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LUYN THI I HC VT L 2014- 2015 Thy Lm Phong

Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 1

Sn c

TNG HP CC CNG THC C BN V CNG THC GII NHANHVT L 12 - LUYN THI I HC 2014

CHNG II: SNG C

Bi 1: Sng C v S Truyn Sng C.Cch hnh thnh:

+ Xy ratrong mi trng vt cht (rn, lng, kh)

+ Dao ng trong mi trng c lc lin kt n hi:

1 phn t dao ngLin

Kt Cc phn t k cn dao ng (ti ch)

+ c im:_ Sng c khng truyn trong chn khng ( khc sng in t)_ Cng ra xa ngun cng tr pha hn.

Khi nim v phn loi:

+ Sng c l s lan truyn dao ng c ( pha dao ng , nnglng )trong mt mi trng vt cht.

+ Sng c gm 2 loi sng ngang v sng dc.

Bc sng - bin sng - tc truyn sng:

+Bc sngk hiu l (c l lm-a), c hai nh ngha:

.L qung ng sng i c trong mt chu kT ( nh vy 1 1 T)

.L khong cch ngn nht gi hai im dao ngcng pha trn mt phng truyn sng.

+ Cng thc tnh bc sng l =v

f

= vT

( Trong v l tc truyn sng, f l tn s sng,T l chu k sng )

Vn tc truyn sngcng tnh bi cng thc v = .f =

T(khng i trong cng mt mi trng).

Vn tc truyn sngtrong mi trng cht rn (dy) cn c tnh bi cng thc v =T

( Trong T l lc cng dy v l h s ma st trn dy kg/m ).

Ch :cn phn bit vn tc truyn sng vn tc dao ng ca cc phn t sng ti mt im.

S ph thuc ca tc truyn sng trong cc mi trng:

Sng ngang Sng dc

_ dng bin dng lch.

_ c phng dao ng vunggcvi phng truyn sng.

_ xy ra trong mi trng

cht rn v b mt cht lng

_dng bin dng nn dn.

_ c phng dao ng trngvi phng truyn sng.

_ xy ra trong mitrng rn,

lng, kh.

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+ Vn tc ph thuc vo nhit mi trng ( T v ), tnh n hi ca mi trng , p sutca mi trng ( p v )

+ Vn tc truyn sng cc mi trng theo th t: v rn> v

lng> v

kh> 0.

+ Vn tc m thanh trong khong 330 v 340 m/s.

Tn s sng - chu k sng - nng lng sng:

+ Cc i lng tn s sng - chu k sng - tn s gc ca sng tha mn:= 2f =2T

+ c bit, khi truyn qua cc mi trng khc nhau, tn s lun khng i( f = const).

+ Nng lng sng cng ging nh nng lng bn c dao ng: E =1

2m2A

2 = const = hng s.

Phng trnh sng mtngun:

+ Gi s ngun O pht sng c phng trnh

u O= Acos2ft.

_ im M cch ngun O mt khong d= OM, nn c

_ Phng trnh sng u

M= Acos

2

ft -2d

.

( Du tr "-" trong biu thc l do im M tr pha so vingun O )

Ch :u Mv d c th khc n v, v d u

M tnh theo (mm) , d tnh theo (cm)

_ Gi s im A v B ln lt cch ngun O nhng khong d 1v d2.

Ta c phng trnh sng ti A v B ln lt l u A= Acos2ft -

2d 1

v u B= Acos2ft -

2d 2

Do pha ban u ca A=2d 1

v B=

2d 2

.

lch phagia hai im A v B l = | A- B| =

2|d 1- d2|

(t AB =|d 1- d2| =

2AB

_ hai im A v B dao ng cng pha= k2AB = k

_ hai im A v B dao ng ngcpha= (2k + 1)AB = (k + 0,5)

_ hai im A v B dao ng vung pha= (2k + 1)

2AB = (k + 0,5)

2

_ hai im A v B dao ng lch pha = (2k + 1)

Nhng lu v kinh nghim khi gii bi tp Sng c :_ Sng nh ln n ln (n - 1)T = ? ( vi T l chu k sng)

_ Sng c n ngn lin tip(n - 1)= ? ( vi l bc sng)

_Khi sng truyn qua cc mi trng khc nhau th tn s sng lun khng i. ( bc sng vtc truyn sng thay i )

_m thanh pht ra (sng m), sng dng trn si giy hay hin tng gioa thoa sng trn mtncu c chung bn cht l sng c.

_ Sng c ch truyn c trong mi trng rn, lng, kh, khng truyn c trong chn khng.

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Bi 2: GIAO THOA SNG C.Khi nim v hin tng:

+ Th nghim:(Kho st vi ngun 2ngun S1v S

2cng pha)

m thoa

Cn rungS 1

S 2

Hnh nh giao thoa sng nc.

+ Do hai ngun S 1v S2xut pht t mt ngun nn chng dao ng cng pha, cng tn sto

ra hai sng kt hp, hnh thnh nn nhng ng hybebol nm hai bn ng trung trc.

_ Cc im cc i khi 2 sng cng pha gp nhau v ngcli.

_ Nu hai ngun S 1v S2cng phath ng trung trc c

bin cc i a max.

_ Nu hai ngun S 1v S2ngc phath ng trung trc c

bin cc i a min.

nh ngha: Giao thoa sng l hin tng trong vnggiao thoa ca hai sng kt hp, ti nhng im xc nhhoc lun tng cng nhau (cc i) hoc lun lm yunhau (cc tiu).

Phng trnh giao thoa sng v ngha lch pha:(Kho st vi hai ngun cng pha)

A min A

max _Gi s hai ngun S1v S

2dao ng cng pha c pt u

S1= U

S2= Acos(t)

M _ im Ml ni tip nhn ca hai ngun sng S 1v S2nn ta c:

N d 1 d 2 u

M= u

MS1+ u

MS2(*)

_ Da vo pt sng mt ngun ta lp phngtrnh u MS1v u

MS2vi:

S 1 O S2 u

MS1= Acos(t -

2d 1

) vu MS2= Acos(t -2d 2

)

_ Dng cng thc lng gic Cosa + Cosb = 2Cos(a + b2).Cos(a - b

2) thay vo

(*) ta c: u M= 2Acos

(d 1- d2)

.cos

t -(d 1+ d

2)

.

t

a = 2Acos

(d 1- d2)

l bin cagiao thoa sng tiM

M=

(d 1+ d2)

l pha cahai sng kthptiMu M= acos(t -

M)

Ta c

MS1=

2d 1 v

MS2=

2d 2 . Xt lch pha= |

MS1-

MS2| =

2|d 1- d2|

(d 1- d

2)

=

2 y chnh l biu thc lin h gia bin v lch pha ca giao thoa sng.

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V tr vn cc i ( ng gn sng ) d 1- d2= ka

max= 2A

V tr vn cc tiu ( ng ng yn, khng dao ng)d 1- d2= (k + 0,5)a

min= 0

iu kin xy ra hin tng giao thoa sng:

+ Phi c 2 sng kt hp to ra t hai ngun kt hp (xem k hin tng, th nghim)

+ Hai ngun kt hpl hai ngun c cng tn s v lch pha khng i theo thi gian.

Bi ton xc nh s im dao ng cc i, cc tiu trn on S 1S2: (Xt 2 ngun cngpha)

+ Ta cd

1+ d

2= S

1S

2d 1- d

2= k

2d 1= S 1S 2+ kd 1= S

1S

2

2+ k2

VV d 1l khongcch d

1> 0

Do M S 1S2d

1< S

1S

2

0 < d 1< S1S

2- S

1S

2< k< S

1S

2

Chng minh tng t ta c s im dao ng cc tiu l - S 1S2< (k + 0,5)< S

1S

2

+ T y cc bn c th xt trng hp 2 ngun ngc pha, vung pha hay lch pha .

Sau y l bng tng hp cch m cc im dao ng vi bin max - min:

CNG PHA NGC PHA VUNG PHA

lch pha

=2(d 1- d

2)

=2(d 1- d

2)

+

=2(d 1- d

2)

+

2

Cc i giao thoa d 1- d2= k d

1- d

2= (k + 0,5) d

1- d

2= (k + 0,25)

Cc tiu giao thoa d 1- d2= (k + 0,5) d

1- d

2= k d

1- d

2= (k + 0,75)

S im dao ngcc i trn S1S

2

- S 1S2< k< S

1S

2 - S

1S

2< (k + 0,5)< S

1S

2 - S

1S

2< (k + 0,5)< S

1S

2

S im dao ngcc tiu trn S1S

2

- S 1S2< (k + 0,5)< S

1S

2 - S

1S

2< k< S

1S

2 - S

1S

2< (k + 0,75)< S

1S

2

Bin tng hpa = 2Acos

2

a = 2Acos

2

-

2

a = 2Acos

2

-

4

Bi ton xc nh s im dao ng cc i, cc tiu trn mt onS 1S2:

+ Ta s dng quy tc " Chn im":

TH1: Ti mt im lp thnh tam gic. (xt 2 ngun S 1v S2cng pha)

A min A

max Yu cu 1: m s im dao ng cc i trn on MS1

M MS 1 - MS2k< S

1S

1- S

1S

2

N d 1 d 2 Yu cu 2: m s im dao ng cc tiu trn on MS2

MS 1 - MS2(k + 0,5)< S

2S

1- S

2S

2

S 1 O S2 Yu cu 3: m s im dao ng cc i trn on MN

MS 1 - MS2k< NS

1- NS

2

TH2: Ti mt im lp thnhhnh ch nht hoc hnh vung. (xt 2 ngun S 1v S2ngcpha)

A B Yu cu 4: m s im dao ng cc i trn on AB

I AS 1 - AS2(k + 0,5)BS

1- BS

2

Yu cu 5: m s im dao ng cc tiu trn on BS 1

S 1 S2 BS

1 - BS

2kS

1S

1- S

1S

2

TH3: m s im dao ng a maxv a

mintrn ng trn (C) hoc elip (E)

Khi ta ch vic nhn i s im tm c nhng cn ch ti 2 ngun S1S2c tha mn khng

Ch : Hai im cc i (cc tiu) lin tip cch nhau

/2

im cc i v cc tiu lin tip cch nhau /4

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Bi 3: SNG DNG V S PHN X SNG DNG.Sng phn x:

+ Hin tng: trong mt mi trng, sng ang truyn m gpvt cn th b phn x. Hin tng trn bao gm sng tiv sng phnxc cng bc sng v tn s f.

+ Vt cn gm c:Cnh: sng tingcpha sngphnxTdo: sng ticng pha sngphnx

Sng dng:

+ Hin tng: l s giao thoaca hai sng kt hpc cng phng nhng chiu ngc nhau. ( hay l sgiao thoa ca sng ti v sng phn x)

_ im lun ng yn: nt sng

_ im lun dao ng vi a max: bng sng.

_ im bng v nt lunxen k, cch u nhau.

_ K/cch gia hai bng (hai nt ) lin tip l /2.

+ Kho st vi cc vt cn:

Vt cn c nh Vt cn t do

Bin a = 2A

cos

2d

+

2

a = 2A

cos2d

Phng trnh u M= acos(t - /2) u

M= acos(t)

iu kin c sngdng l

dy=k.(b sng) = k

2(k Z) ldy=(2k + 1)

4(k Z)

Ta nt v bngsng

x nt= k

2

x bng= (2k + 1)

4

x nt= (2k + 1)

4

x bng= k2

S nt v s bng S nt = k + 1 , S bng = k S nt = S bng = k + 1

ng dng ca hin tng sng dng:

Nh ta bit, chiu di dy lv tn s f rung ca m thoa trn si dy ta c th xc nh c. Nh vovic m cc b sng hnh thnh trn si dy ta tm c bc sng v = .f o tc truyn sng.

Nhng lu v kinhnghim khi gii bi tp Sng dng:

Hai u c nh Mt u c nh mt u t do

_ Nt c ch s l ngc pha vi Q _ Bng c ch s l ngc pha vi Q

_ Nt c ch s chn cng pha vi Q _ Bng c ch s chn cng pha vi Q

_ Tt c cc im bng u vung pha vi Q _ Tt c cc im nt u vung pha vi Q

_ Khong cch gia hai 2 nt hoc 2 im bng lin tip l /2

_ Khong cch gia hai im bng v nt gn nht l /4

i lng sng bin thin ( theo v hoc f):

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+ Bc 1: Tm i lng sng bin thin k

+ Bc 2:Chn k theo bi ( Gii i s hoc dng my tnh cm tay)

V d 1:Hai im M, N cng nm trn mt phng truyn sng lun dao ng cng pha vi nhau cchnhau 9 cm dao ng vi tn s f = 50 Hz. Bit rng vn tc truyn sng trong khong 70 cm/s v 80 cm/s.Tm gi tr ca v = ?

HD gii: ta c lch pha 2 im M,N tha mn = k2MN = kv =f.MN

k=

450

k

Do 70 v 80 70 450k

80 5,6 k 6,4 k = 6 v = 75 cm/s

V d 2: Hai im M, N cng nm trn mt phng truyn sng lun dao ng ngc pha vi nhau cchnhau 10 cm dao ng vi tn s f = 30 Hz. Bit rng vn tc truyn sng trong khong 1,6 m/s v 2,9 m/s.Tm gi tr ca v = ?

HD gii: ta c lch pha 2 im M,N tha mn = (2k + 1)v =2MN.f

2k + 1

Do 1,6 v 2,9 dng my tnh cm tay lp bng TABLE chn nghim k = 1 v v = 2 m/s

V d 3: Dy AB di 100 cm, u B c nh, u A dao ng vi tn s trong khong 58 Hz < f < 63 Hz.

Bit rng vn tc truyn sng trn dy l 10 cm/s. S nt sng trn dy l ?

HD gii: iu kin c sng dng trn dy vi 2 u c nh l AB = k2

= kv

2f

f =kv

2ldng my tnh cm tay lp bng TABLE chn nghim k = 12 s nt = 13

Bi 4: SNG M - CC C TRNG VT L V SINH L CA MSng m v cc khi nim c bn:

+ nh ngha: l sng c lan truyn trong mt mi trng (rn, lng, kh).

+ Ngun m: l ngun pht ra sng m gy ra cm gic m cho con ngi.

+ Cm gic m: khi sng m tc dng ln mng nh th gy ra cm gic m ( ph thuc vo ngun mv tai con ngi ).

Cc c trng vt l ca m:

+ Gm c Tn s m - Cng m - Mc cng m.

_ Tn s m: l tn s ca ngun m.

H m Tai con ngi nghe c Siu m

16Hz 20000 Hz

m c tn s xc nh gi l nhc m( th dao ng l mt ng cong tun hon).

m khng c tn s xc nh gi l tp m( th dao ng l mt ng cong khng tun hon).Cc loi ng vt c th nghec sng h m(f < 16 Hz) nh c voi, hu cao c, voi, rn,... v

sng siu m( f > 20000 Hz) nh c heo, di,... c ng dng trong chn on hnh nh trong y khoa,...

_ Cng m: l nng lng m truyn qua mt n v din tch (S) t vung gc vi phngtruyn trong mt n v thi gian. ( k hiu l I, n v l W/m2 )

I =PS

=P

4R2

Vi P l cng sut ca ngun m, S l din tch ca sng cu.

R l khong cch t im n ngun m.

T y ta nhn xt thy cng m I t l nghch vi bnh phng khong cch R2 I1

I 2= R

22

R12

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_ Mc cng m: l i lng so snh mt m vi m chun. K hiu l L, n v l Bel (B) hoc

-xi-ben (dB) L = lgII o

(B) = 10lgII o

(dB)( Vi lg : l logarit thp phn c s 10)

V I o= 10-12

W/m2 1000 Hz

Ngng nghe: l mc cng m nh nht ( L min) m tai bt u c cm gic m.

Ngng au: l mc cng ln nht ( L max) = 10 W/m2 ng vi L = 130 dB m tai bt u

c cm gic au.

Tai con ngi nghec trong khong 0 dB

130 dB

Ting nicon ngi c tn s t 200 Hz 1000 Hz

Cc c trng sinhl ca m:

+ gm c cao ca m- to ca m - m sc.

_ cao ca m: c trng cho cm gic " thanh " hay " trm " cam. (ph thuc vo tn s f).

_ to ca m: ty thuc vo mc cng m m gy ra cm gic m. ( ph thuc vo L v f)

Bng gii thiu to ca mt s m

Ting ni th thm 20 dBTing ni chuyn bnh thng 40 dB

Ting nhc to 60 dB

Ting n rt to ngoi ph 80 dB

Ting n ca my mc nng trong cng xng 100 dB

Ting st 120 dB

Ting ng c phn lc cch 4m 130 dB

_ m sc: l i lng c trng cho s khc nhau ca " cm gic m " ( ph thuc vo bin vtn s m) v c biu th qua th dao ng m.

Ngun nhc m:L cc dng pht ra sng m, in hnh nh dy n, ng so, hp cnghng m,

+ i vi dy n: ( hai u c nh)

Th l dy n= n. (b sng) = n2

f = nV

2l dy n (n = 1,2,3,...)

Nu n = 1 f1=V

2l dy nl ha m bc 1( tn s m c bn)

Nu n = 2 f2= 2V

2l dy n= 2f1l ha m bc 2.

+ i vi ng so( mt u h, mt u kn)

Th l

dy n= n. (b sng) +

1

2(b sng) = (2n + 1)4f = (2n + 1)

V4l dy n(n = 0,1,2,3,...)

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Nu n = 0 f1=V

4l dy nl ha m bc 1( tn s m c bn)

Nu n = 1 f2= 3V

4l dy n= 3f1l ha m bc 3

Nu n = 2 f3= 5f1l ha m bc 5 Ha m

trong ng so lun l ha m bc l.

Hnh 1: Hnh nh so nhn t bn ngoi.

Hnh 2: L trng bn trong ng so.

+ i vi hp cng hng( thuc v mt nhc c )

_ Hp cng hng c chc nng khuch im c bn v mt s ha m khc.

Nhng lu v kinh nghim khi gii bi tp v Sng m:

+ Ngun m S pht sng m ng hng, mi trng khng hp th m. im A, B ln lt cchngun m S nhng khong R A, R

Bc mc cng m tng ng l L

A, L

B. Tm mi quan h gia

cng m IA, IBvi R

A, R

BvL

A, L

B? ( Gi s L

A> L

B).

Ta c L = 10LgI

I ov I =

P

S=

P

4R2 v log

ab - log

ac = log

a

b

c

Xt L A- LB= 10Lg

I AI o

- 10LgI BI o

= 10LgI AI B

L A- LB= 10Lg

I AI B

Mt khcI AI B

=RBRA2

L A- LB= 10Lg

I AI B

= 10lgRBRA2

= 20lgRBRA

( do log ab = log ab)

Nh vy ta c L A- LB= 10Lg

I AI B

= 20lgRBRA

+ Ngun m S pht sng m ng hng, mi trng khng hp th m. im A, B ln lt cch ngun

m S nhng khong R

A, R

Bc mc cng m tng ng l L

A, L

B. Nu I

A=10n

I

Bth L

A- L

B= ?

Ta cL A- LB= 10Lg

I AI B

= 10Lg10n = 10n ( do logaa = 1)

Nh vy nuI AI B

= 10n LA- L

B= 10niu ny c ngha l nu cng m tng 10

n ln

th mc cng m tng 10n ln.

+i vi quan h gia c trng sinh l v c trng vt l ca m, ta c nhn xt:

_ cao ca m ch ph thuc vo tn s. (f)

_ to ca m ph thuc mc cng m v tn s( L v f)

_m sc ph thuc bin v tn s m(

a v f)_ Cc c trng sinh l ca m ( cao, to, m sc) u ph thuc vo tn s.

CHC CC EM N TPHIU QU V T KT QU CAO NHT TRONG KTHI TUYN SINH I HC 2014- 2015

[email protected] - FB : Lamphong Windy

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