| = amax = e2A = 2ax mvA. + Lc ko v:F = ma= - kx. + Quo chuynngca vtdao ngiuhal mton thngc chiudiL = 2A. * Phng php gii: +tmccilngctrngcamtdaongiuhakhibitphngtrnhdaonghocbitmts ilngkhccadaongtasdngcccngthclinquannnhngilngbitvilng cn tmrisuyra v tnhi lngcn tmtheo yucu cabi ton. +tmccilngcadaongiuhatimtthiimtchotathaygitrcatvophngtrnh linquan tnhilng. Lu:Hmsinvhmcoslhmtunhonvichuk2tnnkhithayt vo nu c gc ca hm sin hoc hmcos l mts lnhn2t thta b ica gc mts chncat d bm my. +tmthiimmx,v,ahayFcmtgitrcthnothtathaygitrnyvophngtrnhlin quanv giiphngtrnhlnggic tmt. Lu:ngstnghim:vihmsinthlythmgcbvigctmc,cnvihmcosthly thmgcivinvnhhmsinvhmcoslhmtunhonvichuk2tngbstcch nghim.Cngngdnghim:Cncvoducaccilnglinquanloibthnghimkhng phhp. * Bi tp minh ha: 1.Phngtrnhdaongcamtvtl:x=6cos(4tt+ 6t)(cm),vixtnhbngcm,ttnhbngs.Xc nh li, vntc v giatc ca vtkhit = 0,25 s. 2.Mtvtnhkhilng100gdaongiuhatrnquothngdi20cmvitnsgc 6 rad/s. Tnh vntc cc i v giatc cc i ca vt. 3.Mtvtdaongiuhotrnquodi40cm.Khivtrclix=10cmvtcvntc 20t 3 cm/s. Tnhvntc v giatc cc ica vt. 4.Mtchtimdaongiuhovichuk0,314svbin8cm.Tnhvntccachtimkhi n i qua vtrcn bngv khin i qua vtrc li 5 cm. 5.Mtchtimdaongtheophngtrnh:x=2,5cos10t(cm).Vothiimnothphadao ng t gi tr 3t? Lcy li, vntc, giatcca vtbngbao nhiu? 6.Mtvtdaongiuhaviphngtrnh:x=5cos(4tt+t)(cm).Vtiquavtrcnbngtheo chiudngvo nhngthiimno? Khi lnca vntc bngbao nhiu? 7.Mtvtnhckhilngm=50g,daongiuhaviphngtrnh:x= 20cos(10tt + 2t) (cm). Xc nh lnvchiuca cc vc t vntc, giatc v lc ko v tithiimt = 0,75T. 8.Mtvtdaongiuhatheophngngangvibin2 cmvvichuk0,2s.Tnhlncagia tc ca vt khin c vntc 10 10cm/s. 9.Mtvtdaongiuhaviphngtrnh:x=20cos(10tt+ 2t)(cm).Xcnhthiimu tin vt i qua vtrc li x = 5 cm theochiungcchiuvichiudngk t thiimt = 0. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 5 10.Mtvtdaongiuhaviphngtrnh:x=4cos(10tt- 3t)(cm).Xcnhthiimgnnhtvn tc ca vt bng20t 3cm/s vangtngk t lct = 0. * Hng dn gii v p s: 1. Khi t = 0,25 s thx = 6cos(4t.0,25 +6t) = 6cos67t= - 3 3 (cm); v = - 6.4tsin(4tt + 6t) = - 6.4tsin67t= 37,8 (cm/s);a = - e2x = - (4t)2. 3 3 = - 820,5 (cm/s2). 2. Ta c: A = 2L= 220= 10 (cm)= 0,1 (m);vmax = eA = 0,6 m/s;amax = e2A = 3,6 m/s2. 3. Ta c: A = 2L= 240= 20 (cm);e = 2 2x A v= 2t rad/s; vmax = eA = 2tA = 40t cm/s;amax = e2A = 800 cm/s2. 4. Ta c: e = 314 , 014 , 3 . 2 2=Tt= 20 (rad/s). Khix = 0 thv = eA = 160 cm/s. Khix = 5 cmthv = e2 2x A = 125 cm/s. 5. Ta c: 10t = 3t t = 30t (s). Khi x = Acos3t = 1,25 (cm);v = - eAsin3t= - 21,65 (cm/s); a = - e2x = - 125 cm/s2. 6. Khii quavtrcn bngthx = 0 cos(4tt + t) = 0 = cos(2t). Vv > 0 nn4tt + t = - 2t + 2kt t = - 38 + 0,5k vik e Z. Khi |v|= vmax = eA = 62,8 cm/s. 7. Khit = 0,75T = 0, 75.2te= 0,15 s thx = 20cos(10t.0,15 + 2t) = 20cos2t = 20 cm;v =- eAsin2t = 0; a = - e2x = - 200 m/s2; F = - kx = - me2x = - 10 N; a v F uc gi tr m nn gia tc v lc ko v uhngngcvichiudngca trcta . 8. Ta c: e = 2Tt = 10t rad/s; A2 = x2 + 22ve= 2 22 4v ae e+ |a| = 4 2 2 2A v e e = 10 m/s2. 9. Ta c: x = 5 = 20cos(10tt + 2t) cos(10tt + 2t) = 0,25 = cos(0,42t).Vv0;lynghim"+"khiv00;lynghim"+" khiv < 0); vis = ol (o tnhra rad); v lli; vntc tithiimt = 0. + Phngtrnhdao ngcacon lc nc thvitdi dngli gc: o = o0cos(et + ); vis = ol; S0 = o0l (o v o0 tnh ra rad). * Phng php gii: Davocciukinbitonchovcccngthclinquantmraccgitrc thca tn s gc, bin v pha banu ri thayvo phngtrnhdao ng. Lu:Saukhigiimtsbitoncbnvdngnytartramtsktlundnggiinhanhmts cu trc nghimdngvitphngtrnhdao ng: +Nukovtracchvtrcnbngmtkhongnorithnhthkhongcchchnhlbindao ng.Nuchngcthigianlcthvtth:=0nukovtra theo chiu dng; = t nu ko vt ra theo chium. +Nutvtrcnbngtruynchovtmtvntcndaongiuhathvntcchnhlvntc cci,khi:A= maxve,(conlcn S0 = maxve). Chn gc thi gian lc truyn vn tc cho vt th: = - 2t nuchiutruynvntc cngchiuvichiudng; = 2t nuchiutruynvntc ngcchiudng. * Bi tp minh ha: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 10 1.Mtconlclxothngnggmmtvtckhilng100gvlxokhilngkhngngk,c cng40N/m.Kovtnngtheophngthngngxungphadicchvtrcnbngmton5cmv thnhchovtdaongiuho.ChntrctaOxthngng,gcOtrngvivtrcnbng;chiu dnglchiuvtbtuchuynng;gcthigianlcthvt.Lyg=10m/s2.Vitphngtrnhdao ngca vt. 2.Mtconlclxogmvtnngkhilngm=400g,lxokhilngkhngngk,ccngk=40N/m.Kovtnngracchvtrcnbng4cmvthnh.Chnchiudngcngchiuvichiu ko, gc thigianlcth vt.Vitphngtrnhdao ngca vtnng. 3.Mtconlclxockhilngm=50g,daongiuha trn trc Ox vi chu k T = 0,2 s v chiu di quolL=40cm.Vitphngtrnhdaongcaconlc.Chngcthigianlcconlcquavtrcn bngtheochium. 4.Mtconlclxotreothngnggmmtvtnngkhilngmgnvolxokhilngkhngng k,ccngk=100N/m.Chntrctothngng,gctotivtrcn bng, chiu dng t trn xung.Kovtnngxungphadi,cchvtrcnbng5 2 cmvtruynchonvntc20t 2 cm/s theochiuttrnxungthvtnngdaongiuhovitns2Hz.Chngcthigianlcvtbtu dao ng.Cho g = 10 m/s2, t2 = 10. Vit phngtrnhdao ngca vt nng. 5.Mtconlclxogmmtlxonhccngkvmtvtnhckhilngm=100g,ctreo thngngvomtgicnh.TivtrcnbngOcavt,lxogin2,5cm.Kovtdc theo trc ca l xoxungdicchOmton2cmritruynchonvntc40 3 cm/stheophngthngnghng xungdi.ChntrctoOxtheophngthngng,gctiO,chiudnghnglntrn;gcthi gianl lcvt bt u dao ng.Lyg = 10 m/s2. Vitphngtrnhdao ngca vt nng. 6.Mtconlcncchiudil=16cm.Koconlclchkhivtrcnbngmtgc90rithnh.B quamimast,lyg=10m/s2,t2=10.Chngcthigianlcthvt,chiudngcngchiuvichiu chuynngban u cavt. Vitphngtrnhdao ngtheoli gc tnhra rad. 7.MtconlcndaongiuhavichukT=2s.Lyg=10m/s2,t2=10.Vitphngtrnhdao ngcaconlctheolidi.Bitrngtithiimbanuvtcligco=0,05radvvntc v = - 15,7 cm/s. 8.Mtconlcncchiudil=20cm.Tithiimt=0,tvtrcnbngconlcc truyn vn tc 14cm/stheochiudngcatrcta.Lyg=9,8m/s2.Vitphngtrnhdao ng ca con lc theo li di. 9.Mtconlcnangnmyntivtrcnbng,truynchonmtvntcv0=40cm/stheophng ngangthconlcndaongiuha.Bitrngtivtrcligco=0,1 3 radthncvntc v=20cm/s.Lyg=10m/s2.Chngcthigianllctruynvntcchovt,chiudngcngchiuvi vntc ban u. Vitphngtrnhdao ngca con lctheoli di. 10.ConlcndaongiuhavichukT= 5ts.Bitrngthiimbanu con lc v tr bin,c bin gc o0 vi coso0 = 0,98. Ly g = 10 m/s2. Vitphngtrnhdao ngcacon lc theoli gc. * Hng dn gii v p s: 1. Ta c: e =mk= 20 rad/s; A =222220 20200) 5 ( + = +evx = 5(cm); cos = 550=Ax= - 1 = cost = t. Vy x = 5cos(20t + t) (cm). 2. Ta c: e =mk= 10 rad/s; A =222220 201004 + = +evx = 4 (cm);cos = 440=Ax= 1 = cos0 = 0.Vyx = 4cos20t (cm). 3. Ta c: e =Tt 2= 10t rad/s; A = 2L= 20 cm;cos = Ax0= 0 = cos(2t); vv < 0 = 2t. Vy:x = 20cos(10tt +2t) (cm). July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 11 4. Ta c: e = 2tf = 4t rad/s; m = 2ek= 0,625 kg; A = 220 20evx + = 10 cm; cos = Ax0 = cos(4t); v v > 0 nn = - 4t. Vy:x = 10cos(4tt - 4t) (cm). 5. Ta c: e = 0lgA= 20 rad/s; A = 220 20evx + = 4 cm;cos = Ax0 = 42 = cos(32t); vv < 0 nn = 32t. Vy:x = 4cos(20t + 32t) (cm). 6. Ta c: e = lg = 2,5t rad/s; o0 = 90 = 0,157 rad; cos =000oooo = = - 1 = cost = t. Vy:o = 0,157cos(2,5t + t) (rad). 7. Ta c: e = Tt 2 = t; l = 2eg = 1 m= 100 cm;S0 = 222) (eovl += 5 2cm;cos = 0Sl o= 21 = cos(4t); vv < 0 nn = 4t. Vy:s = 5 2 cos(tt + 4t) (cm). 8. Ta c: e = lg = 7 rad/s; S0 = ev = 2 cm; cos = 0Ss= 0 = cos(2t);vv > 0 nn = - 2t.Vy:s = 2cos(7t- 2t) (cm). 9. Ta c S20= 220ev= s2 + 22ev = o2l2 + 22ev = 42 2eog + 22ev e = 2 20v vgo= 5 rad/s; S0 = e0v = 8 cm;cos = 0Ss= 0 = cos(2t); vv > 0 nn = - 2t. Vy:s = 8cos(5t - 2t) (cm). 10. Ta c: e = Tt 2= 10 rad/s; coso0 = 0,98 = cos11,480 o0 = 11,480 = 0,2 rad; cos = 0oo = 00oo = 1 = cos0 = 0. Vy:o = 0,2cos10t (rad). 4. Cc bi ton lin quan n th nng, ng nng v c nng ca con lc l xo. * Cc cng thc: + Thnng:Wt = 21kx2 = 21kA2cos2(e + ).+ ngnng:W =21mv2 =21me2A2sin2(e+) =21kA2sin2(e+ ). Thnngvngnngcaconlclxobinthintunhonvitnsgce=2e,vitnsf=2fv vichuk T= 2T. +Trongmtchukc4lnngnngvthnngcavtbngnhaunnkhongthigianlintipgiahai lnngnngv th nngbngnhaul 4T. + C nng:W = Wt+ W =21kx2 + 21mv2 = 21kA2 = 21me2A2. * Phng php gii: tmccilnglinquannnnglngcaconlctavitbiuthclinquannccilng bitv ilngcn tmt suyra v tnhi lngcn tm. * Bi tp minh ha: 1.Mtconlclxocbindaong5cm,c vn tc cc i 1 m/s v c c nng 1 J. Tnh cng ca l xo, khilngca vt nngvtn s dao ngcacon lc. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 12 2.Mtconlclxoccngk=150N/m v c nng lng dao ng l W = 0,12 J. Khi con lc c li l2 cm thvntc ca n l 1 m/s.Tnhbin vchuk dao ngca con lc. 3.Mtconlclxockhilngm=50g,daongiuha trn trc Ox vi chu k T = 0,2 s v chiu di quo lL = 40 cm. Tnh cnglxo v c nngcacon lc. 4.Mtconlclxotreothngnggmmtvtnngckhilngmgnvolxockhilngkhng ngk,ccngk=100N/m.Kovtnngxungvphadi,cchvtrcnbng5 2 cmvtruyn chonvntc20t 2 cm/sthvtnngdaongiuhovitns2Hz.Chog=10m/s2,t2=10.Tnh khilngca vtnngv c nngca con lc. 5.Mtconlclxodaongiuha.Bitlxoccng36N/mvvtnhckhilng100g.Lyt2 = 10. Xc nhchuk v tns binthintunhonca ngnngcacon lc. 6.Mtconlclxockhilngvtnhl50g.Conlcdaong iu ha theo phng trnh: x = Acoset. Csaukhongthigian0,05sthngnngvthnngcavtlibngnhau.Lyt2=10.Tnhcng ca lxo. 7.Mtconlclxogmlxonhvvtnhdaongiuhatheophngngangvitnsgc10 rad/s. Bitrngkhingnngvthnngcavtbngnhauthvntccavtclnbng0,6m/s.Xcnh bin dao ngca con lc. 8.Mtvtnhdaongiuhatheophngtrnh:x=10cos(4tt- 3t)cm.Xcnhvtrvvntcca vtkhingnngbng3 lnth nng. 9.Mtconlclxodaongiuhavitnsgce=10rad/svbinA=6cm.Xcnhvtrv tnh lnca vntc khithnngbng2 lnngnng. 10.Conlclxogmvtnhckhilngm=400gvlxoccngk.Kchthchchovtdaong iuhavicnngW=25mJ.Khivtiquali-1cmthvtcvntc-25cm/s.Xcnhcng ca lxo v bin cadao ng. * Hng dn gii v p s: 1. Ta c: W = 21kA2 k = 22AW= 800 N/m;W = 21mv2max m = 2max2vW= 2 kg;e = mk= 20 rad/s; f = te2= 3,2 Hz. 2. Ta c: W = 21kA2 A = kW 2= 0,04 m= 4 cm. e =2 2x A v= 28,87 rad/s; T =et 2 = 0,22 s. 3. Ta c: e = Tt 2= 10t rad/s; k = me2 = 50 N/m;A = 2L= 20 cm;W = 21kA2 = 1 J. 4. Ta c: e = 2tf = 4t rad/s; m = 2ek= 0,625 kg; A =220 20evx + = 10 cm;W =21kA2 = 0,5 J. 5. Tns gc v chuk cadao ng:e = mk= 6t rad/s; T = et 2= 31s. Chukvtnsbinthintun honca ngnng:T = 2T = 61s; f = '1T= 6 Hz. 6.Trongmtchukc4lnngnngvthnngbngnhaudokhongthigianlintipgiahailn ngnngv thnngbngnhaul 4T T = 4.0,05 = 0,2 (s); e = Tt 2= 10t rad/s; k = e2m = 50 N/m. 7. Khingnngbngthnng:W = 2W hay 21me2A2 = 2.21mv2 A =2ev= 0,06 2m = 6 2cm. 8. Ta c: W = Wt + W = Wt + 3Wt = 4Wt 21kA2 = 4. 21kx2 x = 41A = 5cm. v = e2 2x A = 108,8 cm/s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 13 9. Ta c: W = Wt + W = Wt +21Wt = 23Wt 21kA2 = 23. 21kx2

x = 32A = 4,9 cm; |v|= e2 2x A = 34,6 cm/s. 10. Ta c: W = 21kA2 = 21k(x2 + 22ev) = 21k(x2 + kmv2) =21(kx2 + mv2) k = 222x mv W = 250 N/m. 5. Con lc l xo treo thng ng v con lc l xo t trn mt phng nghing. * Cc cng thc: + Con lc lxo treo thngng:Al0 = kmg; e =km= 0gl A. + Con lc lxo t trnmtphngnghing:Al0 = sin mgko; e = km = 0sin gl oA. + Chiudicc i ca lxo:lmax = l0 + Al0 + A. Chiudicc tiuca lxo: lmin = l0 + Al0 A. + Lc n hicc i,cc tiu:Fmax = k(A + Al0), Fmin = 0 nu A > Al0; Fmin = k(Al0 A) nu A < Al0. + lnca lc n hitivtrc li x: Fh=k|Al0 +x|nuchiudnghngxung;Fh=k|Al0 -x| nuchiudnghngln. * Phng php gii: +Ccbitonvvitphngtrnhdaongthchintngtnhconlclxotnmngang.Trng hpconlclxotreothngngtnsgccthtnhtheocngthc:e = 0gl A; cn con lc l xo t trn mtphngnghingthtn s gcc th tnhtheocngthc:e = 0sin gl oA. +tmmtsilngtrongdaongcaconlctavitbiuthclinquannccilngbitv ilngcn tmt suyra v tnhi lngcn tm. * Bi tp minh ha: 1.Mtconlclxogmmtqunngkhilng100g,lxoccng100N/m,khilngkhng ng ktreothngng.Choconlcdaongvibin5 cm. Ly g = 10 m/s2; t2 = 10. Xc nh tn s v tnh lc n hicc i, lc nhicc tiuca lxo trongqutrnhqu nngdao ng. 2.Mtconlclxotreothngng,udicmtvtmdao ng vi bin 10 cm v tn s 1 Hz. Tnh ts gialcn hicc tiuv lc nhicc i cal xo trongqu trnhdao ng.Lyg = 10 m/s2. 3.Mtconlclxotreothngngcvtnngckhilng100g.Kchthchchoconlcdaongtheo phngthngngththyconlcdaongiuhavitns2,5Hzvtrongqutrnhvtdaong, chiudicalxothayitl1=20cmnl2=24cm.Xcnhchiuditnhincalxovtnhlc n hicc i, cc tiuca lxo trongqutrnhdao ng.Lyt2 = 10 v g = 10 m/s2. 4.Mtconlclxotreothngngdaongiuhavichuk0,4s;bin6cm.Khivtrcn bng, l xo di44 cm.Ly g= t2 (m/s2). Xc nhchiudi cc i,cc tiuca l xo trongqu trnhdao ng. 5.Mtconlclxotreothngnggmlxocchiuditnhin20cm,cng100N/m, vt nng khi lng400g.Kovtnngxungphadicchvtrcnbng6cmrithnhchoconlcdaongiu ha.Lyg=t2(m/s2).Xcnhlncalcnhical xo khi vt cc v tr cao nht v thp nht ca quo. 6.Mtconlclxogmqucukhilng100ggnvolxokhilngkhngngkccng 50N/mvcditnhin12cm.Conlccttrnmtphngnghingmtgcosovimtphng ngangkhi l xo di11 cm. B qua mast. Lyg = 10 m/s2.Tnhgc o. 7.Mtconlclxottrnmtphngnghinggco=300sovimtphngnmngang.vtrcnbng lxoginmton5cm.Kchthchchovtdaongthnsdaongiuhavivntccci 40cm/s.Chntrctatrngviphngdaongcavt,gctativtrcnbng,gcthigiankhi vti qua vtrcn bngtheochiudng.Vitphngtrnhdao ngca vt. Lyg = 10 m/s2. 8.Mtconlclxogmvtnngckhilngm=500g,l xo c cng k = 100 N/m, h c t trn mtphngnghingmtgco=450sovimtphngnmngang,gicnhphatrn.Nngvtlnnv July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 14 trmlxokhngbbindngrithnh.Bquamast.Lyg=10m/s2.Chntrctatrngvi phngdaongcavt,gctativtrcnbng,chiudnghngxungdi,gcthigianlcth vt.Vitphngtrnhdao ngca vt. * Hng dn gii v p s: 1. Ta c: e = mk= 10t rad/s; T = et 2= 0,2 s; f= T1 = 5 Hz;W = 21kA2 = 0,125 J;Al0 = kmg= 0,01 m = 1 cm;Fmax = k(Al0 + A) =6 N; Fmin = 0 v A > Al0. 2. e = 2tf =0lgA Al0 = 2 24 fgt= 0,25 m= 25 cm;Fmax = k(Al0 +A). Al0 > A Fmin = k(Al0 - A) ) () (00maxminA l kA l kFF+ A A= = 73. 3. Ta c: 2A = l2 l1 A = 21 2l l = 2 cm;e = 2tf = 5t rad/s; Al0 = 2eg = 0,04 m = 4 cm; l1 = lmin = l0 + Al0 A l0 = l1 - Al0 + A = 18 cm; k = me2 = 25 N/m; Fmax = k(Al0 + A) = 1,5 N; Al0 > A nn Fmin = k(Al0 - A) = 0,5 N. 4. Ta c: e = Tt 2 = 5t rad/s; Al0 = 2eg= 0,04 m = 4 cm;lmin = l0 + Al0 A = 42 cm; lmax = l0 + Al0 + A = 54 cm. 5. Ta c: e = mk = 5t rad/s; Al0 = 2eg= 0,04 m = 4 cm;A = 6 cm = 0,06 m. Khi v tr cao nht l xo c chiu di:lmin = l0 + Al0 A = 18 cm, nn c bin dng |Al| = |lmin l0| = 2 cm = 0,02 m |Fcn| = k|Al| = 2 N. Khi vtrthpnhtlcn hit gitrcc i:|Ft n| = Fmax = k(Al0 + A) = 10 N.6. Ta c: Al0 = l0 l = 1 cm= 0,01 m;mgsino= kAl0 sino= mgl k0A = 21 o = 300. 7. Ta c: e = 0sinlgAo = 10 rad/s; A = emaxv = 4 cm; cos = Ax0 = 0 = cos(2t); v v0 > 0 nn = - 2t rad. Vy:x = 4cos(10t- 2t) (cm). 8. Ta c: e = mk = 10 2rad/s; Al0 = kmg o sin= 0,025 2m = 2,5 2cm; A = Al0 = 2,5 2cm; cos = Ax0 = AA = - 1 = cost = t rad. Vy: x = 2,5 2 cos(10 2 t + t) (cm). 6. Tm cc i lng trong dao ng ca con lc n. * Cc cng thc: + Tns gc;chukv tns: e = gl; T = 2tgl v f= lgt 21. + Thnng:Wt = mgl(1- coso). ngnng:W =21mv2 = mgl(coso - coso0). + C nng:W = Wt + W = mgl(1- coso0). + Nuo0 s 100 th:Wt = 21mglo2; W = 21mgl(20o -o2); W =21mgl20o ;o v o0 tnhra rad. Thnngv ngnngca conlc nbinthintunhonvie = 2e; f = 2f; T =2T. + Vntc khii qua li gco: v =) cos (cos 20o o gl . July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 15 + Vntc khii qua vtrcn bng(o = 0):|v| = vmax =) cos 1 ( 20o gl . + Nuo0 s 100 th:v =) (2 20o o gl ; vmax = o0 gl ; o, o0 tnh ra rad. + Sc cngca sidy khii qua li gco: To = mgcoso +lmv2 = mg(3coso- 2coso0). TVTCB = Tmax = mg(3 - 2coso0); Tbin = Tmin = mgcoso0. Vio0 s 100: T = 1 + o20 - 23o2; Tmax = mg(1 + o20); Tmin = mg(1- 202o). *Phngphpgii:tmmtsilngtrongdaongcaconlcntavitbiu thc lin quan n cc ilng bitv i lngcn tmt suyra v tnhilngcntm. * Bi tp minh ha: 1.Tinicgiatctrngtrng9,8m/s2,conlcndaongiuhovichuk 72ts. Tnh chiu di, tn s v tns gc ca dao ngca con lc. 2.cngmtnitrnTritconlcncchiudil1daongvichukT1=2s,chiudil2dao ngvichukT2=1,5s.Tnhchukdaongcaconlcnc chiu dil1 + l2 v con lc n c chiu dil1 l2. 3.Khiconlcncchiudil1,l2(l1>l2)cchukdaongtngnglT1,T2tinic gia tc trng trngg=10m/s2.Bittini,conlcncchiudil1+l2cchu k dao ng l 2,7; con lc n c chiudi l1 - l2 c chu k dao ngl0,9 s. TnhT1, T2 vl1, l2. 4.TrongcngmtkhongthigianvcngmtnitrnTritmtconlcnthchinc60dao ng.Tngchiudicanthm44cmthtrongkhongthigian,conlcthchinc50daong. Tnhchiudi vchuk dao ngban u ca con lc. 5.Tinicgiatctrngtrngg=9,8m/s2,mtconlcnvmtconlclxodaongiuhavi cngtns.Bitconlcncchiudi49cm,lxoccng10N/m.Tnhkhilngvtnhcacon lcl xo. 6.Tinicgiatctrngtrngg,mtconlcn dao ng iu ha vi bin gc 0 nh (0 < 100). Ly mc th nng vtr cn bng. Xc nhvtr (li gc ) m th nngbng ng nngkhi: a) Con lc chuynng nhanhdn theo chiudng v vtr cn bng. b) Con lc chuynng chmdn theo chiudng v pha vtrbin. 7.Mtconlcngmmtqucunhkhilngm=100g,treovousidydil=50cm, mt nicgiatctrngtrngg=10m/s2.Bquamimast.Conlcdaongiuhavibingc o0=100=0,1745rad.Chngcthnngtivtrcnbng.Tnhthnng,ngnng,vntcvsccng ca sidy ti: a) Vtrbin.b) Vtrcn bng. * Hng dn gii v p s: 1. Ta c: T = 2tgl l = 224tgT= 0,2 m;f = T1= 1,1 Hz;e = Tt 2= 7 rad/s. 2. Ta c: T2+ = 4t2gl l2 1 + = T21 + T22 T+ = 2221T T + = 2,5 s; T- = 2221T T = 1,32 s. 3. Ta c: T2+ = 4t2gl l2 1 + = T21 + T22 (1); T2+ = 4t2gl l2 1 = T21 - T22 (2)T (1) v (2) T1 =22 2 ++T T= 2 s; T2 =22 2 +T T= 1,8 s; l1 =2214tgT= 1 m; l2 =2224tgT= 0,81 m. 4. Ta c: At = 60.2tgl = 50.2tgl 44 , 0 + 36l = 25(l + 0,44) l = 1 m;T = 2tgl = 2 s. 5. Ta c: mklg= m = gk l.= 500 g. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 16 6. Khi W = Wt th W = 2Wt 21mlo20= 221mlo2 o = 20o. a) Con lc chuynngnhanhdn theo chiudng t v tr bino = - o0 n v tr cn bngo = 0: o = -20o. b) Con lc chuynng chmdn theo chiudng t vtr cn bngo = 0 n v tr bino = o0: o = 20o. 7. a) Tivtrbin:Wt = W = 21mgl20o =0,0076 J; W = 0; v = 0; T = mg(1- 22oo) = 0,985 N. b) Tivtrcnbng:Wt= 0; W = W = 0,0076 J; v = mWd2= 0,39 m/s;T = mg(1+ o20) = 1,03 N. 7. S ph thuc ca chu k dao ng ca con lc n vo cao v nhit . S nhanh chm ca ng h qu lc s dng con lc n. * Cc cng thc: +Nucaoh,nhittconlcncchu k: T = 2tgl; cao h, nhit t con lc n c chu k T=2t'hlgthtac: 2tRhTT A+A=A o ;viAT=T-T;Ah = h - h ; At = t - t; o l h s n di ca dy treoconlc;R=6400kmlbnknhTrit.Vinghmgiysdngconlcn:khiAT>0th ngh chychm,khiAT < 0 thngh chynhanh. + Thigianchysai mingym(24 gi):At = '86400 .TT A. *Phngphpgii:tmmtsilnglinquannsphthuccachukdaongcaconlc nvocaosovimttvnhitcamitrngtavitbiuthclinquanncc i lng bit v ilngcn tmt suyra v tnhi lngcn tm. * Bi tp minh ha: 1.Trnmttnicgiatctrngtrngg=10m/s2.Mtconlcndaongvi chu k T = 0,5 s. Tnh chiudicaconlc.Nuemconlcnylncao5kmthndaongvichukbngbaonhiu(ly n 5 ch s thpphn).Cho bn knhTrit lR = 6400 km. 2.Ngitaamtconlcntmttlncaoh=10 km. Phi gim di ca n i bao nhiu % chukdao ngca n khngthayi. Bitbn knhTrit R = 6400 km. 3.MtconlcndaongtiimAcnhit25 0CvtiaimBcnhit 10 0C vi cng mt chuk.HisovigiatctrongtrngtiAthgiatctrngtrngtiBtnghaygimbaonhiu%?Choh s ndi ca dytreo conlc lo = 4.10-5 K-1. 4.Mtconlcnghcthcoilconlcn.nghchyng mc ngang mt bin. Khi a ng h lnnhnicao4000mthnghchynhanhhaychychmvnhanhchmbaolutrongmtngym? Bitbn knhTrit R = 6400 km. Coi nhit khngi. 5.Qulcnghcthxemlmtconlc n dao ng ti mt ni c gia tc trng trng g = 9,8 m/s2. nhit15 0CnghchyngvchukdaongcaconlclT = 2 s. Nu nhit tng ln n 25 0C thnghchynhanhhaychmbaolutrongmtngym.Chohsndicathanhtreoconlc o = 4.10-5 K-1. 6.Conlccamtnghqulcccoinhmtconlcn.Khitrnmttvinhitt=27 0C thnghchyng.Hikhianghnylncao1kmsovimttththnhitphilbao nhiunghvnchyng?BitbnknhTritlR=6400kmvhsndicathanhtreocon lcl o = 1,5.10-5 K-1. * Hng dn gii v p s: 1. Ta c: l = 224tgT= 0,063 m;Th = TR h R += 0,50039 s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 17 2.Tac:T = 2tgl= 2t''gl=> l = gg' l =) (h RR+2l = 0,997l. Vy phi gim di ca con lc 0,003l, tc l0,3% di can. 3. Ta c: TA = 2tAAgl= 2tAB A Bgt t l )) ( 1 ( +o= TB = 2tBBgl gB = gA(1 + o(tA tB) = 1,0006gA.Vygiatc trngtrngtiB tng0,06% so vigiatc trngtrngtiA. 4. Ta c: Th = R h R +T = 1,000625T > T nnngh chychm. Thigianchmtrongmtngym:At = hhTT T ) ( 86400 = 54 s. 5.Tac:T=T ) ' ( 1 t t +o =1,0002T>Tnnnghchychm.Thigianchmtrongmtngyml: At = ') ' ( 86400TT T = 17,3 s. 6. ngh vnchyngthchuk ca con lc cao h v trnmtt phibngnhauhay: 2tgl= 2thhgt t l )) ( 1 ( +o th = t - oggh 1 = t - o21|.|

\|+h RR =6,2 0C. 8. Con lc n chu thm cc lc khc ngoi trng lc. * Cc cng thc: +Nungoilccngcasidyvtrnglc,qunngcaconlcncnchuthmtcdngcangoi lc Fkhngithta c thcoi conlc c trnglc biukin: ' P = P+ F v giatc rit do biukin: ' g= g+ mF. Khi:T = 2t' gl. +Cclcthnggp:Lcintrng F =q E ;lcquntnh: F =-ma;lcyacsimet(hngthng ngln)c ln:F = mtvmvg. + Cc trnghp c bit: F cphngngangthg= 2 2) (mFg +;vtrcnbngmilchsoviphngthngngmtgco vitano = FP. Fc phngthngnghnglnthg = g- mF ; vt chulcy acsimet:g = g(1- mtv) Fc phngthngnghngxungthg = g + mF. + Chuk ca conlc ntreo trongthangmy: Thangmyngynhoc chuynngthngu:T = 2tgl. Thangmyilnnhanhdn u hoc i xungchmdn u vigiatc ahngln:T = 2ta g l+. Thangmyilnchmdn u hoci xungnhanhdn u vigiatc ahngxung:T = 2ta g l. * Phng php gii: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 18 tmchukdaongcaconlcnkhiconlcnchuthmlctcdngngoitrnglctavitbiu thctnhchukcaconlcntheogiatcritdobiukinvsosnhvichukcaconlcn khi con lcchchutc dngcatrnglc suyra chuk cn tm. * Bi tp minh ha: 1.Mtconlcntreotrongthangmynicgiatctrngtrng10m/s2.Khithangmyngyncon lcdao ngvichuk 2 s. Tnhchuk dao ngcacon lc trongcc trnghp: a) Thangmyi lnnhanhdn u vigiatc 2 m/s2. b) Thangmyi lnchmdnu vigiatc 5 m/s2. c) Thangmyi xungnhanhdnu vigiatc 4 m/s2. d) Thangmyi xungchmdn u vigiatc 6 m/s2. 2. Mt con lc n c chiu di dy treo 50 cm v vt nh c khi lng 0,01 kg mang in tch q= + 5.10-6 C, ccoilintchim.Conlcdaongiuhatrongin trng u m vectcng in trng c ln E = 104 V/m, hng thng ng xungdi. Ly g =10 m/s2. Xc nh chu k dao ng ca con lc. 3.Treoconlcnvotrnmtttinicgiatctrngtrngg=9,8m/s2.Khitngynthchuk daongiuhacaconlcl2s.Tnhchukdaongcaconlckhitchuynngthngnhanhdn u trnngnmngangvigiatc 3 m/s2. 4.MtconlcncchukdaongT=2s.Nutreocon lc vo trn mt toa xe ang chuyn ng nhanh dnutrnmtngnmngangththyrngvtrcnbngmi,dytreoconlchpviphngthng ngmtgco = 300. Cho g =10 m/s2. Tm gia tc ca toa xe v chu k dao ng mi ca con lc. 5.Mtconlcngmqucuckhilngring=4.103kg/m3.khittrongkhngkhndaong vichukT=1,5s.Lyg =9,8 m/s2. Tnh chu k dao ng ca con lc khi n dao ng trong nc. Bit khi lngring ca nc ln = 1 kg/l. * Hng dn gii v p s: 1. Khi thangmyngynhoc chuynngthngu:T = 2tgl. a)Khithangmyilnnhanhdnu a hngln,lcquntnhF ma = hngxung,giatcrit do biuking = g + a nnT = 2ta g l+ T = Ta g g+= 1,83 s. b) Thangmyi lnchmdnu:T = Ta g g= 2,83 s. c) Thangmyi xungnhanhdnu:T = Ta g g= 2,58 s. d) Thangmyi xungchmdn u:T = Ta g g+= 1,58 s. 2.Vtnhmangintchdngnnchutcdngcalcintrng F hngttrnxung(cngchiu vivc t cng intrng E ).V F ||E||P P = P + F giatc rit do biukinl g= g + mE q | |= 15 m/s2.Chuk dao ngca conlc ntrongintrngl T = 2t' gl ~ 1,15 s. 3. Trnglc biukintc dnglnvt: ' P = P + qtF ; qtF = - ma ' g = g - a ; v g a g= 2 2a g + ~ 10,25 m/s2. Khit ngyn:T = 2tgl; khit chuynngc giatc:T = 2t' gl

TT'= ' gg T = T' gg = 1,956 s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 19 4. Ta c: tano =PFqt=ga a = gtano = 5,77 m/s2. V a g g =2 2g a += 11,55 m/s2. T = T' gg = 1,86 s. 5.Ta c: n = 1 kg/l = 103 kg/m3. trong nc qu cu chu tc dng ca lc y Acsimet aFhng ln c lnFa = n.V.g = nmgnnc giatc rit do biuking = g- ng = 7,35 m/s2 T = T' gg = 1,73 s. 9. Con lc vt l. * Cc cng thc: + Phngtrnhnglchc: PM = I ; vi o s 100 (o tnh ra rad), ta c:o + d mgIo = 0. + Phngtrnhdao ng:o = o0cos(et + ); vie = d mgI. + Chuk, tns ca con lcvtl:T = 2tdImg; f = 12td mgI. + Con lc vtltngngvicon lc nc chiudil = dIm. * Phng php gii: tmccilnglinquannconlcvtltavitccbiuthclinquanni lngcn tmv cc i lng bitt suyra v tnhi lngcn tm. * Bi tp minh ha: 1.Mtvtrnnhckhilngm=1kgcthdaongiuhavibinnhquanhmttrcnm ngangvitnsf=1Hz.Momenquntnhcavtivitrcquaynyl0,025kgm2.Giatc trng trng nit vtrn l9,8 m/s2.Tnhkhongcch t trngtmca vtrn n trcquay. 2.Mtconlcvtlckhilng2kg,khongcchttrngtmcaconlcntrcquayl100cm,dao ngiuhavitnsgcbng2rad/stinicgiatctrngtrng9,8m/s2.Tnhmomenquntnhca con lcnyi vitrcquay. 3.Mtconlcvtllmtvtrnckhilngm=4kgdaongiuhavichukT=0,5s.Khong cchttrngtmcavtntrcquaycanld=20cm.Lyg=10m/s2vt2 =10.Tnhmomen qun tnhca con lcnyi vitrcquay. 4.Mtconlcvtlckhilng1,2kg,khongcchttrngtmntrcquayl 12 cm, momen qun tnh i vitrcquayl0,03 kgm2.Lyg = 10 m/s2. Tnhchuk dao ngca con lc. 5.Mtthcdi,mnhcchiudi1,5mctreomtu,daongnh mt con lc vt l ti ni c gia tc trngtrngg = 10 m/s2.Lyt2 = 10. Tnhchukdao ngca n. 6.Mtthanhkimloickhilngkhngngk,di64cm,mtchtimckhilng500gcgn vomtuthanh,thanhcthquayquanhtrcnmngangiquauthanhcnli.Lyg=t2 m/s2.Tnh chukdao ngca h. 7.Mtconlcvtlctreotrongmtthangmy.Khithangmyilnnhanhdnuvigiatc 110gth chukdao ngca con lcthayinh thno so vilcthangmyngyn? * Hng dn gii v p s: 1. Ta c: f= 12td mgI d = 2 24 f Imgt = 0,1 m= 10 cm. 2. Ta c: e = d mgI I = 2d mge = 4,9 kgm2. 3. Ta c: T = 2tdImg I = 22d4mgTt= 0,05 kgm2. 4. Ta c: T = 2tdImg= 0,913 s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 20 5. Ta c: T = 2tdImg= 2t2132mllmg= 2t23lg = 2 s. 6. Ta c: T = 2tdImg= 2t2mlmgl= 2tlg= 1,6 s. 7.Thangmyilnnhanhdnunnahngthngngtdiln,dolcquntnh qtF=-ma hngxungcnghngvitrnglcP nngiatc rit do biuking = g+ a = g + 110g = 1110g . Ta c: T = 2tdImg; T =2t' dImg= 2t1011 dImg= T1011. 10. Dao ng tt dn, dao ng cng bc, cng hng. * Cc cng thc: + H dao ngcngbc s c cnghngkhitns f ca lccngbc bngtn s ringf0 h dao ng. +Trongdaongttdnphncnnggimingbngcngcalcmastnnviconlclxodao ngtt dn vibin ban u A, hs ma st ta c: Qungngvt ic n lcdng li:S = gAmgkAe 2 22 2 2= . gimbin sau michuk:AA = kmg 4= 24eg. S dao ngthc hinc:N = mgAmgAkAAe 4 42= =A. Vntc cc ica vt t c khith nhchovt dao ngt vtrbinban u A: vmax =gAkg mmkA22 2 2 + . *Phngphpgii:tmmtsilnglinquanndaongttdn,daongcngbcvs cnghngtavitbiuthclinquannccilngbitvilngcntmtsuyravtnhi lngcn tm. * Bi tp minh ha: 1.Mtconlclxodaongttdn.Csaumichuk,bincangim0,5%.Hinnglngdao ngca con lcb mtisau midao ngtonphnlbao nhiu% ? 2.Mtconlclxoangdaongtt dn. C nng ban u ca n l 5 J. Sau ba chu k dao ng th bin ca n gimi20%. Xc nhphnc nngchuynha thnhnhitnngtrungbnhtrongmichuk.3.Mtconlclxogmvinbinhkhilngmvlxokhilngkhngngkccng160N/m. Conlcdaongcngbcditcdngcangoilctunhonctnsf.Bitbincangoilc tunhonkhngi.Khithayifthbindaongcavinbithayivkhif=2tHz th bin dao ngca vinbi t cc i. Tnhkhilngca vinbi. 4.Mttuhachytrnmtngray,ccchkhong6,4mtrnngraylicmtrnhnhgiach niccthanhray.Chukdaongringcakhungtutrncclxogimxcl1,6s.Tubxc mnh nht khichyvitc bngbao nhiu? 5. Mt con lc l xo gm vt nh khi lng 0,02 kg v l xo c cng 1 N/m. Vt nh c t trn gi c nhnmngangdctheotrclxo.Hsmasttrtgia gi v vt nh l0,1. Ban u gi vt v tr l xo b nn 10 cm ri bung nh con lc dao ng tt dn. Ly g = 10 m/s2. Tnh vn tc cc i m vt t c trong qu trnhdao ng. 6. Mt con lc l xo gm vt nh khi lng 0,2 kg v l xo c cng 20 N/m. Vt nh c t trn gi c nhnmngangdctheotrclxo.Hsmasttrtgiagivvtnhl0,01. T v tr l xo khng b bindng,truynchovtvntcbanu1m/sththyconlcdaongttdntronggiihnn hi ca l xo. Ly g = 10 m/s2. Tnh lnca lc n hi cc i ca l xo trongqu trnhdao ng. * Hng dn gii v p s: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 21 1. Ta c: AAAA A '1' == 0,05 AA'= 0,995. 2' '|.|

\|=AAWW= 0,9952 = 0,99 = 99%, do phn nng lng ca con lcmti saumidao ngton phnl1%. 2.Tac:W= 21kA2.Sau3chukbindaongcaconlcgim20%nnbincnli:A= 0,8A, cnnglc:W= 21kA2 = 21k(0,8A)2=0,64.21kA2=0,64.W.Phncnngchuynhathnhnhit nngtrongbachuk:AW= W - W = 0,36.W= 1,8 J. Phn c nng chuyn ha thnh nhit nng trong 1 chuk:W A = 3W A= 0,6 J. 3.Bincadaongcngbctccikhitnscalccngbcbngtnsringcaconlc: f = f0 = mkt 21 m= 2 24 fkt = 0,1 kg = 100 g. 4.Tubxcmnhnhtkhichukkchthchcangoilcbngchukringcakhungtu:T=T0= vL

v = 0TL= 4 m/s= 14,4 km/h. 5.ChntrctaOxtrngvitrccalxo,gctaO(cnglgcthnng) ti v tr l xo khng bin dng,chiudnglchiuchuynngcaconlclcmibungtay.Vtttclnnhttrong 41chu kutin.Gixllicavtrvtttccci(x 0. 2.7 Chn cu tr li ng. Chu k dao ng l : A. S dao ng ton phn vt thc hin c trong 1s B. Khong thi gian vt i t bn ny nbn kia ca qu o chuyn ng.C. Khong thi gianngn nht vt tr liv tr ban u. D. Khong thi gianngn nht vt tr litrng thi ban u. 2.8 Chn cu tr li ng.A. Dao ng ca h chu tc dng ca lc ngoi tun hon l dao ng t do.B. Chu k ca h dao ng t do khng ph thuc vo cc yu t bn ngoi.C. Chu k ca h dao ng t do khng ph thuc vo bin dao ng. D. Tn s ca h dao ng t do ph thuc vo lc ma st. 2.9 Mt vt dao ng iu ha theo phng trnhx = 5cos(10tt + 4t ), x tnh bng cm,t tnh bng s. Tn s dao ng ca vt lA.10Hz B. 5Hz. C. 15HZD. 6Hz 2.10 Motchatiemdaoongieuhoatheophngtrnh( ) t x t 2 cos 5 = cm, chu k dao ong cua chat iem la: A. T = 1 s. B. T = 2 s. C. T = 0,5 s.D. T = 0,25 s. 2.11 Cng thc no sau y biu din s lin h gia tn s gc e, tn s f v chu k T ca mt dao ng iu ha ? A. e = 2tf = 1T.B. e = tf= Tt.C. T = 1f = 2etD. e = 2tT = 2ft. 2.12 Mt vt dao ng iu ha theo phng trnhx = 10cos( 10tt + 3t ), x tnh bng cm,t tnh bng s. Tn s gcv chu k dao ng ca vt lA. 10t(rad/s); 0,032s. B. 5t(rad/s) ; 0,2s. C.10t(rad/s); 0,2s. D. 5t(rad/s) ; 1,257s. 2.13 Mt vt dao ng iuha, n thc hin c 50 dao ng trong 4 giy.Chu k dao ng ca vt l A. 12,5 sB. 0,8 sC. 1,25 sD. 0,08 s 2.14 Mtvtckhilngmtreovo l xo c cng k. Kch thch cho vt dao ng iu ha vi bin 3cm th chu k dao ng ca n l T = 0,3s. Nu kch thch cho vt dao ng iu ha vi bin 6cm th chu k dao ng ca con lc l xo l A. 0,3 s B. 0,15 s C. 0,6 sD. 0,423 s 2.15 Cho con lc l xo dao ng iu ho theo phng thng ng Ti ni c gia tc trng trng l g. v tr cn bng l xo dn lAl. 1. Tn s gc ca vt dao ng lA. 2kme t = B. 12kmet= C. gle =AD. mke =2.Chu k dao ng ca con lc c tnh bng cng thc A..g21T A t= B..g2 T At = C..km21Tt= D..mk2 T t =July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 25 2.16 Mtconlcl xo treo thng ng. Qu cu c khi lng 100g. Khi cn bng, l xo dn ra mt on bng 4cm. Cho con lc dao ng theo phng thng ng. Ly g =t2 (m/s2). Chu k dao ng ca con lc l A. 4s. B. 0,4s C. 0,07s. D. 1s. 2.17 Mt vt khi lng m treo vo u di ca mt l xo, u trn ca l xo c gi c nh. Khi h cn bng l xo c chiu di hn chiu di ban u 1 cm. Ly g = 10 m/s2 chu k dao ng ca vt l A. 0,1 sB. 0,2 sC. 0,3 sD. 0,4 s 2.18 Vt c khi lng m gn vo l xo c cng k, dao ng iu ha c tn s gcA. 2kme t = B. 12kmet= C. kme = D. mke =2.19 Mt con lc l xo c treo thng ng ni c gia tc trng trng g = 10m/s2. Vt nngckhilngmvdaongiuhatheophngthngngvitnsgce= 20rad/s.Trongqutrnhdaong,chiudilxobinthint18cm n 22cm. L xo c chiu di t nhin0 lA. 17,5cm. B. 18cm. C. 20cm. D. 22cm. 2.20 Vt c khi lng m gn vo lo xo c cng k, dao ng iu ha c chu k. A. T = 2mkt B. T = 2kmtC. T =mkt 21D. T = kmt 21 2.21Chu k dao ng iu ha ca con lc l xo t l thun vi A. khi lngm.B. cng k ca l xo. C. cn bc hai vi khi lngm. D. cn bc hai vi cng k ca l xo.2.22Vt gn vo l xo c cng k, dao ng iu ha c tn s t l A. thun vi cng k.B. nghch vi cng k.C. thun vi cn bc hai vi cng k.D.nghchvicnbchaivicngk. 2.23 Mt con lc l xo c cng k, nu tng khi lng ca vt ln 2 ln th chu kA. tng ln 2 ln.B. gim2 ln.C. tng2ln.D.gim 2ln 2.24 Con lac lo xo dao ong ieu hoa, khi tang khoi lng cua vat len 4 lan th tan so cua vat la: A.Tang len4 lan.B. Giam i 4 lan. C.Tanglen2lan.D. Giam i2 lan. 2.25 Mtconlclxockhilngbngmdaongvichuk T. chu k con lc gimi mt na th: A. Gim khi lng i 2 ln.B. Gim khi lng i 4 ln. C. Tng khi lng ln4 ln.D. Tng khi lng ln2 ln. 2.26 Conlacloxogomvatm=100gvaloxok=100N/m,(lay 2t =10)dao ong ieuhoa vi chu k : A. T = 0,1 s. B. T = 0,2 s. C. T = 0,3 s. D. T = 0,4 s. 2.27 Mot con lac lo xo dao ong ieu hoa vi chu k T = 0,5 s, khoi lng cua qua nang la m = 400 g, (lay 2t = 10). o cng cua lo xo la: A. k = 0,156 N/m.B. k = 32 N/m. C. k = 64 N/m. D.k= 6400 N/m. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 26 2.28 MtconlclxodaongtheophngthngngvichukT=0,4s.Cho g=2t (m/s2). gin ca l xo khi vt v tr cn bng lA. 0,4cm ;B. 4cm ;C. 0,1m ;D. 10cm 2.29 Mt con lc l xo c cng k v vt c khi lng m dao ng iu ha . Khi khi lng ca vt l m = m1 th chu k dao ng l T1 , khi khi lng ca vt l m = m2 th chu k dao ng l T2 . Khi khi lngca vt l m = m1 + m2th chu k dao ng lA. 1 21TT T=+B. 1 2T T T = +C.2 21 2T T T = +D.1 22 21 2TTT T + 2.30 Mt con lc l xo c cng k v vt c khi lng m dao ng iu ha . Khi khi lng ca vt l m = m1 th chu k dao ng l T1 = 0,6s , khi khi lng ca vt l m = m2 th chu k dao ng l T2 = 0.8s . Khi khi lng ca vt l m = m1 + m2th chu k dao ng lA. T = 0,7s B.T = 1,4s C.T = 1s D.T =0,48s 2.31 Con lc l xo gm l xo c cng k v vt c khi lng m dao ng vi chu k 0,4 s. Nu thay vt nng m bng vt nng c khi lng m gp i m. Th chu k dao ng ca con lc bng A. 0,16sB. 0,2sC. 0,4. 2 sD. 0, 42s 2.32 Mt con lc l xotreo thng ng c chiu di t nhin l0 = 30 cm, dao ng iu ha ti ni c g = 10 m/s2 vichu k dao ng ca vt T = 0,628 s. Chiu di ca l xo ti v tr cn bng c gi tr no sau y? A. 40 cmB. 30 cm C. 31 cmD. 30,1 cm 2.33 Mt l xo c khi lng khng ng k c chiu di t nhin l0 c treo vo im O c nh. Nu treo vo l xo vt c khi lng m1 = 100g vo l xo th chiu di ca n l l1 = 31 cm. Treo thm vt c khi lng m2 = 100g th di ca l xo l l2 = 32 cm. cng ca l xo l A. 200 N/mB. 100 N/mC. 160N/m D. 50 N/m2.34 Chukdaongiuhacaconlcncchiudiltinicgiatctrng trng gA. 12lTg t= .B. 12gTl t=C.2gTlt = D. 2lTgt =2.35 Con lac n gom vat nang khoi lng m treo vao si day l tai ni co gia toc trong trng g, dao ong ieu hoavi chu k T phu thuoc vao: A. l va g. B. m va l.C. m va g. D. m, l va g. 2.36 Ti mt ni xc nh, chu k dao ng ca con lc n t l thun vi A. chiu di con lc.B. gia tc trng trng. C. cn bc hai chiu di con lc.D. cn bc hai gia tc trng trng. 2.37 Ti cng mt v tr, nu chiu di con lc n tng 4 ln th chu k dao ng iu ho A. Tng 2 ln.B. Gim 4 ln.C. Tng 4 ln. D. Gim 2 ln. 2.38 Vi nhng dao ng vi bin nh ca con lc n, mun tn s dao ng tng gp i th chiu di ca con lc July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 27 A. tng 2 ln.B. Gim hai ln.C. tng 4 ln. D. Gim 4 ln. 2.39 Ti mt ni xc nh, Chu k ( tn s) dao ng iu ha ca con lc n ph thuc vo A. t s gia trng lng v khi lng ca con lc. B. bin dao ng. C. khi lng ca vt D. pha dao ng ca vt. 2.40 Ti mt ni xc nh, tn s dao ng ca con lc n t l nghch vi A. chiu di con lc.B. gia tc trng trng. C. cn bc hai chiu di con lc.D.cnbchaigiatctrng trng. 2.41 Ti mt ni xc nh, tn s gc dao ng ca con lc n t l thun vi A. chiu di con lc.B. gia tc trng trng. C.cn bc hai chiu di con lc.D.cnbchaigiatctrng trng. 2.42 Mt con lc n, gm hn bi c khi lng nh m v mt si dy khng gin c chiu dil = 1m, dao ng ti ni c gia tc trng trng g = 10 m/s2. Chu k dao ng ca con lc l A. 0,1 s B. 0,2 sC. 1 s D. 2 s 2.43 Mtconlcncchiu dildao ng ti ni c gia tc trng trng g = 10m/s2, vi chu k T = 0, 2 s. Chiu di con lc c gi tr bng. Ly 2t= 10. A. 1mB. 1 cmC.10 cmD. 1mm 2.44 Mot con lac n co chu k dao ong T = 4s, thi gian e con lac i t VTCB en v tr co li o cc ai la: A. t = 0,5 s.B. t = 1,0 s. C. t = 1,5 s. D. t = 2,0 s. 2.45 ni ma con lac n em giay (chu k 2s ) co chieu dai 1m, th con lac n co o dai 3m se dao ong vi chu k la: A. T = 6 s.B. T = 4,24 s.C. T = 3,46 s.D. T = 1,5 s. 2.46 Mt con lc n c chu k dao ng T = 3s, thi gian con lc i t v trc li x = A/2n v tr c li x = A l A. t = 0,250sB. t = 0,375sC. t = 0,500sD.t = 0,750s 2.47 Mt con lc n c chiu di l1 dao ng iu ha vi chu kT1 = 1,5s. Mt con lc n khc c chiudi l2 dao ng iu ha c chu k l T2 = 2 s. Ti ni , chu k ca con lc n c chiu di l = l1 + l2 s dao ng iu ha vi chu k l bao nhiu? A. T = 3,5 sB. T = 2,5 s C.T = 0,5 s D.T= 0,925 s 2.48 Hai con lc n c chu k T1 = 2,5s v T2 = 2s. Chu k ca con lc n c chiu di bng hiuchiu di ca hai con lc trn l: A. 1,5s. B. 1,0s. C. 0,5s. D. 3,25s. 2.49 Mtconlcncdytreodi20cmdaongiuhovibingc0,1rad.Cho2g = 9,8m/ s . Khi gc lch dy treo l 0,05rad th vn tc ca con lc l: A.0,2m/sB.0,2m/sC. 0,14m/sD.0,14m/s July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 28 Vn Tc - Gia Tc Trong Dao ng iu Ha 2.50 Trongdaoongdieuhoa) cos( e+ = t A x ,vantocbienoiieuhoatheo phng trnh: A.) cos( e+ = t A v .B. ) cos( e e + = t A v . C. ) sin( e+ = t A v . D.) sin( e e + = t A v . 2.51 Vn tc ca mt vt dao ng iu ha c ln t gi tr cc i ti thi im t. Thi im y c th nhn gi tr no trong cc gi tr sau : A. Khit = 0.B. Khi t = T4.C. Khi t = T. D.Khivtqua VTCB. 2.52 Trong dao ng iu ho, gi tr cc i ca vn tc l A.. A Vmaxe =B. . A V2maxe = C.A Vmaxe = D. . A V2maxe =2.53 Mt vt dao ng iu ha vi chu k T = 3,14s v bin A = 1m. Ti thi im vt qua VTCB vn tc ca vt c gi tr l : A. 0,5m/s.B. 1m/s.C. 2m/s.D. 3m/s. 2.54 Vt c khi lngm = 0,1kg gn vo l xo c cng k = 40N/m. Dao ng iu ha c bin A = 10cm. Vn tc ca vt qua v tr cn bng l A.20cm/s.B.100cm/s.C.200cm/s.D.50cm/s 2.55 Mtvtthchindaongiuhatheophngox vi phng trnh x = 2cos( 4t +3t), vi x tnh bng cm , t tnh bng s . Vn tc ca vt c gi tr ln cc lA.2cm/s.B.4cm/s.C.6cm/s.D. 8cm/s. 2.56 Mt vt dao ng iu ha dc theo trc Ox, ti v tr cn bng vn tc ca vt c ln 40 cm/s, chu k dao ng 0,2tgiy. Bin dao ng ca vt c ln bng. A. 0,4 mB. 0,04 mC. 4 mD. 40 m 2.57 Mot con lac lo xo gom qua nang co khoi lng 1kg va mot lo xo co o cng 1600N/m. Khi qua nang v VTCB, ngi ta truyen cho no vat toc ban au bang 2 m/s. Bieno dao ong cua qua nang la: A. A = 5 m.B. A = 5 cm.C. A = 0,125 m.D. A = 0,125 cm. 2.58 Trong dao ng iu ho, vn tc tc thi ca vt dao ng bin i A. Cng pha vi li.B. sm pha 4tso vi li .C. Ngc pha vi li . D. sm pha 2tso vi li . 2.59 i vi mt cht im dao ng iu ha vi phng trnh: x = Acos(et + 2t) th vn tc ca n A. bin thiniu ha vi phng trnhv = Acos(t + t). B. bin thiniu ha vi phng trnhv = Acos(t + 2t).C. bin thiniu ha vi phng trnhv = Acos(t). D. bin thiniu ha vi phng trnhv = Asin(t + 2t). 2.60 Motvat dao ong ieu hoa theo phng trnh( ) t x t 4 cos 6 = cm, van toc cua vat tai thi iem t = 7,5s la: A. v = 0 B. v = 75,4 cm/s. C. v = -75,4 cm/s.D. v = 6 cm/s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 29 2.61 th biu din s bin thin ca vn tc theo li trong dao ng iu ha c hnh dng no sau y A. ng parabol.B. ng trn.C. ng elp.D. ng hyperbol. 2.62 thbiudin s bin thin ca gia tc theo li trong dao ng iu ha c hnh dng no sau y A. ng parabol.B. ng trn.C. ng thng.D.on thng. 2.63 Xt mt vt dao ng iu ha c phng trnh x = Acos(et 3t). Vn tc ca vt c ln cc i khiA. t = 0.B. t = T4.C. t = T12. D. t = 5T12. 2.64 Mt vt dao ng iu ha c phng trnh x = 4cos5tt (cm). Thi im u tin vt c vn tc bng na ln ca vn tc cc i l : A. 1130s.B. 730s.C. 16s.D. 130s. 2.65 Trongdaoongieuhoa) cos( e+ = t A x ,giatocbienoiieuhoatheo phng trnh: A. ) cos( e+ = t A a . B. ) cos(2 e e + = t A a . C. ) cos(2 e e + = t A a . D.) cos( e e + = t A a . 2.66 Li v gia tc ca mt vt dao ng iu ha lun bin thin iu ha cng tn s vA. cng pha vi nhau B. lch pha vi nhau 4t C. ngc pha vi nhau D. lch pha vi nhau 2t 2.67Trong dao ng iu ha, gia tc bin i : A. cng pha vi vn tcB. sm pha 2t so vi vn tc C. ngc pha vi vn tcD. Tr pha 4t so vi vn tc. 2.68Gia tc ca vt dao ng iu ho bng khng khi A. Vt v tr c li cc i.B. Vn tc ca vt t cc tiu. C. Vt v tr c li bng khng.D.Vtvtrcphadao ng cc i. 2.69 Motvatdaoongtheophngtrnh( ) t x t 4 cos 6 = cm,gia toc cua vat tai thi iemt = 5s la: A. a = 0.B. a = 947,5 cm/s2. C. a = -947,5 cm/s2.D. a = 947,5 cm/s. 2.70 Trong dao ong ieu hoa: A.Gia toc bien oi ieu hoa cung pha so vi van toc. B.Gia toc bien oi ieu hoa ngc pha so vi van toc. C. Gia toc bien oi ieu hoa sm pha 2tso vi van toc. D.Giatoc bien oi ieu hoa cham pha 2tso vi van toc. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 30 2.71 Trong dao ng iu ha gia tc bin iA. cng pha vi li.B. ngc pha vi li C. sm pha 2tso vi li .D. Chm pha 2tso vi li . 2.72 Mt vt dao ng iu ha c phng trnh x = Acos ( ) t e + , gia tc ca cht im l: A. a= 2e xB. Maxa= A2eC. Gia tc a v li x lun ngc pha nhau.D. C ba cutrn u ng. 2.73 Trong dao ng iu ha, gia tc ca vt A. tng khi li tng.B. gimkhi li gm. C. khng i.D. lun gimkhi li thay i. 2.74 Gia tc ca mt cht im dao ng iu ha bng khng khi: A. li cc iB. li cc tiu bng 0 C. vn tc cc i hoc cc tiu.D. Vn tc bng khng. 2.75 Xt mt vt dao ng iu ha c phng trnh x = Acos(et 3t). Gia tc ca vt c ln cc i khiA. t = 0.B. t = 5T12.C. t = T12. D. t = T4. 2.76 Mtvt dao ng iu ha c phng trnh x = Acos2Ttt. Thi im u tin gia tc ca vt c ln bng na gia tc cc i nhn gi tr l : A. t = 5T12.B. t =T6.C. t = T12. D.t = T4. 2.77 MtvtDHvichukT=1s.thiimphadaong l 34t, vt c vn tc v 4 2 cm/ s t = . Ly t2 = 10. Gia tc ca vt thi im cho c gi tr no : A. 0,8 2 (m/s2). 2 (m/s2).C. 0,8 3 (m/s2). 0,8 3 (m/s2). 2.78 Chn cu tr li ng. Khi vt dao ng iu ha th : A. Vct vn tcvv vct gia tcal cc vct hng s. B. Vct vn tcvv vct gia tcai chiu khi vt qua v tr cn bng. C. Vct vn tcvv vct gia tcahng cng chiu chuyn ng ca vt. D. Vct vn tcvhng cng chiu chuyn ng ca vt, vct gia tcahng v v tr cn bng. 2.79 Mt cht im dao ng iu ha khi qua v tr bin th vn tc v gia tc l: A. v = 0;a = A2e B.v= Ae ;a= A2eC.v= Ae; a = 0.D. v = 0; a = 0. 2.80 Mt cht im dao ng iu ha khi qua v tr cn bng th vn tc v gia tc l: A. v = 0;a = A2e B.v= Ae ;a= A2eC.v= Ae; a = 0.D. v = 0; a = 0. 2.81 Khini v dao ng iu ha ca mt cht im, pht biu no sau y l ng : A. Khicht im qua v tr cn bng, n vn tc cc i v gia tc cc tiu.B. Khi cht im qua v tr bin, n vn tc cc tiu v gia tc cc i.July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 31 C. Khi cht im qua vtr cn bng, n vn tc cc i v gia tc cc i. D. A v B. 2.82 Mt vt DH vi phng trnh x= 5cos4tt(cm). Li v vn tc ca vt sau khi n bt u chuyn ng c 5s nhn gi tr no sau y ? A. 0cm ; 20t.cm/s.B. 5cm ; 0cm/s.C.-5cm ; 0cm/s.D. 0cm ; -20t.cm/s. 2.83 Trong dao ong ieu hoa cua chat iem, chat iem oi chieu chuyen ong khi: A. Lc tac dung oi chieu. B. Lc tac dung bang khong. C. Lc tac dung co o ln cc ai. D. Lc tac dung co o ln cc ai. 2.84 Chn cu tr li sai. Lc tc dng vo cht im dao ng iu ha : A. C biu thc F = - kx.B.Clnkhngitheo thi gian. C. Lun hng v v tr cn bng.D.Binthiniuhatheothi gian. 2.85 Chn cu tr li sai.A. Vn tc ca vt dao ng iu ha c gi tr cc i khi qua v tr cn bng. B. Khi qua v tr cn bng, lc hi phc c gi tr cc i. C. Lc hi phc tc dng ln vt dao ng iu ha lun hng v v tr cn bng.D. Lc hi phc tc dng ln vt dao ng iu ha lun hng v v tr cn bng. 2.86 Nuchngctotrngvivtrcnbngththiimt,biuthcquang h giabin A (hay xm), lix, vn tc v v tn s gce ca cht im dao ng iu ho l A..vx A222 2e+ = B.. v x A2 2 2 2e + =C..xv A222 2e+ = D.. x v A2 2 2 2e + =2.87 .Mt vt dao ng iu ha c chu k T = 0,2 s,bin 5cm. Tc ca vt ti li x3cm = +l:A.40tcm/s.B.20tcm/s.C.30tcm/s. D.50tcm/s 2.88Mt vt dao ng iu ha c tn s f= 5Hz,bin 10cm.Li ca vt ti ni c vn tc 60tcm/s l A.3cmB.4cmC.8cm D.6cm H Thc c Lp 2.89 Mt dao ng iu ha c m t bi phng trnh x = Acos(et + ). H thc lin h giabin A , li x, vn tc gc e v vn tc v l : 2.90 A. A = x2 + ve.B. A2 = x2 ve.C. A2 = x2 22ve.D. A2 =x222ve. 2.91 Cngthclinh gia bin A, li x, vn tc v v tn s gce trong dao ng iu ha c dng: A. v2 = e2(A2 + x2). B. v2 = e2A2 x2.C. v2 = e2(A2 - x2). D.v2= 22 2 x - Ae. 2.92 Mt dao ng iu ha vi tn s gc 10 rad/s. Ti v tr vt c li 4 cm th vn tc ca vt c gi tr 30 cm/s. Bin dao ng ca vt bng. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 32 A. 3 cmB. 7 cmC. 25 cmD. 5cm 2.93 Mtvtdaongiuhaviphngtrnhx=Acos(et 2t).Bitrngtrong khong 1/60(s) u tin , vt i t v tr cn bng v t c li x = A 32 theo chiu dng ca trc Ox. Ti vtr c li x = 2cm vn tc ca vt v = 40t 3 cm/s. Tn s gc v bin dao ng ca vt ln lt l bao nhiu ? A. 40t(rad/s) ; 4cm.B. 30t(rad/s) ; 2cm.C. 20t(rad/s) ; 4cm.D. 10t(rad/s) ; 3cm. Phng Trnh Dao ng iu Ha 2.94 Mot chat iemdao ong ieuhoa theo phng trnh |.|

\|+ =2cos 3ttt x cm, pha dao ong cua chat iemtai thi iemt = 1s la: A.t (rad). B.t 2 (rad). C. t 5 , 1 (rad).D.t 5 , 0 (rad).2.95 Mot vat dao ong ieu hoa theo phng trnh( ) t x t 4 cos 6 = cm, toa o cua vat tai thi iemt = 10s la: A. x = 3 cm B. x = 6 cmC. x = -3 cm D. x = -6 cm 2.96 et.Pha ban u ca dao ng bng bao nhiu? A. 0.B. 2t. C. t.D. 2t.2.97 Cho mt vt dao ng iu ho vi phng trnh x = 4cos(10t + t) (cm). Thi im vt qua v tr c li x = 2 2 cm ln th nht l A. t = 403(s).B. t = 401(s). C. t = 405(s). D.t = 407(s). 2.98 Cho mt vt dao ng iu ho vi phng trnh x = 4cos(10t + t) (cm). Thi im vt qua v tr c li x = -2 2 cm ln th nht l A. t = 403(s).B. t = 401(s). C. t = 405(s). D.t = 407(s). 2.99 Phngtrnhchuynngcavtcdngx=Asin(et)b.Chn pht biu ng : A. Vt dao ng iu ha xung quanh v tr cn bng c ta x = 0.B. Vt dao ng iu ha xung quanh v tr cn bng c ta x = b.C. Vt dao ng iu ha xung quanh v tr cn bng c ta x =b. D. Chuyn ng ca vt khng phi dao ng iu ha. 2.100 Trong cc phng trnh sau phng trnh no khng biu th cho dao ng iu ha : A. x = 5costt 1.(cm). B.x=3tcos(100tt 6t)(cm).July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 33 C. x = 2sin2(2tt 6t)(cm). D.x=3sin5tt3cos5tt. (cm). 2.101 Phng trnh dao ng ca vt c dng x = Asin2(et 4t). Chn kt lun ng. A. Vt dao ng vi bin A/2.B. Vt dao ng vi bin A. C. Vt dao ng vi bin 2A.D. Vt dao ng vi phan ban u l t/4. 2.102 Phng trnh dao ng ca vt c dngx = asinet acoset. Bin dao ng ca vt l: A. a/2.B. a.C. a 2 .D. a 3 . 2.103 Mt vtdao ng iu ho trn on thng MN di 8 cm vi tn sf = 5 Hz, lc t = 0 vt i qua v tr cn bng theo chiu dng. Phng trnh dao ng ca vt l A. x = 8cos(10tt -2t) (cm).B. x = 4cos(5tt -2t) (cm). C. x = 4cos(10tt -2t) (cm). D. x = 4cos(10tt +2t) (cm). 2.104 Mt vtdao ng iu ho trn on thng MN di 8 cm vi tn sf = 5 Hz, lc t = 0 vt i qua v tr cn bng theo chiu m. Phng trnh dao ng ca vt l A. x = 8cos(10tt -2t) (cm).B. x = 4cos(5tt -2t) (cm). C. x = 4cos(10tt -2t) (cm). D. x = 4cos(10tt +2t) (cm). 2.105 Vt dao ng iu ha c phng trnh: x = Acos(t + 2t). Chn Gc thi gian ti A. li x = - A.B. li x = +A.C. qua VTCB dng. D. qua VTCB m. 2.106 Mt vt dao ng iu ha vi bin A = 6cm, tn s f = 2Hz.Khi t = 0 vt i qua li cc i. Phng trnh dao ng ca vt l : A. x = 6cos(4tt 2t)cm.B. x = 6cos4tt(cm). C. x = 6cos(2tt)cm.D. x = 6cos(4ttt)cm. 2.107 Mt vt dao ng iu ha x = 4cos(2t 3t)cm. Gc thi gian c chn vo lc : A. Vt c li x = 2cm v i theo chiu dng.B. Vt c li x = 2 2 cm v i theo chiu m. C. Vt c li x = 2cm v i theo chiu m. D. Vt c li x = 2 2 cm v i theo chiu dng. 2.108 Mt con lc l xo nm ngang, ko vt theo phng ngang sang phi n v tr cch v tr cn bng 8cm ri th nh cho vt dao ng. Chu k dao ng ca vt T = 2s. Chn gc ta ti v tr cn bng, chiu dng hng sang phi, gc thi gian lc i qua im cch v trcn bng 4cm ln th nht.Phng trnh dao ng ca vt lA. x = 8 cos (tt + t/3) cmB. x = 8 cos (tt + 5t/6) cm C. x = 8 cos (2tt - t/3) cm D. x = 8cos (2tt - 7t/6) cm July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 34 2.109 Conlclxodaongtheophngthngng.Thi gian vt i t v trthp nht nvtrcaonhtcchnhau10cml1,5s.ChntrcOxthngng,gcOlvtrcn bng,chiudnghngln,gcthigianvtvtrthpnht.Phngtrnhdao dng ca vt l A. x = 5 cos ( 32tt + t ) (cm)B. x = 20 cos ( 34tt ) (cm) C. x = 10 cos ( 34tt - 2t) (cm)D. x = 5 cos ( 32tt + 2t) (cm) 2.110 Mt cht im dao ng dc theo trc Ox c phng trnh dao ng l x = 10cos(2tt +3t)(cm).Tithi im t1 vt c li x1 = 6cm v ang chuyn ng theo chiudng th sau 0,25s vt c li l : A. 6cm. B. 8cm. C. 9cm. D. -8cm. 2.111 Mt cht im dao ng dc theo trc Ox c phng trnh dao ng l x = 10cos(2tt +3t) (cm). Ti thi im t1 vt c li x1 = 6cm v ang chuyn ng theo chium th sau 0,25s vt c li l A. -6cm.B. 8cm. C. 1cm. D. -8cm. Tnh Qung ng Vt i c Trong Khong Thi Gian t 2.112Mt cht im dao ng iu ha c bin A, tn s gc le. Sau thi gian t = 4T tnh t v tr cn bng vt i c qung ng l A. AB.2A C.4A. D.2A 2.113 Mt vt dao ng iu ha vi phng trnh x = 4cos(t2t)cm. Thi gian vt i t v tr cn bng n v tr c li x = 2cm l : A. 610s.B. 6100s.C. 16s.D. 136s. 2.114 Mt vt dao ng iu ha vi phng trnh x = 4cos(5t3tqua v tr x = 1cm trong giy u tin l : A. 4.B. 6.C. 7.D. 5. 2.115 Mtvt dao ng iu ha vi phng trnh x = 4cos(5t56t)cm. Sau khong thi giant = 4,5s vt i c qung ng l : A. 179,5cm.B. 180cm.C. 181,5cm. D. 182cm. 2.116 Mtcon lc l xo gm mt l xo c cng k = 100 N/m v vt c khi lng m = 250 g, dao ng iu ho vi bin A = 6 cm. Chn gc thi gian t = 0 lc vt qua v tr cn bng. Qung ng vt i c trongs10tu tin l A. 9 cm.B. 24 cm.C. 6 cm. D. 12 cm 2.117 Mtvtthchindaongiuhatheophngoxviphngtrnhx=10cos( 20tt), vi x tnh bng cm , t tnh bng s .1. Thi gianngn nht khivt i t VTCB n li x = 5cm l A.1( )60s .B. 1( )30s .C. 1( )120s .D. 1( )100s . 2.Thi gianngn nht khivt i t x = 10cmn li x = 5cm l July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 35 A.1( )60s .B. 1( )30s .C. 1( )120s .D. 1( )100s2.118 Phng trnh dao ng ca con lc l xo l :x = Acostt( x = cm ; t = s) Thi gian qu cu dao ng t v tr cn bng n v tr bin l : A.1sB.0,5s C. 1,5sD.2s 2.119 Mt vt dao ng iu ha dc theo trc Ox, Khi qua v tr cn bng vt c vn tc v0 = 40tcm/s tn s dao ng f = 5 Hz. Chn gc thi gian lc vt c li x =- 2 cm v ang vo v tr cn bng. Phng trnh dao ng ca vt l: A. x = 4cos(10t t + 6t) cmB. x = 4cos(10t t + )3t) cm C. x = 4cos (10t t - 6t) cmD. x = 4cos(10t t - )3t) cm 2.120 Phng trnh vn tcca mt vt dao ng iu ha c dng v = eAcoset.Kt lun no ng ? A. Gc thi gian l lc cht im qua v tr cn bng theo chiu dng.B. Gc thi gian l lc cht im qua v tr cn bng theo chiu m. C. Gc thi gian l lc cht im c li x =A. D. Gc thi gian l lc cht im c li x =A. 2.121 Cnngcavtdaongiuha W = 3.105J. Lc cc i tc dng ln vt bng 1,5.103N. Chu k dao ng T = 2s v pha ban u =3t. Phng trnh dao ng ca vt l : A. x = 0,04cos(tt 3t)cm. B.x=0,02cos(tt 3t)cm. C. x = 0,4cos(tt 3t)cm.D.x=0,2cos(tt 3t)cm. 2.122 Mtconlc l xo treo thng ng gm qu nng c khi lng m = 1kg v l xo c cng k =1600N/m. Khi qu nng v tr cn bng, ngi ta truyn cho n vn tc ban u 2m/shng thng ng theo chiu dngxung di. Chn gc thi gian l lc truyn vn vn tc cho vt. Phng trnh dao ng ca vt l A. x = 0,5cos 40t (m)B.x= 0,05cos(40t +2t) (m) C. x = 0,05cos(40t- )2t (m)D.x= 0,05 2 cos40t ( m) 2.123 Mt vt dao ng iu ha vi bin A = 10cm, tn s f = 2Hz. thi im t = 0 vt chuynngngc chiu dng. thi im t = 2s vt c gia tc a = 8 3 m/s2. Ly t2 = 10. Phng trnh dao ng ca vt l : A. x = 10cos(4tt 6t)cm.B.x=10cos(4tt 56t)cm. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 36 C. x = 10cos(4tt 6t)cm.D.x = 10cos(4tt 76t)cm. 2.124 Mt con lc l xo treo thng ng gm mt nng c khi lng m = 80g, mt l xo c cng k v c khi lng khng ng k, tn s dao ng ca con lc f = 4,5Hz. Trong qu trnh dao ng, di ngn nht ca l xo l 40 cm v di nht l 56 cm. Chn gc ta ti vtrcnbng, chiu dng hng xung, lc t = 0 l xo ngn nht. phng trnh dao ng ca con lc l A. x = 8 2 cos(9t t - 2t) cm B.x=8cos(9t t+ 2t) cm C. x = 8cos(9t t - 2t) cmD.x= 8cos(9t t+t) cm 2.125 Mt vt nng khi lng m treo vo l xo c cng k theo phng thng ng lm ti ni c g = 10 m/s2, lm l xo gin thm mt onl A = 10 cm. T v tr cn bng ngi ta kovtthngngxungmtkhongbng3cmri bung nh khng vn tc u cho vt daong.Chngcthigianlcbungvtchiudnghnglntrn.Phng trnh dao ng ca vt l A. x = 3cos(10t +2t) cmB. x = 3cos (10t + ) t cm C. x = 5cos (10t -t ) cm D. x = 5cos(10t -2t) cm 2.126 Mt vt nng khi lng m treo vo l xo c cng k theo phng thng ng lm ti ni c g = 10 m/s2, lm l xo gin thm mt onl A = 10 cm. T v tr cn bng ngi ta ko vt thng ng xung mt khong bng 3 cm ri truyn cho vt vn tc v = 0,4 m/s. Chn gcthigian lc vt c li x = 2,5 ang hng theo chiu dng . Phng trnh dao ng ca vt l A. x = 3cos(10t +2t) cmB. x = 3cos10t cm C. x = 5cos(10t +3t) cmD. x = 5cos(10t -3t) Lc Ko V - Lc n Hi 2.127Mt con lcl xo treo thng ng ti v tr cn bng chiu di con l xo c gi tr 40 cm. Kch thch cho vt dao ng iu ha dc theo 0x phng thng, chiu dng hng ln. Chiu di ca l xo khi vt c ta x = +3 cm l A. 43 cmB. 40,3 cm C. 37 cmD. 33,7 cm 2.128 Mtcon lc l xo treo thng ng ti v tr cn bng chiu di con l xo c gi tr 30 cm. Kch thch cho vt dao ng iu ha dc theo 0x phng thng, chiu dng hng ln. Chiu di ca l xo khi vt c ta x = -2 cm l A. 32 cmB. 30,2 cm C. 28 cmD. 3,20 cm 2.129 Lc lm vt dao ng iu ha theo phng ngang c gi tr cc i l A. Fmax = kA.B. Fmax = k (A - A ).C. Fmax = 0. D. Fmax = k A . 2.130 Mt con lc l xo c cng l k treo thng ng, u trn c nh, u di gn vt. Gigincalxokhivtvtrcnbngll.Choconlcdaongiuhatheo July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 37 phng thng ng vi bin lA . Lc n hi ca l xo c ln cc i trong qu trnh dao ng lA. F = kA.B. F = 0. C. F = k( l-A ).D. F = k(A + l).2.131 Mt vt c khi lngm = 0,1kg dao ng iu ha vi chu k T = 1s. Vn tc ca vt qua v tr cn bng l v0 = 31,4cm/s. Ly t2 = 10. Lc hi phc cc i tc dng ln vt c gi tr l : A. 0,2N.B. 0,4N.C. 2N.D. 4N. 2.132 Mt vt c khi lng m = 50g dao ng iu ha tn on thng MN di 8cm vi tn s f = 5Hz, vt i qua v tr cn bng theo chiu dng. Lyt2 = 10. Lc gy ra chuyn ng ca cht im thi im t = 1/12(s) c ln l : A. 1N.B.3 N.C. 10N.D. 10 3 N. 2.133 LxokhivtvtrcnbnglAl.Choconlcdaongiuhotheophng thng ng vi bin l A (A > Al ). Trongqu trnh dao ng lc n hi ca l xo c ln nh nht l A. Fmin = kAl.B. Fmin = kA.C. Fmin = 0.D.Fmin= k(A -Al ). 2.134 Mtconlclxo c cng l k treo thng ng c bin 5cm. Ti VTCB l xo dn 2,5cm. Lc n hi c ln nh nht l A. Fmin = 5N.B. Fmin = 5NC. Fmin = 0.D. Fmin = 7,5N2.135 ConlacloxongangdaoongvibienoA=8 cm, chu k T = 0,5 s, khoi lngcuavatnanglam=0,4kg (lay 2t = 10). Gia tr cc ai cua lc an hoi tac dung vao vat la: A. Fmax = 525 N. B. Fmax = 5,12 N.C. Fmax = 256 N. D.Fmax= 2,56 N. 2.136 Treo qu cu c khi lng m vo l xo ti ni c gia tc trng trng g. Cho qu cu dao ng iu ho vi bin A theo phng thng ng. Lc n hi cc i ca l xo c xc nh theo cng thc : A. Fhmax = mg.B. Fhmax = kA.C. Fhmax = kA + mg.D.Fhmax = mg - kA. 2.137 C nng ca mt cht im dao ng iu ho t l thun vi A. Bnh phng bin dao ng.B. Li ca dao ng.C. Bin dao ng. D. Chu k dao ng. 2.138 C nng ca con lc l xo xc nh bng cng thc. Chn cu sai A. 21 me2A2 B. 21 k A2C. 21 kx2 D. 21mv2 +21 kx2 2.139 Chn cu tr li ng. Nng lng ca mt vt dao ng : A. Gim 4 ln khibin gim 2 lnv tn s tng 2 ln. B. Gim 4/9 lnkhitn s tng 3 ln v bin gim 9 ln. C. Gim 25/9 ln khitn s tng 3 ln v bin gim 3 ln. D. Tng 16 lnkhi bin tng 2 ln v tn s tng 2 ln. 2.140 Phng trnh dao ng iu ha ca vt c dng x = Acoset. Kt lun no sau y sai : A. Phng trnh vn tc ca vt v = eAsinet. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 38 B. ng nng ca vt W = 12me2A2cos2(et). C. Th nng ca vt Wt = 12me2A2sin2(e ).D. C nng W = 12me2A2. 2.141 Chn pht biu sai khini v nng lng ca h dao ng iu ha : A. C nng ca t l vi bnh phng bin dao ng. B.Trong qu trnh dao ng c s chuyn ha gia th nng v ng nng v cng ca lc ma st. C. C nng ton phn c xc nh bng biu thc W = 12me2A2.D. Trong sut qu trnh dao ng, c nng ca h c bo ton.2.142 Phat bieu nao sau ay la khongung ? A.Congthc 221kA E = chothaycnangbangthenangkhivatcoliocc ai. B.Cong thc 2max21mv E =cho thay c nang bang ong nang khi vat qua v tr can bang. C.Cong thc 2 221A m E e =cho thay c nangkhong thay oi theo thi gian. D.Cong thc 2 22121kA kx Et= = cho thay the nang khong thay oi theo thi gian. 2.143 Vt dao ng iu ha chuyn ng hng v v tr cn bng, th nng ca vt A. tng.B. gim.C.khng i.D.lctng,lc gim. 2.144 Conlclxodaongiu ha trn trc 0x, c phng trnh : x = Acos(et + ). ng nng (th nng) ca vt A. bo ton trong sut qu trnh dao ng.B.t l vi tn s gc e. C. bin thiniu ha vi tn s gc eD.binthintunhonvitns gc 2e. 2.145 Chn cu tr li ng. Nng lng ca vt dao ng iu ha : A. Binthiniu ha theo thi gianvi chu k T. B. Binthintun hon theo thi gian vichu k T/2. C. Bng ng nng ca vt khi qua v tr cn bng. D. Bng th nng ca vt khiqua v tr cn bng.2.146 Mt vt dao ng iu ha vi phng trnhx = 5cos(10tt) (cm) th th nng ca n bin thintun hon vi tn s:A. 2,5 Hz B. 5 HzC. 10 HzD. 18 Hz 2.147 Qu nng gn vo l xo t nm ngang dao ng iu ha c c nng l 3.10-5 J v lc n hi l xo tc dng vo vt c gi tr cc i l 1,5.10-3 N. Bin dao ng ca vtl A. 2 cm.B. 2 m.C. 4 cm.D. 4 m. 2.148MtconlclxoccnngW=0,9Jvbin dao ng A = 15cm. Hi ng nng ca con lc ti li x = -5cm l bao nhiu. A. 0,8J.B. 0,6J. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 39 C. 0,3J.D.Khngxcnhcvchabitcng ca l xo. 2.149 Mt vt gn vo mt lxo c cng 100N/m,dao dng iu ho vi bin 5cm. Khivt cch v tr cn bng 3cm th n c ng nng l A.0,125J.B. 0,09J.C. 0,08J.D. 0,075J. 2.150 Vtnngckhilng100g,daongiuhavi vn tc v = 10t cost t (cm/s). Lyt2 =10. Nng lng ca vt bng A.0,005JB.0,05J C.0,5J D. 5J 2.151Mt vt c khi lng m = 1kg dao ng iu ha vi chu k T = 5ts. Bit nng lng dao ng ca vt l 0,02J. Bin c gi tr no ? A. 6,3cm.B. 4cm.C. 2,25cm.D. 2cm. 2.152 Conlacloxogomvatm=400gvaloxok=40N/m. Ngi ta keo qua nang ra khoiv tr can bang mot oan 4 cm roi tha nhe cho no dao ong. C nang dao ong cua con lac la: A. E = 320 J.B. E = 6,4.10-2 J.C. E = 3,2.10-2 J.D.E= 3,2J. 2.153 Mt con lc l xo thc hin c 5 dao ng trong thi gian 10 s, vn tc ca vt nng qua v tr cn bng c ln8tcm/s. V tr vt c th nng bng 13ln ng nng cch v tr cn bng; A. 0,5 cmB. 2 cmC. 4 cmD. 2 2cm 2.154 Mt con lc l xo dao ng iu ha c phng trnh x = Acos(et+2t), c c nng l E 1.Th nng ca vt ti thi im t l A. Et = Esin2(et+2t)B. Et 2sin E t e =C.2ostE Ec t e = D. Et = Ecos2(et+2t) 2. ng nng ca vt ti thi im t l A.E = Esin2(et+2t)B. E 2sin E t e =C. Et 2os Ec t e = D. Et = Ecos2(et + 2t) 2.155Conlclxodaongiuhaviphngtrnhx=Acoset.Vtnngckhi lngm. Khi vt m qua v tr c li x =2A th ng nng ca vt c gi tr l : A. 83me2A2. B. 41me2A2. C. 81me2A2. D. 21me2A2. 2.156 mt thi im, vn tc ca vt dao ng iu ha bng 20% vn tc cc i, t s giang nng v th nng ca vt l A. 24B. 1/24C. 5D. 0,2 2.157 Mtconlclxockhilng l m, dao ng iu ha vi bin A, nng lng dao ngl E. Khi vt c li x = A/2 th vn tc ca vt c gi tr l July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 40 A. 2EmB. 2EmC. EmD. 32Em 2.158 Mt vt thc hin dao ng iu ha c phng trnhx = 10cos(4tt +2t). (cm) vi t tnh bng giy.ng nng ( th nng) ca vt bin thin vi chu kA.0,50s.B. 0,25s.C.1,00s.D.1,50s 2.159Dao ng ca con lc l xo c bin A, nng lng l E0 . ng nng ca qu cu khiqua li x = 2A l A.034E B.02EC. 04E D. 03E

2.160Dao ng ca con lc l xo c bin A v nng lng l E0 . Li x khi ng nng bng 3 ln th nngl A. 4Ax = B.2Ax = C. 22Ax = D. 24Ax = 2.161Vt c khi lngm = 100g, tn s gc e = 10t(rad/s), bin A = 5cm.Cho 210. t =Nng lngdao ng ca vt l A. 12,5J.B. 0,125J.C.1250J.D.1,25J 2.162Vtckhilngm,gnvo l xo c cng k = 100N/m.Dao ng iu ha c bin A = 5cm. Nng lng dao ng ca vt l A. 12,5J.B. 0,125J.C.1250J.D.1,25J 2.163 Mt con lc n dao ng iu ha vi bin gc 6. Khi ng nng ca con lc gp hai lnth nng th gc lch ca dy treo so vi phng thng ng l: A. 2B. 2C. 3,45D. 3,45 2.164 Con lc n c chiu di = 1m, g =10m/s2, chn gc th nng v tr cn bng. Con lc dao ng vi bin 0 = 90. Vn tc ca vt ti v tr ng nng bng th nng l: A. 9 2 cm/sB. 9 5 m/sC. 0,43m/sD. 0,35m/s 2.165 Mt vt c khi lng m = 0,1kg gn vo l xo dao ng iu ha theo phng ngang c tn s f = 5 Hz, bin 5cm.Cho t2 = 10. 1. cngk ca l xo l A. 75N/m.B.1N/m.C.50N/m. D.100N/m. 2. Lc n hi ln cc i trong qu trnh dao ng l A.500N.B.100N.C. 5N.D.2N 3. Nng lng trong qu trnh dao ng l. A. 12,5J.B.0,125J.C.1250J.D.1,25J Loi 2: TNG HP CC DAO NG IU HO 2.166 Haidao ong ieu hoa cung pha khi o lech pha gia chung la: A.t n 2 = A(vi neZ).B.( )t 1 2 + = A n(vi neZ). C.( )21 2t + = A n(vi neZ). D.( )41 2t + = A n(vi neZ). 2.167 Haidao ong ieu hoa nao sau ay c goi la cung pha? A. cm t x|.|

\|+ =6cos 31ttvacm t x|.|

\|+ =3cos 32tt . July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 41 B.cm t x|.|

\|+ =6cos 41tt vacm t x|.|

\|+ =6cos 52tt . C. cm t x|.|

\|+ =62 cos 21ttvacm t x|.|

\|+ =6cos 22tt . D. cm t x|.|

\|+ =4cos 31ttvacm t x|.|

\| =3cos 32tt . 2.168 Mtvtthchinngthihaidaongiuhacngphng,cngtns: x1=A1sin(et+ 1) v x2 = A2sin(et + 2).Bin dao ng tng hp l A. A = 2 21 2 1 2 2 12 os( ) A A A A c + + B. A = 2 21 2 1 2 2 12 os( ) A A A A c + C. A = A1 + A2 + 2 A1A2 cos (2 - 1) D. A = A1 + A2 + 2 A1A2 cos (2 - 1)2.169 Mt vt thc hinng thi haidao ng iu ha cng phng, cng tn s:x1=A1sin(et + 1) v x2 = A2sin(et + 2). Pha ban u ca dao ng tng hp l A. tg =1 1 2 21 1 2 2sin sinos osA AAc A c +B. tg = 1 1 2 21 1 2 2sin sinos osA AAc A c ++ C. tg = 1 1 2 21 1 2 2sin sinos osA AAc A c +D. tg = 1 1 2 21 1 2 2os ossin sinAc A cA A ++ 2.170 Bin dao ng tng hp A ca hai dao ng iu ha cng phng , cng tn s , khc bin c pha ban u vung gc l : A. A = A1 + A2B. A = A1 A2 C. A = 2 21 2A A D.A = 2 21 2A A + 2.171 Bindaongtnghpcahaidaongiuhacng phng , cng tn s , cng bin A , c lch phat/3 l : A.A = A 2 B.A = A 3C. A = A2D. A = 32A 2.172 Motvatthchienongthihaidaoongieuhoacungphng,cung tanso co bien o lan lt la 8 cm va 12 cm. Bien o cua dao ong tong h p la: A. A = 2 cm.B. A = 3 cm.C. A = 5 cm. D. A = 21 cm. 2.173 Hai dao ng iu ha cng phng c phng trnh ln lt l: x1 = 4cos100t (cm) v x2 = 3cos(100t +2t) (cm). Dao ng tng hp ca hai dao ng c bin l: A. 5cm. B.7cm. C. 1cm D.3,5cm. 2.174 Hai dao ng iu ha cng phng c phng trnh ln lt l: x1 = 10cos100t (cm) v x2 = 3cos(100t) (cm). Dao ng tng hp ca hai dao ng c bin l: A.5cm. B.7cm. C. 1cm D.13cm. 2.175 Mt vt thc hin:x1 = 2cos(10tt) (cm);x2=2cos(10tt +3t) (cm).Bin dao ng tng hp ca vt l A.5cm. B.7cm. C. 2 3 cm D.3,5cm July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 42 2.176 Haidao ng iu ha cng phng, cng tn s, c bin A1 = 3cm v A2 = 4cm v lch pha l 1800 th bin dao ng tng hp bng bao nhiu ? A.5cmB.3,5cmC.7cmD.1cm 2.177 Mtvtthchinngthihaidaongiuha,cngphng,cngtnsc phng trnh :( ) ( )1 2x 3cos 4 cm; x 3cos 4 cm3t ttt t| |= + = |\ .. Bin v pha ban u ca dao ng tng hp l A. 3 3 ;3cm t. B.2 3 ;6cm t C.3 3 ;6cm t D.2 ;6cm t 2.178 Motchatiemthamgiaongthihaidaoongieuhoacungphng, cung tan sox1 = sin 2t (cm)vax2 = 2,4 cos 2t (cm). Bien o cua dao ong tong hp la : A. A = 1,84 cm.B. A = 2,60 cm.C. A = 3,40 cm.D. A = 6,67 cm. 2.179 Phat bieu nao sau ay la ung: A. Trong dao ong tac dan, mot phan c nang a bien oi thanh nhietnang.B.Trong dao ong tac dan, mot phan c nang a bien oi thanh hoa nang.C.Trong dao ong tac dan, mot phan c nang a bien oi thanh ien nang.D.Trong dao ong tac dan, mot phan c nang a bien oi thanh quang nang.2.180 Nhn nh no sau y sai khi ni v dao ng c hc tt dn? A. Trong dao ng tt dn c nng gim dn theo thi gian. B. Lc ma st cng ln th dao ng tt dn cng nhanh. C. Dao ng tt dn l dao ng c bin gim dn theo thi gian.D. dao ng tt dn c ng nng gim dn cn th nng bin thin iu ha. 2.181 Pht biu no sau y l ng ? A. Dao ng duy tr l dao ng tt dn mta lm mt lc cn ca mi trng i vi vt dao ng. B. Trong dao ng tt dn, mt phn c nng bin i thnh ho nng. C. Dao ng duy tr l dao ng tt dn m ngi ta tc dng ngoi lc vo vt dao ng cng chiu vi chiu chuyn ng trong mtphn ca tng chu k. D. Trong dao ng tt dn, mt phn c nng bin i thnh quang nng.2.182 Dao ng ca ng h qu lc l: A.Dao ng duy tr.B.Dao ng cngbc.C.Dao ng tt dn.D.S cng hng. 2.183 Chn cu sai. Dao ng cng bc l dao ng A. chu tc dng ca ngoi lc bin thintun hon. B. c tnh iu ha. C.c bin ch ph thuc vo bin ca lc cng bc. D. c tn s bng tn s ca lc cng bc. 2.184 Khi ni vmt h dao ngcng bc giai on n nh, pht biu no di y l sai? A. Bin ca h dao ng cng bc ph thuc vo tn s ca ngoi lc cng bc. B. Bin ca h dao ng cng bc ph thuc bin ca ngoi lc cng bc. C. Tn s ca h dao ng cng bc lun bng tn s dao ng ring ca h. D. Tn s ca h dao ng cng bc bng tn s ca ngoi lc cng bc. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 43 2.185 Pht biu no sau y l ng. A. Bin dao ng cng bc khng ph thuc vo pha ban u ca ngoi lc tun hon tc dng ln vt. B. Bin ca dao ng cng bc khng ph thuc vo bin ngoi lc tun hon tc dng lnvt. C. Hin tng cng hng ch xy ra vi dao ng tt dn. D. Bin ca dao ng cng bc khng ph thuc vo h s cn (ca ma st nht) tc dng lnvt. 2.186 Phat bieu nao sau ay la ung: AHientng cong hng ch xay ra vi dao ong ieu hoa.B.Hientng cong hng ch xay ra vi dao ong rieng. C.Hientng cong hng ch xay ra vi dao ong tac dan. D.Hientng cong hng ch xay ra vi dao ong cng bc.2.187 Pht biu no sau y l khng ng ? A. iu kin xy ra hin tng cng hng l tn s gc lc cng bc bng tn s gc dao ng ring. B. Tn s ca dao ng cng bc bng tn s ca lc cng bc.C. Chu k ca dao ng cng bc bng chu k ca lc cng bc.D.iukin xy ra hin tng cng hng l bin lc cng bc bng bin dao ngring. 2.188 Khixy ra hintng cng hng c th vt tip tc dao ng A. vi tn s bng tn s dao ng ring.B.vitnsnhhntnsdao ng ring. C. vi tn s ln hn tn s dao ng ring.D. m khng chu tc dng ca ngoi lc. 2.189 Pht biu no sau y l sai ? A. Dao ng tt dn l dao ng c bin gim dn theo thi gian.B. Dao ng cng bc l dao ng chu tc dng ca mt ngoi lc bin thin tun hon.C. Khi cng hng xy ra, tn s dao ng cng bc ca h bng tn s ring ca h daong . D. Tn s ca dao ng cng bc lun bng tn s ring ca h dao ng.2.190 Nhn xt no sau y l khng ng. A. Dao ng tt dn cng nhanh nu lc cn ca mi trng cng ln. B. Dao ng duy tr c chu k bng chu k dao ng ring ca con lc . C. Dao ng cng bc c tn s bng tn s ca lc cng bc.D. Bin ca dao ng cng bc khng ph thuc vo tn s lc cng bc. 2.191 Mt ngi xch mt x nc i trn ng, mi bc i di 40 cm th nc trong x sng snh mnh nht. Chu k dao ng ring ca nc trong x l 0,25 s. Vn tc ca ngi l: A. 3,6 m/sB. 4,2 m/sC. 4,8 m/sD. 5,76 m/s 2.192 Mt chic xe my chy trn ng lt gch, c cch khong 9m trn ng li c rnh nh. Chu k dao ng ring ca khung xe trn cc l xo gim xc l 1,5s. Hi vi tc bao nhiuth xe b xc mnh nht? A. 6m/s. B. 24 km/h. C. 9m/s.D. 13,5m/s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 44 i.I.CCCUTRCNGHIMTRONG THITNPT ST-C2.1.Mtvtnhdaongiuhatheomttrccnh.Phtbiunosauy ng? A. Qu o chuyn ng ca vt l mt on thng. B. Lc ko v tc dng vo vt khng i. C. Qu o chuyn ng ca vt l mt ng hnh sin. D. Li ca vt t l vi thi giandao ng. ST-C2.2.Mt con lc l xo gm mt l xo khi lng khngng k, mt u c nh v mtugnvimtvinbinh.Conlcnyangdaongiuhatheophngnm ngang. Lc n hi ca l xo tc dng ln vinbi lun hngA. theo chiu chuyn ng ca vin bi.B. theo chiu m quy c.C. v v tr cn bng ca vin bi.D. theo chiu dng quy c. ST-C2.3.Mt con lc l xo gm vt nh khi lng 400g, l xo khi lng khng ng k vccng100N/m.Conlcdaongiuhatheophngngang. 2=10. Dao ng ca con lc c chu k l A. 0,8s.B. 0,4s.C. 0,2s.D. 0,6s. ST-C2.4.Mtconlc n gm qu cu nh khi lng m c treo vo mt u si dy mm, nh, khng dn, di 64cm. Con lc dao ng iu ha ti ni c gia tc trng trng g. 2 (m/s2). Chu k dao ng ca con lc l A. 1,6s. B. 1s.C. 0,5s.D. 2s. ST-C2.5.Mt con lc n gm mt hn bi nh khi lng m, treo vo mt si dy khng gin, khi lng si dy khng ng k. Khi con lc n ny dao ng iu ha vi chu k 3 s th hn bi chuyn ng trn mt cung trn di 4 cm. Thi gian hn bi i c 2 cm k t v tr cn bng l A. 0,25 sB. 0,5 s C. 1,5 s D. 0,75 s ST-C2.6.Biu thc tnh chu k dao ng iu ha ca con lc vt l l T =mgd12t ; trong nh xuyn qua vt, m v g ln lt l khi lng ca con lc v gia tc trng trng ti ni t con lc. i lng d trong biu thc l B. khong cch t trng tm ca con lc n ng thng ng C. chiu di ln nht ca vt dng lmcon lc. D. khi lng ringca vt dng lmcon lc. ST-C2.7. tnh bng cm, t tnh bng s). Ti thi im t = 5s, vn tc ca cht im ny c gi tr bng A. 5cm/s. C. -D. 0 cm/s. ST-C2.8. ca cht im tiv tr cn bng c ln bng A. 4 cm/s. B. 8 cm/s. C. 3 cm/s. D. 0,5 cm/s. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 45 ST-C2.9.Mt con lc l xo gm mt l xo khi lng khng ng k, cng k, mt u cnhvmtugnvimtvinbinhkhilng m. Con lc ny ang dao ng iu ha c c nngA. t l nghchvi khi lngm ca vinbi.B. t l vi bnh phng chu k dao ng.

C. t l vi bnh phng bin dao ng.D. t l nghchvi cng k ca l xo. ST-C2.10.Haidaongiuhacngphng,cphngtrnhx1=1cos( )3A ttt +v x2= )32cos(2tt t A hai dao ngA. cng pha.B. ngc pha.C. lch pha /3. D. lch pha /2. ST-C2.11.Chohaidaongiuhacngphngcccphngtrnhlnltl x1= )4cos( 3tt tv x2= )4cos( 4tt+ t . Dao ng tng hp ca hai dao ng ny c bin l A. 7cm.B. 12cm.C. 5cm.D. 1cm. ST-C2.12.Chohaidaongiuhacngphngcccphngtrnhlnltl x1= )6cos( 4tt tv x2= )2cos( 4tt t . Dao ng tng hp ca hai dao ng ny c bin l A. 8cm.B. 3 4cm.C. 2cm.D.2 4 cm. ST-C2.13.Dao ng tt dn A. c bin gim dn theo thi gian.B. lun c li. C. c bin khng i theo thi gian. D. lun c hi. ST-C2.14.Mt h dao ng chu tc dng ca ngoi lc tun hon F = F0cos 10tth xy ra hintng cng hng. Tn s dao ng ring ca h phi l A. 10 Hz.B. 10 Hz.C. 5 Hz.D. 5 Hz. II.CC CU TRC NGHIM TRONG THI I HC ST-C2.15.Mt vt dao ng iu ha c phng trnh x = Acos(et + ). Gi v v a ln lt l vn tc v gia tc ca vt. H thc ng l : A. 2 224 2v aA + =e e.B. 2 222 2v aA + =e eC. 2 222 4v aA + =e e.D. 2 222 4aAve+ =e. ST-C2.16.Mt con lc l xo dao ng iu ha. Bit l xo c cng 36 N/m v vt nh c khi lng 100g. Lyt2 = 10. ng nng ca con lc bin thin theo thi gian vitn s. A. 6 Hz.B. 3 Hz.C. 12 Hz.D. 1 Hz. ST-C2.17.Mt con lc l xo c khi lng vt nh l 50 g. Con lc dao ng iu ha theo mttrccnhnmngangviphngtrnhx=Acoset.Csaunhngkhongthigian 0,05 s th ng nng v th nng ca vt li bng nhau.Ly t2 =10. L xo ca con lc c cng bng A. 50 N/m.B. 100 N/m.C. 25 N/m.D. 200 N/m. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 46 ST-C2.18.Ti mt ni trn mt t, mt con lc n dao ng iu ha. Trong khong thi gian At, con lc thc hin 60 dao ng ton phn; thay i chiu di con lc mt on 44 cm th cng trong khong thi gianAt y, n thc hin 50 dao ng ton phn. Chiu di ban u ca con lc l A. 144 cm.B. 60 cm.C. 80 cm.D. 100 cm. ST-C2.19.Mt vt dao ng iu ha c ln vn tc cc i l 31,4 cm/s. Ly3,14 t = . Tc trung bnh ca vt trong mt chu k dao ng lA. 20 cm/sB. 10 cm/sC. 0.D. 15 cm/s. ST-C2.20.Tinicgiatctrngtrng9,8m/s2,mtconlcn v mt con lc l xo nm ngang dao ng iu ha vi cng tn s. Bit con lc n c chiu di 49 cm v l xo c cng 10 N/m. Khi lngvt nh ca con lc l xo l A. 0,125 kgB. 0,750 kgC. 0,500 kgD. 0,250 kg ST-C2.21.Mt vt dao ng iu ha theo mt trc c nh (mc th nng v tr cn bng) thA. ng nng ca vt cc i khi giatc ca vt c ln cc i.B. khi vt i t v tr cn bng ra bin, vn tc v gia tc ca vt lun cng du.C. khi v tr cn bng, th nng ca vt bng c nng. D. th nng ca vt cc i khi vt v tr bin. ST-C2.22.Mtconlclxogmlxonhvvtnhdaongiuhatheophng ngang vi tn s gc 10 rad/s. Bit rng khi ng nng v th nng (mc v tr cn bng ca vt) bng nhau th vn tc ca vt c ln bng 0,6 m/s. Bin dao ng ca con lc lA. 6 cmB.6 2 cmC. 12 cmD. 12 2 cm ST-C2.23.Chuynngcamtvtltnghpcahaidao ng iu ha cng phng. Haidao ng ny c phng trnh ln lt l 1x 4cos(10t )4t= + (cm) v 23x 3cos(10t )4t= (cm). ln vn tc ca vt v tr cn bng l A. 100 cm/s.B. 50 cm/s.C. 80 cm/s.D. 10 cm/s. ST-C2.24.Khini v dao ng cng bc, pht biu no sau y l ng? A. Dao ng ca con lc ng h l dao ng cng bc. B. Bin ca dao ng cng bc l bin ca lc cng bc. C. Dao ng cng bc c bin khng i v c tn s bng tn s ca lc cng bc.D. Dao ng cng bc c tn s nh hn tn s ca lc cng bc. Cu 1:Chncu ngnhttrongcc cusau : A. Con lcl xo gmmtvt nngtreo vo lxo. B. Con lcngmmtvt nngtreo vomt sidy khngginc khilngkhngngk. C. Con lcl xo gmmtvt nngc khilngm treo vo lxo c cngl k. D. Tt c u ng. Cu 2:Chncu ngtrongcc cu sauy: A. Bin dao ngca con lcl xo chphthuccch chngc thigian. B. Chu k con lcnphthucbin khidao ngvibin nh. C. Tns ca h dao ngt do phthucbin July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 47 D. Chuynngcacon lc nxemldao ngt do ti1 vtrxc nh. Cu 3:Chncu ng: A. Nng lngca dao ngiuha binthintheothigian. B.Nnglngdaongiuhacahqucu+lxobngngnngcaqucukhiquavtr cn bng. C. Nng lngca dao ngiuha chphthucc imcah. D. Khibin ca vtdao ngiuha tnggp ithnnglngcah gimmtna.Cu 4:Timtaimtrnmttrng: A. Dao ngca conlc lxo v dao ngca con lcnkhngphildao ngt do. B. Dao ngcacon lc lxo v dao ngca con lcnl dao ngt do. C.Daongcaconlclxoldaongt do cn dao ng ca con lc n khng phi l dao ng t do. D.Daongcaconlclxokhngphildao ng t do cn dao ng ca con lc n l dao ng t do. Cu 5: iuno sauy l ng: A. Chu k dao ngnhcacon lc nt lthunvig . B. Chu k con lclxo t lnghchvik . C. Chu k conlc l xo v con lc n u ph thuc vo khi lng vt. D. Chu k conlc n khi dao ng nh khng ph thuc vo khi lng hn bi treo vo dy treo. Cu 6: Trongdao ngiuho ca conlc lxo, c nngcan bng: A. Tngngnngvth nngca vtkhiqua mtvtrbt k. B. Thnngca vt nngkhiqua vtrcn bng. C. ngnngcavt nngkhiqua vtrbin. D. C A,B,C u ng. Cu 7: Mt con lcl xo dao ngiuho, c nngton phnc gitrlW th: A. Tivtrbindao ng:ngnngbngW.B. Tivtrcn bng:ngnngbngW. C. Tivtrbt k:thnnglnhnW.D. Tivtrbt k:ngnnglnhnW. Cu 8: Trongdao ngcacon lc lxo, nhnxtno sau y lsai: A. Chu kringchphthucvo c tnhca h dao ng. B. Lc cn camitrngl nguynnhnlmcho dao ngtt dn. C. ngnngli lngkhngbo ton. D. Bin dao ngcngbc chphthucvo bin ca ngoilc tunhon. Cu 9: Bin camtcon lc lxo thngngdao ngiuho: A. L xmax. B. Bngchiuditia trchiudi vtrcn bng. C. L qungngi trong 41 chukkhivtxutphtt vtrcn bnghoc vtrbin. D. C A,B,C, u ng. Cu 10: Khithayi cchkchthchdao ngcacon lc lxo th: A.v A thayi, fvekhngi.B.v W khngi, T vethayi. C. , A, fve u khngi.D. , A, f veu thayi. Cu11:Mtconlclxocngktreothngng,utrncnh,udi gn vt. dn ti v tr cn bngll A .Choconlcdaongiuhotheophngthngngvibinxm( l xmA < ).Trongqu trnhdao nglctc dngvo imtreo c lnnhnhtl: A.0 = F .B.) (mx l k F A = .C.) (mx l k F + A = .D.l k F A = . July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 48 Cu12:Mtconlclxocngktreothngng,utrncnh,udi gn vt. dn ti v tr cn bngll A .Choconlcdaongiuhotheophngthngngvibinxm( l xmA > ).Trongqu trnhdao nglccc i tc dngvo imtreo c lnl: A.l kx FmA + = .B.) (mx l k F + A = .C.) (mx l k F A = .D. mx l k F + A = . Cu13:Khignqunngckhilngm1vomtlxo,thyndaong iu ha vi chu k T1. Khi gn qunngckhilngm2volxo,ndaongvichukT2.Nugnngthim1vm2 cng vo l xo , thchuk dao ngca chngl: A. 2 21 2T T T = + B. 2 21 2T T T = + C. 1 22T TT+=D. 1 2T T T = +Cu14:Mtlxoccngbanulk,qucukhilngm.Khi gim cng 3 ln v tng khi lng vtln2 lnthchuk mi: A. Tng6ln.B. Gim6 ln. C. Khngi.D. Gim6ln. Cu15:Mtvtckhilngmctreovoumtlxo.Vtdaongiuhavitnsf1=12Hz. Khitreo thmmtgiatrng10 m g A =thtns dao nglf2 = 10Hz.Chncu ng: A. m = 50gB. k = 288N/mC. T = 0,23sD. Tt c u sai Cu16:Nucngkcalxovkhilngmcavttreoulxoutnggpi th chu k dao ng ca vts thayi nhth no ? A. Tng2 lnB. Khngthayi C. Tng2 lnD. Gim2ln Cu17:Haiconlclxoccngcngk.Bitchukdaong 2 12T T= .Khilngcahaiconlc lin h vinhautheocngthc : July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 49 A. 2 14m m=B. 1 2m m = /4C. 2 12m m=D. 2 12m m=Cu18:Choconlclxodaongkhngmasttrnmtphngnghing1gco sovimtphngnm ngang,utrncnh,udignvtm,lxoccngk.Khiqu cu cn bng, gin ca l xol A , giatc trongtrngl g.Chukdao ngl: A. lkTA= t 2 B. glTA= t 2 C. otsin2glTA= D. glTotsin .2A=Cu19:Mtconlclxodaongiuhokhngmasttrnmtphngngang,lxoccngk,khi lngqu cu l m,bin dao ngl xm. Khngnhno sau y lsai: A. Lc n hicc i c ln mkx F = . B. Lc n hicc tiu0 = F. C. Lc n hicc i c ln) ( l x k FmA = , vil Al gincal xo tivtrcn bng. D. Lc phchibnglc n hi. Cu 20: Mt con lcl xo gmqu cu khilnglm v lxo c cngk. Khngnhno sau y lsai: A. Khilngtng4 lnthchuktng2 ln. B. cnggim4 lnthchuk tng2 ln. C. Khilngtng4 lnngthi cngtng4 lnthchuk khngi. D. cngtng4 lnthnnglngtng2 ln. Bi tp trcnghim: Dng 1: CC I LNG C TRNG CACON LC LXO Cu1:Mtconlclxoccng mNk 150 = vcnnglngdao ng l 0,12J. Bin dao ng ca n l: A. 0,4m.B. 4mm.C. 0,04m.D. 2cm. Cu2:Mtconlclxotreothngngvdaongiuhovitns4,5Hz.Trongqutrnhdaong chiudi lxo binthint 40cmn 56cm.Ly 210smg = . Chiudi r nhinca n l: A. 48cm.B. 46,8cm.C. 42cm.D. 40cm. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 50 Cu3:Mtlxokhilngkhngngk,treovomtimcnh,cchiuditnhinl0.Khitreovt m1 = 0,1kg thn di l1 = 31cm.Treo thmmtvt m2 = 100g th dimill2 = 32cm. cng k ll0 l: A. 100mN v 30cm.B. 100mN v 29cm.C. 50mN v 30cm.D. 150mN v 29cm. Cu4:Mtlxokhilngkhngngk,cchiuditnhinl0,cngktreovomtimcnh. Nutreomtvtm1=50gthnginthm0,2cm.Thaybngvtm2=100gthndi20,4cm.Chnp n ng: A. l0 = 20cm,k = 200mN.B. l0 = 20cm,k = 250mN.C. l0 = 25cm,k = 150mN. D. l0 = 15cm,k = 250mN.Cu5:Conlclxotreothngngdaongiuhotheophngtrnh:) )(220 cos( 2 cm t xt+ = .Chiudi tnhincalxolcm l 300 = .Ly 210smg = .Chiuditithiuvtiacalxotrongqutrnhdao ngl: A. 30,5cm v 34,5cm. B. 31cm v 36cm.C. 32cm v 34cm.D. Tt c u sai. Cu6:Mtlxocng mNk 80 = .Trongcngmtkhongthigiannhnhau,nutreoqucukhi lngm1thnthchin10daong,thaybngqucukhilngm2thsdaonggimphnna.Khi treo c m1 v m2 thtn s dao nglHzt2. Tmkt qung: A.kg m 41 =vkg m 12 = . B.kg m 11 =vkg m 42 = .C.kg m 21 =vkg m 82 = . D.kg m 81 =vkg m 22 = .Cu7:Mtlxockhilngkhngngk,chiuditnhincm l 1250 = treothngng,udic qucum.Chngctotivtrcnbng,trcOxthngng,chiudnghngxung.Vtdaong viphngtrnh:) )(62 cos( 10 cm t xtt = . Ly 210smg = . Chiudi lxo thiimt0 = 0 l: A. 150cm.B. 141,34cm.C. 133,66cm.D. 158,66cm. July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 51 Cu8:Mtconlclxodaongtheophngngangvichiudiquol14cm,tnsgc sradt e 2 = . Vntc khipha dao ngbngrad3t l: A. scmt 7 .B. scm3 7t .C. scm2 7t .D. scm37t. Cu9:Mtconlclxotreothngngcvtnngkhilngg m 500 = ,cng mNk 25 = angdao ngiuho.Khivntccavtl scm40 thgiatccanbng 23 4scm.Bindaongcavt l: A. 4cm.B. 4 2 cm.C. 20 3 cm.D. 8cm. Cu10:Mtvtkhilngg m 400 = treovomtlxocng mNk 160 = .Vt ang dao ng iu ha theophngthngngvibin10cm.Vntccavttitrungimcavtrcnbngvvtrbinc lnl: A. sm3 .B. scm3 20 .C. scm3 10 . D. sm2320 . Dng 2: CHU K DAO NG CA CON LCL XO Cu1:Mtvtnngtreovomtlxo,daongtheophngthngng.Nuvtckhilngm1th vt cchukdaongl3s.Nuvtckhilngm2 thvtcchukdaongl4s.Hichukdaong ca vtl bao nhiukhi vtc khi lng bngtnghaikhilngtrn? A. 7s.B. 5s.C.s712.D. C A,B,C u sai. Cu2:Mtvtnnggnvolxotreothngnglmlxodnramton0,8cm.Lyg=10m/s2.Chu k dao ngt do ca vtnnggnvo lxo l: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 52 A. 0,178s.B.1,78s.C. 0,562 s.D. 222 s. Cu3:Mtvtckhilngmtreovomtlxoccngk.Kchthchchovtdaongiuhavi binl5cmthchukdaongcavtlT=0,4s.Nukchthchchovtdaong vi bin dao ng l10cm thchukdao ngca n c thnhngitrno di y: A. 0,2s .B.0,4s.C. 0,8s .D. Mt gitrkhc. Cu4:Mtconlclxodaongiuhatheophngthngngviphngtrnh ) )(6cos( 10 cm t xte+ = .Trongqutrnhdaong,tsgialnnhtvnhnhtcalcnhical xo l 37. Cho g = 10m/s2.Chu k dao ngca con lcl : A. T = 2 (s)B. T = 6 (s)C. T = 1 (s)D. T = 4 (s) Cu4:Khignqucukhilngm1volxothndaongvichukT1.Khignqucuckhi lngm2volxotrnthndaongvichukT2=0,4s.Nugnngthi hai qu cu vo l xo th n dao ngvichuk T = 0,5s.VyT1 c gitrl: A. s T321 = .B.s T 3 , 01 = .C.s T 1 , 01 = .D.s T 9 , 01 = . Cu5:Mtlxoccngk.Lnltgnvolxoccvtm1, m2, m3 = m1 + m2, m4 = m1 m2 vi m1 > m2 . Ta thychuk dao ngcacc vttrn lnltlT1, T2, T3 = 5s , T4 = 3s . T1 , T2 c gitrl: A. T1 = 8s vT2 = 6s.B. T1 = 2,82s vT2 = 4,12s. C. T1 = 6s vT2 = 8s.D. T1 = 4,12s vT2 = 2,82s. Cu6:Mtvtckhilngg m 160 = treovomtlxothngngthchukdaongiuhl2s. Treo thmvo lxo vt nngc khilngg m 120'=thchuk dao ngca hl: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 53 A. 2s.B.7 s.C. 2,5s.D. 5s. Cu7:Mtlxoccngk=80N/m,lnlttreohaiqucuckhilngm1,m2volxovkch thchchochngdaongththy:trongcngmtkhongthigianvtm1thchinc10daong,trong khim2chthchinc5daong.NutreochaiqucuvolxothchukdaongcahlT= 1,57s = 2t s. Him1 v m2 c gitrl: A. m1 = 3kg vm2 = 2kg .B. m1 = 4kg v m2 = 1kg . C. m1 = 2kg v m2 = 3kg .D. m1 = 1kg vm2 = 4kg . Cu8:Mtvtkhilngmcgnlnltvohailxoccngk1, k2th chu k ln lt l T1 v T2. BitT2 = 2T1 v k1 + k2 = 5N/m . Gitrca k1 v k2 l: A. k1 = 3N/m vk2 = 2N/m .B. k1 = 1N/m v k2 = 4N/m . C. k1 = 4N/m v k2 = 1N/m .D. k1 = 2N/m vk2 = 3N/m . Dng 3: PHNG TRNH DAO NG CACON LC LXO Cu1:Mtconlclxogmqucug m 300 = , mNk 30 = treovomtim c nh. Chn gc to vtrcnbng,chiudnghngxung,gcthigianllcvtbtudaong.Koqucuxungkhi vtrcnbng4cmritruynchonmtvttcbanu scm40 hngxung.Phngtrnhdaongca vtl: A.) )(210 cos( 4 cm t xt = .B.) )(410 cos( 2 4 cm t xt+ = . C.) )(410 cos( 2 4 cm t xt = .D.) )(410 cos( 4 cm t xt+ = . Cu 2: Con lcl xo c thnhhnhv. Phngtrnhdao ngca vtl: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 54 A.) ( 10 cos 4 cm t x t = .B.) ( 5 cos 8 cm t x t = . C.) )(25 cos( 4 cm t xt = .D.) )(25 cos( 4 cm t xt+ = . Cu3:Mtconlclxotreothngngc cng mNk 7 , 2 = , qu cukg m 3 , 0 = . T v tr cn bng ko vtxung3cmricungcpmtvntc scm12 hngvvtrcnbng.Lyt0=0tivtrcnbng. Phngtrnhdao ngcavt l: A.) )( 3 cos( 5 cm t x t = .B.) ( 3 cos 5 cm t x = . C.) )(43 cos( 5 cm t xt+ = .D.) )(23 cos( 5 cm t xt = . Cu4:Khitreoqucumvomtlxothnginra25cm.Tvtrcnbngkoqucuxungtheo phngthngng20cmribungnh.Chnt0=0llcvtquavtrcnbngtheochiudnghng xung.Ly 210smg = . Phngtrnhdao ngcavt c dng: A.20cos(2 )( )2x t cmtt = .B.) ( 2 cos 20 cm t x t = . C.45cos(2 )( )2x t cmtt = .D.) ( 100 cos 20 cm t x t = . Cu5:Mtconlclxotreothngnggmvtnngg m 250 = ,cng mNk 100 = ..Kovtxung dicholxogin7,5cmribungnh.ChntrcOxthngng,chiudnghngln,gctov trcn bng,t0 = 0 lcthvt. Ly 210smg = . Phngtrnhdao ngl: A.) )(220 cos( 5 , 7 cm t xt = . B.5cos(20 )( ) x t cm t = + . C.) )(220 cos( 5 cm t xt+ = .D.) )(210 cos( 5 cm t xt = . Cu6:Mtlxocngk,uditreovtg m 500 = ,vtdaongvicnng10-2J.thiimban u n c vntc sm1 , 0v giatc 23sm . Phngtrnhdao ngl: x(cm) t(s) 0,1 O July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 55 A.4cos(10 )( )2x t cmt= + .B.) ( cos 2 cm t x = . C.2cos(10 )( )6x t cmt= .D.2cos(20 )( )6x t cmt= + . Cu7:Mtlxoutncnh,uditreomtvtkhilngm.Vtdaongiuhotheophng thngngvitnsHz f 5 , 4 = .Trongqutrnhdaong,chiudilxothoiukincm l cm 56 40 s s . Chngctovtrcnbng,chiudnghngxung,gcthigianllclxongnnht.Phng trnhdao ngcavt l: A.8cos(9 )( ) x t cm t t = + .B.) )(29 cos( 16 cm t xtt+ = . C.) )(25 , 4 cos( 8 cm t xtt = .D.) )(29 cos( 8 cm t xtt = . Cu8:Mtconlclxotreothngnggmmtqunngckhilngkg m 1 = vmtlxo c cng l mNk 1600 = .Khiqunngvtrcnbng,ngitatruynchonvntcbanu sm2 hngthng ngxungdi. Chngc thigianllctruynvntc cho vt. Phngtrnhdao ngca vtl: A.) ( 40 cos 5 , 0 m t x = .B.) )(240 cos( 05 , 0 m t xt+ = . C.0, 05cos(40 )( ) x t m t = + D.) ( 40 cos 2 05 , 0 m t x = . Cu9:Conlclxogmqunngckhilngkg m 4 , 0 = vmtlxoccng mNk 40 = tnm ngang.Ngitakoqunnglchkhivtrcnbngmtonbng12cmvthnhchondaong.B quamast.ChntrcOxtrngviphngchuynngcaqunng,gctalvtrcnbng,chiu dngtheohngko vt,gc thigianl lcvt qua vtrcn bnglnth nht.Chn p nsai: A. srad10 = e .B.cm xm12 = .C. 2t = .D.) )(210 cos( 12 cm t xt =July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 56 Cu10:Lxockhilngkhngngk,cng mNk 100 = ,utrncnh,uditreovtc khilngg m 400 = .Kovtxungdivtrcnbngtheophngthngngmtonbngcm 2 v truynchonvntc scm5 10 ndaongiuha.B qua ma st. Gc ta l v tr cn bng, chiu dnghngxung,gcthigianllcvtlix=+1cmvdichuyntheochiudngOx.Phng trnhdao ngcavt l: A.) )(310 5 cos( 2 cm t xt = .B.) )(310 5 cos( 2 cm t xt+ = . C.) )(310 5 cos( 2 2 cm t xt+ = .D.) )(310 5 cos( 4 cm t xt+ = . Cu11:TreovoimOcnhmtucamtlxockhi lng khng ng k, di t nhin l0 = 30 cm.udicalxotreomtvtM,lxoginmtonbng10cm.Bquamilccn.Ly 210smg = .NngvtMlnvtrcchOmtkhongbng38cmritruynchonmtvntcbanu hngxungbng scm20 .Chnchiudnghngxung.Gctativtrcnbng,gcthigianl lccungcp vntc ban u. Chn p n ng: A. srad10 = e .B.cm xm2 2 = .C.) )(410 cos( 2 2 cm t xt+ = .D. A v B ng. Cu12:Mtlxockhilngkhngngk,utrncnh,uditreomtvtckhilng80g. Vtdaongiuhatheophngthngngvitns4,5Hz.Trongqutrnhdaongdingnnht calxol40cmvdinhtl56cm.Ly 210mgs= .GctalVTCB,chiu dng hng xung, t = 0 llclxo ngnnht.Phngtrnhdao ngl: A.) )(29 cos( 2 8 cm t xtt = .B.) )(29 cos( 8 cm t xtt+ = . C.) )(29 cos( 8 cm t xtt = .D.8cos(9 )( ) x t cm t t = + . Cu13:Mtconlclxockhilngcavtm=2kgdaongiuhatrntrcOx,ccnngl J W 18 , 0 = .Chnthiimt0=0lcvtquavtrcm x 2 3 = theochiumvtithnngbngng nng.Phngtrnhdao ngca vt l: July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 57 A.) )(452 5 cos( 6 cm t xt+ = . B.) )(42 5 cos( 6 cm t xt+ = . C.) )(45 cos( 6 cm t xtt+ = . D.) )(455 cos( 6 cm t xtt+ = . Cu14::Mtconlclxo,gmmtlxoccng mNk 10 = ckhilngkhngngkv mt vt c khilngm=100gdaongiuhodctheotrcOx.Thiimban u c chn l lc vt c vn tc sm1 , 0 v giatc 21sm .Phngtrnhdao ngca vt l: A.) )(310 cos( 2 cm t xt+ = .B.) )(310 cos( 2 cm t xt = . C.) )(310 cos( 2 cm t xt = .D.2 cos(10 )( )4x t cmt= .Cu15:Mtconlclxockhilngcavtm=2kg,daongiuhocnnglngdaongl J W 125 , 0 = .Tithiimbanuvtcvntc sm25 , 0 vgiatc 225 , 6sm .Phngtrnhdaong ca vtl: A.) )(425 cos( 10 cm t xt+ = .B.) )(425 cos( 10 cm t xtt = . C.) )(425 cos( 2 cm t xt+ = .D.) )(425 cos( 2 cm t xt = . Cu16:Mtconlclxodaongtheophngthngng.Trongqutrnhdaong,lxocchiudi binthint48cmn58cmvlcnhiccicgitrl9N.Khilngcaqucul400g.Chn gcthigianllcqucuiquavtrcnbngtheochiumcaquo.Cho 2210smg = = t .Phng trnhdao ngcavt l: A.) ( 5 cos 5 cm t x t = .B.5cos(5 )( )2x t cmtt = + .July 22, 2011 TI LIU CHNG IINHM HC L 360* HOCNHOM360.HNSV.COM TMail:vietan16@yahoo.com Page 58 C.) )( 5 cos( 5 cm t x t t = .D.) )( 5 cos( 5 cm t x t t+ =Cu18:Mtconlclxoccng mNk 100 = khilngkhng n