Vật lý 12 tóm tắt lý thuyết và bài tập trắc nghiệm tập 1 (p1)

ThS V O M A N H H U N G VAT LY 12 TOM TAT LY THUYET VA BAI TAP TRAC NGHIEM (TAP 1) THJ Vlf N TINH BINHTHUAN N H A X U A T B A N D A I H Q C SU" P H A M T P H C H I M I N H VAT LY - TOM TAT LY THIIY£T VA BAI TAP TBAC WGHIEM - TAP NHA XUXT BAN DAI HpC SIT PHAM TP Hb CHf MINH LOflNOIDAU 280 An Di/ang Vi/ang, Phw&ng 4, Qugn 5, TP Ho Chi Minh Di?n thogi: (08) 38 301 303 - Fax: (08) 39 381 382 Email: nxb&hcmuo.edu.vn Than chao quy thay c6, cac bac phu huynh va cac em hoc sinh lop 12! Website: httD://nxb.hcmuD.edu.vn/ Nh3m givip cac em hpc sinh c6 them tai lieu ciing c6 kien thuc va ren Vft ly 12 T6m tit ly thuyet B^i tgp trie nghifm luyen ky nang giai cac bai tap Vat l y dS chuan bi cho cac ky thi t6t nghiep Trung hoc Ph6 thong, tuyen sinh Dai hpc va Cao dSng, chiing toi da smi tam va bien soan bo sach Tom tat ly thuyet va b^i tap trSc nghif m Vat ly 12 (Tgpl) Bo sach gom hai tap: Tac gia: ThS Vd Mgnh Hung T a p 1: Dao dong ca Song ca va Dong di?n xoay chieu T a p 2: Dao dgng di?n tir Song anh sang, Lugmg t u anh sang va Chju trach nhi$m xuat ban: Hat nhan nguyen tu Glim doc-T6ng blen tip PGS TS NGUYIN KIM HSNG M o i tap sach dugc chia lam nhi§u chirong M o i chuong dago phan loai theo timg van de trpng tam ma hoc sinh can n i m vGng Trong timg \kn Bien tgp ngi dung: &k, noi dung l y thuyet dugc tom tat mpt each ngSn ggn nhSt, sau la bai LY QUdC BINH tap mau c6 giai chi tiat va mgt s6 cau trac nghifm lien quan Sau moi chuong, CO khoang 100 cau trdc nghi?m de hpc sinh on tap tung vSn de dugc neu Bien tgp ky thugt: Trong bg sach Tom t§t ly thuyet va bai tap trSc nghifm Vat ly 12 C N G TY TNHH TMDV VHP CO suu tam mot so cau t r i e nghiem t u cac da thi tot nghiep Trung hoc Pho thong, tuyan sinh Dai hgc va Cao d i n g cua Bg Giao dye va Dao t^o de cac Tnnh bay va minh hga: em hgc sinh t h u sue C N G TY TNHH T M D V V H P Mac du da c6 gang bian soan ky, nhung chdc chan se khong tranh kh6i nhung sai sot Mong quy dgc gia thong cam va gop y da i k tai ban dugc Si>o ban in: hoan thien hem C N G TY TNHH TMDV VHP Chung t o i hy vgng bg sach se dong gop mgt phdn quan 3ng qua trinh hgc t§p cua cac em hgc sinh Chuc cac em hgc t|ip tot! In 1000 cuon kho 16 x 24 cm tgi Cong ty TNHH MTV In an Mai Thjnh Ou-c Dang ky ke hogch xuat b^n so 845-2012/CXB/02-168/DHSPTPHCM Quyt djnh xuat ban so 500/QD-NXBDHSP ky ng&y 06 thing 09 n3m 2012 In xon va nop lu'u chieu thing 12 nSm 2012 ' Tac gia: T h S V Manh Hung VAT tY - TOM TAT LY THUY£T VA BAI TAP TRAC NEHIEM - TAP MVC LUC L6IN6IDAU C H U O N G 1: D A O D O N G C O H Q C V4n de 1: Dao dpng di^u hda Vin de 2: Th(n gian, pha ban d^u dao d^ng di^u h6a Vin dk 3: Con iSc 16 xo D I 4: DOng nSng, thi nang ca nSng Vdn De 5: Lyre d^n h6i - Lye k€o \h Van E>k 6: Con I5c don D^ 7: T6ng hfjfp dao dOng Vin De 8: Dao dOng tat dlln wk dao dpng cuSng biic Bhi t§p trac nghif m t6ng hgrp Chuomg Dip in Chuang CHl/ONG 2: S6NG C O Vdn d^ 1: Djnh nghia s6ng ca d^ 2: Qud trhih truyen s6ng - DO l?ch pha - Phuong trlnh truy^n s6ng Vin 6k 3: Giao thoa song vdi ngu6n k^t hijrp cung pha V4n ih 4: S6ng dung via dl5:S6ngam Bhi t§p trac n h§ t6ng h(jrp Chuong gim Dip &n Chuong CHL/ONG 3: D N G D l ^ N X O A Y CHIEU Vkn de 1: C&c bilu thurc ca bin win ik 2: Do?ui m^ch kh6ng phSn nhinh d^ 3: Cong sudt - Hf s6 c6ng sudt Vdn dl 4: M?ch cpng hudng win de 5: CuOn d&y khOng thu^ ckm Van d^ 6: Bien thien R, L, C V ^ d^ 7: Phuong phip giin 6h Vecta Vin dh 8: M&y phdt difn, dOng co difn, m&y biln thi, truyk t&\n nang Bai t§p trac nghifm tong hgrp Chuong Dip an Chuong Trang CHIfflme : DAO D$WG Ctf HOC CHirONQ I DAO DONG CO HOC VAN DE 1: DAO DONG DIEU HOA Dao dong dieu hda la chuyen dgng c6 phuong trinh tudn theo qiiy lugt sin hogc cosin theo thai gian Phu'ong trinh dao dong X = Acosia)t + (p) o A: Bien dao dpng (L = 2A: chieu dai quy dao) o CO: Tan so goc (rad/s) 6> = 2nf = 2n o f: Tan so (Hz) So dao dong tren mot gidy o T: Chu ki (s) Thai gian thuc hien mgt dao dong o (p : Pha ban d^u (phu thuoc vao each chon goc thai gian) o x: Li do, dai cua vat so vai vi tri can bang Vsln toe, gia toe: a Van t6c: v = x' = - Acosin(cot + (p)= Acocos(a)t + + ^) ^ IV Inax = Aco X = (tai VTCB) ^ v = khix = ±A (tai vi tri bien) b Gia t6c: a = v' = x" = co^Acos(a)t + cp + rr) = - co^x ^ a Lax = to^A X = ± A (tai vi tri bien) / a = X = (tai VTCB) c Mdilienhegiuaa^a;^ vaVjnax^ ,.w , ^ ,^^^^ d Cong thuc doc lap: V,2 max = Aamax CD CO CO CHKflM61:DAB OflNECdHqC VAT lY 1? - T O M T A T lY THttYgT V A B A I T A P T B A C H E H I E M - T A P Nhan xet: • Li X, van tSc v va gia tSc a biin thien dieu hoa ciing chu ki va tan so • Van (6c v sam pha hem li x mot gdc - • Gia tdc a sam pha han van toe v mot gocj^ • Gia tSc a ngugrc pha vai li dg x Bai tap mau: Mot vat dao dong di^u hoa voi phuong trinh x = 5cos Ant + n cm, t tinh bang giay Goc toa tai vi tri can bSng a Xac djnh bien do, chiSu dai quy dao, tin so goc, tan s6, chu ki, pha ban dau b Tinh li x tai thai diem t = 0,25 s c Viet bilu thuc eua van t6c v, gia toe a theo thai gian t d Xac dinh cae gia tri van toe, gia t6e eua ehdt diim tai thai dilm t = 0,25s c Tinh Ian van t6e eua chSt diem qua vi tri can bSng va Ian gia t6e ciia chit dilm a vi tri bien f Tinh van toe chit diem c6 li x = 3cm g Tinh li chat dilm c6 van toe 127r em/s h Tinh van toe chdt dilm c6 gia t6c a = 4871^ cm/sl Giai: x = 5cos Ant H— em a -4 = 5cm; L = 2A- 10cm; a> = 4rr r a d / s / = ^ = IHz.T = = 0,5s,(p = -^rad b t = 0,25s -> X = cos (47r 0,25 + f) = 0: Vi tri can bSng c 17 =-207r sin (^4jrt+ ^ ) c m / s , a - - S O T T ^ COS {\nt +1) cm/s^ d i; = - 207r sin (47r 0.25 + f) = 207r cm/s, a e- = - S O T T ^ COS (47r Vynax = i4aj 025 = 207r cm/s; +1) a^ax = = = ^Orr^ cm/s^ CHI/fl«G1: DAB fi^K VAT LY 12 - T D M T A T L Y THUVfT «A B A I T A P T B A C W E K I E M - T A P => V = ±0)7^42"^ = ±4n-V52 - 32 CtfHpC = +167r cm/s cm V = ±0) , (487r2)2 a A^~ —4 = —±4n ' - - ( ^ ; ^ ) = ±16n:cm/s Bai tap Cau 1: Chon phat bieu dung noi ve vat dao dong diSu hoa A Van toe va li ngugc pha B Van toe va gia t6c cung pha C Li va gia toe vuong pha nha D Van toe va gia toe vuong pha • Cau 2: Van t6e ciia chat dilm dao dong diSu hoa c6 Ian cue dai A li eua chat diem eo Ian cue dai B li eua chat diem bang khong C gia toe eua chat diem c6 Ian cue dai D pha cua dao dpng e\rc dai i Cau 3: Mot vat dao dpng dieu hoa vai phuong trinh x = 6cos(4;:r)em, ehilu dao cua vat la: dai quy A 4cm B 12cm C 4m D 6m Cau 4: Mot chdt dilm dao dpng dieu hoa vai phuang trinh x - 5cos(4;rt) cm S6 dao dpng vat th\rc hi?n duprc s la: § A 10 B.20 C.5 D 100 Cau 5: Mot chat diem dao dpng dieu hoa vai phuang trinh x = 5cos(2;zf) gm, li dp cua chat dilm tai thai diem t = 1,5 s la: A 1,5cm B.-5cm C 0cm D.-6cm VAT lY CHIfttWe 1: BAO D$l»6 Cd HgC THUYfTVA I Z - T I J M T A T LY BAI TAP TBAC WSHlgM - TAP Cau : V^n toe cua mpt ehat diem dao dong dieu hoa c6 bieu thuc V = 50sin(l0t) cm/s, bien dp dao dpng eua chat d i l m la A 2,5 em B 5cm C 5m D 40cm Cau 7: (De thi tot nghiep nam 2009) Mot chat dilm dao dong dieu hoa tren true Ox theo phuong trinh x = 5cos(4;Tt) (x tinh bSng cm, t tinh bSng s) Tai thai dikm t = 5s, van toe cua chdt diem c6 gia tri bSng A 5cm/s B 20;Tcm/s C.-20;rem/s VAT LY 1Z - TdM TAT LY THUYfl VA B A I TAP TRAC UGHIgM - CHifams : BAD oOwGCdHqc TAP Cau : Mpt vat nho dao dpng dieu hoa vofi phuong trinh l i dp X = Scos (107rt)(x tinh bang em, t tinh bang s) Khoang cdch tir vj tri can bang den vi tri ma van t6c c6 dp Ian 40;i cm/s la • A ± c m B.±3cm C 3em Cau : Mpt vat nho dao dpng dieu hoa vai phuong trinh l i dp :t = Scos (107rt),(x tinh bang cm, t tinh bang s) G6c tpa dp tai vi tri can bang Dp Ian v ^ toe cua vat v$t c6 l i dp 4cm la D Oem/s A SOTT cm/s A 16;r m/s D 6n cm/s A 0,1 s D 974,5cm/sl B I s C s D 0,5 s D 8;rem/s C m/s Cau 15: Mpt dao dpng dieu hoa tren quy dab dai 40 cm Khi a vj tri x = 10 cm v$t C O van t6c ZOTTVS cm/s Chu ki dao dpng cua vat la: B 4;rcm/s C 16 ;r cm/s B 3n C 60n cm/s Cau 8: Mot chat diSm dao dong dieu hoa tren true Ox theo phuong trinh x = 4eos(47rt) (x tinh bing cm, t tinh bang s) G6e toa dp tai vi tri can bang Do Ion van t6c cua chk dikm vat qua vi tri can bang D ± c m V A N D E 2: T H O f I G I A N , P H A B A N D A U T R O N G D A O Cau : Mot vat dao dong dieu hoa vai phuong trinh x = 6cos(4;7?)cm, gia D O N G D I E U H O A toe cua mpt vat tai thai diem t = 5s la: a Thai gian dao dgng dieu hda dugrc xdc dinh b&ng sa sau A -947,4cm/s^ B 947,4cm/sl T Cau : Mot chat di6m dao dpng dieu hoa vai chu ky 0,5;r(s) va bien dp 2cm Chon goc toa tai vi tri can bang, \ky = 10 Gia toe cua chat diem tai vi tri c6 l i Ian nhat eo dp Ian A.4cm/sl B 16cm/sl C 32cm/sl D 0,5cm/sl B 10cm C 15cm D 20cm -A CO + - T = ^2 A' CO ^ a^ C - v + - = A ' CO CO B D a' v^ - T + CO^ co^ 2 CO V + a CO = A^ T ^/— VTCB 2 =r- -t— " V " T T 13 la b Pha ban dau tp: (phirmg trinh dao dgng: x = Acos^Oit + (p)) Cau : (De t h i dai hoc nam 0 ) Mpt vat dao dpng diSu hoa c6 phuong trinh x = Acos(cot + 9) Gpi v va a Ian lupt la van toe va gia toe cua vat H? thiie dung la: A ^ A Cau 1 : Vat dao dong dieu hoa vai van t6c c\rc dai la 20 ; T cm/s va gia t6c cue dai la 4m/s^ LSy TT^ = 10 thi bien dp dao dong ciia vat la: A 5cm T Goc thai gian: t = Vat qua v i tri x = X Q : COSO) , = — A Chieu duomg: v > ^ sincp < 0; chieu am: v < -> sincp > c Co cp = > tr^ng thai ban dau (vgt & dau va dt chuyen theo chieu nao) x(t 0) xo v(t = 0) =vo '\c thai gian liic vat qua X Q va theo chilu J duomg (vo > 0) hoac theo chilu am (vo < 0) WAT It 12-TOM TAT lY THUYgl VA BAI TAP TRAC BEHlgM-TAPI CHlfdNGI: DAB g p G CdHQC tfAT LY 12 - TJM TAT IX THUYfT VA BAI TAP TRAC H6HIEM - TAP CHtftfUE 1: B A O D Q M G C K H Q C a G C thai gian luc vat qua v i t r i can bang theo chidu duang C h u y: • Trong / T vat di dugc quang duomg A b G C thai gian luc vat qua v i tri can bang theo chieu am • Trong vat di dugc quang duomg A c Goc thai gian luc vat dang a v i tri bien duang • Trong vat di dugc quang ducmg hkng A neu xuSt phat tir vi tri bien hoac vi tri can bSng Giai: Bai tap mau: (x tinh bang cm va t tinh bang s) a Tai thai diem t = 0, vat chuyen dgng nhu the nao ? b Tinh thai gian ngSn nhat U t u luc t = dgn k h i vat di dugc quang duong 4cm c a t , > O.Xo b t = 0, 0, XQ = < = d Trong khoang thai gian 5s kk tir luc t = vat di dugc quang COSCp =^=0 A sincp < COSCp = V = sln(p C t = 0, +A,Vo XQ = d t = e t = tvr luc t = +1) a t = O.Xf) = Ovkvo cm f t = = -AOncm/s = ^ , V Q > = » coscp h s = A = c m => Vat d i tir v i t r i can bang den v i t r i bien 0,05 s t = 257 -» s = 25 X 4/1 = 0 c m = m so lan chat diem qua v j tri c6 x=2cm ^ , la 10 X = 20 (trong chu k i chat diem qua V T C B hai Ian) Bai 2: Xac djnh pha ban dau (p cac truang hgp sau (Goc tga dg tai vi tri can bang) -1 A => q) = n ^ = cosq) = - C A ^ = - 2=^(p sinq) > = - ^ Cau : Bi6u thuc l i dg theo thai gian ciia mgt vat dao dgng diSu hoa la x = Acos (at + (p) Chgn dieu kien ban d i u nhu thS nao d l phuomg trinh cua no 0,2 0,2 sinq) = + ^ , l o < => r ' " " CO dang dan gian la x = Acos e - = — = = » t = lOT 7" Bai tap T 0,Xo ^ < 0: Vat qua v} t r i canbSng theo chieu am (ngugc chieu duomg) d - = — = 25 = ,n sinq) < G i a i : x = 4cos (lOnt = -= 0,^0 _ sinq) = duong la bao nhieu? e T i n h s6 Mn vat qua l i dg x = 2cm khodng thai gian 2s kg n = _ £o _ COS(p = -J = A -A,Vo 0,Xo > => {cos(p = T = ( Sau khoang thai gian ngSn nhat la bao nhieu d l vat c6 l i dg x = -2cm =^t va chuygn dgng cung f Goc thai gian luc vat a v i tri +A/2 va dang chuyen dgng vk v i t r i can bang • Trong - T quang duomg ngSn nhat ma vat di dugc la A Mot vat dao dgng diSu hoa c6 phucmg trinh x = 4cos(107rt +1) e Goc thcri gian luc vat qua v j t r i ^ chigu duang • Trong i T quang duomg dai nli4t ma vat di dugc la A Bail: d Goc thofi gian luc vat a v i tri bien am {(oi) A Goc tga dg trung v o i v j t r i can bang B t = vat qua v i tri can bang va dang chuyen dgng theo chihu duong C t = luc vat bat dau chuyen dgng t u bien duomg D t = vat qua v i t r i can bang va dang chuyen dgng ngugc chieu duong CHIfdNG : DAB flflNE Cfl HgC VAT lY 1Z - TDM T A T LY THUYfT VA B A I TAP TRAC HGHIEM-TAPI Cau 2: Vat dao dpng dieu hoa c6 quy dao la 2cm, liic t = dugc chon la luc qua chu di qua vi tri c6 l i dp x = 0,5cm va dang chuyen dgng theo chieu duong Pha ban dhu cua vat la: IZ-TOMTAT THUYfTVA B A I TAP TBAC NGHIfM - TAP CHIflflllB : DAO OfiNG Cd HQC Cau 9: Mpt 15c 16 xo c6 dp cung k =100N/m, vat nang khoi luong m = 250g, dao dpng dilu h6a vai bien dp 4cm hiy to = luc vat a vi trf bien thi quang duong vat di dupe thai gian 7r/40 (s) dau tien la A 12cm B 8cm C 4cm D 24cm Cau 10: Thai gian ngSn nhSt de mpt vat dao dpng dieu h6a vai phuang Cau 3: (De thi cao dSng nam 2009) Mgt chat dilm dao dong dieu hoa c6 phuong trinh van t6c la v = 4;rcos2;rf(cm/s) Goc tpa dp a vj tri can bang M6c thai giandugc chon vao luc chat diem c6 l i dp va van toe la: A x = 2cm;v = trinh x = Acos(;rt) cm di tir vj tri xi = — den vj tri X2 = A la: A i s B x = 0; v = 4;rcm/s C B.is C.is > D.is D x =0; v = ATT cm/s X = -2cm; v = Cau 4: Mot \hc 16 xo dao dong theo phuang ngang vai phuang trinh: (cm) Thai gian ngin nhat tir luc t = den thai diem vat X = lOcos V CO VATIY li -5cm la V A N D E 3: C O N L A C L O X O • Mgt Id xo CO dQ cung k, rriQt dau dirfrc treo vao rngt diim c6 dink ddu Igi treo vgt c6 khSi luQng m Bo qua khoi hf(mg cua Id xo • A B C.^s 12 J).-s Cau 5: Mot iSc 16 xo dao dong diSu h6a vai phuang trinh x = cos(47rt) (cm) Quang duong vat di dupe thai gian 3s ke tir liic t = la A 16cm B 3,2m C 6,4cm D 0,96m Cau 6: Mot vat dao dong d'lhu h6a vai phuang trinh x = Acosnt (cm) G6c tpa dp tai vi tri can bang Thai gian ngin nhat de vat di tir vi tri bien vh vi tri can bang la: A 2s B Is C.0,5s D 0,25s Tdn sd goc, chu ki, tdn sd: k_ T = 2n m B Sau thai gian T/2, vat di dupe quang duong 2A C Sau thai gian T/4, vat di dupe quang duong A f = ± k_ ' In m 4t Con l i e 16 xo treo thkng dung (Goc tQa tai vj tri can bang, chieu duomg huangxuSng) •1 A Ae' Cau 7: (De thi cao dang nam 2009) Khi noi vh mot vat dao dpng diSu hoa CO bien dp A va chu ky T, vai m6c thai gian t = la luc vat a vi tri bien, phat bieu nao sau day la sai: A Sau thai gian T/8, vat di dupe quang duong 0,5A k V •0 (VTCB) X Dp bien dang cua 16 xo a VTCB: mg D Sau thai gian T, vat di dupe quang duong 4A Cau 8: Mot vat dao dpng dilu h6a vai chu ki T Neu chpn goc thai gian t = luc vat qua vi tri can bang, thi nua chu ki dau tien, van toe cua vat bang khong a thai diem A T/4 B.T/2 C.T D 3T/4 ^ Chi^u dai cua 16 xo khl v^t a vi tri can bang: = /Q + ^ ^ Chieu dai cua 16 xo vat a vj tri x bat ki: |/ = /o+A/ + x V A T LY 12 - TtM TAT LY THIIYgT V A BAI T A P TBAC NGNIEM - * ^ CHKdMG 1: DAO OQNGCdHOC ^ Chieu dai cue dai: T u ( l ) v a (2): / _ ax *~ mm Cau 1: Mot lac 16 xo dao dong diSu h6a vai chu ky 0,5 s, vat nang c6 kh6i luong 400g (lay TT^ = 10) Dp cung cua 16 xo la (2) CHtflfllBI: DAI Blue rtHjC B a i t a p 3; (1) Chieu dai toi thieu: tfAT lY 12 - TOM T A T LY THUYfT V A BAI T A P TBAC WBHIgM - T A P A.0,156N/m ^min B 32N/m C 64N/m D 6400N/m Chu y: Cac cong thiic tren dSu ap dung cho lac 16 xo nam ngang nhung A/ = B a i t a p m a u : Mot iSc 16 xo c6 chiSu dai ban dSu h = 20cm treo thing dung, dau dual gin vat c6 kh6i lucmg m =100 g, dp cung k = 100 N/m LSy g = lOm/s^, TT^ « 10 Bo qua mpi l\rc can a Tinh dp bien dang cua 16 xo va chieu dai cua 16 xo vat a vj tri can bSng Cau 2: (De thi cao dSng nSm 2007) Mpt lie 16 xo g6m vat c6 khoi lupng m va 16 xo c6 dp cung k khong doi, dao dpng dieu hoa N6u kh6i lugmg m = 200 g thi chu ki dao dpng cua lie la s Da chu ki lie la s thi khoi lupnng m bftng A 200 g B.lOOg C.50g D 800 g Cau 3: Mpt lie 16 xo gom vat c6 kh6i lupng m va 16 xo c6 dp cung k, dao dpng dieu h6a N I U tSng dp cung k len l i n va giam khoi lugrng m di Ian thi tan so dao dpng cua vat se b Xac dinh tan so goc, chu ki, tan so A tang l i n B giam ISn c Tir vi tri can bSng keo vat xu6ng theo phuorng thing dung mot doan 5cm r6i tha nhe Goc tpa dp tai vj tri can b3ng, chilu ducmg la chieu keo vat, Goc thoi gian luc tha vat V i l t phuomg trinh dao dpng cua v^t nang C g i a m l i n D.tang4ian d Tinh chieu dai cua 16 xo vat c6 v ^ toe 40Tr cm/s va dang a phia tren vi tri can bang e Tinh chieu dai c\rc dai, chieu dai cue tieu cua 16 xo Giai: m = 100^ = 0,1^5, a Ai = ^ h 0) = = ^ k = lOON/m, = 0,01m = \cm,U ^ = IOTT rad/s, lo + M T = 0,2 s,f = = 21cm A 1,4s SHz c Phuong trinh dao dpng: x = Acosicot + (p) d A^ = x^+-,=^x (p = 0-* x = cos(107rt) cm = ± A^-^= A 160cm/s ±3cm V i vat n i m tren vj tri can bSng nen x = -3 cm l = l^^ e + x = 21 + ( - ) = 18cm lmax = lcb +A 16cm = 21 +5 = 26cm, Imln = lcb-A B,4cm C 4>/3cm D 10>/3cm = 21-5 B.2s C.2,8s D.4,0s CSu 6: Mpt lie 16 xo c6 kh6i lupng m = 0,4kg gin vao dSu 16 xo c6 dp cung 40N/m Nguai ta keo qua nang khoi vj tri can bing mpt doan 4cm roi tha nhe cho n6 dao dpng Van t6c c\rc dai cua vat nang la: m A = 5cm, w = IOTT rad/s, A 16cm Cau 5: Khi gin qua nang mi vao mpt 16 xo, no dao dong vai chu ky Ti = 1,2s Khi gan qua nSng m vao 16 xo do, no dao dpng vai chu ky Tj = 1,6s Khi g i n dong thai mi v^ m v^o 16 xo thi dao dpng cua chung la: = 20cm = Cau 4: Mpt lie 16 xo gom 16 xo c6 dp cung 20 N/m va vien hi c6 khoi lupmg 0,2 kg dao dpng dilu hoa Tai thoi dilm t, van t6c va gia toe cua vien bi Ian lupt la 20 cm/s va 2^ m/s^ Bien dp dao dpng cua vien bi la = B.80cm/s C.40cm/s D.20cm/s CSu 7: Mpt lie 16 xo gom mpt qua n3ng c6 khoi lupmg 1kg va mpt 16 CO dp cung 1600N/m Khi qua nSng a vi tri can bing, nguod ta truySn cho mpt van t6c ban dau bing 2m/s Bien dp dao dpng cua qua nSng la A 5m B.5cm C 0,125m D 0,125cm CHlflfHei:DAOD()HECtfHqC WAT i t K - TJM TAT I * THDYfT VA BAI TAP THAC WEHigM - W Cau : (De thi cao ding nam 2009) Mpt idc 16 xo treo thSng dung dao dong dieu hoa voi chu ky 0,4s Khi vat a vi tri can bang, 16 xo dai 44cin Lay g = ( m/s^) ChiSu dai t\ nhien cua 16 xo la: A 36cm B 40cm C 42cm D 38cm Cau 9: Mpt idc 16 xo treo thing dung va dao dpng dieu hoa voi tan s6 f = 4,5 Hz Trong qua trinh dao dpng chieu dai cua 16 xo bien thien tir 40 cm den 56 cm Lay g = 10 m/s^ Chigu dai tu nhien cua 16 xo la A 46,8 cm B 48 cm C.40cm D 42 cm Cau 10: Mot vat dao dong dieu hoa qua vi tri can bang v§t c6 van t6c V = 20 cm/s Gia toe e\rc dai cua vat la amax = 2m/s^ Chgn t = la luc vat qua vi tri can bang theo chieu am cua tr\ic to? dp Phuorng trinh dao dpng cua vat la: A x = 2cos(10t) em B x = 2eos(10t + 7c) em C.x = 2eos(10t-7i/2)em D x = 2cos(10t + 7i/2) cm V A N D E 4: D O N G N A N G , T H E N A N G V A C O N A N G 4i DgngnSng: Wa =-wv^ =-7ri4Vsin^(fflr+9?)=-fc4^sin^(fflr+^) !2 2 The nang: W, = ^fcc^= ^ iU^cos^ {cot + (p) ^ C»nang: W = + W, = -^^4' =^mo)'A^ Chuv: • • • • vAl LY 12 - TOM TAT LY THUYgT VA BAI TAP TBAC HEHIgM - TAP CHtftfHGlrBAO flOWGCUHqC Bai tap mau: Mot lac 16 xo treo vai vat nang m = lOOg dang dao dong dieu h6a voi phuang trinh x - 2cos(10nt)cm G6c thi nang va g6c tpa dp tai vi tri can bang Lay g = n^ = 10 m/s^ Tinh: a Dp cung k ciia 16 xo THLf VIEN TINHBINHTHUAN b Tan so cua dpng nang c Co nang cua lac d The nang va dpng nang t = s e Dpng nang cua lac vat nang qua vi tri c6 li dp x=lcm f Ti s6 dpng nang va thi nang vat c6 van t6c v = lOn cm/s g Khoang each tu vi tri can bang din vi tri c6 li dp ma dpng nang bang Ian the nang Giai; m = 100^ = OMg.x = 2cos(10nt)cm a = m j = 0,L(107r)2 = lOON/m fdQngnSng = "^fdaodqug = 2.5 = lOHz W = \kA^ = \ 100.0,022 = 0,027 d t = I s -> X = 2cm -> M^t = J/f^^ = 0,02/, W^^W -Wt = Q e WA = W-Wt= -^kA" -\kx'' = 0,02 - i 100.0,012 = 0,015, f = \mv^ = io,l(0,l7r)2 = 0,005/ -» M^^ = 0,015/ ^ = i g = 3Wt W = 4Wt =^ Wt = => \x\ 2,5 cm Bai tap ; Cau 1: Mpt vat dao dpng digu hoa theo phuang trinh: x = lOcos (4nt + n/2) cm Dpng nang cua vat b i k thien vai tan s6 la A.4Hz B.2Hz C IHz D 6Hz Cau 2: Mpt vat dao dpng dieu hoa theo mpt tryc e6 dinh (m6c thi nang a vi tri can bang) thi W = Wt(rruxx)/3 cos(5;r/)(c/w) v^ = sVJcosf5;r/ ](c/«) 3J CHliflHB1:DAOD0HeCdHqG I * 12 - T J M TAT iV THUYft VA BAI TAP TBAC WfiMieM - T*LL Bien dao dong t6ng h(?p cua hai dao dpng tren bang A cm B cm cm D S cm Cau 114: (De thi d^i hQC nSm 2010) Dao dgng t6ng h