Backup of Cach Giai Nhanh Vat Ly

8/6/2019 Backup of Cach Giai Nhanh Vat Ly

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ti:Phng php, th thut gii nhanh cc dng trc nghim vt l 12

T L Cng ngh. Trng THPT chuyn L Qu n Trang 1

S GIO DC & O TO BNH NH TRNG THPT CHUYN L QU N

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SNG KIN - KINH NGHIM

TN TI:

PHNG PHP,TH THUT GII NHANH CC DNG TRC NGHIM

VT L 12

Ngi thc hin: Nguyn Trn Cng n v:Trng THPT chuyn L Qu n

Quy Nhn, thng 5/2010


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TN TI : PHNG PHP, TH THUT GII NHANH CC DNG

TRC NGHIM VT L 12

A. M U 1. L do chn ti:

Vt l l mn khoa hc c bn nn vic dy vt l trong trng ph thngphi gip hc sinh nm c kin thc c bn, trng tm ca b mn, mi quan hgia vt l v cc mn khoa hc khc vn dng cc quy lut vt l vo thc tini sng.

Vt l biu din cc quy lut t nhin thng qua ton hc v vy hu ht cckhi nim, cc nh lut, quy lut v phng php ca vt l trong trng phthng u c m t bng ngn ng ton hc, ng thi cng yu cu hc sinhphi bit vn dng tt ton hc vo vt l tr li nhanh, chnh xc cc dng bitp vt l nhm p ng tt cc yu cu ngy cng cao ca cc thi TNPT vTSH.

Vn t ra l vi s lng ln cc cng thc vt l trong chng trnhPTTH lm sao nh ht vn dng, tr li cc cu hi trong khi thi trc nghimph ht chng trnh, khng trng tm, trng im, thi gian tr li mi cu hi qungn, (khng qu 1,5 pht ) nn vic suy lun v chng minh cc cng thc cnvn dng l bt kh thi.

V vy chng ti chn ti: Nh ti thiu cc cng thc c bnv cc cngthc c tnh tng qut nht ca chng trnh v a ra cc phng php, th thutvn dng nhm gii quyt nhanh, chnh xc cc cc dng bi ton trong chngtrnh.2. Nhim v ca ti Gii hn ti.

a. Nhim v ca ti: + Ch ra cc cng thc c bn, trng tm, tng qut nht trong chng trnh vt llp 12 thuc tng chng vi s lng ti thiu hc sinh d nh nht. + Ch ra cc mi quan h trc quan ca cc i lng vt l, phng php, ththut s dng cc cng thc ny gii nhanh nht, chnh xc nht cc bi tp. +Thng qua ti rn luyn, pht trin t duy, tnh sng to ca hc sinh.

b. Gii hn ti: Ni dung, kin thc trong chng trnh vt l 12 vi ti ny ta xt 3 phn: + ng trn lng gic. +Tng hp dao ng iu ha cng phng, cng tn s. + Giao thoa sng c. c. Hng pht trin ti: Ni dung, kin thc nghin cu tip theo ca ti. + Dng gin vecto trong bi ton in xoay chiu. + Cc cng thc tnh nng lng, ng lng trong chng vt l ht nhn. + Mt s th thut ca cc chng cn li. 3. Phng php tin hnh .+ Tm hiu, c, phn tch, tng hp cc ti liu trn mng internet. + Tng hp t kinh nghim ging dy ca bn thn v hc hi kinh nghim gingdy ca cc ng nghip trong cc t tp hun chuyn mn, bi dng thay schgio khoa.4. C s v thi gian tin hnh nghin cu ti. ti hnh thnh trong qu trnh ging dy ti trng chuyn L Qu n, trongcc t bi dng chuyn mn v tp hun thay sch gio khoa, k t nm 2008.


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B. NI DUNG PHN MT

NG TRN LNG GIC

1. M t tnh trng s vic hin ti : S lng cng thc yu cu hc sinh nh vn dng trong chng dao ng

c rt nhiu ch tnh phn t m, bt buc l 16 cng thc nhng vi s lngcc cng thc cng ch gii quyt c cc cu hi rt c bn, khng th giiquyt c ht cc dng bi tp t ra ca chng ny.

phn dao ng kin thc ton lin quan l cc cng thc lng gic vgii cc phng trnh lng gic y l kh khn ln i vi a s cc loi itng hc sinh k c hc sinh kh gii v rt hay st nghim bi tnh lp li cahm tun hon.

Hin ti trn ng trn lng gic ch s dng mt trc cosin cho phngtrnh dao ng x = Acos( t + ) (trc Ox) v cc dng ton chng ny thngcn c vo cc d kin bi ton cho t phng trnh dao ng dng x

=Acos( t+ ), tm chu k, tn s, ng i, khong thi gian i t to x1 n to x2, tm vn tc, gia tc ti mt thi im no , khong thi gian l xonn, gin * Nhn thy mt s nhc im ca phng php ny khi lm trc nghim:

S kh khn cho hc sinh khi gp phi loi cu hi d kin bi ton khngcho phng trnh dao ng dng li x = Acos(t + ) m cho dng vn tc tcthi v = - Asin( t + ) hoc cho dng gia tc tc thi a = - 2Acos( t + ). Lcny hc sinh b ng khng th biu din hm (v) v hm (a) trn ngtrnlng gic. Mun biu din c trn ng trn lng gic th phi t hm(v),(a) vit li dng hm (x) bng cch ly tch phn bc nht hm vn tc (v) hoc

bc 2 hm gia tc (a) y l cch rt kh khn cho hc sinh. Nu mun trnh iuny th phi nh hm vn tc (v) sm pha hn li (x) 1 gc / 2 , cn hm gia tc(a) ngc pha vi hm li (x) v gii cc phng trnh lng gic lin quan iuny mt nhiu thi gian, cha mun ni chnh xc vi a s hc sinh l rt thp.

Khng th nh ht cc cng thc, cc mi quan h phc tp ca cc ilng c hc, v thiu tnh trc quan, thiu mi quan h gn b gia cc hintng vt l nn thng tr li sai cc cu hi d c bn nht.

Sau y, chng ti xin trnh by mt phng php khc rt trc quan, thhin c mi quan h gia cc i lng nhm gip cc em hc sinh v h trgio vin trong vic gii nhanh nht, chnh

xc nht cc dng ton v dao ng c. 2. M t ni dung gii php mi : a. C s l thuyt :

Dao ng iu ho c biu dinbi hm sin (cosin) + Li tc thi : x = Acos( t + x )+ Vn tc tc thi: v =- Asin( t + x)+ Gia tc tc thi: a =- 2Acos( t + x)b. Gii php mi:

Biu din c ba hm li (x), vntc (v) v gia tc (a) trn cng mt ngtrn lng gic nh sau: + Li : x = Acos( t + x ) l hm cosin=> cng chiu trc cosin c hng (+) t tri sang phi vi bin l xmax = A

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+ Vn tc: v = x= - Asin( t + x) l hm tr sin => ngc chiu trc sin nn c hng (+) hng t trn xung vi bin Vmax = A iu ny tng ng vi hm v = Acos( t + V) vi / 2v x + Gia tc: a = v= - 2Acos( t + x) = - 2 x l hm tr cosin (ngc hm x) => ngc chiu trc cos c hng (+) t phi sang tri vi bin amax = 2 A iu ny tng ng vi hm a = 2Acos( t + a) Vi / 2

a x v

Thng qua cch biu din ny ta thy mt s im c bit v vng c bitv mi quan h v pha ca li (x), vn tc (v), gia tc (a) cng nh vic khai thccc kin thc l thuyt lin quan v dao ng iu ha, cc dng nng lng cadao ng iu ha c th hin mt cch trc quan trn hnh v vi mt vi v dsau :+ Bn im c bit: - V tr bin dng I: ( xmax = A ; v = 0 ; a = - 2 A)=>Th nng cc i, ng nng cc tiu

- V tr cn bng II: ( x = 0 ; v = - A ; a = 0 )=>Th nng cc tiu, ng nng cc i - V tr bin m III: ( x = -A ; v = 0 ; a max = 2 A )=>Th nng cc i, ng nng cc tiu - V tr cn bng IV: ( x = 0 ; V max = A ; a = 0)=>Th nng cc tiu, ng nng cc i

Vy chu k dao ng tun hon ca hm ng nng v hm th nng ca

dao ng iu ha ch bng chu k T ca hm li (x), khong thi gian ngnng (th nng) t cc i thnh cc tiu hay ngc li l chu k T ca hm li (x). + Bn vng c bit: Vng 1: x>0 ; v0 v th nng gim, ng nngtng. Vng 2: x chuyn ng chm dn theo chiu (-) v a.v0=> chuyn ng nhanh dn theo chiu (+) v a.v>0 v th nng gim, ng nngtng. Vng 4: x>0 ; v>0 , a chuyn ng chm dn theo chiu (+) v a.vvn tc (v) sm pha hn li (x) mt gc/ 2 , tr pha hn gia tc (a) mt gc

/ 2 =>gia tc (a) sm pha hn vn tc (v) mtgc / 2 , ngc pha vi li (x)

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3. Chng minh tnh kh thi ca gii php mi: Sau y l chng minh thy r cc u im ca phng php v th

thut gii nhanh cc dng ton ca phn dao ng c thng qua tnh trc quan vs lin h mt thit gia cc mi quan h cu li (x), vn tc (v) v gia tc (a),ng nng, th nng, c nng Cu 01 . Mt vt dao ng iu ho theo phng trnh x = 6cos(2 t)cm, vn tcca vt ti thi im t = 7,5s l: A. v = 0cm/s. B. v = 75,4cm/s. C. v = - 75,4cm/s. D. v = 6cm/s.Gii: Dng trc Ox biu din (Hnh1): lc ban uvt v tr I sau thi gian t = 7,5s vt quaymt gc t =2 .7,5 = 15 lp li 7,5 vngn v tr III => c vn tc v = 0, chn A Cu 02. Mt vt dao ng iu ho theophng trnh x = 6sin (4 t + / 2 )cm, gia tcca vt ti thi im t = 5s l: A. a = 0cm/s 2 B. a = 946,5cm/s 2.

C. a = - 946,5cm/s2

D. a = - 946,5cm/s.Gii: Dng trc Ox biu din (Hnh1). cho hmx dng sin cn chuyn sang cos c dng x = 6cos (4 t )cm => ban u vt v tr Isau thi gian t =5s vt quay 1gc t = 4 .5 = 20 lp li 10 vng n v tr c. => c gia tc a =- 2 A = - 947,5cm/s 2 , chn C Cu 03. Mt cht im dao ng iu ho theo c phng trnh vn tcv=10 cos(2 t + / 2 ) cm/s, to ca cht im ti thi im t = 1,5s l A. x = 1,5cm. B. x = - 5cm . C. x= + 5cm. D. x = 0cm.Gii: Dng trc Ov biu din (Hnh1): lc ban u vt v tr I sau thi gian t = 1,5s vtquay 1 gc t =2 .1,5 = 3 lp li 1,5 vng n v tr III => c to x =- 5cm, chn B Cu 04: Vn tc ca mt vt dao ng iu habin thin theo thi gian theo phng trnh v =2 cos(0,5 t /6) m/s). Vo thi im no sauy vt qua v tr c li x=2cm theo chiudng ca trc ta . A. 8/3s B. 2/3s C. 2s D. 4/3sGii: Dng trc Ov biu din (Hnh 2) lc ban u vt v tr V sau thi gian t vt quay 1 gc =0,5 . t = /3 v c li x= 2cm, bin A= 4 cmv chuyn ng theo chiu (+) n v tr VI=> mt thi gian t = 2/3s, chn B Cu 05. Mt vt dao ng iu ho vi bin A= 4cm v sau thi gian t =3s vt i c qungng 24cm, chn gc thi gian l lc vt i qua

v tr c li x =23

cm theo chiu dng.Phng trnh vn tc ca vt l A. v = 4cos(3 t - / 3 )cm./sB. v = 4 cos( t + / 3 )cm ./sC. v = 4cos(3 t + / 6 )cm/s.

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D. v = 4 cos( t - / 6 )cm/sGii: Dng trc Ox biu din (Hnh 3) lc ban u vt v tr V hm vn tc c to ban u l + / 3 , bin vn tc A = 4 cm/s v chu k T = 2s ( t = 3s i cqung ng 6A mt thi gian 1,5T) => hm v = 4 cos( t + / 3 )cm ./s , chn B Cu 06: Vt dao ng iu ha c phng trnh vn tc v = 2cos(0,5 t- /6) cm/s.Vo thi im no sau y vt qua v tr c li x = 2cm theo chiu dng ca trcta . A. 6s B. 2s C. 4/3s D. 14/3s Gii: Dng trc Ox biu din (Hnh 4) lc ban uvt v tr V bin A = 4 cm/s vVmax =A =2 sau thi gian t vt n v tr VIc li x = 2cm theo chiu (+) v chu k T = 4snn thi im t = T/6 +kT => t = 14/3s , chn D Cu 07: Mt con lc l xo treo thng ng,u di c vt m=0,5kg, phng trnh daong ca vt l a =100cos t (cm/s 2). Ly g =10 m/s 2. Lc tc dng vo im treo vo thiim 0,5 (s) l A. 5N B. 1 NC. 5,5N D. Bng 0 Gii: Dng trc Oa biu din (Hnh 5) lc ban uvt v tr III, chu k T = 2s nn sau thi gian t= 0,5s vt v tr IV l v tr cn bng. Lc tcdng vo im treo l lc n hi => F = k l0 = mg = 5 N , chn A Cu 08: Mt l xo cng k treo thng ng,u trn c nh, u di gn vi vt m, ly g= 10 m/s 2. Chn trc Ox thng ng, chiudng hng xung. Vt dao ng viphng trnh v = 20 cos(5 t - 2 /3)cm/s. Thi im vt qua v tr l xo b gin 2cm ln u tin l A. 1/30s B. 11/30s C. 1/25s D. 1/10s Gii: Dng trc Ov biu din (Hnh 5) lc ban u vt v tr V, ti v tr cn bng l xogin 0 2 4( )

mg gl cm

k nn ti v tr l xo gin

2cm vt c li x =-2cm v tr VI (do chiu (+)hng xung) => gc quay 5 .t = / 2 => t = 1/10 (s), chn D Cu 09: Mt vt dao ng iu ha vi biu thcli x=4cos(5 / 6 0,5 t ), trong x tnh bng

(cm), t tnh bng (s).Vo thi im no sau yvt s qua v tr x=2 3 (cm) theo chiu m catrc ta ?

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Hnh 4

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A. 6(s) B. 3(s) C. 2/3(s) D. 4/3(s)Gii: Dng trc Ox biu din (Hnh 6). cho hm x dng cosin c nhng 0 cnchuyn sang dng x=4cos( 0,5 5 / 6t )cm, c chu k T = 4s. Ban u vt v tr Vsau thi gian vt quay 1 gc t = 0,5 .t = ( v vt v tr VI ) =>t = 2(s) p n 2 + kT (s), chn A Cu 10 : Mt vt dao ng iu ha dc theo trc Ox vi phng trnh:v=24 cos(4 t+ /6) cm/s. Qung ng vt i c t thi im t1=2/3 (s) nthi im t2= 37/12 (s) lA. 141cm B. 96cm C.234cm D. 117cm Gii: Dng trc Ov biu din (Hnh 7). Hm c A =6cm ban u vt v tr V, ti thi imt1 =2/3(s) vt quay 1 gc 1

24

3 n v tr VI.

Trong khong thi gian 2 129

12t t t s quay

c gc 29 24 912 3

ngha l lp li 4,5

vng ri n v tr I c tng qung ng i ngvi 4,5 Chu k cng thm T/3 chu k na. => s =18A+0,5A+1A=19,5x6 =117 (cm), chn D 4. Mt s vn lin quan v vn dng:

a. Vn lin quan: Khi hc sinh bit cch s dng vng trn lng gic trn vi 3 trc to

tng ng Ox, Ov, Oa th c th vn dng gii cc bi ton v dao ng int vi cc mi quan h gia dao ng c v dao ng in t nh sau:

S tng t gia dao ng c v dao ng in t

i lng c

i lng in t Dao ng c Dao ng in t

x q x + 2x = 0 q + 2q = 0v i k

m

1

LC

m L x = Acos( t + ) q= q 0cos( t + )

k 1C

v = x=- Asin( t+ ) i=q=- q0sin( t+ )F u 2 2 2( )v A x

2 2 20 ( )

iq q

R F = -kx = -m 2x 2qu L qC

W W t (WL) W =1

2mv 2 W t =

1

2Li2

W t W (WC) W t =1

2kx2 W =

2

2

qC

Nh vy chng ta c th thay th: +Trc (Ox) thnh trc (Oq) hay (OuC)+Trc (Ov) thnh trc (Oi) +Trc (Oa) thnh trc (OuL)

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Ta s dng vng trn lng gic gii cc dng ton tm: Chu k, thi im, in tch, dng in, in p gia hai u t, hai u cun dy,cc gi tr ca cc hm nng lng nh nng lng in, nng lng t

Tng t nh dao ng c ta cng rt ra c cc im c bit, cc vngc bit cng nh mi quan h ca cc i lng mt cch trc quan thng quamt s v d sau: + Bn im c bit:

- V tr I: ( qmax = Q 0 ; i = 0 ; u L = - 0QC

= - L 2 Q0 )

=>Nng lng in cc i, nng lng t cctiu - V tr II: ( q = 0 ; i= - 0Q ; u L = 0 )=>Nng lng in cc tiu, nng lng t cci

- V tr III: (q =-Q 0 ; i = 0 ; u L max = 0QC

= L 2 Q0 )

=>Nng lng in cc i, nng lng t cctiu - V tr IV: ( q = 0 ; i max = 0Q ; u L = 0)=>Nng lng in cc tiu, nng lng t cc i

Vy chu k dao ng tun hon ca hm nng lng in v hm nnglng t ca dao ng in t ch bng chu k T ca hm in tch (q), khongthi gian nng lng in (nng lng t) t cc i chuyn thnh cc tiu hayngc li l chu k T ca hm in tch (q). + Bn vng c bit: Vng 1: q>0, i0 => Nng lng in gim, nng lng t tng Vng 4: q>0, i>0, u L Nng lng in tng, nng lng t gim +Mi quan h v pha ca in tch (q), cng dng in tc thi (i), inp trn hai u cun dy (u L ):

Qua hnh v thy c mi quan h v pha ca in tch (q), cng dngin tc thi (i), in p trn hai u cun dy (uL): / 2i q v

/ 2 Lu q i .

=> cng dng in tc thi (i) sm pha hn in tch (q) hay in p trn haiu t in (uC) mt gc / 2 , tr pha hn in p trn hai u cun dy (uL) mtgc / 2 .=> in p trn hai u cun dy (uL) sm pha hn cng dng in tc thi (i)mt gc / 2 . ngc pha vi in tch (q).

b. Phn dnh cho hc sinh vn dng, tnh ton tr li : Cu 01 . Mt mch dao ng l tng (LC). Dao ng in t ring (t do) camch (LC) c chu k 2,0.10-4 s . Nng lng in trng trong mch bin thin tunhon vi chu k l: A. 4,0.10 -4 s. B. 0,5.10 -4 s . C. 1,0.10 -4s . D. 2,0.10 -4 s. p n C.Cu 02.

Mt mch dao ng in t (LC) l tng gm cun cm thun c t

cm 5 H v t in c in dung 5 F. Trong mch c dao ng in t t do.Khong thi gian gia hai ln lin tip m in tch trn mi bn t in c lncc i l A. 5 .10 -6s. B. 2,5 .10 -6s. C . 10 .10 -6s. D. 10 -6s.

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p n A Cu 03. Mt mch dao ng in t t do gm mt cun dy c h s t cm L =0,636H v t in C = 0,255nF. Bit t in c tch in n hiu in th U0.Thi gian ngn nht nng lng in trng ca t gim t cc i n 0 l: A. t = 8.10 -5 (s). B. t = 4.10 -5(s). C. t = 2.10 -5(s). D. t = 10 -5(s). p n C Cu 04. Mt mch dao ng (LC) l tng. Bit in p tc thi gia hai bn t c

biu thc u = 60sin(104

t + /6)(V). Cun dy thun cm c h s t cm L = 1mH.Khong thi gian ngn nht m in tch ca t in tng t mt na cc i mn mt na cc i dng l A. t = 0,5.10 - 4 (s). B. t = 10 - 4 /3 (s). C. t = 0,67.10 - 4 (s). D. t = 10 - 4 /6 (s). p n B Cu 05. Mt mch dao ng l tng (LC) gm cun dy thun cm c L = 0,2mHv t in c in dung C = 8pF. Bit ban u t c cung cp mt nng lngW = 0,25.10 -3mJ. Chn gc thi gian lc dng in qua cun dy c gi tr cc i.Biu thc in p gia hai bn t in l? A. u = 250 cos (25.10 6t - /2)(V). B. u = 250 cos(25.10 6t + )(V).C. u = 250 cos (25.10 6t)(V). D. u = 220 cos (25.10 6t)(V). p n A Cu 06. Mt mch dao ng in t l trng c tn s dao ng l 0,5kHz, t inc in dung C = 1 F. in p cc i gia hai bn t l U0 = 100V. Chn gc thigian l lc q = - 3 Q0/2 th biu thc in tch ca t theo thi gian l A. q = 10 -4 cos (100 t + /2)(C). B. q = 10-4cos (100 t + /6)(C).C. q = 0,01cos (1000 t - /6) (mC). D. q = 0,1cos (1000 t + 7 /6)(mC) p n D Cu 07. Mt mch dao ng l trng (LC). in p hai bn t l u = 5sin10 4 t(V),

in dung C= 0,4( F). Biu thc cng dng in trong khung l: A. i = 2.10 -2 cos (10 4 t ) (A) B. i = 2.10 2 sin(10 4 t + ) (A)C. i = 2cos (10 4 t + /2 ) (A) D. i = 0,02cos(10 4 t- ) (A) p n A Cu 08. Mt mch dao ng (LC) l tng in tch trn t bin thin theo phngtrnh q = Q 0cos(7000t + /3) (C), vi t o bng giy. Thi im ln u tin nnglng in trng trong t in bng nng lng t trng trong cun dy l A. 74,8 s B. 14,96 C. 112,22 s D. 186,99 s p n C Cu 09.

Mt mch dao ng in t gm mt cun dy thun cm L = 4H v mtt in c in dung C. Trong qu trnh dao ng, cng cc i qua cun dyl 12mA. Khi cng dng in qua cun dy l 4mA th nng lng in trng t in l A. 3,2mJ. B. 0,256mJ. C. 0,320mJ. D. 0,288mJ p n B Cu 10. Mt mch in RLC ni tip c dng in i = I0sin(2 ft + )(A). in lngchuyn qua tit din thng ca dy dn trong 0,5 chu k, k t lc dng in b trittiu lA. I0 /2 f (C). B. 2I 0 / f (C) C. I 0 / f (C) D. I0 /4 f (C).

p n C 5. u im:Vic s dng vng trn lng gic cng mt lc vi 3 trc Ox, Ov, Oagip c ho chng ta c thun li nhiu vn sau:


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Th nht: Trnh c cc kin thc ton hc cao cp nh o hm, tchphn lm cho vt l khng b ton hc ha y cng l phn m a s hc sinh mi i tng u gp kh khn.

Th hai: Cng thng qua hnh v ny ta c th xc nh ngay lp tc cc gitr tng ng ca vn tc (v) v gia tc (a) khi bit li (x) hoc tm thy vn tc(v), ta (x) khi bit gia tc hoc ngc li ti mt thi im (t) hoc (t +t ) no. Gii quyt hu ht cc dng ton v dao ng iu ha nh thc trng nu.

iu ny nu tnh ton bng phng php i s rt lu v thng b sai. Th ba: Cng nh phn dao ng c hc phn dao ng in t t do mch dao ng l trng (LC) ta c th ch ngay c cng dng in tc thi(i) khi bit in tch (q) trn hai bn t, d dng tm thy ngay chu k dao ng tunhon ca nng lng in, nng lng t ca mch dao ng c th gii quythu ht cc dng ton ca chng ny theo yu cu ca thi tt nghip v ihc hin nay.

Th t: Khng cn nh nhiu cc cng thc c th ca to , vn tc, giatc ca phn dao ng iu ha v in tch trn t in, in p hai u t, inp hai u cun dy cng dng in trong mch dao ng l tng

(LC)Phng php ny lin kt c mt cch c h thng, trc quan cc miquan h c tnh cht tng qut, trng tm ca phn dao ng c, dao ng in t t hiu c bn cht trong qu trnh nhm tr li nhanh, chnh xc nhtcc cu hi dng l thuyt v bi tp theo yu cu ca bi. 6. Nhc im v khc phc:

Khi mi s dng ng trn lng gic mt s hc sinh cn hiu lm l: + Gin vc t ca hm (x), hm (v), hm (a) cho nn dn n kt qu sai l chohm (v) tr pha hn hm (x) [ hm (i) tr pha hn hm (q)] mt gc / 2 trnhtrng hp ny khi tnh lch pha ca cc hm s ta c theo nh ngha

12 1 2 t tr li hm no sm (tr) pha hn. + Trong mch in xoay chiu khng phn nhnh (RLC) hc sinh vn cho hm (i)sm pha hn hm (uC) mt gc / 2 cn lu cho hc sinh bit iu ny ch xy rakhi in tr thun R = 0( ).

CY CU S TNG HP DAO NG


12/26



PHN HAI TNG HP DAO NG IU HO IU HO CNG PHNG,

CNG TN S 1. M t tnh trng s vic hin ti:

Hin ti tng hp hai dao ng iu ho cng phng cng tn s nh sau: x1= A1cos( t+ 1) v x 2=A2cos( t + 2) ta c mt dao ng iu ho cngphng cng tn s x=Acos( t+ ).Trong : 2 2 21 2 1 2 2 12 os( ) A A A A A c

v 1 1 2 21 1 2 2

sin sintan

os os

A A A c A c

vi 1 2 (nu 1 2 )

* Nu = 2k (x1, x2 cng pha) AMax = A1 + A2* Nu = (2k+1) (x1, x 2 ngc pha) AMin = A1 - A2 => Tng qut bin dao ng : A1 - A2 A A1 + A2

Khi bit mt dao ng thnh phn x1 = A 1cos( t + 1) v dao ng tng hpx=Acos( t+ ) th dao ng thnh phn cn li l x2 = A2cos( t + 2).Trong : 2 2 22 1 1 12 os( ) A A A AAc

1 12

1 1

sin sintan

os os

A A Ac A c

vi 1 2 ( nu 1 2 )

Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cngtn s x1 = A1cos( t + 1, x2 = A2cos( t + 2) th dao ng tng hp cng l daong iu ho cng phng cng tn s:x = Acos( t + ).Chiu ln trc Ox v trc Oy trong h xOy. Ta c: 1 1 2 2os os os ... x A Ac Ac A c v 1 1 2 2sin sin sin ... y A A A A

2 2 x y A A A v tan

y

x

A

A vi [ Min, Max]Hoc song song vi cch trn th ngi ta biu din gin Fresnel t

tm bin A v pha ban u * Nhn thy mt s nhc im ca phng php ny khi lm trc nghim:

Mt nhiu thi gian biu din gin vct, i khi khng biu din cvi nhng bi ton tng hp t 3 dao ng tr ln, hay tm dao ng thnh phn.

Ta thy vic xc nh bin A v pha ban u ca dao ng tng hptheo phng php Frexnen l rt phc tp v d nhm ln khi thao tc nhp my i vi cc em hc sinh, thm ch cn phin phc ngay c vigio vin.

Vic xc nh gc hay 2 tht s kh khn i vi hc sinh bi v cngmt gi tr tan trong bi ton vt l lun tn ti hai gi tr ca v d tan =1 th

= / 4 hoc 3 / 4 vy chn gi tr no cho ph hp vi bi ton. Sau y, chng ti xin trnh by mt phng php khc nhm gip cc em

hc sinh v h tr gio vin kim tra nhanh c kt qu bi ton tng hp daong trn. 2. M t ni dung gii php mi :

a. C s l thuyt: Nh ta bit mt dao ng iu ho os x Ac t

+ C th c biu din bng mt vect quay A c di t l vi bin A v tovi trc honh mt gc bng gc pha ban u .+ Mt khc cng c th c biu din bng s phc di a*= Ae j ( )t v ccdao ng cng tn s gc c tr s xc nh nn thun tin trong tnh ton ngi


13/26



ta thng vitvi quy c a*= Ae j trong my tnh CASIO fx- 570ES k hiudi dng m lA .+ c bit gic s c hin th trong phm vi :-180 0< 180 0 hay < rt ph hp vi bi ton tng hp dao ng iu ho.

Nh vy vic tng hp cc dao ng iu ho cng phng, cng tn sbng phng php Frexnen cng ng ngha vi vic cng cc s phc biu din

ca cc dao ng . b. Gii php mi: (Cc thao tc vi my tnh CASIO fx 570ES )Chn ch mc nh ca my tnh:

+ tnh dng to cc :A . Bm my tnh nh sau: 3 2SHIFT MODE + tnh dng to cc: a + ib. Bm my tnh nh sau: 3 1SHIFT MODE

thc hin cc php tnh v s phc th ta phi chnMode ca my tnh dng Complex (dng s phc) pha trn mn hnh xut hin chCMPLX . Tabm my nh sau: 2 MODE

ci t n v o gc (Deg, Rad, Gra) cng c tc dng vi s phc. Nutrn mn hnh hin th k hiuD th ta phi nhp cc gc ca s phc c n v ogc l . Nu mn hnh hin th k hiuR th ta nhp cc gc vi n vi rad. Chnch ny c th bm my nh sau:SHIFT 3 MODE l chn ch tnh theo ,cn bm my SHIFT 4 MODE l chn ch tnh theo rad. Kinh nghim cho thynhp vi n v nhanh hn n v rad nhng kt qu sau cng cn phi chuynsang n v rad i vi nhng bi ton cho theo n v rad.

nhp k hiu gc ca s phc ta n SHIFT .V d: Dao ng 6 os / 3 x c t s c biu din vi s phc6 60 hay 6 / 3

ta nhp my nh sau: - Ch tnh theo (D) : 6 SHIFT 60 s hin th l 6 60 .

- Ch tnh theo rad (R): 6 SHIFT ( :3) s hin th l6 / 3 . 3. Chng minh tnh kh thi ca gii php mi:

a. tm dao ng tng hp ta thc hin php tnh cng: Cu 1: Dao ng tng hp ca hai dao ng iu ha cng phng, cng tn s x 1 = a 2 .cos( t+ /4)(cm), x 2 = a.cos( t + ) (cm) c phng trnh dao ngtng hp l

A. x = a2

.sin( t +2 /3)(cm) B. x = a.sin( t + /2)(cm)C. x = 3a/2.sin( t + /4)(cm) D. x = 2a/3.sin( t + /6)(cm)Gii: Tin hnh nhp my: Ch tnh (D). Tm dao ng tng hp

2 45 + 1 180 . H in th:1 90 , chn B Cu 2: Ba dao ng iu ha cng phng, cng tn s c phng trnh ln ltl x 1 = 4 cos( t - /6) (cm) , x 2 = 5cos( t - /2) cm v x 3=3cos(20t+2 /3) (cm).Dao ng tng hp ca 3 dao ng ny c bin v pha ban u l A. 4,82cm; -1,15 rad B. 5,82cm; -1,15 rad

C.4,20cm; 1,15 rad D.8,80cm; 1,15 radGii: Tin hnh nhp my: Ch tnh rad (R). Tm dao ng tng hp 4 / 6 5 / 2 3 2 / 3 . Hin th:4.82... 1,15... chn A


14/26



b. tm dao ng thnh phn ta thc hin php tnh tr: Cu 3: Mt cht im dao ng iu ho c phng trnh dao ng tng hpx=5 2 cos( t + 5 /12) vi cc dao ng thnh phn cng phng, cng tn s lx 1 =A1 cos( t + 1 ) x 2 =5cos( t+ /6 ), pha ban u ca dao ng 1 l: A. 1 = 2 /3 B. 1 = /2 C. 1 = /4 D. 1 = 3 / Gii: Tin hnh nhp my: Ch tnh rad (R). Tm dao ng thnh phn 5 2 5 /12 5 / 6 . Hin th:5 2 / 3 , chn A

Cu 4: Mt vt ng thi tham gia 3 dao ng cng phng, cng tn s cphng trnh dao ng: x

1= 2 3 cos(2 t + /3) cm, x

2= 4cos(2 t + /6) cm v

phng trnh dao ng tng hp c dng x = 6cos(2t - /6) cm. Tnh bin daong v pha ban u ca dao ng thnh phn th 3: A. 8 cm v - /2 . B. 6cm v /3. C. 8cm v /6 . D. 8cm v /2.Gii: Tin hnh nhp my: Ch tnh rad (R). Tm dao ng thnh phn th 3

6 / 6 2 3 / 3 4 / 6 . Hin th :8 / 2 , chn A * Lu : + Khi thc hin cc php tnh m kt qu php tnh c hin th c th di dngi s a+bi . Tc l cha mc nh dngA . Hoc c dng A cn chuynqua dng a + bi. Ta phi chuyn kt qu ny v li dng cn thit. Bng cch: - Chuyn t dng to cca + bi sang dng to cc A : SHIFT 2 3 v d: 8 SHIFT ( :3) hin th: 4+4 3 i ta cn chuyn sang dng A bm

SHIFT 2 3 s c kt qu l : 8 / 3 - Chuyn t dng to ccA sang dng to cca + bi : SHIFT 2 4 v d: 8 SHIFT ( :3) hin th : 8 / 3 ta cn chuyn sang dng a+bi bmSHIFT 2 4 s c kt qu l : 4+4 3 i

+ Theo kinh nghim th cn chn ch mc nh theo dng to ccA biton nhanh hn, v thc t trong phn tng hp dao ng cha cn thit s dngdng cc. 4. Mt s vn lin quan v vn dng:

a.Vn lin quan: Hin ti trn mng inter net c ti liu hng dn cc thao tc s dng vimy tnh CASIO fx 570MS nhng y l loi my c cu hnh yu hn my tnhCASIO fx 570ES (c php mang vo phng thi) m chuyn ny cpn. Mt khc kt qu him th caCASIO fx 570MS v bin A ri sau lgc lch phi thng qua mt bc tnh na, cn my tnhCASIO fx 570EShin th ng thi.

b. Phn dnh cho hc sinh vn dng, tnh ton, tr li: Cu 01: Mt vt thc hin ng thi hai dao ng iu ho cng phng, cng tns x1=cos(2 t + )(cm), x 2 = 3 .cos(2 t - /2)(cm). Phng trnh ca dao ngtng hp A. x = 2.cos(2 t - 2 /3) (cm ) B. x = 4.cos(2 t + /3) (cm)C. x = 2.cos(2 t + /3) (cm) D. x = 4.cos(2 t + 4 /3) (cm) p n A


15/26



Cu 02: Cho hai dao ng iu ha cng phng, cng tn s x1= 3cos(5 t+ /2) (cm) v x 2 = 3cos( 5 t + 5 /6)(cm). Phng trnh dao ng tng hp l A. x = 3 cos ( 5 t + /3) (cm). B. x = 3 cos ( 5 t + 2 /3) (cm).C. x= 2 3 cos ( 5 t + 2 /3) (cm). D. x = 4 cos ( 5 t + /3) (cm) p n B Cu 03: Dao ng tng hp ca hai dao ng iu ho cng phng, cng tn sx1=cos(10 t+ / 3 )(cm) v x 2 = 2cos(10 t + )(cm). Phng trnh dao ng tnghp A. x = 2 cos(10 t + 4 )(cm) B. x = 3 cos(10 t + 5 6 / )(cm) C. x = 2cos(10 t + 2 )(cm) D. x = 2 3 cos(10 t + 4 )(cm) p n B Cu 04: Mt vt thc hin ng thi hai dao ng iu ha cng phng, cng tns theo cc phng trnh: x

1= - 4sin( t ) v x

2= 4 3 cos( t) cm. Phng trnh

ca dao ng tng hp A. x

1= 8cos( t + /6) cm B. x

1= 8sin( t - /6 ) cm

C. x1 = 8cos( t - /6 ) cm D. x1 = 8sin( t + /6 ) cm p n A Cu 05: Mt vt tham gia ng thi hai dao ng iu ho cng phng, cng tns c phng trnh ln lt l x1 = 2.sin(10t - /3) (cm), x 2 = cos(10t + /6) (cm) (to bng giy). Xc nh vn tc cc i ca vt. A. 5 (cm/s) B. 20 (cm) C. 1 (cm/s) D. 10 (cm/s) p n D Cu 06: on mch xoay chiu co in tr thun , cun dy thun cam va tinmc ni tip. B la mt im trn AC v i u AB = cos100 t (V) v uBC = 3cos(100 t-

/2 ) (V). Tm biu thc in p uAC.A.

AC Cosu 2 2 (100 t) V B. AC cosu 2 100 t / 3 V C. AC osu 2c 100 t / 3 V D. AC osu 2c 100 t / 3 V p n D Cu 07: Mt vt ng thi thc hin 3 dao ng cng phng, cng tn s cphng trnh dao ng ln lt l : 1x 2 3cos 2 t / 3 cm ,

2x 4cos 2 t / 6 cm v 3x 8sin 2 t cm . Gi tr vn tc cc i ca vt vpha ban u ca li dao ng tng hp ln lt l: A. 12 (cm/s) v / 3 (rad) B. 16(cm/s) v / 6 (rad)C. 16cm/s v / 6 (rad) D . 12(cm/s) v - / 6 (rad) p n D Cu 08: Mch in xoay chiu ba pha mc sao c dy trung ho. Cng dngin tc thi qua cc dy pha l 5 2 os(100 )1 c t i (A); 8 3 os(100 2 /3)2i c t (A)v 3 5 os(100 2 /3)3i c t

(A). Cng dng in tc thi qua dy trung ho lA. 6,97 os(100 2,05)i c t (A) B. 5,97 os(100 2,05)i c t (A)C. 6,97 os(100 1,09)i c t (A) D. 5,97 os(100 1,09)i c t (A) p n A Cu 09:

Mt vt thc hin ng th i 4 dao ng iu hoa cung phng , cng tns co cc phng trnh: x1 = 3sin( t + ) cm, x 2 = 3cos t (cm), x 3 = 2sin( t + ) cm,x4 = 2cos t (cm). Phng trnh dao ng tng hp cua vt . A. 25 cos( / 4) x t cm B. 5 2 cos( / 2) x t cm


16/26



C. 5cos( / 2) x t cm D. 5cos( / 4) x t cm p n A Cu 10: Mt vt thc hin ng th i 4 dao ng iu hoa cung phng , cng tns co cac phng trnh x1= 3 cos t 2 (cm), x 2 = 3 3 cos(

22

t )(cm),

x3=6cos(3

42

t ) (cm) , x 4= 6cos(

3

22

t )(cm). Phng trnh dao ng tng hp

ca vt.A. x=6cos(

3

42

t )(cm) B. x = 6cos(

3

22

t )(cm)

C. x = 12cos(3

42

t ) (cm) D. x= 12cos(

3

t )(cm)

p n A 5. u im:

Th nht: Thc hin nhanh c bi ton tng hp vinhiu dao ng vpha ban u ca cc dao ng c th c tr s bt k . iu ny c minhchng trong lp bi dng SGK nm 2008-2009 ca c hai t pha bc tnh vv nam tnh bi ton s 2 v thi lng nu tnh bng phng php gin Fresnel mt 15 pht cn gii bng phng php s dng my tnh mt khong 2pht

Th hai: L phng php ti u v c th ni l duy nht tnh cc daong tng hp t 3 hoc 4 dao ng thnh phn tht nhanh v chnh xc.

Th ba: Khi tnh ton bng hm phc th gi tr ca l chnh xc, duy nhtcn tnh theo hm tan ta phi chn nghim, ngoi ra cn tn rt nhiu thao tc.6. Nhc im v khc phc:

Do hc sinh khng c trang b l thuyt v s phc nn vic dng my

tnh ban u c th gp rc ri m khng bit cch khc phc. (v d nh MODE,ch Deg, Rad, ). Nhng thao tc my nm ba ln ri s quen, v cng khngcn thit bit my tnh thc hin tnh ton hm phc nh th no.

Tc thao tc ph thuc nhiu vo cc loi my tnh khc nhau, khngdng cho cc loi my tnh c cu hnh yu hn. (Nhc im ny, gio vin cth khc phc d. Nhng vi hc sinh, cha c my tnhfx 570ES c th muagi khong 250.000 ng ).

Khi tr v ch tnh c bn thng qun khng chn li ch tnh bnhthng 1 MODE . Nu qun iu ny th kt qu tnh ton cc php tnh c bnca cc bi ton tip theo s b sai cn lu iu ny.

HIN TNG SNG


17/26



PHN BA GIAO THOA SNG C

1. M t tnh trng s vic hin ti : Hin ti sch gio khoa cung cp kin thc v giao thoa nh sau:

a) Hai ngun dao ng cng pha, cng bin : Giao thoa ca hai sng pht ra t hai ngun sng kt hp S1, S 2 cch nhau

mt khongl . Xt im M cch hai ngun ln lt d1, d 2. Phng trnh sng ti haingun cng pha c dng 1 2 Acos(2 )u u u ft + Phng trnh sng ti M do hai sng t hai ngun truyn ti:

11 Acos(2 2 ) M

d u ft

v 22 Acos(2 2 ) M

d u ft

+ Phng trnh giao thoa sng ti M:u M = u 1M + u 2M => C lch pha: 1 22 1 1 2

22 M

d d d d

=> C bin : 2 os 2 M A A c

- Nu dao ng cc i: 1 2 0, 1, 2, 3...d d k k

- S ng hoc s im (khng tnh hai ngun): l lk

- Nu dao ng cc tiu: 1 2 0,5 0, 1, 2, 3...d d k k

- S ng hoc s im (khng tnh hai ngun): 1 12 2

l lk

b) Nu hai ngun dao ng thnh phn ngc pha nhau ta m rng

thm: - Nu dao ng cc i: d1 d2 = (k+0,5) 0, 1, 2, 3...k - S ng hoc s im (khng tnh hai ngun): 1 1

2 2l l

k

- Nu dao ng cc tiu: d1 d2 = k 0, 1, 2, 3...k -S ng hoc s im (khng tnh hai ngun): l lk

* Nhn thy mt s nhc im ca phng php ny khi lm trc nghim: Vn rt kh khn hin nay l cc cu hi, bi tp trc nghim yu cu

gii cc bi ton tng qut khi dao ng khng cng pha, ngc pha m lch pha

nhau mt gc bt k. Cng nh vic tng hp hai sng khng cng bin . iuny bt buc hc sinh phi lm li bi ton t u mt nhiu thi gian, cha chcchnh xc.

Sau y, chng ti xin trnh by mt phng php khc bng cch yu cuhc sinh cn nh 2 cng thc c bn nhng tng qut nht ca chng ny. Tuys lng cng thc khng nhiu nhng n c th gii quyt hu ht cc dng tonca phn giao thoa sng c khng nhng p ng tt cho cc bi thi tt nghip mc cc bi thi tuyn sinh i hc. 2.M t ni dung gii php mi :

a. C s l thuyt: Giao thoa ca hai sng pht ra t hai ngun sng kt hp S1, S 2 cch nhaumt khongl . Xt im M cch hai ngun ln lt d1, d 2.

+ Phng trnh sng ti 2 ngun l 1 1 1A cos(2 )u ft v 2 2 2A cos(2 )u ft + Phng trnh sng ti M do hai sng t hai ngun truyn ti:


18/26



11 1 1A cos(2 2 ) M

d u ft

v 22 2 2A cos(2 2 ) M

d u ft

+ Phng trnh sng ti M do s tng hp hai sng:u M = u 1M + u 2M b. Gii php mi: Ta c th tm hm sng u M thng qua tm bin A v pha ban u M bng

phng php s dng my tnh casio fx 570ES khi bit c c gi tr t hm (u1M )v (u 2M ) nh phn hai trnh by.

Ta c th rt ra 2 cng thc cn nh gii cc dng ton phn sng c: - lch pha ca hai sng t hai ngun n M l:

2 1 1 22

M M M d d

(1) vi 2 1

=> hiu ng i ca sng t hai ngun n M l: 1 2 ( ) 2 M d d

(2)

i vi hc sinh ch cn nh cc cng thc (1), (2) bng cch hc thuc (1)suy ra (2), khi tnh ton cn lu vi d kin bi ton cho v yu cu ca bi.

Cn ch : - 2 1 l lch pha ca hai sng thnh phn ca ngun 2 so vi ngun 1 - 2 1 M M M l lch pha ca hai sng thnh phn ti M ca ngun 2 so vingun 1 do sng t ngun 2 v ngun 1 truyn n - C th tnh bin bng cng thc: 2 2 21 2 1 2 M2 os A A A A A c

Vi bi ton tm s ng dao ng cc i, cc tiu hoc theo yu cu no ca bi ton gia hai im M, N cch hai ngun ln lt l d1M, d 2M, d 1N, d 2N.Lc ta t dM= d 1M - d 2M ; dN = d 1N - d 2N v gi s dM < dN Th ta c:

dM < 1 2 ( ) 2 M d d

< dN. Vi s gi tr nguyn ca k tha mn biu thc

trn l s ng cn tm. 3. Chng minh tnh kh thi ca gii php mi:

chng minh ta vn dng 2 cng thc trn tr li cc cu hi sau:Cu 01: Ti hai im A v B trn mt cht lng c hai ngun pht sng dao ngtheo phng thng ng vi cc phng trnh ln lt l u1 = a 1cos(50 t + /2) vu2 = a 2cos(50 t + ). Tc truyn sng trn mt cht lng l 1 (m/s). Mt im Mtrn mt cht lng cch cc ngun ln lt ld1 v d 2. Xc nh iu kin Mnm trn ng cc i? (vi k l s nguyn) A. d 1 - d 2 = 4k + 2 (cm) B. d 1 - d 2 = 4k + 1 (cm)C. d 1 - d 2 = 4k - 1 (cm) D. d 1 - d 2 = 2k - 1 (cm)Gii:Bc sng 100 4( )

25v

cm f

. Dng cng thc (2) 1 2 ( ) 2 M d d

V M

nm trn ng cc i nn 2 M k => (d1- d 2) =4

(2 / 2) 4 12

k k

=>chn C

Cu 02: Trn mt nc c hai ngun pht sng kt hp S1 v S 2, dao ng theocc phng trnh ln lt l: u1 = a 1cos(50 t + /2) v u 2 = a 2cos(50 t). Tc truyn sng ca cc ngun trn mt nc l 1 (m/s). Hai im P, Q thuc h vngiao thoa c hiu khong cch n hai ngun l PS1 - PS 2 = 5 cm, QS 1- QS 2 = 7cm. Hi cc im P, Q nm trn ng dao ng cc i hay cc tiu? A. P, Q thuc cc i B. P, Q thuc cc tiu C. P cc i, Q cc tiu D. P cc tiu, Q cc i


19/26



Gii: Bc sng 100 4( )

25

vcm

f . Dng cng thc (1) 1 2

2 M d d

=> 2 .5 24 2P

=2k => im P thuc cc i

=> 2 .7 3 (2 1)4 2Q

k

=> im Q thuc cc tiu=> chn C Cu 03: Trn mt nc hai ngun sng A v B dao ng iu ho theo phngvung gc vi mt nc vi phng trnh u1 = u 2 = acos(10 t). Bit tc truynsng 20(cm/s), bin sng khng i khi truyn i. Mt im N trn mt nc chiu khong cch n hai ngun A v B tho mn AN- BN = 10 cm. im N nmtrn ng ng yn .. k t trung trc ca AB v v . A. th 3- pha A B. th 2 - pha A C. th 3 - pha B D. th 2 - pha BGii:

Bc sng 20 4( )5

vcm

f . Dng cng thc (1) 1 2

2 M d d

=> 2 .10 0 5 (2 1) 24 N

k k

=> im N nm trn ng ng yn th 3 v pha B v d1> d 2 => chnC Cu 04: Ti hai im A v B trn mt cht lng cch nhau 10 cm c hai ngun phtsng kt hp dao ng theo phng trnh u1 = acos(10 t), u 2 = bcos(10 t + ). Tc truyn sng trn mt cht lng 20 (cm/s). Tm s cc tiu trn on AB. A. 5 B. 6 C. 4 D. 3Gii:

Bc sng20

4( )5

vcm

f

. Dng cng thc (2) 1 2( )

2 M d d

V M nm trn ng cc tiu nn (2 1) M k =>(d 1-d2) = 4

(2 1) ) 42

k k

m AB < d 1-d 2 < AB nn ta c -2,5 < k < 2,5 c 5 im cc tiu => chn A Cu 05: Hai ngun sng kt hp A, B cch nhau 21cm dao ng theo cc phngtrnh u 1= acos(4 t), u 2 = bcos(4 t + ), lan truyn trong mi trng vi tc 12(cm/s).Tm s im dao ng cc i trong khong AB A. 7 B. 8 C. 6 D. 5Gii:

Bc sng12

6( )2v

cm f . Dng cng thc (2) 1 2 ( ) 2 M d d

V M nm trn ng cc i nn 2 M k => (d 1-d2) =6

(2 ) 6 32

k k

m AB < d 1-d 2 < AB nn ta c - 3 < k < 4 c 6 im cc i=> chn C Cu 06: Ti hai im A v B trn mt cht lng cch nhau 15 cm c hai ngun phtsng kt hp dao ng theo phng trnh u1 = acos(40 t), u 2 = bcos(40 t + ). Tc truyn sng trn mt cht lng 40 (cm/s). Gi E, F l hai im trn on ABsaocho AE = EF = FB. Tm s cc i trn on EF. A. 7 B. 6 C. 5 D. 4Gii: Bc sng 40 2( )

20

vcm

f . Dng cng thc (2) 1 2 ( ) 2 M d d


20/26



V M nm trn ng cc i nn 2 M k => (d 1-d2) =2

(2 ) 2 12

k k

m AB/3 d1-d 2 AB/3 nn ta c -2 k 3 c 6 im cc i (s d c dubng v EF nm gia AB)=> chn B Cu 07: Ti hai im A v B trn mt nc cch nhau 16 cm c hai ngun phtsng kt hp dao ng theo phng trnh u1 = acos(30 t), u 2 = bcos(30 t + /2).Tc truyn sng trn mt nc 30 (cm/s). Gi E, F l hai im trn on AB saocho AE = FB = 2 cm. Tm s cc tiu trn on EF. A. 10 B. 11 C. 12 D. 13Gii:

Bc sng 30 2( )15

vcm

f . Dng cng thc (2) 1 2 ( ) 2 M d d

V M nm trn ng cc tiu nn (2 1) M k

=> (d 1-d2) = 2

(2 1) / 2) 2 0,52

k k

m 12 d1- d 2 12

ta c -6,25 k 5,75 c 12 im cc tiu (s d c du bng v EF nm trongon gia AB)=> chn C Cu 08: Trn mt nc c hai ngun kt hp A v B cch nhau 8 cm, dao ngtheo phng trnh ln lt l u1 = acos(8 t), u 2 = bcos(8 t). Bit tc truyn sng4cm/s. Gi C v D l hai im trn mt nc sao cho ABCD l hnh ch nht ccnh BC = 6cm. Tnh s im dao ng vi bin cc tiu trn on CD. A. 8 B. 9 C. 10 D. 11 Gii:

Bc sng 4 1( )4

vcm

f .

Dng cng thc (2) 1 2 ( ) 2 M d d

V M nm trn ng cc tiu nn(2 1) M k

=> (d 1-d2) = 1

(2 1) 0) 0,52

k k

m 4 d1-d2 4 nn ta c - 4,5 k 3,5 c 8 im cc tiu=> chn A Cu 09: Trn mt cht lng c hai ngun kt hp A v B cch nhau 8 cm, dao ng

theo phng trnh ln ltl u 1 = acos(8 t), u 2 = bcos(8 t + ). Bit tc truynsng 4 cm/s. Gi C v D l hai im trn mt cht lng m ABCD l hnh ch nhtc cnh BC = 6 cm. Tnh s im dao ng vi bin cc i trn on CD. A. 8 B. 9 C. 10 D. 11 Gii:Bc sng 4 1( )

4v

cm f

.

Dng cng thc (2) 1 2 ( ) 2 M d d

V M nm trn ng cc i nn 2 M k

=> (d 1-d2) =

1(2 ) 0,5

2k k

m 4 d1-d 2

4 nn ta c -3,5 k 4,5 c 8 im cc tiu=>

10cm

A B

CD

8cm

6cm

10cm

A B

CD

8cm

6cm


21/26



chn A Cu 10: Hai ngun S1 v S 2 dao ng theo cc phng trnh u1 = a 1cos(90 t)cm,u2=a 2cos(90 t + /4)cm trn mt nc. Xt v mt pha ng trung trc caS 1S 2 ta thy vn bc n i qua imM c hiu s MS1-MS 2 = 13,5 cm v vn bc n + 2(cng loi vi vn n) i qua imM' c MS1-MS2 = 21,5 cm. Tm tc truynsng trn mt nc, cc vn l cc i hay cc tiu? A. 25cm/s, cc tiu B. 180 cm/s, cc tiu C. 25cm/s, cc i D. 180cm/s, cc i Gii: Xt (d 1-d 2) = MS 1-MS 2 = 13,5 = n v (d1- d 2) = MS1-MS2 = 21,5 = (n+2) ta c: 2 =8 => = 4 (cm) vy v = 4 45 180( / ) f x cm s

Dng cng thc (1) 2 1 1 22

M M M d d

=> 2 .13,5 6,5 ( 0,5)4 4 M

k

vy cc vn l cc tiu=> chn B

4. Mt s vn lin quan v vn dng: a.Vn lin quan: Mt trong nhng cu hi v giao thoa rt thng gp trong cc thi vt l

l tm s cc i (gn sng) v cc im cc tiu (khng dao ng nu 2 ngun ccng bin ). y l vn khng kh, nhng khi gii quyt ta thng nhm lnti v tr ca ngun l mt cc i (hay cc tiu) dn n s nghim thng d 2.

Rt nhiu sch bi tp ang bn th trng cng thng nhm ln nhtrn, gy khng t hoang mang cho ngi c, v mi tc gi li c cch gii quytkhc nhau cng mt vn .

Chng ti ng vi quan im Ti v tr ca ngun trong hin tng giao thoa (v sng dng) khng th l cc i hay cc tiu vi l do sau: + n gin ta xt 2 ngun A, B trn mt nc dao ng cng pha c phngtrnh sng u A= u B = acos2ft. Ngha l bin dao ng ti ngun la.+ Ti im M trn mt nc ni hai sng cng pha s xut hin cc i vi bin 2a , v nu hai sng ngc pha th bin bng 0 tc cc tiu hay ng yn. + Th v th khng gian u = f(x) tc hnh nh mi trng vo mt thi im nhtnh trn mt nc ta quan st c nh sau:

+ Gi A l ngun, M1 l im cc tiu, M2 l im cc i nu ta k mt ngthng (1) // Ox qua M 2 th ng ny cch Ox mt khong 2a cn ng thng(2)// Ox qua A s cch Ox mt khong 1a nh vy d thy ngun A ch c thnm trn (2) ngha l ngun A khng th trng im cc i M2 hay im cc tiuM1.+ Mt khc trong th khng gian (Oxu) th chu k chnh l bc sng = M1M3 tacng d dng chng minh dc theo Ox ngun A c bin l a cch M1 mt

khong d = /12 v nu n nm trong M1 th s cch M2 mt khong d= /6 (gingnh thi gian i t x = A/2 n O hoc t x = A/2 n x = A trong dao ng iuha)

Th l r l ngun A khng th l cc i hay cc tiu nh vy tm scc i trn AB ta nn lm nh sau:

A

M 3

M 2

M 1 x

1 u

2


22/26



- AB < 1 2 ( ) 2 M d d

< AB => (khng th bng) => Tm c bao nhiu gi tri k tha mn khong h trn l c by nhiu cc i. + Thng thng n gin ha vn ngi ta thng dng cu xem nh ngun rt gn cc tiu (tc l nt trong sng dng) v thc ra /12 l khng ngk so vi AB khi tm s cc tiu ta c th cho:- AB 1 2 ( )

2 M d d

AB =>Ngha l khong k tm c l khong kn!

+ i vi cc bi gii khi tm cc i li cho:-AB 1 2 ( ) 2 M d d

AB

tc cng nhn ngun trng cc i th kt qu s d 2 gi tr khi gp trng hpAB chia c mt s nguyn. Rt may mn l dng y thng rt t gp c lngi ra s phi bn ci nhiu chng?

b.Phn dnh cho hc sinh vn dng, tnh ton, tr li : Cu 01: Trn mt cht lng, ti A v B cch nhau 6 cm c hai ngun dao ng kthp uA = u B = 0,5 sin100 t (cm). Tc truyn sng v = 60 cm/s. Ti im M trnmt cht lng cch A, B nhng khong d1= 4,2 cm, d 2 =1,8 cm thuc vn cc ibc A. k = 0. B. k =1. C. k =3. D. k =2. p n D Cu 02: Hai ngun kt hp A,B cch nhau 16cm ang cng dao ng vung gcvi mt nc theo phng trnh x = a sin50 t (cm). C l mt im trn mt ncthuc ng cc tiu, gia C v trung trc ca AB c mt ng cc i. BitAC= 17,2cm. BC = 13,6cm. S ng cc i i qua cnh AC l A. 16 ng B. 6 ng C. 7 ng D. 8 ng p n D Cu 03: Hai im M v N trn mt cht lng cch 2 ngun O1O 2 nhng on lnlt l O1M =3,25cm, O 1N =33cm , O 2M = 9,25cm, O 2N= 67cm, hai n gun daong cng tn s 20Hz, tc truyn sng trn mt cht lng l 80cm/s. Hai imny dao ng th no A. M ng yn, N dao ng mnh nht. B. M dao ng mnh nht, N ng yn. C. C M v N u dao ng mnh nht. D. C M v N u ng yn. p n DCu 04: Ti hai im A n B trn mt nc dao ng cng tn s 16Hz, cng pha,cng bin . im M trn mt nc dao ng vi bin cc i vi MA = 30cm,MB = 25,5cm, gia M v trung trc ca AB c hai dy cc i khc th tc truyn sng trn mt nc l A. v = 36cm/s. B. v =24cm/s. C. v = 20,6cm/s. D. v = 28,8cm/s. p n B Cu 05: Cho 2 ngun pht sng m cng bin , cng pha v cng tn s f =440Hz, t cch nhau 1m. Hi mt ngi phi ng u khng nghe thy m. Cho tc ca m trong khng kh bng 352m/s. A. 0,3m k t ngun bn tri. B. 0,3m k t ngun bn phi. C. 0,3m k t 1 trong hai ngun D. Ngay chnh gia, cch mi ngun 0,5m p n C Cu 06: Ti hai im S1, S 2 cch nhau 10cm trn mt nc dao ng cng tn s50Hz , cng pha cng bin , tc truyn sng trn mt nc 1m/s. Trn S1S 2 c bao nhiu im dao ng vi bin cc i v khng dao ng tr S1, S 2 : C.. im dao ng vi bin cc i v .. im khng dao ng. A. 9 - 9 . B. 11 - 10 . C. 10 - 11 . D. 9 - 10 . p n D


23/26



Cu 07: Ti hai imA v B cch nhau 8m c hai ngun m kt hp c tn s m440Hz, tc truyn m trong khng kh l 352m/s. Trn AB c bao nhiu imnghe to v bao nhiu im nghe nh: C .. im m nghe tr A, B v. im nghe ... A. 21 - nh - 18 - to. B. 19 - to - 20 - nh. C. 19 - nh - 20 - to. D. 21 - to - 20 -nh. p n C Cu 08: Hai im A, B trn mt nc dao ng cng tn s 15Hz, cng bin vcng pha, tc truyn sng trn mt nc l 22,5cm/s, AB = 9cm. Trn mt ncquan st c bao nhiu gn li tr A, B ? A. c 13 gn li. B. c 11 gn li. C. c 10 gn li. D. c 12 gn li. p n B Cu 09: Ti hai im A v B (AB = 16cm) trn mt nc dao ng cng tn s50Hz, cng pha, tc truyn sng trn mt nc 100cm/s . Trn AB s im daong vi bin cc i l A. 13 im B.15 im . C. 16 im . D. 14 im . p n B Cu 10: Hai im M v N (MN = 20cm) trn mt cht lng dao ng cng tn s50Hz, cng pha, tc truyn sng trn mt cht lng l 1m/s . Trn MN s imkhng dao ng l A. 18 im. B. 19 im. C. 21 im. D. 20 im. p n D 5. u im:

Th nht: Nh vy vi 2 cng thc cn nh (1) l 1 22

M d d

v

cng thc (2) l 1 2 ( ) 2 M d d

ta c th vn dng gii tt cc bi ton tm

s cc i, cc tiu, lch pha ra sao ca nhiu dng bi ton m dao ng cahai ngun khng nht thit phi cng pha, ngc pha nh sch gio khoa. Th hai: Gii quyt hu nh trn vn tt c dng bi tp ca chng trnh

giao thoa sng c nu bit kt hp cng thc (1), (2) vi phng php tm bin v pha ban u ca dao ng tng hp bng my tnh CASIO fx 570ES thkhng cn phi nh nhiu cc cng thc c th.

Th ba: Rn luyn c kh nng t duy t hc cho hc sinh thng quabn cht qu trnh truyn sng ta c th gii cc bi ton giao thoa nh sng nucoi khong vn (i): 2i= 6. Nhc im v khc phc: -Vic hc sinh nh c cng thc (1), (2) lc u l kh khn v vy cn thchnh vn dng cng nhiu cng tt.

S GIAO THOA SNG


24/26



PHN M T TH HIN MC TRIN KHAI CA TI

+ Trong nm hc 2008 2009: ti c chng ti vn dng trong vic ging dy lp 12L, 12T, 12Si vi ktqu im thi tnh trung bnh nh sau:

BNG TNH IM BNH QUN MN L (Theo s liu thng k ca nh trng lu vn phng)

Lp Thi tt nghip nm 2008-2009 Thi i hc nm 2008-200912L 9,55 8,9012T 9,61 8,8312Si 9,30 x

+ Trong nm hc 2009 2010: ti c chng ti vn dng trong vic ging dy lp 12L, 12T, 12Si, 12A1.Bi kim tra gm 20 cu hi phn mt vng trn lng gic trong thi gian 30pht kt qu im c thng k phn loi nh sau:

BNG TNH IM CC BI KIM TRA (Theo s liu kho st ca ti )

Lp S lng (%)Km

(0 =>2.0)Yu

(2.5=>5.0)Trung bnh(5.5=>6.0)

Kh(6.6=>7.5)

Gii (8.0=>10)

12L 23 00%0

0%3

13,04%2

8,70%18

78,26%

12T 25 00%0

0%5

20,00%3

12,00%17

68,00%

12Si 30 00%1

3,33%7

23,33%6

20,00%16

53,34%

12A1 40 12,50%2

5,00%10

25,00%9

22,50%18

45,00%

Bng tng hp:

i tng Slng (%)

Km(0

=>2.0)

Yu (2.5=>5.0)

Tr. bnh(5.5=>6.0)

Kh(6.6=>7.5)

Gii (8.0=>10)

Kho stban u

130 43,08%

107,69%

3224,62%

2116,15%

6348,46%

Kt qu pdng

118 10,85%

32,54%

2521,19%

2016,95%

6958,47%

+ Nhn xt: Qua theo di kt qu nhn thy s hc sinh c tip xc, bi dng thng

qua ti c cch gii thnh tho hn, m bo thi gian lm bi theo yu cu,cu tr li chnh xc hn, t kt qu im kh cao.


25/26



C. KT LUN 1. Khi qut cc kt lun cc b:

Trong khun kh chuyn ny, chng ti mong mun gip cho hc sinhnh ti thiu nhng kin thc c bn, c tnh tng qut ca chng trnh vt l 12 cc chng dao ng c, sng c, dao ng in t. ng thi cung cp chocc em mt s phng php v th thut nhm gii quyt nhanh, chnh xc ccdng ton trong chng trnh theo yu cu ca cc thi TNPT v TSH. Vi chuyn ny chng ti cn lu dnh phn cho hc sinh t rn luyn,vn dng cc phng php v th thut hc sinh t chim lnh tri trc v phthuy tnh c lp sng to, t c th suy ngh tm ti phng php ring ca bnthn, y cng l mc tiu rn luyn gio dc hc sinh theo hng Thy thit k,Tr thi cng. 2. Li ch v kh nng vn dng:

Bn thn ti p ng tt cho yu cu v lm bi trc nghim vi mc chtr li nhanh, chnh xc, loi b c yu t ton hc phc tp ca phn lnggic, o hm, tch phn.... C th tng phn ta thy:

Phn mt v vng trn lng gic n th hin rt r tnh cht trc quan giacc mi quan h v li , vn tc, gia tc cng nh ng nng, th nng hoc phn dao ng in t th in tch, in p hai u t in, hai u cun dy,cng dng in cng nh nng lng in, nng lng t tng thi im,v pha dao ng, v tn s dao ng, v s tng hoc gim ca cc i lng, vgi tr ca cc i lng trong cng thi im ....

Phn hai tng hp dao ng cng phng cng tn s bng phng phps dng my tnh casio fx- 570 ES rn luyn hc sinh thao tc nhanh, chnhxc trong vic s dng my tnh cm tay, c th coi y l phng php duy nhtv mt nhanh, vi chnh xc cao.

Phn ba v giao thoa sng c vi hai cng thc va c bn nhng c tnhtng qut m ti nu ra c th gii quyt hu ht cc dng ton v giao thoasng c hin ti va nhanh va chnh xc.

Chng ti mong mun chuyn ny c ng dng hiu qu vo thc tinkhng nhng trng chuyn L Qu n m cn p dng rng ri cho cctrng khc vi i tng hc sinh i tr, nht l trong vic bi dng hc sinhchun b tham gia cc k thi tt nghip v tuyn sinh i hc. Nhng c th dokinh nghim cn thiu, khng th trnh khi nhng thiu st. Chng ti rt mongnhn c kin ng gp t pha ng nghip v hc sinh chuyn nyngy cng hon hin hn. Chng ti xin chn thnh cm n. 3. xut- kin ngh:

Kin ngh b mn vt l cn t chc cc cuc hi tho chuyn mn, tp trungv phng php c kt nhng kinh nghim qu bu ca cc thy c gioging dy trong ton tnh Bnh nh, t ph bin rng ri cn b, gio vin,v hc sinh hc tp, vn dng vo thc tin cho b mn vt l ngy cngmnh hn.

Quy Nhn ngy 10 thng 5 nm 2010


26/26


PHN NHN XT NH GI CA HI NG KHOA HC

TRNG THPT CHUYN L QU N .............................................................................................................................

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