JOURNAL OF SCIENCE OF HNUE
Educational Sci., 2009, VCJI. 54, No. 8, pp. 23-27
C H E TAO T H I E T BI T H I N G H I E M
T 6 N G H O P H A I D A O D Q N G D I E U H O A D U N G T R O N G
D A Y H O C VAT LI 12 d T R U C J N G T H P T
Dudng Xuan Quy
Trudng Dai hoc Su pham Hd Ndi
M d dau
Trong chudng trinh Vat If 12 d trirdng THPT, kiin thQe vi tdng hdp hai dao dpng diiu hda la npi dung quan trpng ma hpc sinh (HS) cin chiim linh. Di thue hien tdt nhat qua trinh nay, viec td chQc hoat dpng hpc cho HS cin di theo tiin trinh sau:
ThQ n h i t , hudng d i n d i HS giai quyit bai toan If thuyit [1, 2, 5]: Mot vdt thuc hien ddng thdi hai dao ddng diiu hda ciing pliuang, cung tdn sd, cd phuang trinh chuyen ddng la Xi = Ai cos{u}t + ^\) ua X2 = A2Cos{u}t -|- (^2)- Tim phuang trinh dao ddng tdng hap? Vdi cac kiin thQe da dupe trang bi (phuPng phap biin ddi lupng gidc hoac phuong phdp vec td quay Frex-nen), HS se tim dupe phupng trinh chuygn dpng tdng hdp co dang: x = Acos{ujt -+- gj), trong dd bign dp A va goc pha ban d i u (/? dupc xac dinh theo cae cdng thQe:
. 0 A-y r. A , , N AlSin^Pl + A2sing:)2
A<) = Ai + Al + 2A1A2 cos(v32 - ^\) va isxmp = — ^—.
Ax c o s 9 , -\- A2S\nip2
ThQ hai, td chQc tao diiu kien cho HS thuc hien viec kiim nghiem cac kit qua thu dupc tQ viec giai bai tddn If thuyit. Dieu nay lam cho kiin tliQc ma HS chiim linh dQpc trd ngn vQng chdc, qua dd ren luyen va phat trign dupe d eac em ndng lUc sang tad [3].
Tuy nhign, trdng thQc t i day hpc d trudng T H P T cua nudc ta tQ trudc din nay h i u nhu chUa cd thiet bi tign dung ndd cd t h i giup dap Qng dQpc ygu ciu thQ hai nay. Vi vdy, chung toi xin trinh bdy d day viec chi tao thiit bi tong hop hai dao dpng diiu hda dung trong day hpc mon Vdt li 12 b trudng T H P T .
2. Noi dung nghien ciJu 2.1. Chi tao thiet bi
Thiit bi dupc chi tac gdm ba bp phdn cP bin.
* Bg phdn tga ra hai dao dgng dieu hda
Duong Xuan Quy
Hinh 1. Bp phan tgo dao dpng dieu hda Gid dp kep cd gdn hai thanh thep
gidng nhau dat thdng dQng (1); hai thanh sdt gidng nhau (2) (hai eon ldc vat ll) dung d i tao ra hai dao dOng cung chu ki va g i n dung la dao dpng diiu hda, hai con ldc nay cd true quay dang dao tua vao hai tay do gdn vdi khdp ndi da ndng (3) d i cd t h i thay ddi chiiu cao ciia chung; d i u tren ciia hai con lac cd chd d i gdn hai thanh nhdm nhe, cQng, gidng nhau (4) dung lam ddn bdy d i truyin dao dpng ciia hai con ldc, qua hai thanh truyin, vao bp phan thue hien viec tdng hop hai dad dOng (Hinh 1).
* Bp phdn tong hdp dao dpng Bp phan nay c6 chQc nang tao ra ddng thdi hai dao dOng thanh p h i n va dao dpng tdng hpp cua chung. Bp phan nay cd ciu tao nhu sau:
- Mot gid dd hinh chu nhat bdng nhom va gd ddn de gan vao dd cac mde cd dinh bang day kim loai. Cae Id cua mde kim Idai nay cd dudng kfnh Idn hon dudng kfnh ciia cac dng nhua se ludn qua dd mot chut. C i n cd 6 mde nhu vay.
- Hai thanh AB va CD lam bdng cac dng nhQa (tao tQ cac rupt but bi). Mdi thanh nay tua trgn hai mde cd dinh va cd t h i trUpt d i dang tren dd (hai thanh nay each nhau cd 10 cm). Mot trong hai d i u ciia mdi thanh (A va D) dupc chpn la chd lign kit vdi he thdng thanh truyin - don bay gan vdi hai con ldc de nhan dao dpng tQ chung. Khi cho hai con ldc dao dpng thi thanh AB va CD se trirpt tren vdng race va thuc hien dao dpng ngang (Hinh 2).
- Hai dng nhQa MN va PQ (each nhau 8 cm) cd nhiem vu tao ra hai dao dpng thanh phin. Mudn vay, cin phai cho mot d i u cua mdi thanh tQa' qua mde cd dinh, d i u kia ludn qua mot mde (hoac chdt) di dOng gdn tren mdi thanh trupt AB va CD.
Hai thanh MN vd PQ se ddng vai trd la cac don bay dao dpng theo hai thanh AB vd CD. Khi do hai trung diim Di va D2 cua cac thanh nay se la hai dao dpng thanh p h i n cin khao sat. Dd thi li dp ciia hai dao dpng nay se dupe hai but gan vdi Dl va D2 ve Ign bang ghi dd thi.
- Ong nhua UV lam nhiem vu tao ra dao dpng tdng hpp cua hai dao dpng thanh phin. Di thuc hien dUdc dieu nay, can cho hai d i u dng tua tren hai mde di dpng gan tren hai thanh AB va CD.
Hinh 2. Bp phan tong hdp dao dpng
Che tao thiet bi thi nghiem tdng hap hai dao dong diiu hda diing trong day hoc Vdt li 12...
Trung digm D3 cua dng (cung cd gdn but) se thuc hien dao dpng tdng hpp cua hai dao dpng Di va D2 (Hinh 3).
M U
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D,
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D3 D,
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D N
M6c c6 dinh M6c di dong
Hinh 3. Sd do hp phan tong hdp dao dpng
* Dg cac digm Di, D2, D3 thdng hang, ta gdn vao phfa dudi va gin sdt vdi diim gdn but cua cac dng (MN, PQ, UV), nhQng dean dng nhd, ngdn. Mpt spi day, gdn vdi gid cd dinh, cdng ngang ludn qua cae doan dng do se dinh hudng cho chung trUpt tren cung dudng thdng.
Chung toi dung cdc dng nhua nhd, nhe (loai rupt but bi Hdng Ha hoac Thign Long) di he thdng nay ft tigu hao nang lupng dao dpng cua hai con ldc, lam cho dao dpng tdt d i n cham hpn.
Cd t h i mo ta boat dpng cua hg thdng nay nhu sau: Khi cho con ldc bgn trdi dich sang trai (giQ cd dinh con ldc bgn phii) thi thanh AB bi ddy vi ben phai. Luc nay do lign kit qua vdng mde, cdc thanh MN vd UV se bi di chuyin lam cho eac digm Dl va D.j cUch sang phai mpt doan aj. Niu luc nay ta cho con ldc bgn phai dich sang trai thi thanh CD bi ddy vi bgn phai. Diim D2 bi dich vi ben phii mpt doan 02, nhu vay diem D3 lai bi keo sang phai mpt doan 02. Dp dich tdng cong cua D3 se la ai-l-a2. Niu kfch thfch hai con lac dao dpng cung bien dp thi ai = 02 = a va dp dich cua D3 la 2a.
Neu ta cho con lac ben phai dich sang phai thi thanh CD dich sang trdi. Kit qua la diim D2 va D3 dich mpt doan 02 vi ben trai, do dd dp dich ciia D3 se la
loi — 02!, dp dich nay se bdng 0 niu Oi = 02-
Hinh 4- Bp do thi dao dpng tong hdp hai dao dpng ciing pha
Dudng Xuan Quy
Nhu vay, niu ban d i u ddng thdi kfch cho hai con ldc bdt d i u dao dpng tQ cung mpt phfa (hai dao dpng cung pha cung t i n sd) thi bien dp dao dOng tdng hpp dupc tdng cudng. Con niu ban d i u kfch cho hai con lac dao dpng ngupc phfa (hai dao dpng ngupc pha) thi bien dp dao dpng tdng hpp bi giam di.
* Bp phdn ghi dd thi li dp cua cdc dao dpng phu thupc thdi gian D i ghi dQpc dd thi dao dOng, trQdc
h i t ta gdn cac but nam cham nhd vao eac diim Di, D2, D3. Khi cdc diim nay dang thuc hien dao dQng thi ta cho mpt bing bpt sdt chay diu bgn canh cac but nam cham [4]. Cdc dudng cong mat sdt do but nam ehdm keo ra phia ngoai cua mat bang se biiu diin dd thi li dp eua cacdao dpng phu thupc thdi gian (Hinh 4).
2 . 2 . T i i n h a n h t h i n g h i e m - Ddt hai con lac len gia dd va gan he thdng tdng hpp hai dao dpng Ign do; lien kit hai con ldc vdi he thdng tdng hpp dao dOng; bd tri he thdng ghi dao dpng d vi trf phu hpp d i cd t h i ghi dupe dd thi li dp. Chu y la he thdng dupc coi la hoat dpng dupc khi ta thQ cho hai eon ldc ddng thdi dao dpng thi hai con ldc keo he thdng hoat dpng trpn tru va du lau
(khoang tren 10 dao dpng), ddng thdi Hinh 5. D6 thi dao dpng tdng hdp hai cae but nam cham ve dupc cac dudng ^ " ^ ^ ^ " 9 ngUdc pha
qui dao dao dpng du rd ma d i u but khong cham vao mat bang.
- Di thuc hien viec tdng hpp hai dad dpng cung pha, ta keo hai con ldc vg cung mpt phfa sau dd tha nhe ddng thdi. Ngay sau dd, nhanh chdng bat dign cho md to keo bang bpt sdt d i thu dupc cdc dd thi dao dpng (Hinh 4).
- Viec tdng hpp hai dao dpng ngupc pha dupe thuc hien bang each kec hai con ldc vi hai phfa ngupc nhau rdi t h i nhe ddng thdi va cung thuc hien tUPng tu, ta se thu dupc cac dd thi dao dpng (Hinh 5).
2 . 3 . C a c phi:fofng a n siJ d u n g t h i i t b i t r o n g d a y h o c
- Thuc hien viec kiem nghiem cac kien thQc ve tong hpp hai dao dpng dieu hoa cung t i n sd:
+ Dao dOng tdng hpp cd chu ki ( t i n sd) bang chu ki ciia hai dao dpng thanh
Che tao thiet hi thi nghiem tdng hap hai dao dong dieu hoa dung trong day hoc Vdt li 12...
phin.
-f Bign dp dao dpng tdng hpp cua hai dao dpng cung pha la tdng cua hai bign dp dao dOng thanh p h i n .
-i- Bign dp dao dpng tdng hpp eua hai dao dong ngupc pha la hieu ciia hai dao dpng thanh p h i n .
- Thuc hien vige khio sat (dinh tfnh) sU tdng hpp hai dao dpng khac t i n sd Cd the chpn hai con ldc khde nhau d i tao ra hai dao dpng khae t i n sd. Qua thi nghiem cho HS t h i y dao dpng tong hdp Id 'mpt dao dpng t u i n hodn pliQc tap.
TQ dd HS cung cd t h i hiiu: Mpt dao dpng phQc taj) co t h i la tdng hpp tQ nhiiu dao dpng dieu hda don giin. Day la If do ma trudc h i t p h i i xet dao dpng diiu hoa ddn giin, tad ra tQ cac he If tudng.
3. Ket luan
Thiit bi thf nghiem tdng hpp hai dad dpng diiu hda la thiit bi dupc ehi tao tQ cdc vdt Iigu dg kiim, re tien, viec chi tao va sU dung khong cpid phQc tap, thdi gian thao tdc nhanh. Thiit bi nay da dQpc chung toi thu nghiem nhiiu lin cung nhu da tien hanh thQc nghiem sQ pham. Ddy la thiit bi thich hpp d i thuc hien A'iec day hpc Vat If d trirdng trung hpc phd thong.
TAI LIEU T H A M K H A O
[1] Nguyin T h i Khdi (Tdng chii bign), 2005. Vd,t li 12. Nxb Giao due. Ha Npi.
[2] Sdch gido vien - bd 1. Nxb Giao due, Hd Npi, 2005.
[3] Ludng Duygn Binh (Tdng chu bign). Vdt li 12 - Sdch gido khoa thi diem.
Nxb Gidd due Ha Ndi.
[4] Sdch gido vien - bd 2. Nxb Gidd due. Ha Npi, 2005.
[5] Nguyin DQc Thdm ( Chu bign ), 2002. Phuang phdp day hoc vdt li d trudng phd thdng. Nxb Dai hpc Su pham. Ha Npi.
[7] DQPng Xudn Quy, 2008. Xdy dung vd sii dung thiit In con lac Id xo ngang trong day hoc dao ddng ca - Vdt li 12 THPT. Tap chf Giao due. Ha Npi.
[8] David Haliday, Robert Resnick, Jearl Walker, 2002. Co sd vdt li, tap 2, tai Heu dich. Nxb Gido due.
A B S T R A C T
Manufacturing experimental e q u i p m e n t
t o s y n t h e s i z e harmonious oscillation used in 12 grade P h y s i c s teaching programme in high schools
The report describes the manufacturing experimental equipment to synthesize harmonious oscillation. The equipment manufactured by easy to find cheap mate- rials, not excessively complicated to manufacture or to use, with fast manipulated time which are suitable for teaching Physics in high schools.
