Ứng dụng phần mềm marlab mô phỏng hiện tượng vl

8/2/2019 ng dng phn mm Marlab m phng hin tng VL

1/20

S gio dc v o to thanh ha

trng THPT Thch thnh 3

-------- --------

tn SKKN

ng dng phn mm matlaBm phng cc hin t- ng vt l

H v tn tc gi: nguyn tt thnhChc v: Gio vin

n v cng tc: T L - Ha - Cng ngh

SKKN thuc mn: Vt l

SKKN thuc nm hc: 2007 - 2008


2/20

1

A. t vn :

I. li M u

Ngy nay cng ngh thng tin bng n trn ton th gii. Vicngdng tinhc vo mi lnh vc: K thut, khoa hc, ngh thut, vn ho - x hi trn ton

cu l iu tt yu. Trc nhng pht trin mnh m ca khoa hc k thuti

hi cc nh gio dc phi i mi khng nhng ni dung dy hc m cn c v

phng php dy hc. Khi my tnh tr thnh mt cng cc lc th vic

gii cc bi tp cng nh lp trnh m phng cc hin tng vt l sem li

hiu qu cao trong dy hccc trngi hc, cao ng, cc trng trung hcph thng hay trong nghin cu vt l.

Hin nay, c nhiu phn mm c thit k, xy dng phc v trong

cng vic gio dc, dy hc. Nhng cc phn mm ny th ng cng nhc, kh

thay i c ni dung, cc thng s hoc nu c thay i c th cng b gii

hn nn vic s dng cc phn mm ny trong dy hc cng b hn ch. V vy

ti tm hiu, nghin cu v ng dng phn mm MATLAB m phng cc

hin t ng vt l phc v trong cng tc ging dy.

y, tc gi s dng Matlab nh mt cng c c lc m phng v

tnh ton cc bi ton vt l rt a dng, gm nhiu phn khc nhau ca vt l

hc nh c hc, nhit hc, in hc, quang hc, vt l nguyn t, ht nhn, thin

vn hc

ng dng phn mm Matlab gii cc bi ton vt l s tr nn ph bin

rng ri. Cng vi vic s dng Notebook trong MATLAB to ra cc gio n

in t hon thin hn, phc v cng tc dy - hc vt l.

II. Thc trng ca vn nghin cu:

1. Thc trng

Qua mt nm ging dy, ti thy hc sinh hc vt l kh khn do t c

thc hnh. Mt khc ph ng php dy hc ca gio vin vn ch a c ch

trng. V vy vic hiu bn cht cc hin t ng vt l ca hc sinh b hn ch.


3/20

2

2. Kt qu, hiu qu ca thc trng trn

T thc trng trn khincc em kh tip thu bi hc v cng ngi hc mn

vt l hn. cng vic t hiu qu tt hn, ti mnh dn ci tin ni dung,

phng php:

ng dng phn mm Matlab

m phng cc hin t ng vt l

B. Gii quyt vn :

I. Cc gii php thc hin

1. Gii thiu phn mm MATLAP

Matlab (MATrix LABoratory) l mt b ch ng trnh phn mm ln

ca cng ty MathWork,l ngn ng c tnh nng cao trong tnh ton khoa hc,

k thut. B lnh rt mnh ca Matlab cho php gii quyt cc bi ton vi

cc i tng khc nhau l i l ng v h ng, vector, cc h phng trnh

tuyn tnh, phi tuyn, hay cc xu k t vi kt qu nhanh chng v chnh xc.

Matlab cho php x l d liu, biu din ha mt cch linh hot, n gin

v chnh xc trong khng gian hai chiu cng nh ba chiu, gip ng i s dng

c th quan st kt qu mt cch trc quan v a ra gii php tt nht. c

tch hp cng vi mt s ngn ng lp trnh thng dng khc nh C, C++,

Fortran, Java lm cho nhng ng dng ca Matlab c th c chuyn i d

dng sang nhng ngn ng .

Matlab v ang c s dng rng ri trn nhiu lnh vc t cc lnh

vc k thut chuyn ngnh nh in, in t, iu khin t ng cho n cc

ngnh x l ton chuyn dng nh thng k, k ton... nhiu n c trn ton th

gii.


4/20

3

- Ccphp ton bnh thng c th thc hin ngay ca s chnh v kt

qu s hin dng di sau php tnh.

- Nu mun lu ni dung php ton (hay mt ch ng trnh) th lm trong

ca s lp trnh. N s c l u li dng m-file. Ni dung file c th chnh sa

ty .

- Cc hnh v, m phng s c hin ra ca s hnh (Figure).

MATLAB c th m t trong khng gian hai hoc ba chiu. ca s hnh

c th phng to, thu nh hnh nh, c th tm dng hnh, xoay hnh nh d i cc

gc nhn khc nhau v c th chuyn hnh nh sang cc dng nh c ui thngth ng nh : .gif, .bitmap, .JPEG

Ca schnh

Hin ktqu

Ca s lp trnh,

l u vn bn

Hnh nhm phng


5/20

4

2.Mt s hm trong MATLAB

a) Nhm cc hm ha

TT Tn hm Ni dung

plot(x,y, k )v th 2 chiu vi x v y l cc mng s,

k l kiu mu, kiu ng nt.

plot3(x,y,z, k )v th 3 chiu vi x, y v z l cc mng s,

k l kiu mu, kiu ng nt.

axis([xmin,xmax,ymin,ymax]) Thit lp cc gi min, max ca h trc ta

title( st ) Ghi tn ca figure ha l st

xlabel( st ) Tn nhn cho trc x

ylabel( st ) Tn nhn cho trc y

legend( st1 , st2 )Ghi ch chi tng ng trong th ln l t

l st1, st2,...

b) Nhm cc hm tnh ton

TT Tn hm Ni dung

sin(x) tnh sinx

cos(x) tnh cosx

tan(x) tnh tgx

abs(x) tnh tr tuyt i ca s thc v tnh argument ca s phc x

round(x) Lm trn x n gi tr nguyn gn nht

c) Cc lnh iu khin v cc php ton:

if

else

end

Hm kim tra iu kin . Nu tha mn th thc

hin nhm lnh . Nu khng tha mn th thc

hin nhm lnh .

for a=m:p:n

end

Vng lp vi s b c xc nh. Cho bin a chy t m

n n vi b c nhy p, ng vi mi gi tr a thc hin

lnh While

end

Vng lp vi b c khng xc nh. My thc hin lnh

khi no iu kin cn ng v thot khi

khi sai.

+ , - , * , / Cng, tr, nhn, chia ma trn.

+ , - , .* , ./ Cng, tr, nhn, chia mng


6/20

5

0 5 10 15 20 25 300

5

10

15

20

25

30

35

40

alpha=0

alpha=30

alpha=45

alpha=60

alpha=90

II. Cc gii php t chc thc hin

1. M phng hin t ng c hc

Bi: Chuyn ng ca vt bnm

VD1: Tmt nh thp cao H ngi ta nm mt hn xung t vi vn tcv theophng hp vi mt phng ngang mt gc . Vi = 0, 30, 45, 60,

90. B qua sc cn ca khng kh.

Nhp cao H = 30m, v = 10 m/s (vi H v v khng b gii hn).

Ta c thnh hnh v:

Da vo hnh v ny ta thy tm

xa cc i khi gc nm bng 45.

VD2: Ta nm mt vt c khi lng 0.1kg cao 15m vi vn tc ban u

30m/s vi gc nm 30. Bit rng lc ma st t l vi vn tc vi h sbng

0.15kg/s v gia tc trng hp dn bng 9.8m/s2

a. V th x, y, vx, vy theo thi gian v quo ca vt t khi nm n khi

chm t.

b.Nu khng c ma st th vt s i xa hn bao nhiu?

c. V qu o ca vt khi c ma st v khi khng c ma st trn cng mt

hnh?

d.Xc nh thi im v v tr khi vt t ti cao cc i

e.Xc nh gc nm ti u vt c th xa nht

f. V hnh nh chuyn ng ca vt.


7/20

6

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

16quy dao cua qua bong tu khi nem den khi cham dat

0 10 20 30 40 50 60 70 80 90 1000

5

10

15

20

25

30

quy dao khi khong co ma sat

quy dao khi co ma sat

0 2 4

5

10

15

t

x1(t)

0 2 4

0

5

10

15

20

t

y1(t)

0 2 4

0

5

10

15

20

t

vx1(t)

0 2 4

-5

0

5

10

t

vy1(t)

0 10 200

10

20

x

y

Quy dao khi co ma sat

VD3: Nm mt qu bng c khi lng 0.1kg cao 10m vi vn tc banu 25m/s, gc nm nghing vi phng nm ngang mt gc bng 30 theo

phng vung gc vi mt bc tng cao thng ng, cch v tr nm 15m.

Bit rng lc ma st t l vi vn tc, h s t l bng 0.1 kg/s v gia tc

trng hp dn bng 9.8m/s2, qu bng va chm n hi vi tng

a.Xc nh thi im qu bng khi chm tng?

b.Xc nh cao ca qu bng khi chm tng?

c. Gc ny ca qu bng khi chm tng?d.Xc nh thi im qu bng chm t

e. Khi chm t, qu bng cch tng mt khong bng bao nhiu?

f. Tnh v v qu o ca qu bng t khi nm ti khi chm t?

th x, y, vx, vy theo thi gian v qu o

ca vt t khi nm n khi chm t.

Qu o ca vt khi c ma st vkhi khng c ma st.


8/20

7

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8Dao dong dieu hoa

Ep(t)

Ek(t)

Et(t)

0 1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5Dao dong dieu hoa

x(t)

v(t)

Bi:Va chm n hi v va chm mm

0 50 100 150 200 250

-100

-80

-60

-40

-20

0

20

40

60

80

100

va cham dan hoi m1=1;m2=2;v2=0;v1=25; cham mau vang tren do thi la diem va cham

0 50 100 150 200 250-100

-80

-60

-40

-20

0

20

40

60

80

100

va cham mem m1=1;m 2=2;v2=0;v1=25; cham mau vang tren do thi la diem va cham

Bi:H dao ng

VD1:Xt dao ng iu ho ca mt con lc l xo vi h s n hi k=1.5, khi

lng m=1kg. T v tr ban u ca con lc l x0 = 1cm, ta th cho con lc dao

ng. Hy:

a.Biu din th ca to v vn tc ca con lc theo thi gian

b.Biu din th ca ng nng, th nng v nng lng theo thi gian

c.Biu din mt phng nng lng

d.Biu din ng ng tr nng lng trn mt phng pha

VD2: Cho mt h 100 dao ng t tuyn tnh l nhng ht c khi lng bng nhau

m=1, gn vi nhau bng cc l xo c h s n hi bng nhau k=10, khong cch

gia cc ht a=1, cc dao ng t chu tc dng ca lc ma st t l vi vn tc ca

ht vi h s t l =0.1. Ti t=0 cc ht ng yn x i(0)=0, vi(0)=0. Ngoi lc tc

dng ln ht th nht bng Fc=b cos(wet), vi b=0.5, we=1.

a. Tnh v v ly ca cc dao ng t th 1, 45, 80, 100 theo thi gian b. Tnh v v vn tc ca cc dao ng t th 1, 45, 80, 100 theo thi gian


9/20

8

0 10 20 30 40 50 60 70 80 90 100

-0.05

0

0.05do thi ly do cua he dao dong tu khi t=10s

0 10 20 30 40 50 60 70 80 90 100-0.05

0


0 10 20 30 40 50 60 70 80 90 100-0.05

0


0 10 20 30 40 50 60 70 80 90 100

-0.05

0


0 10 20 30 40 50 60 70 80 90 100-0.1

0

0.1do thi van toc cua he dao dong tu khi t=10s

0 10 20 30 40 50 60 70 80 90 100-0.05

0


0 10 20 30 40 50 60 70 80 90 100-0.05

0


0 10 20 30 40 50 60 70 80 90 100

-0.05

0


-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

Dao dong dieu hoa

0 1 2 3 4 5 6 7 8 9 10-2

-1

0

1

2Dao dong dieu hoa

(t)

'(t)

0 1 2 3 4 5 6 7 8 9 10-2

-1

0

1

2Dao dong tat dan

(t)

'(t)

0 1 2 3 4 5 6 7 8 9 100

5

10

15Dao dong bao toan

Et1(t)

Ed1(t)

E1(t)

0 1 2 3 4 5 6 7 8 9 100

5

10

15Dao dong tat dan

Et2(t)Ed2(t)

E2(t)

c. Tnh v v ly ca h dao ng t khi t=10, 40, 80, 100

d. Tnh v v vn tc ca h dao ng tkhi t=10, 40, 80, 100

Bi:Cho con lc c di l dao ng trong trng hp dn

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

Dao dong tat dan


10/20

9

-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Bi: Chuyn ng ca hnh tinh

Hnh tinh chuyn ng di tc dng H sao kp ch hp dnln nhauca lc hp dn gia mt tri v hnh tinh

2. M phng hin t ng nhit

Bi:Tnh ton p sut v nhit ca h ng hc phn t v biu din cc il- ng trn th theo thi gian.

Kt qu ta thu c nh sau:

Bi:Biu din s ht trong hp lc ban u v sau khi h tin ti trng thi cn

bng. V th biu din s ht na tri ca hp theo thi gian.Kt qu ta thu c l:

-4 -3 -2 -1 0 1

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

x

y


11/20

10

Ban u cc ht c phn b Sau mt khong thi gian h tin ti cn bng,

v pha bn tri ca hp s ht na tri s n nh.

Kim tra tnh xc sut gia l thuyt theo phn b Maxcell_Boltmann:(v) = kT

mv

evkT

m222

32

)(2


12/20


13/20

12

Bi:Xc nh in trngv in th ti nhng im bn trong(rR) qu cu. Biu din bng th kt qu .

0 0.5 1 1.5 2 2.5 3

x 10-3

0

0.5

1

1.5x 10

-3

r (m)

E(

N/C)

R

Ben trong Ben ngoai

0 0.5 1 1.5 2 2.5 3

x 10-3

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4x 10

-6

r (m)

V(

J/C) Ben trong Ben ngoai

Bi:Trng in sinh bi mt in tch chuyn ng u.

-20 -15 -10 -5 0 5 10 15 20

-20

-15

-10

-5

0

5

10

15

20

X-axis

Y-axis

Bi : Th t vect ca si dy thng di.

Kho st mt si dy dn c chiu di 2L, dng i0 chy qua.

a) Xc nh th t vct dc theo ng thng i qua tm ca ng dy dn.

Gi s L di sao cho c th b quahiu ng ca cc u dy.

b) V trng vct B do dy di v cng trong trng hp hai v ba chiu.


14/20

13

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1tu truong cua soi day dai vo cung trong khong gian 2 chieu

truc x

truc

y

-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1

truc x

tu truong cua soi day dai vo cung trong khong gian 3 chieu

truc y

truc

z

Bi :Kho st chuyn ng ca ht tch in trong t trng khng i

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1

0

1

2

0

0.5

1

1.5

2

2.5

y

x

z


15/20

14

Bi :Kho st mt ht tch in chuyn ng trong t trng u khng i v

in trng bin i theo thi gian. Cho in trng hng dc theo trc x v

c dng Ex = E0cos( t). T trng B khng i v hng dc theo trc z, Ti

thi im t = 0 ht ng yn ti gc to .

4. M phng hin t ng quang hc:

Bi : S khc x qua mt hoc hai lp bn mng.

Bi : Vt sng AB - c t song song vi mn v cch mn mt khong c nh

L. Mt thu knh hi t c trc chnh qua im A v vung gc vi mn - c di

chuyn gia vt v mn. M phng s to nh qua thu knh khi vt dch

chuyn.


16/20

15

Bi :M phng s lan truyn ca sng in t.

Bi :Mt ngun sng n sc pht ra nh sng c b- c sng= 0,6m. Chiunh sng trn vo hai khe h hp song song cch nhau a = 1mm v cch ungun sng. Trn mt mn nh t song song v cch mt phng cha hai khemt on D = 1m, ta thu - c mt h thng vn giao thoa.M phng hnh nhgiao thoa.


17/20

16

Bi :M phng hnh giao thoa nu s khe l 2; 3; 4 hoc 5 bit khong cch

cc tia bng 10 ln b- c sng.

Kho st h vn khi khong cch a gia

hai khe thay i. a = 1; 2; 3; 4 mm.

Ph c ng cc vn giao thoa quahai khe h hp song song.


18/20

17

Bi :M phng h vn nhiu xt tr- c l trn truyn sng bn knh r.

Bi :M phng h vn nhiu x vi s khe N = [1, 2, 3, 5]


19/20

18

Bi : V th nng l ng bc x theo cng thc Planck

Planck coi l ng t nng l ng ng vi tn s v c tr s bng h.

h = 6,625.10-34 (J.s) gi l hng s Planck. Cng thc Planck nh sau:

1

125

2

,

KT

hcT

e

hc

Cng thc ny c nghim ng vi mc chnh xc cao mi nhit thc nghim

v mi b c sng o c.

V th nng l- ng bc xT, vi T = 300K; 400K; 500K

5. Ngoi ra trong bi ging in t cn c th s dng link

C th link bi ging n cc file dng khc nh : cc on phim, cc flat ..

Gio vin c th t to bng k thut lp trnh hoc ln mng copy cc ch ng

trnh c sn.


20/20

C. kt lun:

Thc t sau khi p dng ph ng php dy hc mi ti thu c kt qu

sau:

1. Kt qu nghin cu (Tnh hiu qu so vi kt qu c)

+ Tr c khi p dng ph ng php mi:

Hc sinh hiu bi, hiu bn cht ca

hin t ng vt l60%

Hc sinh thch hc Vt l 30%

+ Sau khi p dng ph ng php mi:

Hc sinh hiu bi, hiu bn cht cahin t ng vt l

95%

Hc sinh thch hc Vt l 80%

Vi cng c Matlab, chng ta khng ch thu c cc kt qu nhanh chng

v chnh xc, m cn c th v d dng cc th v hnh nh chuyn ng ca

cc vt mt cch trc quan, sinh ng, trong khi vic v cc th bng tay tn

rt nhiu cng sc v thi gian.Vi ti ny, chng ta c th pht trin Matlab to nn mt cng c

cho vic ging dy v hc tp vt l ca hc sinh.

2. Kin ngh, xut

My mc ch l phng tin gip cho bi ging hay hn, sinh ng hnsong n khng l tt c. Hiu qu tit hc vn tp trung vo vai tr ca ngithy. Ngi thy khng ch l ngi truyn th kin thc m cn phi bit cchdn dt HS tham gia tch cc bi ging nh th no v kt qu l phi xem HSlnh hi ctri thc bao nhiu. p dng c vo vic son ging bng gion in t th i hi phi c n chiu Projecter, my vi tnh, p dng ng lotth mi lp hc cng u phi c trang b.

ng dng phng php ging dy mi ny t hiu qu ch thc thnn p dng dn dn t cp hc nh n ln. iu s gip cc em c th tipcn c phng php ging dy mi, ng thi cc GV nn c phiu hc tp

pht cho cc em HS h thng li bi hc

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