Tài liệu Bí quyết ôn luyện thi đại học đạt điểm tối đa vật lý tập 1 - lê văn vinh

ThS. L EVAN VINH G I A O V I E N C H U Y E N L U Y E N T H I D A I HOC Bl QUYET ON LUYEN THI DAI HOC DAT DIEM TO'I DA VAT L I TAP 1 THEO TtfNG o n j Y f i N D 6 vA GiAi a n TI6T Bien soon theo cau true de thi cua Bo Giao Due va Dao too (Tdi ban tan thA nhcU c6 sita chSa va bo sung) - Tuyen tap cac bai toan cd ban, hay la va kho - Tong hgfp cac phiiang phap giai nhanh nhat khi lam bai trac nghiem - Sach danh cho hoc sinh luyen thi Dai hoc - Cao dang NHA AUAT BAN DAI H O C QUOC GIAH A NQI NHr^ x u n r B A N D R I H O C O U O C c m Hf^ N O I 16 H d n g C h u o i - H a i B d TrUng - H d N p i D i e n t h o a i : B i e n t a p - C h e b a n : (OA) 3 9 7 1 4 8 9 6 : H d n h c h i n h : C04) 3 9 7 1 4 8 9 9 ; T o n g b i e n t a p : ( 0 4 ) 3 9 7 1 4 8 9 7 L a m g i de c6 d u o c d i e m 10 m o n Vat ly? De thoi, neu d o la d i e m 10 t r o n g t h i tot nghiep. Vay t h i cao dang t h i sao! Co de n h u tren khong? N o i c h u n g la cung Fax: (04) 3 9 7 1 4 8 9 9 k h o n g k h o l a m . The con d i e m 10 trong de t h i dai hoc t h i sao! C o k h o l a m Chiu trdch nhiem khong? T a i day chac nhieu ban da c6 cau tra l o i r o i ! D a so nhieu ban cho rang xuat ban dieu nay la qua k h o va k h o n g the l a m duac. T u y nhien ben canh d o cijng c6 Gidm doc - Tong bien tap : TS. PHAM THj TRAM nhieu ban cho rang cijng k h o n g c6 g i kho l a m va hoan toan c6 the l a m dugc neu CO n h i r n g b i quyet o n luyen t h i bo ich truoc k h i buoc vao p h o n g thi. C u o n Bien CONG TY KHANG VIET Che ban Trinh sach nay la d a n h cho cac ban n h u the va ciang danh cho cac ban c6 nit.:-,i t i n la THU LUYEN tap CONG TY KHANG VIET bay bia m i n h se l a m d u o c dieu m o n g uoc do. Co so de t i n vao dieu nay la t r o n g k h u o n kho cua de t h i k h o n g c6 cau nao qua k h o n h u ben de t h i toan. Vay cai k h o 6 day la gi? D o chinh la t h o i gian l a m bai. T r o n g de t h i c6 tong cong 50 cau n h u n g chi c6 90 p h i i t v i the chia ra t h i 1 cau chi c6 108 giay (1,8 p h i i t ) . Vay b i Tong phdt hdnh vd dot tdc lien ket xuat ban: C O N G T Y TNHH MTV DjCH V g VAN HOA K H A N G V I | T Dia ch?: 71 Dinh Tien Hoang - P.Da Kao - Q.1 - TP.HCM Dien thoai: 08. 39115694 - 39105797- 39111969 - 39111968 Fax: 08. 3911 0880 Email: khangvietbookstore©yahoo.com.vn Website: www.nhasachkhangvlet.vn quyet nao d e g i a i bai toan k h o khan 6 tren? Bi quyet d a u tien de c6 d i e m 10 la phai bie't phan loai cap d o cau hoi va sau do de l a m truoc, k h o l a m sau. Bi quyet t h u hai la phai lam duoc n h u n g cau k h o dugc p h a n loai 6 tren trong thai gian cho phep. De l a m dugc dieu nay d o i hoi cac ban phai giai dugc that nhieu bai tap va phai phan ra t u n g chuyen de cu the de bie't bai toan giai quyet theo h u o n g nao. N h u vay de giai quyet cac bai tap thugc d a n g k h o nay cho nhanh va c h i n h xac nhat la n h i i n g g i cuon sach l a m dugc. V i the'de thuc hien dugc b i quyet t h u hai t h i can l a m dugc n h i i n g viec sau: 1. Su d u n g thanh thao m o i lien h? giira dao d o n g dieu hoa v o i chuyen dgng tron deu va cong cu de l a m dugc viec tren chinh la vong tron lugng giac. K h i SACH L I E N K E T sir d u n g v o n g t r o n l u g n g giac, cac bai toan n h u ve pha, thai gian, quang Bf Q U Y E T ON LUYEN THI DAI HQC DAT DIEM TOI DA VAT U - d u o n g ... dugc giai quyet true quan de hieu va it t o n t h o i gian (thi trac nghiem TAP 1 can dieu nay nhat). 2. Phai s u d u n g tot kien thuc h i n h hgc phang de t i m d o I o n cac dai l u g n g M a so: 1 L - 5 3 1 D H 2 0 1 3 Ma so ISBN: vecto. T r o n g c h u o n g dao d g n g co t h i can n a m va ap d u n g d u g c d i n h ly h a m so 978-604-934-857-0 In 2.000 cuon, kho 1 6 x 2 4 c m sin va cosin de t i m bien d o va pha ban dau cua dao d g n g tong h g p . Ben - • c h u o n g d o n g dien xoay chieu la can nhat v i cac h i n h phang 6 day dugc ve t u Tai: Cty TNHH MTV IN AN MAI THjNH DLfC phuang phap vecta trugt. D e giai tot cac bai tap t r o n g c h u o n g nay p h u o n g Dia c h i : 7 1 , Kha Van ^ a n , P. Hiep Binh Chanh, Q . Thu Dufc, TP. H o Chi M i n h So xuat b a n : 2033 - 201 3/CXB/03 - 2 8 4 / D H Q C H N ngay 31/12/2013. Quyet d i n h xuat b^n so: 5 4 3 L K - T N / Q D - N X B D H Q G H N , cap ngay 31/12/2013 In xong va n o p lOu c h i e u quy I n a m 2 0 1 4 phap vecto t r u g t k h o n g the k h o n g bie't t o i . ' 3. Sir d u n g thanh thao may t i n h cam tay (Fx 570 ES hoac Fx 570 ES PLUS). T r o n g t i n h toan b i n h t h u o n g ma cac ban k h o n g c6 thao tac n h a n h tir chie'c ma> tinh thi se mat kha nhieu thoi gian. Ngoai ra, chirc nang ciia cac chie'c may t i n t vugt nay con xa su mong dgi cua chiing ta, no c6 the giai ne'u dua ve chedo phuc. dugc nhieu bai toan ^hdn 1: DAD B O N G CII^U HdA XO LY T H U Y E T 4. Nam chac cau true de thi de on diing trpng tarn, khong lang phi thoi gian I. D A O D O N G C O 1. Thendolddaodongca? on nhiing kien thiic khong ra thi. Ne'u cac ban hpc ben ban co ban hoac khong muon hoc chuong chuyen dong ciia vat ran ben ban nang cao thi cac ban nen Dao dpng co la chuyen dpng qua lai quanh mpt v i tri dac bi§t gpi la vj tri can bang. chQn ban co ban ma thi. De thi cho chiing ta chon mpt trong hai phan nhung chiing ta nen chpn ngay tix dau l u y f n thi chii khong de vao phong thi moi 2, Dao dong tudn hodn chpn. Ne'u cac ban chac chan chpn ban co ban (da sochpn ban nay) thi cac ban khong hgc cac kien thiic sau: nguyen chuang chuyen dpng cua vat ran, nguyen - Dao dpng tuan hoan la dao dpng ma trang thai chuyen dpng ciia vat dupe lap lai nhu cij (vi tri cii va huang cu) sau nhiing khoang thoi gian bang nhau. - Dao dpng tuan hoan don gian nha't la dao dpng dieu hoa chuong tix vi mo den vT mo. Khong hoc phan hieu ling Doppler, con lac vat ly> cac bai toan ve mac hinh sao, tam giac trong dong dien xoay chieu 3 pha, cac bai toan ve thoi gian dao dong, so dao dpng, quang duong vat dao dpng dupe trong dao dpng tat dan, cac cong thiic Einstein trong chuong lupng t u anh II. P H l / O N G T R I N H C U A D A O D O N G D I E U HOA sang, cac bai toan ve thay doi chieu dai, thoi gian trong thuye't tuong doi, cac 1. Vtdu bai toan giao thoa anh sang bang cac dung cu quang hpc. Cac phan nay chi ra - Gia su M chuyen dpng ngupc duong van toe goc la co, P la hinh chieu bi phan tam va mat thoi gian. ciia M len Ox. 5. Nho cong thiic. Chiing ta khong the diing cong thiic nao ciing d i chiing y chieu ben ban nang cao v i the chiing ta phai loai bo ngay cac kien thiic nay de khong / cot -fj Tai thoi diem t = 0, M co tpa dp goc (p minh ma tot nha't la trong qua trinh hpc hay chiing minh va nho no de lam bai Sau thoi diem t, M co tpa dp goc (ro.t + cp) t^p moi nhanh dupe. Khi do: OP = x => diem P co phuong trinh la: 6. Phai biet tinh nham cac bieu thiic tinh co ban nhu cos—= 0;cos —= —... 2 3 2 nhfmg cai nay nham nhanh hon may tinh nhieu. Phai thanh thao may tinh cam tay nhung dung de phu thupc vao no. 7. T u tin va kien tri on luy^n, nha't dinh cac ban se thanh cong. Tac gia X = OMcos(cot + (p) - Nhd Sdch Khang Viet xin trdn trgng gi&i thi^u tai Quy doc gid vd xin lang nghe tnoi y kien dong gop decuoh sdch ngdy cdng hay han, boich han. - Email: [email protected] Do ham cosin la ham dieu hoa nen diem P dupe gpi la dao dpng dieu hoa Dao dpng dieu hoa la dao dpng trong do l i dp ciia vat la mpt ham cosin (hay sin) ciia thoi gian. 3. Phuong trinh - Cty T N H H Mpt thanh vien - Dich V u Van Hoa Khang Vi?t. Tel: (08) 39115694 - 39111969 - 39111968 - 39105797 - Fax: (08) 39110880 ' ' 2. Dinh nghia ve: 71- D i n h Tien Hoang, Phuang Dakao, Qu|n 1, TP H C M . Dat A = O M t a c 6 : x = Acos(co.t + (p) Trong do A, w, cp la hang so Le Van Vinh Thuxingtti V A CDN LAC L d 4. Phuang trinh x = A cos(co.t + (p) gpi la phuong trinh ciia dao dpng dieu hoa * A la bien dp dao dpng va la li dp cue dai ciia vat. (A > 0). * (co.t + (p) la pha ciia dao dpng tai thoi diem t * cp la pha ban dau tai t = 0 ((p > 0;(p = 0;(p < 0) Chuy a) Diem P dao dpng dieu hoa tren mpt doan thang co the coi la hinh chieu ciia diem M chuyen dpng tron deu len duong kinh la dogn thang do. b) Ta quy uoc chon true x lam goc de tinh pha aia dao dong va chieu tang cua pha tuong ling voi chieu tang ciia goc MOP trong chuyen dong tron deu. I I I . CHU K I , T A N SO, T A N SO GOC CUA DAO D Q N G DIEU HOA 1. Chu ki va tan so Khi vat tro ve vi tri cii, huong cii thi ta noi vat thuc hi^n 1 dao dong toan phan. * Chu ki (T): ciia dao dong dieu hoa la khoang thai gian de vat thuc hien mpt dao dong toan phan. Don vj la s * Tan so (f): cua dao dong dieu hoa la so' dao dong tuan hoan thuc hien trong mgt s. Don vi la 1/s hoac Hz. Neu keo vat khoi vj tri can bang va buong ra vat se dao dong quanh vi tri can bang, gii>a hai vj tri bien V I I . K H A O SAT DAO D Q N G CUA CON LAC LO XO VE M A T D Q N G LI/C HQC Xet vat 6 li dp x, 16 xo gian mot doan Al = x. Luc dan hoi F = - k A l Tong luc tac dung len vat: F = -kx •k Theo dinh luat I I Niu ton: a = x m Dat co^ = k/m => a + co^x = 0 Trong dao dong dieu hoa o) dupe gpi la tan so goc. Giiia tan so'goc, chu ki va tan so c6 moi lien he: Tan so goc: (» = . — m = 27lf T IV. V A N TOC VA GIA TOC CUA DAO D Q N G DIEU HOA 1. Van toe * Chu ki: * Luc keo ve - Van to'e eiing bie'n thien theo thoi gian * Tai * Tai X = ±A thi v = 0 X = 0 thi V = vmax = to.A 2. Gia toe Gia to'c la dao ham ciia van to'e theo thoi gian a = v ' = x" = -co^A cos(a)t + c})) a = -LoH * Taix = Othia = 0 * Tai X = ±A thi a = am.ix = co^A V. DO T H I CUA DAO D Q N G DIEU HOA Do thi ciia dao dpng dieu hoa vai cj) = 0 c6 dang hinh sin nen nguai ta con gpi la dao dpng hinh sin. V I . CON LAC L 6 XO Con lie 16 XO gom mot vat nang m gan vao 1 dau ciia 16 xo eo dp eung k va khoi lupng khong dang ke. Dau eon lai eiia 16 xo eo dinh. Con lac CO 1 v i tri can bang ma khi ta tha vat ra vat se dung yen mai. F M. Luc huong ve vi tri can bang gpi la luc keo ve. Luc keo ve c6 dp Ion ti le Van toe la dao ham ciia l i dp theo thoi gian: V = x' = -coA sin(a)t + cj)) .j't!:i Vay dao dpng eiia con lac 16 xo la dao dpng dieu hoa. 2. Tan so goc CO = ' voi li dp va gay gia toe cho vat dao dpng dieu hoa. V I I I . K H A O SAT D A O D Q N G CUAA LO XO VE M A T N A N G L l j Q N G 1. Dong nang cua con lac Id xo: W^j = ^ mv^ 2. The nang eua con idc Id xo: Wt = - kx - . T The'nang va dpng nang ciia con lac 16 xo bien thien dieu h6a voi chu ki — 3. Co nang eua con idc Id xo. Su bdo toan eo nang 2 2 2 2 Co nang ciia con lac ti le voi binh phuong voi bien dp dao dpng Co nang ciia con lac 16 xo dupe bao toan neu bo qua mpi ma sat. N h a n x e t q u a n t r o n g : dis'dat diem cao phan nay thi hi quyet dau tien la cdc ban phdi su dung dwgc duang tron htgng gidc thuan thuc vd sau day la hi quyet thu nhat. ' ''''' "' 5 S a u d a y l a p h u o n g phap khac rat t r y c quan, the hien d u g c mo'i q u a n he Chuyen de 1 giira cac dai l u g n g n h a m g i i i p cac ban t r o n g vi#c giai n h a n h nha't, c h i n h xac nha't cac dang toan ve dao d o n g co, song co, d o n g d i e n xoay chieu va dao D U O N G T R 6 N LUgrNG GlAC TRONG DAO D O N G Bltu H6K d g n g t r o n g mach L C . co the n o i rang: p h u o n g phap d i i n g d u o n g t r o n l u g n g I. KHO KHAN KHI G I A I BAI TAP: giac quye't d i n h I o n den viee dau hay rot ciia cac ban. So l u g n g cong t h u c yeu cau cac ban n h o van d y n g t r o n g c h u o n g dao d g n g II. PHl/CNG PHAP CO rat n h i e u chi t i n h p h a n to d a m , bat bugc la 16 cong t h i i c n h u n g v o i so' Bieu d i e n ca ba h a m l i d g (x), v a n toe l u g n g cac cong thuc d o cung chi giai quye't d u g c cac cau h o i rat co ban, (v) va gia toe (a) va k h i can ta co the k h o n g the giai quye't d u g c he't cac dang bai tap dat ra cua c h u o n g nay. O bieu d i e n lire tren c i i n g m g t d u o n g phan dao d g n g kie'n thuc toan lien quan la cac cong t h u c l u g n g giac va giai cac p h u o n g t r i n h l u g n g giac day la kho khan I o n do'i v o i da so' cac ban ke ca cac ban kha g i o i v i ra't hay sot n g h i ^ m b o i t i n h lap lai cua h a m t u a n hoan. t r o n l u g n g giac n h u sau: + L i dp: X = A cos(cot + cpx) la h a m cosin D u o n g t r o n l u g n g giac v o i vi^c lien he giira chuyen d g n g t r o n deu v o i dao => c i i n g chieu true cosin co chieu (+) d g n g d i e u hoa se giai quye't n h i i n g kho khan tren m o t each de dang. tir trai sang p h a i v o i bien d g la => x^^^^ = A H i e n tai tren d u o n g t r o n l u g n g giac da so' chi s u d u n g m o t true cosin cho + p h u o n g t r i n h dao d g n g x = A cos(cot + (p^) (true Ox) va cac dang toan c h u o n g V a n toe tijfc thai: v = -A(osin(cot + cpx) la h a m trir sin => ngugc chieu true sin nen co chieu (+) h u o n g tir tren xuo'ng v o i bien d g nay t h u o n g can c i i vao cac d\x kien bai toan cho t u p h u o n g t r i n h dao d g n g v^ax^^-'^- d a n g X = A cos(cot + (Px), de t i m chu k i , tan so', d u o n g d i , k h o a n g t h o i gian D i e u nay tuong duong voi ham v = Acocos(cot + (Pv) voi de d i t u toa d g xi den toa d o xi, t i m van toe, gia toe tai m g t t h o i d i e m nao do, khoang t h o i gian 16 xo nen, gian ... T u y nhien se kho khan cho cac ban k h i gap phai loai cau h o i dv! kien bai + => ngugc chieu true cos co h u o n g (+) tir phai sang trai v o i bien d o a^^ =cf^A toan k h o n g cho p h u o n g t r i n h dao d g n g dang l i do x = Acos((ot + cp^) ma cho dang v a n toe tue t h a i v =-A(osin(cot + (Px)hoac cho d a n g gia toe tiic D i e u nay t u o n g d u o n g v o i ham: a = a)^Aeos(cot + (Pa) v o i (pg = (p^, + ^ = (px + Ji t h o i a = -(a'^Acos((ot + (Px) • Luc nay hau he't cac ban deu b i d g n g k h o n g the bieu d i e n h a m (v) va h a m (a) tren d u o n g t r o n l u g n g giac. T h o n g qua each bieu d i e n nay ta tha'y mpt so diem dac bi?t, vung dac biet M u o n bieu d i e n d u g c tren d u o n g tron l u g n g giac t h i p h a i t u h a m (v), (a) va m o i q u a n he ve pha ciia l i d g (x), van toe (v), gia toe (a) cung n h u viee vie't lai d a n g h a m (x) bang each lay tich phan bac nha't h a m v a n toe (v) hoac khai thae cac kien thuc l y thuye't lien quan ve dao d g n g dieu hoa, cac dang bae 2 h a m gia toe (a) day la each rat kho k h a n cho cac bah v i sang hge k y 2 nang l u g n g ciia dao d g n g dieu hoa dugc the hien m g t each true q u a n tren cac ban m d i d u g c hge n g u y e n h a m va tich phan. Vay giai phap nao c6 the h i n h ve v o i m g t vai v i d u sau: giai quye't cac k h o k h a n neu tren? + Bon vi tri dac biet: N h i e u y kien cho r i n g : N e u m u o n tranh dieu nay t h i p h a i n h o h a m van toe • V i tri bien duong I: (x^^x = A ; v = 0; a = -co^A) => The'nang cue d a i , d g n g nang cue tieu (v) som pha h o n l i d g (x) 1 goc ^ , con h a m gia to'c (a) ngugc pha v o i h a m l i d g (x) t u y nhien, vi^c giai cac p h u o n g t r i n h l u g n g giac lien q u a n dieu nay m a t n h i e u t h o i gian, chua m u o n noi d g chinh xac v o i da so cac ban la rat 6 G i a toe tuc thoi: a = -a)^Aeos((ot + cpx) la h a m trir cosin (ngugc h a m x) - V i tri can bang I I : (x=0; v = -coA; a=0) tha'p. K h o n g the n h o he't cac cong thuc, cac m o i quan he p h u c tap cua cac => The'nang cue tieu, d o n g nang eye dai dai l u g n g co hge, v i thie'u t i n h true quan, thie'u m o i q u a n h ^ gSn bo giira cac V i tri bien am I I I : h i ^ n t u g n g vat ly nen t h u o n g tra l o i sai cac cau h o i d u co ban nha't. (x = - A ; v = 0 ; a^^^x =co^A) I I I . CAC DANG TOAN D a n g l : X a c d j n h c a c dai lUcfng li do, v a n toe, g i a t o e t a i thdi d i e m t. => T h e n a n g cue dai, d o n g nang cue tieu - V j t r i can b i i n g I V : ( x = 0 ; v^-,^=wA; a = 0) => Tlie'nang cue tieu, d o n g nang cue dai Kei luan: Vay dm ki dao dgn^ titan hoan cua ham dgn^ nang va ham thenang cua dao dong dieu hoa chi bang ^ chu ki T cua ham li do (x), khodng thai gian dedong nang (the nang) tie cur dai thanh cite tieu hay ngugc lai la ~ chu ki T cua ham li ga v i DU MAU: V I d u 1 : M o t vat dao d o n g dieu hoa theo p h u o n g t r i n h x = 6eos(27tt)cm, van toe ciia vat tai t h o i d i e m t = 7,5s la: A. V = Ocm/s. B. v = 75,4cm/s. C. V = -75,4cm/s. D . v = 6cm/s. ^hdn tich vd huong ddn gidi do (x). + .. t\. Bon vung dae biet: D u n g true Ox bieu dien : V u n g l : (x>0; lue ban dau vat 6 v j t r i I sau t h o i gian v < 0; a < 0) => vat chuyen d o n g nhanh dan theo chieu (-) v i a.v > O v a the nang giam, t = 7,5s vat quay m o t goc: d o n g nang tang. cot = 271.7,5 = 1571 lap lai 7,5 v o n g V u n g 2: (x < 0; v < 0; a > 0) den v i t r i I I I => co van toe v = 0. => vat chuyen d o n g cham dan theo chieu (-) v i a . v < 0 va the nang tang, C h p n dap an A a II o IV d o n g nang g i a m . V u n g 3: (x < 0 ; V v > 0; a > 0) => vat chuyen d o n g nhanh dan theo chieu (+) v i a.v > 0 va the nang giam, d o n g nang tang. V u n g 4: ( x > 0 ; v>0; d o n g nang g i a m . Mo'i quan he vepha eua li do (x), van toe (v),gia toe (a): Qua h i n h ve nhan tha'y d u g c m o i quan he ve pha ciia ham l i do (x), van toe (v) va gia toe (a) la: 9v = gia toe (a) som pha h o n van toe (v) m o t goc ^ va s o m pha h o n l i do (x) m o t goc 7t hay ngugc pha v o i l i do (x) 2y cm. A . a = Ocm/s2 B. a = 946,5em/s2. C. a = -947,5cm/s2 D . a = -946,5cm/s2. 'Phdn tich vd hu&ng ddn gidi D i i n g true Ox bieu dien De cho h a m x dang sin can chuyen sang cos c6 dang: x = 6cos(47it) cm (Pa = (p^ + ^ - (p, + 71 => van toe (v) sam pha han l i d o (x) m o t goc Vl d u 2: M o t vat dao d o n g dieu hoa theo p h u o n g t r i n h x=6sin gia toe ciia vat tai t h o i d i e m t = 5s la: a < 0) => vat chuyen d o n g cham dan theo chieu (+) v i a . v < 0 va the nang tang, + ; => ban dau vat 6 v j t r i I sau thoi gian t = 5s vat quay 1 goc o)t - 4n.5 - 20n lap lai 10 v o n g den v j t r i cij. => CO gia toe a = -co^A = - 9 4 7 , 5 ^ s C h g n dap an C Vl d u 3 : M o t chat d i e m dao d o n g dieu hoa theo t c6 p h u o n g t r i n h van toe v = 107reos 27rt + - em/s, toa dp ciia chat d i e m tai thoi d i e m t = 1,5s la V 2J _ A . x = l,5cm. B. x = -5em. C. x= + 5cm. D . x = 0cm. 9 Cty n V H H M T V D W H JOtang 'Phdn tich vd hitang dan gidi Bl/ofC 3 : Bieu d i e n dao d p n g dieu hoa tren d u o n g t r o n . Vat d i tir v i t r i X j = 5cm Bien d p A = CO den X2 t u o n g u n g v o i m p t chuyen d p n g t r o n deu d i tir M deh N v a i v a n 271 D u n g true O v bieu d i e n : toe goc CO, ban k i n h la A . L i i c ban d a u vat a v i t r i I sau t h o i gian CO = - A = - 5cm. m C h p n dap an B Vl d u 4 : Van toe cua m p t vat dao d o n g d i e u hoa bien t h i e n theo t h o i gian theo p h u o n g t r i n h v = 27:cos 0 , 5 7 : t - - (cm/s). Vao t h a i d i e m nao sau 6y v i DU Vi d u MAU: 1 : Vat dao d p n g d i e u hoa v o i p h u o n g t r i n h x = Acos(cot + cp) (cm). T i n h t h a i gian ngan nhat vat d i t u : A a) V j t r i can bang den v i t r i x = — . day vat qua v i t r i c6 l i d p x=2cm theo chieu d u o n g cua true tpa d p . A.8/3S B.2/3S ^hdn Bien d o A = ^ ^ CO = — = 0,5Tt C. 2s D . 4/3s tkh vd hii&ng dan gidi 4cm b) V i t r i can bang den v i t r i x = c) V j t r i can bang den v j t r i x = ^hdn (POv sau t h a i gian t v a t quay 1 goc 2 A^f3 tich vd hu&ng ddn gidi A =-7 a) K h i vat d i tir v j t r i can bang den ^ = y , o ^Ov = t u o n g u n g v o i vat chuyen d p n g tren d u o n g t r o n t u M den N d u o c m p t goc a = cot = 0,57rt = — v i c6 l i d p x = 2cm, 3 Acp n h u h i n h ve ben. bien d p A = 4 cm va c h u y e n d o n g theo chieu {+) den v j t r i V I De thay: sin Acp = ^ => Acp = => mat t h o i gian t = 2/3s. C h g n dap an B => K h o a n g t h o i gian ngan nhat de vat d i t u Dang 2: Xac d j n h t h d i g i a n v a t d i tuT vj t r i x j den v i t r i x j : Phi/cTng p h a p g i a i : Cho p h u o n g t r i n h dao d p n g vat c6 dang: X = Acos(cot + (p)cm BUdc 1 : (v;>0; Xac d i n h v j t r i X j tren d u a n g Bl/dc 2: Vj<0; hay >0; Xac d i n h v j t r i X2 tren d u o n g V2 <0; la: At = CO d p n g cua hay V2 = 0 ) . AV2 rad. n 6 T 27t 12 T bang den ^ , -, t u o n g u n g v o l vat chuyen T h a i gian v a t d i t u v i t r i X j den X2 la : At = — . lap lai 1,5 v o n g den v i t r i I I I X «, Slide 4 : Xac d j n h goc cp = M O N . t = 1,5s vat quay 1 goc cot = 2 i i . l , 5 = 3n => CO toa d o Vift vat ^ d u p e m p t goc Acp n h u h i n h ve ben. De thay: sin Acp = — => Acp = - rad. 2 4 ^ V I— =i> K h o a n g t h o i gian ngan nhat de vat d i t u VTCB deh x = — — la: Bi qtiyei on luyen thi dai hoc dat diem toi da Vat It, tap l~Le At = Van Vinh Cty TNHH MTV DWH Tit hang nay, ta sc gicii qiii/ct nhi'mg bai tocin vc thai gian trong dao dong dieu hoa mot each nhanh nlid't neii dccho diem di va diem den dqc bict nhu tren. Acp CO 8 2n T c) K h i vat d i tir v i t r i can b^ng den x = BAI TAP VAN DUNG: p , t u o n g l i n g v a i vat chuyen dpng tren d u o n g t r o n t u M den N duoc m o t goc Acp n h u h i n h ve ben. C a U 1: Vat dao d o n g dieu hoa v o i p h u o n g t r i n h x = Acos(cot + cp) (cm). T i n h : a) Thoi gian ngan nhat vat d i t u v i t r i c6 l i do X| = vat d i t u V T C B den x = At = Acp CO n _3_ 2n I • • A b) Thoi gian ngan nhat vat d i tif v i t r i c6 li do la: X2 = 6 t u o n g u n g vcVi vat chuyen d o n g tren theo chic?Li am. d) Thoi gian ngan nhat vat di tir vj t r i ccS l i dc) x, = -A d u o n g t r o n t u M den N d u o c m o t goc Acp n h u h i n h ve ben. X2 = AV2 den vj t r i c6 l i dc> theo chieu am. De thay: sin Acp = ^ => Acp = ^ rad. 'Phdn tick vd huang dan gidi a) Khoang t h o i gian ngan nhat vat d i tir => K h o a n g t h o i gian ngan nhat de vat d i t u V T C B den x = A la: vi t n X, = AV3 — A den ^2 = y tuong u n g vc>i vat chuyen dcing tren d u o n g _^ ^ tron tir M den N . CO Min t T o m lai: ta c6 bang thoi gian trong dao dong dieu hoa sau: -A den v i t r i c6 l i dc) A X-, = d) K h i vat di t u v i t r i can bang den x = 2 ^\ — ' 2 • AN/2 c) Thoi gian ngan nhat vat d i tir v i t r i c6 l i do x, = - ~ ^ den vj t r i c6 li do 2 Av'3 T Acp At = - den v i t r i c6 l i do A V3 Acp = — rad. ^ 3 => K h o a n g t h o i gian ngan nhat de De thay: sin Acp = Khang Viet o ( 2 A 2 = t Asfd. =1 ~ 6^ T_ 12 >0 + t, 0-^ 2 ; L-1 4 Ta bieu dien cac d i e m tren len true dao d o n g dieu hoa se thay ro h o n T ~A _AS A42 _A o \ T_ 6 •1 L 4 ^ ] r * ' A Ayll A41 T 12 4 1 -x Bf quyei on liiyen ilii dai hoc dat diem tot da Vat It, t^p 1-Le Van Vitth d) Khoang thoi gian ngSn nha't v^t di tu vj tri xi = - A den xj = theo chieu b) Khoang thoi gian ngan nha't vat di tij" vi A ^ tri X j A — den = - —dirong tuongtron imgtixvoi vat chuyen dpngX2 tren M den N. Min t (A A' = t + t 0-+ >0 I2 >—2, I2 y am tuong u n g voi vat chuyen dong tren -A d u o n g tron tu M den N. ^ AV2 ^ = t(~A->0) + t(O^A) + V Aj2] A—•—•— 2 I T 4 ~ 12 ^ 1 2 " 6 Bieu dien len tryc dao dong dao dgng dieu hoa -A O T_ 12 T = —+ — + A 2 L1 2 Ayfl AS A^ A-Jl fT T\ 4 U 8 _A 5T A A-JIAS 6 Tdi day ta da cd cdi nhin mai vedang todn th&i gian trong dao dong dieu hoa, tif day ede bai vequdng duang, toe do trung binh, van toe trung binh... eo the dime gidi qiiyet rat de dang. Ket luan: c) Khoang thoi gian ngan nha't vat di tvr vi tri xi = A>/2 — den X2 = A- Vy 3theo chieu am tuong ung voi vat chuyen dong tren d u o n g tron tu M den N. + t(O^A) + t A>/2 A 7 3 1 = t D a n g 3. B a i t o a n x a c d j n h t h d i d i e m v a t d i q u a v j t r i x d a b i e t ( h o $ c V, a, \Nt, Wd, F) I a n thuT N PHl/QNG P H A P * Trong mot chu ky T (27r) vat di qua x hai Ian neu khong ke den chieu chuyen dpng, neu ke den chieu chuyen dong thi se di qua 1 Ian * Xac dinh Mo dua vao pha ban dau (xo, vo chi quan tarn < 0 hay > 0 hay = 0) * Xac dinh M d u a vao x (hoac v, a, Wi, WJ, F ) 2 - = 1 I fl_ll I I T 12 Bieu dien len true dao dong dieu hoa -A A42 A ^V3 T 8 O A 2 A-Jl A(p Ap d u n g cong thuc t = (0 V Liru y: De ra thuong cho gia tri n nho, con neu n Ion thi tim quy luat de suy ra nghiem t h u N. ^ T 6 T 4 Thai gian dao dong cua vat dugc xdc dinh nhu a tren nhimg hinh ve duai eho edc ban CO cdi nhin true quan hem Cdc loai thucmggap vd cong thiic tinh nhanh - Q u a X k h o n g ke den chieu + N chin: t = ^ ^ ^ T + t2 i h ^hoi gian de vat di qua vi tri x Ian thu 2 ke tu thoi diem ban dau) 15 Bi 4ityet on luySn thi aai ntfcflflfaiem TOi aa var ii, lup I-LB + - N le: t = N - 1 van vmn T + t i (ti t h a i gian de vat d i qua v i t r i x Ian t h u 1 ke t u t h a i N h ^ n xet: each tinh theo cac khoang thoi gian dqt Met tuy trinh bay tren giay thdy nhieu ban nhung thuc tetinh thi rd't nhanh. Khi gidi chiing ta khong can ghi cu the d i e m ban d a u ) md chi viec cong cdc khodng thai gian Iqi thoi. Q u a X ke den chieu (+ ho|c - ) Cung bdi todn tren nhung neu thai diem di qua vat Id rdi l&n thi ta lam nhu vi du sau t = ( N - 1)T + t j ( t i t h o i gian de v a t d i qua v i t r i x theo chieu d a u b a i quy d i n h Ian t h i i 1 ke t u t h a i d i e m ban dau) m vi D U V I dy V l dM 2: M p t v a t dao d p n g dieu hoa v o i p h u o n g t r i n h x = 4cos(47tt + — )cm. 6 T h o i d i e m t h u 2013 vat qua v i t r i x = 2cm. 1 : M p t v a t dao d o n g dieu hoa v o i p h u a n g t r i n h x = 4cos(47it + —) 6 c m . T h o i d i e m t h u 3 v a t qua v i t r i x = 2cm theo chieu d u a n g . A.9/8S B. l l / 8 s C.5/8S ^- -IT' D . 1,5 s Cach 1 : G i a i theo p h u o n g trinh lugrng giac Ta CO x= 2 x = 4cos 4 7 r t + - = 2 6 lv>0^1 V = -167tsin 7l^ 47rt+- 6J = > 4 7 t t + - = — + k27t 6 >0 ^ 24157 C. s 24 3 'Phdn txch vd huang ddn gidi 47tt + — = - + k27: 6 3 t = A + ii (keN) 24 2 ^ ' 47ct + - = — + k27t 6 3 t = - i + Ji ( k e N * ) 8 2^ ' T h o i d i e m t h u 2013 (le) nen ta d u n g cong thuc: t = = > t = - — + — k e N . T h o i d i e m t h u 3 l i n e v o i k = 3=> t = —s 8 2 ^ 8 , Vay t i = d u o n g la qua M 2 . Q u a M 2 Ian t h u 3 u n g v o i vat quay d u g c 2 v o n g (2T) Vat qua v j t r i x = 2cm Ian t h u 2013 la : 6 (qua 2 Ian) v a Ian cuoi cung d i t u M Q den M 2 . ^ ^ t= t=^=Hs 8 Hoac t i n h theo cac khoang t h o i gian dat biet: = 4V3 AV3 . o 4 A ^ = - r - v a XM2 = 2 = - - 2 2 tM o ^ M 2 -VA../3 = 1 +1 6 I (O^-A) + '(-A^O) T _3T_3.0,5_3 4 ^ 4 ^ 1 2 " 4 ~ 4 +t X -A f A k 1 0 1 . . 2 24 2 24 N-1 ^ ^ T + ti 2 " = 2013-1 1 2 1 .- +— = 2^24 12073 24 Cach 2 : S u d v n g duong tron lupng giac Vat qua x = 2 la qua M i va M 2 . -A Vat quay 1 vong (1 chu k y ) qua x = 2 la 2 Ian. Qua Ian t h u 2013 t h i phai quay A^ 0-»— 2y 1 24 — + - = — + - = —(s) C h p n dap an A Mo CO XMO T + ti dau u n g v o i k = 0 (nghiem tren). Pha ban d a u (p = — nen ban d a u v a t 6 v i t r i M Q . V a t qua x = 2 c m theo chieu 3n N-1 V o i ti la t h o i gian de v a t d i qua v i t r i x = 2cm Ian t h u 1 ke tir t h o i d i e m ban C a c h 2 : S u d\ing d u o n g tron luQmg giac Goc quet A(p = 2.2n + D . D a p an khac 27T 271 1, , s) C h u k y dao d o n g : T =: — = — = -- ((s) CO 47t 2 CO 47t Cach 1 : G i a i theo p h u a n g trinh lugtig giac 'Phdn txch vd huang dan gidi x=2 ^ 12061 B. s 24 12073 MAU: 1006 v o n g r o i d i t u M o den M i . M2 Goc quet Acp = 1006.271 + ^ .~8 ^ t . 2 T . ^ , M = ll(s)?«^^''*-; CO 24 24 T { ^ ' ; V;£f^ TiivH BiNH Vvwim C h p n dap an A 16 17 Cty TNHHMTV 2 2 .2 -^0 T j r __T 6 12~12 T 0 5 12073 t = 1006T+ — = 1006.0,5+ — = - ^ = ^ s 12 12 24 ^ m - t 'A = t 2012 vat qua v j t r i c6 v = -87t cm/s. D . 1005,5s B. 1005s C. 2012 s 2T[ Theo bai ra ta c6: v = -167csin(27tt ~ ^ ) 6 = ^ + k2n 6 t=- +k 6 = — + k27t t= l . k 2 6 "^"^ •y +12 la t h o i gian de vat d i qua v j t r i x = 2cm Ian t h i i 2 ke t u t h o i d i e m ban 2 ^ ^ N-2„ ^ 2012-2, 1 t= T + t, = .1 + - = 1005,5 (s) ^ 2 2 - — 6 => ban dau vat 6 M . Vat qua v j t r i can t i m Ian t h u 2 tai P k h i d o = ^^^I^T +1 ^^Y^T +12 = 1005,5T = 1005,5(s) A. -s 8 B. c. 5s —s 24 ^ ' D . 1,5s 8 'Pkan tich vd hu&ng ddn gidi Cach 1: G i a i theo p h u a n g trinh lugng giac I 2^ ^ Vat qua v i t r i x = 2cm Ian t h u 2012 la : 2 Pha ban dau cp = 1 2*2 2 Wd = W t r ^ - m c o ^ A ^ s i n ^ 2 7 r t - ^ = - m o ) A cos 2nt2 2 3^ t. =i + k = - + 0-i(s) 2 — 2 t r i vat d i qua la N va P. t= keN d a u u n g v a i k = 0 (nghiem d u a i ) . Vay => V = Nhan xet: a day ta hieu dien duang tron theo v detinh nhanh han. C a u 2: M o t vat dao d o n g dieu hoa v o i p h u a n g t r i n h x = 8cos(27it - —) cm. 3 T h a i d i e m t h u nhat vat qua v i t r i c6 d o n g nang bang the nang. N-2, Vat qua Ian t h u 2012 (chan) nen ta d u n g cong thuc : t = — ^ — T Voi 2 Qua Ian t h u 2012: Cach 1: G i a i theo p h u a n g trinh lugng giac 27rt — -1 167r vat quay d u o c n u a v o n g nen mat ^2 C h u k y dao d p n g : T = — = — = l(s) 6 -871 T 'Phdn tich ra hu&ng ddn gidi 2ni-^ Theo bai ra: N h i n tren d u a n g t r o n ta thay v j C a u 1: M p t vat dao d o n g dieu hoa vai x = 8cos(27it - —) c m . T h a i d i e m t h i i 6 CO Khang Viet T i n h theo van toe V B A I T A P VAN DUNG: A . 1005,5s + DWH o sm 1-cos 2nt-^ 3 2nt-^ = cos 3 3. 3 1 + cos 4 7 t t - 47tt n 271^ V a y chpn dap an A C a c h 2: S u diing d u a n g tron lug^ng giac + T i n h theo i i d o Ta CO X = A^- = ±4%/3(c m . cos 47tt- 2n^ A . . = 0=>47it 7t , ^ 7 k = — + k7r=:>t = — + — k e [ - l ; « . ) 3 2 24 4 27t T h o i d i e m t h u nhat u n g v o i k = - 1 t = —(s) 24^ ' rf ••^•Ii^,,..• V i v < 0 nen vat qua M i va M2. Qua Ian t h u 2012 C h u y: vi thai gian khong nhan gia tri am nen thai diem thu nhat ung vai gia tri k t h i p h a i quay 1005 v o n g roi d i t u Mo den M2. nho nhat ma lam cho t > 0. Goc que| A(p = 1005.271 + 71 => t = 1005,5 s . 18 19 Biquyet on luyftt thi d^i hgc dat diem tot da Vat It, tap 1 - Le Van Vinh D ? n g 4 . X a c d i n h s o I a n v a t d i q u a x t r o n g thcfi g i a n tu" t i d e n tz ( A t = t 2 - t i ) Cach 2: S u dung duong tron lugng giac x=± Wd = W . A 72 pHl/aNG PHAP » Trong mpt chu ky T (27i) vat di qua x 2 Ian neu khong ke den chieu chuyen dpng, neu ke den chieu chuyen dpng thi se d i qua 1 Ian • CO 4 v i tri M i , M 2 , M a , M 4 tren duang tron. Pha ban dau cp = — nen ban dau vat a Mo 3 Thoi diem dau tien vat qua v i tri W d = W t ling voi vat di t u M o den M 4 . . n Goc quet: A(p = n n ^ A ( p l = — =>t = = —s 4 3 12 CO * Xac dinh M i dua vao ti v a PT x,v ( xi, v i chi quan tam < 0 hay > 0 hay = 0) * Xac dinh M dua vao x (hoac v, a, Wt, W d , F) * A p dung cong thuc A(p = coAt tim so'Ian '' Cdc loai thudng gap va cong thuc tinh nhanh Acp coAt 24 271 ,i = n,p(n + 0,p) 27t Neu khong ke den chieu: N = 2n + N' Cau 3: Mot vat dao dpng dieu hoa voi phuong trinh x = 8cos(7rt - —) cm. N' la so ian di qua x khi tren vong trong lupng giac quay dupe goc 0,p.27t 4 ke tir vi tri ban dau Thoi diem thu 2010 vat qua vj tri c6 dong nang bSng 3 Ian thenang.? Cach 1: giai theo phuong trinh lugng giac Wd = 3Wt ^ sin 2.t-^ = 2 ^ 3 2,tt-^ = - ^ TTt-^ 4 = 3cos^ >cos 4 k27i t = —+ k k e N 12 + k2K t = - —+ k 12 Qua Ian thu 2010 (chin): t = N-2 1 "2 Neu ke den chieu: N = n + N ' N' la so Ian di qua x theo chieu bai toan quy dinh khi tren vong trong lupng giac quay dupe goc 0,p.27t ke t u vj tri ban dau m vi D U M A U : V i d u 1: Cho vat dao dpng dieu hoa theo phuong trinh: keN T + t, = X = 3cos 2010-2 „ 11 12059 -.2 + — = . (s) 12 12 4 A. 4 B. 5 => CO 4 v i t r i tren duong tron M i , M2, M s , M4. Vat qua vj tri x = 1,5 cm ta c6: Qua ian thu 2010 thi phai quay 502 vong 3 cos 47tt-- (moi vong qua 4 Ian) roi di t u M o den M2. Goc quet 4nt-^ =1004Tt + .3 CO 12 . 12 4) = > t = ^ = 1004.11 = l ^ s 20 C. 6 12 D. 7 Cach 1: Giai theo phuong trinh lugfng giac 2 A(p =^ 502.271 + 71 - cm. 'Phdn tick vd hu&ng dan gidi W , = iw=>x = ± ' 47rt-^ 3, So Ian vat di qua v i tri x = 1,5cm trong 1,2s dau tien Cach 2: S u d\ing ducmg tron lugfng giac Wd = 3Wt , ,... 3 = 1,5 => cos = ^ + k2n 3 4 n t - i ^ = - ^ + 127r 3 3 t 2 3 1 k = - + - (keZ) 6 2^ t = l (leZ) Trong 1,2s dau tien tuc: 21 Bi(juyei on luyen 0 < t <1,2<=> thi itai hoc itat dicin 0 < i + ii I - Le V&ti -0,330 Goi M la v i t r i cua vat tai thoi diem t = 1,2s. Goc ma vat quet dugc trong 0,2s la: =^N' = 1 + 1 = 2 ii> N = 2.2 + 2 = 6. n W H mot dao dpng toan phan. Don vi la s Bay gia ta tinh N ' . Ta c6: BOM > AOB => vat di qua A I pHl/aNG PHAP * Chu ki (T): cua dao dpng dieu hoa la khoang thoi gian de vat thuc hi^n T -0,5s => At = 2T + 0,2 => N = 2.2 + N ' AOB = — r a d . , 3 So sanh hai gia trj tren / v^ 2 He thuc doc lap giua van toe va gia toe: A = — j - + CO CO Chieu dai quy dao: S = 2A m vi DU MAU: Vi d u 1: (Trich de thi thu chuyen Nguyen Quang Dieu - Dong Thap Ian 1 nam 2013) Mpt eon lac 16 xo dao dpng dieu h6a theo phuong n3m ngang c6 khoi lupng m = lOOg, dp ciing k = lON/m. Keo vat ra khoi vj tri can bang mpt khoang 2em roi truyen eho vat mpt toe dp 20em/s theo phuong dao dpng. Bien dp dao dpng cua vat la: A. 2^/2 cm. B. V2 cm. C. 4 cm. D. 2 cm. ' ' ' ^hdn tkh vd hit&ng dan gidi Vai bdi todn nay chi can tim dugc tan so goc roi thd vao he thuc doc lap la cd ngay bien do. Tan so goc: co = . t = Vm i ^ . = lOrad / s V0,1 Theo h ^ thuc dpc lap lien h^ giiia li dp va van toe: 22 9 A 22 == Xx 2 2+ + 4 = 2 ' + ^ A Chpn dap an A co^ 10^ = 8=^A = 2V2em 23 V l d u 2: (Trich de thi thu Nam T r ^ c - Nam D i n h Ian 1 nam 2013) todn ve'luc, each gidi chung nhat Id tim hap luc tdc dung vdo vat gay ra gia toe M p t vat dao d o n g d i e u hoa v o i gia toe cue dai amax va toe d p c^c dai Vmax. Tan so dao d o n g la £_ chuyen dong cho vat. Cach giai bai nay n h u sau: ^max g £_ ^max Q f _ ^^-^max -Q £ _ ^max Con lac dao d p n g tren doan thang dai 4cm chinh la q u y dao chuyen d p n g ciia vat S = 2 A = 4 c m , v i the bien dp dao d o n g cua vat la A = 2cm. K h i vat m dao dong, h o p luc cua luc d i ? n t r u o n g va luc dan h o i gay gia toe ^ h d n (icfi pd huang ddn gidi a cho vat. Bdi nay dan gidn chi Id tim moi lien he giua gia toe cue dai vd toe do cue dai. Tai v i t r i bien, vat c6 gia toe cue dai. K h i do ta c6: Fd - Fdh = m.amax Gia toe cue dai: a^^^ = co^A qE - k A = Toe dQ cue dai: v^^^ T u day ta lap t i le la eo ngay lien a„.^^ o^A (oA e6n phia d u a l gan vat m . N a n g m len den v j t r i 16 xo k h o n g bien d a n g roi tha nhe vat dao d o n g dieu h6a theo p h u o n g thang d u n g v a i bien do V l d u 3 : (Trich de thi thu Nghi Lgc 4 - Ngh? A n Ian 1 nam 2013) 2,5cm. L a y g = lOm/s^. T r o n g qua t r i n h dao dong, t r o n g luc ciia m eo eong M o t vat dao d o n g d i e u hoa tren quy dao dai 40em. K h i d o d a i la 10cm vat CO van toe 2071V3 cm/s. Lay TC^ = 10. C h u k i dao d o n g cua vat la: ^hdn C. 0,5s A . 0,41W C. 0,5W D . 0,32W ddn gidi Day Cling la dang todn doc vdo rat la, tuy nhien degidi quyel bdi todn ndy cdc ban chi can nha vecong thiec tinh eong suat ciia luc md da hoc tie cdc lap dual. P^^ = F.v Vai F Id luc tdc dung vdo vat, v Id van toe chuyen dong ciia vat. K h i do d a i vat la lOcm nen l i dp x = ± 1 0 c m Trong dao dong dieu hoa, van toe chuyen dong cua vat Id van toe tiec thai nen eong De cho q u y dao dai S = 2 A = 40 ^ A = 20cm suat ciia luc cung Id eong suat tiec thai. A p d u n g he thuc doc lap lien he giira l i do va van toe Sau day la each giai cu the bai toan nay: 207t73 - = 1=>C0 = - (coAf B. 0,64W E = 2.10^ V / m V l d u 5 : M 6 t eon lac 16 xo c6 do c u n g k = 4 0 N / m dau tren d u p e giir eo d j n h 271 v „ 3 ^ B. I s A = m. — .A » m Chgri dap an A giiia hai dai l u o n g tren „ , , l a = (0 = 27rf =^ f = _ : i ! B a x . . c h p n dap an A A . 0,1s m.w2 = coA Cong suat tuc t h 6 i cua t r p n g luc P „ = P.v = m g v v o i v la van toe cua vat m . = 27trad/s=>T = —= — = l s VA2-X2 720^-10^ « fir 271 V o i m va g la cae hang so nen P^sUax '^'^n\ax= C h p n dap an B V l d u 4 : M o t con lac 16 xo n a m ngang g o m vat nang tich d i f n q = 20|LIC va PcsMax = mgVmax = ^ g A 16 xo CO do c u n g k = l O N / m . K h i vat dang n a m can bang, each d i ^ n , tren — = g A ^ / k ^ (1) = J—A Vm Vm mat ban nhan t h i xua't hien tuc t h o i m o t d i f n t r u a n g deu t r o n g k h o n g Theo bai ra: nang m len den v j t r i 16 xo k h o n g bien d a n g r o i tha nh? nen gian bao q u a n h c6 h u a n g dpe theo true 16 xo. Sau do eon lac dao d o n g bien dp dao d p n g A = Al tren m o t doan thang dai 4cm. D o Ion c u a n g d g d i ^ n t r u a n g E la: ^, m g kA D p bien d a n g cua 16 xo tai VTCB: Al = — = A = > m - — A . 2.10^ V / m . B. 2,5.10^ V / m . ^hdn Thong thuang C . 1,5.10^ V / m . ir„ •; ..W ? j ; , D.lO^V/m. tich vd huang ddn gidi thi bdi todn eo liec la tdc dung vdo vat trong qud trlnh dao dong Thay vao (1): PcsMax = g A V k i ^ = ^^f'Y ^ ^ 40.2,5.10-2 VlO.2,5.10-2 ^ thuang dugc xet vai con Idc dan. Tuy nhien a day bdi todn xet cho con lie Id xo nen nhieu ban se xem day Id dang todn la vd thuang bo qua. Nhin chung thi cdc bdi 24 C h p n dap an C 25 Bi quyei on luy$n thi deii hqc dat diem tot da Vat li, tap 1-Le Van Vinh Fmax V i dM 6 : (Trich de thi thu Quynh L i n i 1 - Ngh# An Ian 1 nam 2013) Treo con \ic 16 xo tren tran cua mpt thang may, khi thang may dung yen thi chu ky cua con lac la T, cho thang may chuyen dgng di xuong nhanh dan deu voi gia toe a = 0,5g (g la gia toe roi ty do) theo phuong thang dung thi chu ky dao dong dieu hoa cua no la T' va A.T' = 2T. B . T ' = T. = 0,5T. C.T' D.T' = T V 2 . ^han tick Pd huang ddn gidi A + Al =^ 76 = - k ^ 75 75 ,„.. Chpn dap an B Nhan xet: khi tim ra Al = 2,25(m) nhieu ban se boi roi vi qua Ion so vai cac dap an decho. Cac ban nen binh tinh khai thdc gid thiet ticp theo ma tinh cho ra ddp so Ndi chung Id hay binh tinh! g' = g - a = g - 0 , 5 g = 0,5g Al Chu ky con lac thang may diing yen: T = 2n^ j— = (A + Al) =^ sau do so sdnh vai ddp an. NeU c6 thi minh da diing con nen sai thi hay xem Iqi. Xuong v i nhanh dan deu a i vi the F^, T nguoc voi P i suy ra Chu ky con lac thang may chuyen dpng: T' = 2n^j^ = 2n / ^ k ( A + Al) T h e o b a i r a : - ^ = -4^ V2T V i the chpn dap an D Ca B A I TAP V A N DUNG: Cau 1 : (Trich de thi thu chuyen Hong Linh Ha tinh Ian 1 nam 2013) Con 15c 16 xo c6 dp cung k va vat nho khoi lupng m c6 the dao dpng khong ma sat tren mat phSng nghieng goe a so voi phuong ngang. 6 vi tri can bang dp bien dang cua 16 xo la Al. Cho gia toe roi t u do tai do la g thi chu ky dao dpng la A.T = 2.pL:. B.T = 2 . I ^ . Ral tiec r^ng day khong phdi la dap an dung ma deym cau, vqy chung ta da sat lam tit dau?. Rieng chu ky con lac Id xo thi chi phu thuoc vao dp cimg cua Id xo va khoi luang am vat nqng ma khong phu thuoc vao vi tri dqt con lac nen khong phu thuoc vao gia toe C.T = 27: D . T = 27i trpng tru'ong. KJji thang may chuyen dong thi gia toe trpng tnecmg cua con lac sS la gia ^gsina VAl toe trong tntang hicu dung thay doi nhimg chu ki/ khong phu thuoc vao gia toe trong d huang dan gidi Thoi gian qua cau di tu vj tri cao nhat den vj tri thap nhat ehinh la di tu bien nay qua bien kia mat — = 1,5 => T = 3(s). Dp bien dang 16 xo tai VTCB: T = 27i Vi the chpn C Cau 2: (Trich de thi thu Nguyen Dinh Chieu - Tien Giang Ian 1 nam 2013) Mpt con lac 16 xo, gom 16 xo nh^ c6 dp cung 50N/m, vat c6 khoi lupng 2kg, dao dpng dieu hoa doc. Tai thoi diem vat co gia toe 75 cm/s^ thi no co van toe 15^y3 (cm/s). Xac djnh bien dp. A.A=6cm. B. A=9em. g 47t^ 10.32 40 = 2,25(m) D. A=10em. (phdn tich ra huang ddn gidi Tan so goe: (o = gT^ _ C. A=5em. = = 5 rad / s thiic dpc lap lien h^ giua van toe va gia toe: v^ a' ' a' ^ Luc dan hoi 6 vj tri thap nhat la lire dan hoi cue dai: F^g^ = k ( A + Al) 27 26 'Phdn tich v>d huang dan gidi => A = = 6cm 20 = 10rad/s 0,2 He thuc dpc lap lien h? giiia van toe va gia toe: Tan so goc: co = ^j— = Chgn dap an A Cau 2: (Chuyen Ha Tinh Ian 1 nam 2013) (I) dieu ki^n kich thich ban dau de con lac dao dpng; (II) chieu dai day treo; Chu ky dao dong nho ciia con lac dan phu thupc vao: A. (II) va (IV). B. (Ill) va (IV). C. (II) va (V). D. (I). ^hdn tkh v>d hu&ng dan gidi Ta bie't rang: chu ky con lac dan dao dong dieu hoa phu thupc vao chieu dai day treo va gia toe trong truong. Chpn dap an C Cau 3: (Trich de thi thu chuyen Tran Phu Thanh Hoa Ian 1 nam 2013) Mot con lac 16 xo thang dung 6 vj tri can bang 16 xo gian mot doan A/ . Neu chieu dai 16 xo dupe cat ngan chi c6n bang 1/4 chieu dai ban dau thi chu ki dao dpng ciia con lac 16 xo bay gi6 la = —+ • (III) bien dp dao dong; (IV) khoi lugng vat nang; (V) gia toe trong truong, •A= 10 v2 - ^ + I (0 — 7 - CO co^A^ = 1 (200V3) 20^ 10^ 10^ = 4cm :'<'• t o r Chpn dap an D Cau 5: (Trich de thi thii chuyen Ha Long Quang Tri Ian 1 nam 2013) Con lie 16 xo CO khoi lupng vat nang la 85g dao dpng dieu hoa, trong 24s thuc hi^n dupe 120 dao dpng toan phan. Lay = 10. Dp cung ciia 16 xo ciia con lac do la y j A. 85N/m. B. lOON/m. C. 120N/m. D. l O N / m . ^hdn tich v>d huang dan gidi Chu ky con lac 16 xo : ^ ^hdn tick m hu&ng dan gidi ^ /m At 4n^n^.m k n At^ Gpi k va k' la dp cung ung voi 16 xo c6 chieu dai / va 1/4. Dp cung va chieu dai 16 xo lien h^ qua cong thuc: kl = k'—=> k' = Ak 4 Dp gian ciia con lac 16 xo c6 chieu dai / tai VTCB: - P « k A l = mg=>-^ = — k g 47t^l20^.0,085 • = 85N/m 24^ Chpn dap an A Cau 6 (Trich de t h i dgii hpc nam 2013): Mot con lac 16 xo c6 khoi lupng vat nho la m j = 300g dao dpng dieu h6a voi chu ki Is. Neu thay vat nho c6 khoi lupng mi bang vat nho co khoi lupng m2 thi con lac dao dpng voi chu ki 0,5s. Gia tri m2 bang A. lOOg B. 150g C.25g D. 75 g ^hdn tich vd hu&ng dan gidi Chu ky con lac 16 xo voi chieu dai //4: Chu ky dao dpng cua con lac co khoi lupng mi va mi Ian lupt la: Tj = 2n Chpn dap an A Cau 4: (Trich de thi thu chuyen Nhu Thanh Thanh Hoa Ian 1 nam 2013) Mot con lac 16 xo c6 dp cung k = 20N/m, khoi lupng m = 0,2kg dao dpng dieu h6a. Tai th6i diem t, van toe va gia toe eua vien bi Ian lupt la 20 cm/s va 2V3 m/s^. Bien dp dao dpng ciia vat nang la: A. l O V S c m . B. 16 cm. C. 4 V 3 e m . 28 D. 4 cm. mj _ Tj 2 m2 = nij. 1 [^2 J = 300. ro,5^ 11J = 75g Chpn dap an D CSu 7: (Chuyen Diic Thp Ha Tinh Ian 1 nam 2013) Hai chat diem dao dpng dieu hoa dpc theo hai duang thang song song vai tryc Ox, canh nhau, cung bien dp va tan so. V i tri can bang ciia chiing xem 29 n h u t r i i n g nhau (cung toa do). Biet rSng k h i d i ngang qua nhau, hai cha't d i e m chuyen d o n g ngugc chieu nhau va deu c6 dp Ian l i do bang m p t nua bien dp. H i e u pha cua hai dao d p n g nay c6 the la gia trj nao sau day: C. 7i; 3 D. 271 ^hdn tich ra hudrng ddn gidi Theobaira: =l i n . ^ =^ = 48cm Dp bien dang ciia 16 xo tai VTCB: 1 (K^A\ — 1 ^ = Al 4n^r = 0,012m = 1,2cm 40.4,5^ . i^r;; v,'; Vay: IQ = l^b - ^1 = 48 - 1 , 2 = 46,8cm V i hai vat dao d o n g dieu hoa ciing • ,• Chpn dap an D bien d p va tan so nen ta bleu dien tren cau m o t v o n g tron n h u h i n h ve. 10: (Trich de thi thu chuyen Phiic Trach H a T i n h Ian 2 nam 2013) Mot con lac 16 xo c6 k= lOON/m, dau tren co dinh c6n dau d u o i gan vat nang H i e u pha ciia hai dao d p n g tren la — . 3 C h p n dap an D i n = 0,4kg. Cho vat nang m dao dpng dieu h6a theo p h u o n g thang d i m g thi thay thai gian 16 xo nen trong mpt chu k i la 0,1s. Cho g= 10 m/s^ « m/s\n dp dao dpng ciia vat la Cau A.8>/3cm 8: (Chuyen D u e T h p H a T i n h Ian 1 nam 2013) B.4cm L 6 xo nen khi A l < A . T h a i gian 16 xo nen k h i vat dao d p n g tir n g u a i ta thie't lap m o t dien t r u o n g nSm ngang, c6 h u a n g t r i i n g v a i true cua - A l - > - A - > - A l (chieu d u a n g chpn thSng d u n g h u o n g x u o n g ) . V i the 16 xo 16 xo, CO c u o n g dp E = 8.10^V/m, k h i do vat d u n g yen 6 v j t r i can bang. nen t r o n g m p t chu ky gap hai Ian t h o i gian vat d i tir - A l - > - A . N g u o i ta d p t ngpt ngat dien t r u o n g . Sau k h i ngat dien t r u o n g vat dao d p n g tn = 2 t ( _ ^ _ _ A ) = 0,1 => t ( _ ^ , ^ _ A ) = 0 , 0 5 ( 5 ) dieu hoa v o i bien dp bang A . 12,5cm. B. 2,5cm; C. 4cm; D . 2cm; Goc vat qu^t d u p e t r o n g thai gian tren: 'Phdn tich m huang ddn gidi cp = « . t ( _ „ _ A ) K h i CO dJQn t r u o n g vat d i r n g yen 6 v i t r i can bSng, k h i d o luc d i e n t r u o n g can bang v o i luc dan hoi (phuc hoi). K h i d o t ngpt ngSt d i e n t r u o n g , vat se bat dau dco d p n g dieu h6a v o i vj t r i bien dp la v j t r i vat can bang cu. Luc cfo: Fdt = F d h M a x «> q E = k A =^ A = qlE k -10 -6 .8.10^ 40 = 0,02m = 2cm C h p n dap an D Cau -Al coscp = -A =^-.t(_^,_A) Al = — =>A A ^ .0,05 Al mg = = cos

d huong dan gidi Dp gian cua 16 xo tai VTCB doi voi eon lac 16 xo dat nghieng mpt goc 37" so A l o 2 = - ^ s i n ( a + Aa) k C a u 1 3 : Trong thang may treo mpt con lac 16 xo c6 do cung 25N/m, vat nang CO khoi lugng 400 g. Khi thang may dung yen ta cho con IMc dao dong dieu hoa, chieu dai con lac thay doi tu 32cm den 48cm. Tai thai diem ma vat 6 vj tri tha'p nhat thi cho thang may di xuo'ng nhanh dan deu voi gia toe ^ = ^ • Lay g = TC" = 10 m/s^. Bien dp dao dong ciia vat trong tru6ng hop nay la B. 19,2 cm. D. 5(rad/s). " . Khi tang them goc nghieng 16° thi dp gian mai tai VTCB la: 5 Chpn dap an A A. 17 cm. C. 15(rad/s). voi phuong ngang. 1 *i • A i mgsina kAloi = m g s m a = > Aloi = j^^ = 100 . =^T = 2 . / ^ = 2 . / l L = I s Vk 37'' so voi phuong ngang. Tang goc nghieng them 16° thi khi can bang 16 xo dpng rieng cua con lac la: A. 12,5(rad/s) B. 10(rad/s). Cac lire tac dung vao vat tai vj tri can bang: Fq, i+Pi Viet = 8cm 12,5cm. Chu ki dao dong rieng ciia con lac 16 xo la B. 7t(s.) D W H Khang ,p = n + 0,p v 6 i n = 0 ; l ; 2 B. 15cm C. 56cm - D. 32cm ^han tick m huang dan gidi Bai todn khong cho thai diem ban dau ti vat d dau, vi theneu ti le thai gian khong lap ti so: ^ = n,p- n + 0,pvoi n = 0; 1; 2 T T Phan tich thanh: At = n — + 0,p— 2 ^2 Vay quang duang vat di la S = S ; + S2 = n.2A + S2 ndm trong truang hap dac Met thi bdi todn se di vdo be tat. Do do ta can xdc dinh ti T S2 la quang duong vat di dugc trong thoi gian 0, p — ke t u v i t r i xi, v i . sotren c6 gid tri nguyen hoac bdn nguyen hay khong. De xac dinh S2 ta dung vong tron lugng giac (goc quay t u v i t r i ban dau Chu ky dao dpng cua vat: T = Is T Ta lap ti so giua khoang thoi gian de cho vai chu ky dao dong ciia vat: t 3 5 - = ^ = 3,5 (so ban nguyen) (p = (o.0,p— = 0,p.7r) C a c h 3: T i m n g a y g o c q u a y , Tu do ta phan tich: t = 3T + 0,5T + voi thai gian ti = 3T => — = 71 = 3.4A = 12A = 48(cm); - + voi thoi gian t2 = 0,5T ^ 8 2 = 0,5.4A = 2A = 8(cm) - S = Si + S2 = 48 + 8 = 56(cm). C h p n d a p a n C - se quay goc nn + 0,pn) khi quay goc nn vat d i duoc quang duang S j = n.2A khi quay goc A(p = 7r.0,p t u v i tri ban dau (xi, vi) ta dua vao vong trgn vay quang duang vat d i dugc la S = S j + S2 = n.2A + S2 DU MAU: V I d u 2 : (Trich dethi thtt chuyen Trdn Phu - Thanh Hoa Idn 1 nam 2013): Mgt con 13c 16 xo gom mgt vat nang m = lOOg mac voi 16 xo c6 dg cung Cach lam: Buac bat huge: tim vi tri ban dau: t = ti tim xi vd vi (chi quan tarn > 0, < 0 hay = 0) = n, p = n + 0, p (nhu vay de d i het thoi gian At tren vong tron TC lugng giac ta tim dugc quang duang di la $2 D a n g 2 . Q u a n g d i T d n g v a t d i du'^c tCr t h d i d i e m t i d e n t 2 . At = t 2 - t i Tu duy loai nay: trong thai gian T/2 (goc quay tren vdng trbn la: n) vat dddh se di duoc quang duang la lA. Ta de xdc dinh quang duong di duoc neu thai gian Id nho han T/2 (goc quay nho han n) dua vdo vdng tron luong gidc 34 - k = 160(N/m) dao dgng dieu h6a giiia hai v i tri bien B va B' quanh VTCB . O (cho BB' = I6V2 (cm)). Tinh quang duang v^t d i chuyen dugc sau 35 t/uygr on m y f » n y t uui mei;i t u i uu vnt unam n, lup i " CtyTNHHM L K V M H VTW/I" khoang thai gian At = — ( s ) , ne'u chpn goc thai gian t = 0 luc v^t d i qua 6,4 B. 151,6 cm C. 66 cm 'PhcLn tick vd hu&ng dan gidi 2 Theo bai ra: t = 0 liic vat di qua VTCB theo chieu duong chinh la diem Mo Tan so goc: ™ , , „ =,/— = 271 Chu ky: T = — 27t CO = 40 rad / s 6,4 20 Tu hinh ve ta thay S 2 = OA = A.cos^ = 8(cm). Vay S = Si + S2 = 143,8(cm). Chpn dap an A Dang 3. Toe dg t r u n g binh cua v a t d i tiT t l i d i d i e m t i den t2: Tie do ta phan tich: t = 3T + 0,125T Vai thoi gian ti = 3T :=> Sj = 3.4A = 96V2(cm); + Vai thai gian t2 = 0,125T t2-ti cp = t j .co = 0,125T. — = - ( r a d ) tir hinh ve ta thay T 4 S2 = O A = A . c o s - = 8(cm). 4 VayS = S i + 5 2 = 143,8 (cm) Cach 2: Tach At theo 2 Ta lap ti so giira khoang thoi gian de cho vai nua chu ky dao dpng cua vat: 7t T T At = 6 - + 0 , 2 5 2 2 2 2.20 Vgy quang duang vat di la S = Sj + m V I DU MAU: Vi du 3: Mot vat dao dpng dieu hoa theo phuong trinh x = 6cos A. 16,9cm/s B. 15,7cm/s C. 20cm/s xo, vo.(vi tri vat qua VTCB theo chieu duong) (cm). cm D. 13,2cm/s Nhan xet: x = -3^/3 = - 6 . — = - A . ^ va x = -3 = -6.1 = - 2 2 2 2 Asll 2 A ^ 2 2 0 r : S 2 la quang duong vat di dupe trong khoang thoi gian 0,25y ke t u v i tri ^ 'Phdn tich vd hic&ng dan gidi A = n.2A + 8 2 = 6.2A + S 2 7rt + 4 Trong khoang thai gian ngan nhat khi vat di t u vi tri c6 li dp x = -3 den vi t r i x =3cm, chat diem c6 toe dp trung binh la A^f3 T 36 vdi S la quang dUdng t i n h nhu" t r e n . vtbl = - + 71 'i VTCB theo chieu duong) 20 1 A(p = 67t + 0,25.7i S2 la quang duong vat quay dupe 0,25.7t ke t u v} tri xo, vo.(vi t r i vgt qua = 3,125 4^ = - ^ = 6,25 = 6 + 0,25 ^6,25 = 6 + 0,25 ^ n n n Vay quang duong vat di la S = S j + S2 = 6.2A + S2 Ta lap ti so' giiia khoang thoi gian de cho voi chu ky dao dpng cua vat: t Tu hinh ve ta thay S 2 = OA = A.cos— = 8(cm). 4 ;«••••• ' »*•„ :;, Vay S = Si + S2 = 143,8(cm) Cach 3: tim ngay goc quay Ta lap ti so'giiia goc quay dupe trong khoang thoi gian de cho voi 7t: 40.^'^ A(p^o)At^__M 71 = — = —s 40 Cach 1: Tach At theo T ^ Goc quay ma vat quay dupe trong khoang thoi gian 0,25 —: D. 132,5 cm Bien do dao dgng: A = - ^ 5 - = 1^2^^ _ g ^ ^ m 2 K / . , n i v V'irf cp = co.O,25l = ^ . 0 , 2 5 l = J r a d VTCB theo chieu duong. A. 143,8 cm I V nvVJI ; ' : . ^ 8 — T 2 A-JIAS 2 2 ^ _ •] ; 37